(4778,176,'<p>Substation Design and Practices</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1652434882961652434882.JPG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','6','Substation Design and Practices','1','3','1','0','1','2022-05-13 15:12:16','2022-05-13 15:12:16'),(4779,176,'<p>Figure 1.1</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1652434993601652434993.JPG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','7','Figure 1.1','1','4','1','0','1','2022-05-13 15:14:13','2022-05-13 15:14:13'),(4780,176,'<p>Substation Design and Practices</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1652435169111652435169.JPG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','8','Substation Design and Practices','1','4','1','0','1','2022-05-13 15:17:13','2022-05-13 15:17:13'),(4781,176,'<p>Figure 3</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1652435831331652435831.JPG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','9','Figure 3','1','4','1','0','1','2022-05-13 15:29:27','2022-05-13 15:29:27'),(4782,176,'<p>Substation Design and Practices - Substation Layout</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1652436204191652436204.JPG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','10','Substation Design and Practices - Substation Layout','1','3','1','0','1','2022-05-13 15:35:12','2022-05-13 15:35:12'),(4783,176,'<p>Figure 4</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1652436400321652436400.JPG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','11','Figure 4','1','4','1','0','1','2022-05-13 15:37:51','2022-05-13 15:37:51'),(4784,176,'<p>Figure 5</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1652436613491652436613.JPG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','12','Figure 5','1','4','1','0','1','2022-05-13 15:42:40','2022-05-13 15:42:40'),(4785,176,'<p>Stay Tuned with edupme</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1652437091351652437091.JPG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','13','Stay Tuned with edupme','1','4','1','0','1','2022-05-13 15:49:05','2022-05-13 15:49:05'),(4786,177,'<p>Welcome to the internship programme.</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655274062691655274062.pdf',NULL,'1','1','Welcome','1','2','1','0','0','2022-06-15 11:51:02','2022-06-15 11:51:02'),(4787,177,'<p>Global SpaceTech ecosystem</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655274219571655274219.jpeg\" style=\"height:609px; width:800px\" /></p>',NULL,'1','2','Ecosystem','1','2','1','0','0','2022-06-15 11:58:51','2022-06-15 11:58:51'),(4788,177,'<p>1. Introduction</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655275154971655275154.PNG\" style=\"height:453px; width:800px\" /></p>',NULL,'1','3','1. Introduction','1','2','1','0','0','2022-06-15 12:09:32','2022-06-15 12:09:32'),(4789,177,'<p>2. Rocket Launch</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655275239731655275239.PNG\" style=\"height:449px; width:800px\" /></p>',NULL,'1','4','2. Rocket Launch','1','2','1','0','0','2022-06-15 12:11:21','2022-06-15 12:11:21'),(4790,177,'<p>3. Milestones</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655275347241655275347.PNG\" style=\"height:447px; width:800px\" /></p>',NULL,'1','5','3. Milestones','1','2','1','0','0','2022-06-15 12:13:11','2022-06-15 12:13:11'),(4791,177,'<p>4. OmSpace Timeline</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655275484941655275484.PNG\" style=\"height:449px; width:800px\" /></p>',NULL,'1','6','4. OmSpace Timeline','1','2','1','0','0','2022-06-15 12:15:22','2022-06-15 12:15:22'),(4792,177,'<p>5. Rocket Exploration</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655275588491655275588.PNG\" style=\"height:448px; width:800px\" /></p>',NULL,'1','7','5. Rocket Exploration','1','2','1','0','0','2022-06-15 12:16:55','2022-06-15 12:16:55'),(4793,177,'<p>6. Partners</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655275668601655275668.PNG\" style=\"height:446px; width:800px\" /></p>',NULL,'1','8','6. Partners','1','2','1','0','0','2022-06-15 12:18:30','2022-06-15 12:18:30'),(4794,177,'<p>7. Our Service and Product</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655275788621655275788.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','9','7. Our Service and Product','1','2','1','0','0','2022-06-15 12:20:06','2022-06-15 12:21:16'),(4795,177,'<p>8. Collaboration: Russia&nbsp;</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655275990671655275990.PNG\" style=\"height:451px; width:800px\" /></p>',NULL,'1','10','8. Collaboration: Russia ','1','2','1','0','0','2022-06-15 12:23:37','2022-06-15 12:23:37'),(4796,177,'<p>9. Collaboration: Australia</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655276094481655276094.PNG\" style=\"height:449px; width:800px\" /></p>',NULL,'1','11','9. Collaboration: Australia','1','2','1','0','0','2022-06-15 12:25:21','2022-06-15 12:25:21'),(4797,177,'<p>10. Collaboration: UK</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655276198211655276198.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','12','10. Collaboration: UK','1','2','1','0','0','2022-06-15 12:27:14','2022-06-15 12:27:14'),(4798,177,'<p>11. Collaboration: India</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655276414531655276414.PNG\" style=\"height:448px; width:800px\" /></p>',NULL,'1','13','11. Collaboration: India','1','2','1','0','0','2022-06-15 12:30:46','2022-06-15 12:30:46'),(4799,177,'<p>12. Our Crew</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655276508161655276508.PNG\" style=\"height:453px; width:800px\" /></p>',NULL,'1','14','12. Our Crew','1','2','1','0','0','2022-06-15 12:32:14','2022-06-15 12:32:14'),(4800,177,'<p>13. Our Crew</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655276642971655276642.PNG\" style=\"height:448px; width:800px\" /></p>',NULL,'1','15','13. Our Crew','1','2','1','0','0','2022-06-15 12:34:35','2022-06-15 12:34:35'),(4801,177,'<p>14. Our advisors</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655276740761655276740.PNG\" style=\"height:449px; width:800px\" /></p>',NULL,'1','16','14. Our advisors','1','2','1','0','0','2022-06-15 12:36:08','2022-06-15 12:36:08'),(4802,177,'<p>15. Contact Us</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655276822601655276822.PNG\" style=\"height:448px; width:800px\" /></p>',NULL,'1','17','15. Contact Us','1','2','1','0','0','2022-06-15 12:37:25','2022-06-15 12:37:25'),(4803,177,'<p>16. Logbook: Page 1</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655277150831655277150.png\" style=\"height:450px; width:800px\" /></p>',NULL,'1','18','16. Logbook: Page 1','1','2','1','0','0','2022-06-15 12:43:11','2022-06-15 12:43:11'),(4804,177,'<p>17: Logbook: Page 2</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/16552772281001655277228.png\" style=\"height:450px; width:800px\" /></p>',NULL,'1','19','17: Logbook: Page 2','1','2','1','0','0','2022-06-15 12:44:14','2022-06-15 12:44:14'),(4805,177,'<p>Quiz Q1: Which of the following is a hypergolic propellant combination used in rocket engines?</p>',NULL,'multi','single','',NULL,'1','20','Quiz Q1','2','1','1','0','1','2022-06-15 14:29:11','2022-06-15 14:48:12'),(4806,177,'<p>Quiz Q2: For a NACA 2415 airfoil of chord length c, which of the following is true?</p>',NULL,'multi','single',NULL,NULL,'1','21','Quiz Q2','3','1','1','0','1','2022-06-15 14:41:58','2022-06-15 14:41:58'),(4807,177,'<p>Quiz Q3: A rocket has an initial mass of 150 kg. After operating for a duration of 10s, its final mass is 50 kg. If the acceleration due to gravity is 9.81 m/s2 and the thrust produced by the rocket is 19.62 kN, the specific impulse of the rocket is&nbsp;&nbsp;</p>',NULL,'multi','single',NULL,NULL,'1','22','Quiz Q3','3','4','1','0','1','2022-06-15 14:44:17','2022-06-15 14:44:17'),(4808,165,'<p>Knolzi 1</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655706867371655706867.jpeg',NULL,'1','16','Knolzi','1','1','1','0','1','2022-06-20 12:04:27','2022-06-20 12:04:27'),(4809,165,'<p>What is Knolzi&nbsp;</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655706962501655706962.pdf',NULL,'1','17','What is Knolzi','1','3','1','0','1','2022-06-20 12:06:02','2022-06-20 12:06:02'),(4810,165,'<p>Knolzi Features</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655707086811655707086.jpeg',NULL,'1','18','Features','1','3','1','0','1','2022-06-20 12:08:06','2022-06-20 12:08:06'),(4811,165,'<p>Knolzi Focus</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1655707681121655707681.jpeg',NULL,'1','19','Focus on Knolzi','1','3','1','0','1','2022-06-20 12:18:02','2022-06-20 12:18:02'),(4812,165,'<p>Knolzi Online</p>',NULL,'single','html','<p>Please click the link below to experience Knolzi</p>\r\n\r\n<p>https://xd.adobe.com/view/971dfb69-0b62-4152-b988-01023b1509d7-c24e/?fullscreen</p>',NULL,'1','20','Online Link','1','4','1','0','1','2022-06-20 12:27:30','2022-06-20 12:27:30'),(4813,165,'<p>Is the measurement of Delivery important and meaningful?</p>',NULL,'multi','single',NULL,NULL,'1','21','Question 2','1','4','1','0','1','2022-06-30 08:53:26','2022-06-30 08:53:26'),(4814,165,'<p>Will it add value if the matrix of&nbsp;level of engagement, attentiveness, correctness &amp; availability is readily displayed on the presenter&#39;s screen?</p>\r\n\r\n<p>&nbsp;</p>',NULL,'multi','single',NULL,NULL,'1','22','Question 3','4','4','1','0','1','2022-06-30 08:57:12','2022-06-30 08:57:12'),(4815,178,'<p>Agenda</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1657004713501657004713.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','1','Agenda','1','2','1','0','0','2022-07-05 12:36:04','2022-07-05 12:36:04'),(4816,178,'<p>Introduction</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1657016899181657016899.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','2','Introduction','1','2','1','0','0','2022-07-05 15:59:08','2022-07-05 15:59:08'),(4817,178,'<p>Basic Functions and Overview</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/16570234601001657023460.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','3','Basic Functions & Overview','1','3','1','0','0','2022-07-05 17:48:53','2022-07-05 17:50:54'),(4818,178,'<p>Basic Terms in Excel</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1657023581851657023581.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','4','Basic Terms in Excel','1','3','1','0','0','2022-07-05 17:50:25','2022-07-07 10:01:45'),(4819,178,'<p>Conditional Formatting</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1657023761451657023761.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','5','Conditional Formatting','1','3','1','0','0','2022-07-05 17:53:14','2022-07-05 17:53:14'),(4820,178,'<p>IF Formula</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1657023845691657023845.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','6','IF Formula','1','3','1','0','0','2022-07-05 17:54:36','2022-07-07 10:01:25'),(4821,178,'<p>Index-Match Formula</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1657023943271657023943.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','7','Index-Match Formula','1','2','1','0','0','2022-07-05 17:56:09','2022-07-05 17:56:09'),(4822,178,'<p>COUNT IF/ COUNT IFs</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1657024042301657024042.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','8','COUNT IF/ COUNT IFs','1','2','1','0','0','2022-07-05 17:59:04','2022-07-06 16:56:28'),(4823,165,'<p>Audio and Video test.</p>',NULL,'multi','html','<p>Select from the&nbsp;following and press &quot;Submit&quot;.</p>',NULL,'1','20','Audio and Video test','1','2','1','0','1','2022-07-06 10:29:20','2022-07-06 10:29:20'),(4824,178,'<p>SUM IF / SUM IFs</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1657106316921657106316.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','9','SUM IF / SUM IFs','1','2','1','0','0','2022-07-06 16:51:47','2022-07-07 10:00:56'),(4825,178,'<p>V Lookup</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1657168194531657168194.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','10','V Lookup','1','2','1','0','0','2022-07-07 10:00:27','2022-07-07 10:00:27'),(4826,178,'<p>PIVOT Table</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1657168365851657168365.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','11','PIVOT Table','1','2','1','0','0','2022-07-07 10:06:30','2022-07-07 10:06:30'),(4827,178,'<p>Demonstration sheet</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1657168675281657168675.xlsx',NULL,'1','12','Demonstration sheet','1','3','1','0','0','2022-07-07 10:07:55','2022-07-07 10:07:55'),(4828,178,'<p>Which if the following cannot be done in MS Excel?</p>',NULL,'multi','single',NULL,NULL,'1','13','Exercise 1','2','1','1','0','1','2022-07-27 15:37:26','2022-07-27 15:37:26'),(4829,178,'<p>Select whether the following statement true or false: &quot;VLOOKUP can look up values from left to right and right to left&quot;</p>',NULL,'multi','single',NULL,NULL,'1','14','Exercise 2','2','3','1','0','1','2022-07-27 15:43:33','2022-07-27 15:43:33'),(4831,178,'<p>What is the use of Pivot Table?</p>',NULL,'user_input','single',NULL,NULL,'1','15','Exercise 3','1','2','1','0','1','2022-07-28 12:37:44','2022-07-28 12:37:44'),(4832,179,'<p>Interaction 1: Agenda</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1659249637761659249637.png\" style=\"height:450px; width:800px\" /></p>',NULL,'1','1','Interaction 1: Agenda','1','2','1','0','0','2022-07-31 12:11:40','2022-07-31 12:17:57'),(4833,179,'<p>Interaction 2: Importance of Problem solving skills</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1659249915321659249915.png\" style=\"height:450px; width:800px\" /></p>',NULL,'1','2','Interaction 2: Importance of Problem solving skills','1','3','1','0','0','2022-07-31 12:17:10','2022-07-31 12:18:27'),(4834,179,'<p>Interaction 2 - Assessment 1: Which of the following is a direct effect of developing problem solving skills?</p>',NULL,'multi','single','',NULL,'1','3','Interaction 2 - Assessment 1','1','4','1','0','1','2022-07-31 12:26:51','2022-07-31 12:59:16'),(4835,179,'<p>Interaction 2 - Assessment 2: State Objectives</p>',NULL,'single','html','<p>Please state your objectives in the&nbsp;form below:</p>\r\n\r\n<p><a href=\"https://docs.google.com/forms/d/e/1FAIpQLSe6oF_FrIke-_aNiIoUMspPkQrIefByqpwdDVDVWKCeoUXPiA/viewform?usp=sf_link\" target=\"_blank\">Click on this link to write your objectives.</a></p>',NULL,'1','4','State Objectives','1','2','1','0','1','2022-07-31 13:09:50','2022-07-31 19:25:26'),(4836,179,'<p>Interaction 3</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1659268877721659268877.png\" style=\"height:450px; width:800px\" /></p>',NULL,'1','5','Interaction 3','1','3','1','0','0','2022-07-31 17:34:27','2022-07-31 17:40:24'),(4837,179,'<p>Interaction 4: Steps in&nbsp;detail</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1659269391471659269391.pptx',NULL,'1','6','Interaction 4: Steps in Detail','1','2','1','0','1','2022-07-31 17:39:51','2022-07-31 17:41:50'),(4838,179,'<p>Interaction 5: Tools and Pitfalls</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1659269775181659269775.pptx',NULL,'1','7','Interaction 5: Tools and Pitfalls','1','2','1','0','0','2022-07-31 17:46:16','2022-07-31 17:46:16'),(4839,179,'<p>Interaction 5 - Assessment 1: Short text answer auto evaluation</p>',NULL,'user_input','html','<p>List benefits of having problem solving skills:</p>\r\n\r\n<p>Ideal answer: &quot;Increased customer satisfaction, market share, profitability and efficientcy. Shorter delivery time and lower costs&quot;</p>',NULL,'1','8','Interaction 5 - Assessment 1: Short text answer evaluation','1','2','1','0','0','2022-07-31 17:49:59','2022-07-31 17:49:59'),(4840,180,'<p>1 Usage of Industrial Robots</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">According to the Robot Institute of America, &ldquo;A robot is a reprogrammable multifunctional manipulator designed to move material, parts, tools, or specialized devices through variable programmed motions for the performance of a variety of tasks.&rdquo; These manipulators have vast amounts of avenues where they can OR they are used. These avenues utilize/install robotic systems based on three distinguishing characteristics that they provide</span></span></p>','','1','1','Usage of Industrial Robots','1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4841,180,'<p>1.1 Working Environments</p>','','single','html','<p>&nbsp;</p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:12pt\"><span style=\"color:#000000\">Industrial Robots come in handy on factory floors with monotonous tasks, specifically, those that do not require skilled manpower OR those that do not extract the optimal throughput available from an active member on the factory floor; i.e tasks that are termed as DULL in layman language.&nbsp;</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:12pt\"><span style=\"color:#000000\">There are certain tasks that are DIFFICULT for humans to execute on&nbsp; a factory floor, for example:</span></span></li>\n	<li style=\"text-align:justify\"><span style=\"font-size:12pt\"><span style=\"color:#000000\">On a continuous production line, where the product is being manufactured in batches; the process flow is well-defined, leading to conveyors being used to keeping up with the high-speed cycle times. On such lines, it is of utmost importance to maintain and keep in check the quality of the parts and assembled products. This task is crucial for any industry as it directly affects their sales. Robots are used to check and discard products that fail to meet the quality standards without tiring and reducing efficiency since human eyes are unable to assess them continuously throughout shifts.</span></span></li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:12pt\"><span style=\"color:#000000\">Some avenues where robots are widely used are the ones that are DANGEROUS and can cause fatalities to human life. These avenues have adapted to using robotic systems on a wide scale which has increased overall throughput and at the same time has made working environments comparatively safer. For example furnace tapping and production of harmful drugs.</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><span style=\"color:#000000\"><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300499971630049997.mp4\">Click here to visualize the topic better</a></span></strong></span></p>\n\n<p>&nbsp;</p>','','1','1.1','Working Environments','1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4842,180,'<p>1.2 Repeatability</p>','','single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300501131630050113.png\" style=\"height:699px; width:600px\" /></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">One of the factors affecting the quality of products is the consistency in its production or final aesthetic. Many tasks like spray painting of automobiles, MIG welding of parts, soldering of PLCs and instrument calibrations require high levels of precision and repeatability when being done on the factory floor. Industrial robots excel at such tasks as they use encoders and motors to repeat and execute movements respectively. This characteristic of REPEATABILITY (defined as the closeness of points achieved by the end-effector of the robot for the same controlled position repeated multiple times under the same prevalent conditions) is why robots have been widely accepted throughout industries.&nbsp;</span></span></li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><span style=\"color:#000000\"><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300516891630051689.gif\">Click here to visualize the topic better</a></span></strong></span></p>\n\n<p>&nbsp;</p>','','1','1.2','Repeatability','1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4843,180,'<p>1.3 Reliability</p>','','single','html','<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">In robotics, reliability is defined in terms of the probability of the robot working correctly under defined working conditions. In industrial applications, reliability is linked to the term- MTTF (Mean Time To Failure) which means the average time of normal working of a robot before it faces a failure and is defined as <img alt=\"&lt;math xmlns=&quot;http://www.w3.org/1998/Math/MathML&quot;&gt;&lt;mi&gt;M&lt;/mi&gt;&lt;mi&gt;T&lt;/mi&gt;&lt;mi&gt;T&lt;/mi&gt;&lt;mi&gt;F&lt;/mi&gt;&lt;mo&gt;&amp;#xA0;&lt;/mo&gt;&lt;mo&gt;=&lt;/mo&gt;&lt;mo&gt;&amp;#xA0;&lt;/mo&gt;&lt;mfenced&gt;&lt;mrow&gt;&lt;mi&gt;P&lt;/mi&gt;&lt;mi&gt;H&lt;/mi&gt;&lt;mi&gt;R&lt;/mi&gt;&lt;mo&gt;&amp;#xA0;&lt;/mo&gt;&lt;mo&gt;-&lt;/mo&gt;&lt;mo&gt;&amp;#xA0;&lt;/mo&gt;&lt;mi&gt;D&lt;/mi&gt;&lt;mi&gt;T&lt;/mi&gt;&lt;mi&gt;D&lt;/mi&gt;&lt;mi&gt;T&lt;/mi&gt;&lt;mi&gt;R&lt;/mi&gt;&lt;mi&gt;F&lt;/mi&gt;&lt;/mrow&gt;&lt;/mfenced&gt;&lt;mo&gt;&amp;#xA0;&lt;/mo&gt;&lt;mo&gt;&amp;#xF7;&lt;/mo&gt;&lt;mo&gt;&amp;#xA0;&lt;/mo&gt;&lt;mi&gt;N&lt;/mi&gt;&lt;mi&gt;R&lt;/mi&gt;&lt;mi&gt;F&lt;/mi&gt;&lt;/math&gt;\" src=\"https://lh3.googleusercontent.com/_9yO_Qc6oLahLYvgLmvBM55ww_RQJoY2fMpurLfOpdv9IwvT_7PB2nSjhGUCy9f-6OwhA8WhHNOT4_-CBEr6mnQcvcJz7obfnC-0nrQewAYpS3or_idBLilIy0wVRvnJbsYrF2kv=s0\" style=\"height:12px; width:253px\" title=\"M T T F space equals space open parentheses P H R space minus space D T D T R F close parentheses space divided by space N R F\" /></span></span></li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Where,</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">PHR- Production Hours of a Robot</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">DTDTRF- Downtime Due To Robot Failure in hours</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">NRF- Number of Robot Failures&nbsp;</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">The reliability of a robotic system is dependent on its ancillary units as well; the following are included when one wants to take into account the RELIABILITY of a cell:</span></span></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Robotic Arm</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Controller</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Gripper and Tools</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Workstation with the workpieces in the robot working area</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Safety interlocks of the cell</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Human operator working on the cell</span></span></li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><span style=\"color:#000000\"><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300517781630051778.mp4\">Click here to visualize the topic better</a></span></strong></span></p>','','1','1.3','Reliability','1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4844,180,'<p>2 Aternatives to industrial Robots</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">In many industrial processes, where a specific task is to be carried out continuously and requires only a simple specific motion, then these tasks can be executed without installing industrial robots as well, these categories can be divided into two parts:</span></span></p>\n\n<p>&nbsp;</p>','','1','2','Aternatives to industrial Robots','1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4845,180,'<p>2.1 Custom Mechanisms</p>','','single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300504881630050488.png\" style=\"height:309px; width:600px\" /></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">This category consists of both, simple (like 4 bar) as well as very complex mechanisms required to complete a process. That is, instead of a pick and place process executed by an industrial robot, if a conveyor and a simple hydraulic/pneumatic cylinder is enough and can push a job in another direction while not harming the product, then this simple yet effective form of mechanism is a cheaper option. Similarly, for a linear welding profile on a job, if a complex mechanism in which a roller conveyor lets the part traverse on while a stationary welding gun mounted on top performs welding based on timer logic, then this is also a very effective substitute in place of a robot. Keeping in mind that such mechanisms which somewhat serve the same purpose as Special Purpose Machinery (SPM) unlike industrial robots, come at the cost of only being able to do one task throughout its product cycle.</span></span></li>\n</ul>\n\n<p>&nbsp;</p>','','1','2.1','Custom Mechanisms','1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4846,180,'<p>2.2 CNC Machines (Computer Numerical Control)</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">CNC machines are generally used in the production of high production low variability parts. These machines are available with many variations from 2-axis to 12-axis and accommodate conventional tools like drills, mills and lathes as well as novel manufacturing tools like plasma cutting, water jet cutting, Electron Beam Machining, Electrochemical machining, Electrical Discharge Machining (EDM), Photochemical machining and Ultrasonic machining. Though CNCs have a certain monopoly in these high production costs, they lag behind industrial robots which can work effectively in low production high variability scenarios. </span></span></p>','','1','2.2','CNC Machines (Computer Numerical Control)','1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4847,180,'<p>3 Types of Industrial Robots</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">The first set of classification of Industrial robots is based on the type of configuration of joints and links as well as mobile and stationary. Industrial robots are broadly divided into static and mobile; these classifications have subsets as well which are elaborated below:</span></span></p>\n\n<p>&nbsp;</p>','','1','3','Types of Industrial Robots','1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4848,180,'<p>3.1 Static Industrial Robots (fixed base robots)</p>','','single','html','<p><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300507271630050727.jpg\" style=\"height:400px; width:600px\" /><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300507471630050747.png\" style=\"height:525px; width:600px\" /></span></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Fixed robots are usually industrial manipulators that operate in specialized environments suited for robotics.</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><strong>Cartesian Robots: </strong>these are the robots whose three principal axes are perpendicular to each other and are linear in nature. These three linear joints, i.e the sliding joints correspond to up-down, in-out and back-forth motions. They have high reliability and precision when it comes to operating and executing tasks in the 3-D space. Examples of the cartesian types are the motions observed in CNC machines and 3D printers.</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><strong>Cylindrical Robots:</strong> these types of robots are simple and similar to cartesian robots with the major difference that they have one rotary and two linear actuators. This makes them have a cylindrical work envelope which can be highly effective for restricted workspaces. These robots are used for pick and place applications as they have a rotary joint that allows a 360-degree reach.</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><strong>SCARA Robots:</strong> </span><span style=\"color:#202122\">The acronym stands for Selective Compliance Assembly Robot Arm or Selective Compliance Articulated Robot Arm. A SCARA robot has two rotary and one prismatic joint. The two rotary joints linked to each other give a human arm type movement and the prismatic joint at the end effector is used for restricted motion in the Z- direction, this restricted motion is why the term &lsquo;selective compliance&rsquo; is used in its nomenclature.</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#202122\"><strong>Delta Robots:</strong> these robots have a unique layout of having 3 joints rooted in a common base which is generally mounted on top (on the roof). All the tree joints have a universal joint at the root and then merge to connect their ends on a triangular end effector. The linkages, that is the arms, are designed to be parallelograms which results in giving the only translational ability to the base of the links, i.e. movement in X, Y and Z directions without any rotational requirement. A Delta robot can be understood as a spatial generalisation of a four-bar linkage. Since the actuators are rooted in the base, the arms can be made of light composite material. As a result of this, the moving parts of the delta robot have small inertia. which allows for high speed and high accelerations. Having all the arms connected to the end-effector increases the robot stiffness, but reduces its working volume.&nbsp;</span></span></li>\n	<li style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#202122\"><strong>Articulated Robots:</strong> </span><span style=\"color:#333333\">Articulated robots are defined as robots containing rotary joints. They can be as simple as a two-axis structure or complex with ten or more axes and are typically powered by servo motors. Most industrial robots have four to six axes, with six axes being the most common. Each joint provides an additional degree of freedom (DoF). An articulated robot begins with a base that is generally vertical to the ground, which contains the first joint. It is further connected to the base through this first revolute joint, and another revolute joint operates perpendicular to the body and connects the shoulder; this chainlike structure continues as necessary. At the end of the robot additional joints are then used to attach the robot wrist and end effector. This robotic structure is designed to closely mimic the human arm.<span style=\"color:#202122\"> </span>Articulated robots provide more dexterity than any other robot type, which is why they are commonly utilized amongst manufacturers.</span></span></li>\n</ul>','','1','3.1','Static Industrial Robots (fixed base robots)','1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4849,180,'<p>3.2 Mobile Robots</p>','','single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300512461630051246.jpg\" style=\"height:338px; width:600px\" /></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#202122\">Mobile robots can move around in their environment and are not fixed to one physical location. Mobile robots can be &quot;autonomous&quot; (AMR - autonomous mobile robot) which means they are capable of navigating an uncontrolled environment without the need for physical/electro-mechanical sensory guidance. Alternatively, mobile robots can rely on guidance devices that allow them to travel a predefined navigation route in relatively controlled space. A mobile robot has three basic environments in which its design must be different because of the difference in mechanics: aquatic (underwater exploration), terrestrial (road vehicles) and aerial (drones). In addition, the classification isn&#39;t strict, for example, there are amphibious robots that move both on and underwater. Robots for these three environments can be further divided into subclasses: terrestrial robots can have legs or wheels or tracks, and aerial robots can be lighter-than-air balloons or heavier-than-air aircraft, which are in turn divided into fixed-wing and rotary-wing.</span></span></li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><span style=\"color:#202122\"><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300518451630051844.gif\">Click here to visualize the topic better</a></span></strong></span></p>','','1','3.2','Mobile Robots','1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4850,180,'<p>4 Selection of Robots</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Selection of Robots consists of multiple factors and moreover, a complex process flow, mentioned below are the factors that are considered in the process of selecting a robot which is further explained in detail.</span></span></p>\n\n<ol>\n	<li style=\"list-style-type:lower-alpha\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Speed</span></span></li>\n	<li style=\"list-style-type:lower-alpha\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Applicable Standards</span></span></li>\n	<li style=\"list-style-type:lower-alpha\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Indoor vs outdoor environments</span></span></li>\n	<li style=\"list-style-type:lower-alpha\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Return on investment</span></span></li>\n	<li style=\"list-style-type:lower-alpha\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Safety</span></span></li>\n	<li style=\"list-style-type:lower-alpha\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Payload</span></span></li>\n	<li style=\"list-style-type:lower-alpha\"><span style=\"font-size:14px\"><span style=\"color:#000000\">End of Arm Tooling</span></span></li>\n	<li style=\"list-style-type:lower-alpha\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Mounting Conditions</span></span></li>\n</ol>\n\n<p><span style=\"font-size:14px\"><span style=\"color:#000000\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300524021630052402.jpg\" style=\"height:700px; width:280px\" /></span></span></p>','','1','4','Selection of Robots','1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4851,180,'<p>4.1 Speed</p>','','single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300623071630062307.png\" style=\"height:299px; width:300px\" /></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">SPEED is considered to be one of the most important factors while selecting an industrial robot; while considering speed, it consists in itself a list of sub-factors ranging from process-specific speed requirements (example: cycle-time) to time lags in electro-mechanical systems.</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300525051630052505.mp4\">Click here to visualize the topic better</a></strong></span></p>\n\n<p style=\"margin-left:48px; text-align:justify\">&nbsp;</p>\n\n<p>&nbsp;</p>','','1','4.1','Speed','1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4852,180,'4.1.1 Speed requirement calculations based on the application of interest','','single','html','<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Required position, speed/velocity and acceleration</span></span></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">This parameter is closely related to each user. The specification table of an industrial arm shows the maximum speed of the model, but we should know that considering the acceleration and deceleration from one point to another, the actual speed will be between zero and the maximum speed and is usually expressed in degrees/second.</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">The requirement of an industrial robot on the factory floor arises mainly because of a few reasons; the need for manpower reduction, reduction in cycle-time, increasing throughput and reducing process inventory. It is clear that all these reasons are at a benefit by increasing the speed of the process/task that is being automated. For example: taking a manual production process of welding has a certain inventory, process flow and cycle time attached to it. Now, in the case of an increase in demand and the want of increasing the weld pool consistency, product parity and overall aesthetic quality, an industrial welding robot is the right choice. To start the analysis, the speed of a manual process is generally known and is based on batch size and the average time taken along with the amount of manpower utilized to complete one product. Once this data is studied, the new speed requirement of the process is based on process deviation (example: opting for automated MIG welding instead of manual TIG welding), order completion time and QC passing time. That is, if our manual welding process took 30 minutes, based on the budget and requirement the cycle-time (and hence the speed) of the robotic welding cell can be decided to be in the range of 5 minutes to 29 minutes. Once the ideal speed is calculated, the acceleration and torque graphs along with space requirements are also studied before pin-pointing the speed. In our example, as welding is a heat treatment and requires a certain amount of curing time, a 5 minute cycle time even if possible should not be opted for.</span></span></li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Required Gripper Speed</span></span></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Once the speed of the robot for the task is decided, it is also essential to decide the gripper speed. The gripper speed is completely dependent on the product the robot is working with and the application. That is, for example: if the task is a simple pick and place, the sensitivity of the product coming in contact with the gripper is important; an iron piece can be transported and gripper at high speeds and high gripping force but at the same time a thin&nbsp;glass film cannot be subjected to high speeds as the gripping force cannot exceed a certain threshold.</span></span></li>\n</ul>','','1','4.1.1',NULL,'1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4853,180,'4.1.2 Time delay caused by a robotic system','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">There is a difference between an ideal cycle time calculated based on the decided speeds of a robot in a robotic cell; this difference occurs because of different time lags/delays termed as latency in robotic systems. This latency is spread throughout many parts of the cell; they are elaborated below:</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Actuator Lag: Time required by the end-of-arm actuator</span></span></p>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">In all actuators, the time difference between when the actuator is commanded to move and when the movement is completed always exists. For example, a pneumatic suction gripper may take 50-500 ms to generate a vacuum depending on the size of the suction cup. A mechanical finger-like gripper may take 10-500 ms depending on the load it carries.&nbsp;</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Mechanical Lag: Movement of the robot</span></span></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">For any robot, there always exists a time difference between when the robot is commanded to move and when the movement is completed. This time is also due to the calculations required by the robot controller to execute inverse kinematics. This time delay, which can also be named mechanical time lag, depends on the design and size of the robot but generally lies in the range of 12-45ms in terms of response time and 10-150ms in terms of tracking time. If large robots with heavy payloads are used, it may even take several seconds to finish a given pick and place task.</span></span></li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Computational Lag: PLC compute time</span></span></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">As explained above, the other factor that adds a delay to the task being carried out by the robot is the PLC compute delay. This is the time required for executing the logic inside the CPU of a PLC. This latency is very much dependent on the resources available within the CPU of the PLC and how complex the calculation is, but for an average task, this lag is in the range of 50&micro;s-250&micro;s for motion control logic, 100ms-1000ms for vision control logic and 1ms-500ms for process control logic.</span></span></li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Sensing and communication delay</span></span></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">This delay is the difference between the time instance the sensor senses a physical quantity (e.g. temperature or pressure) and the instance when the digitized data reaches PLC. This is also known as transport or communication delay. These delays are directly proportional to the distance of communication and the type of communication protocol being used; it ranges from a few microseconds to a few seconds.</span></span></li>\n</ul>\n\n<p>&nbsp;</p>','','1','4.1.2',NULL,'1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4854,180,'4.1.3 Matching required and achievable time','','single','html','<ul>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Once the speed required is obtained and all the factors adding in on the time delays and latencies are also analysed as mentioned above, the matching of time required and time achievable becomes an important part. The fundamental here is to check on available hardware and software which help us bridge the gap OR reduce the latencies as much as possible to obtain the desired cycle time of the process.</span></span></li>\n</ul>\n\n<p>&nbsp;</p>','','1','4.1.3',NULL,'1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4855,180,'<p>4.2 Applicable standards</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">The second factor that is considered while selecting robots are the applicable standards based on the industry where they are used. Mentioned below are four such industries and the standards required there:</span></span></p>\n\n<p>&nbsp;</p>','','1','4.2','Applicable standards','1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4856,180,'4.2.1 Food and Pharmaceutical','','single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300628521630062852.png\" style=\"height:149px; width:150px\" /></p>\n\n<p>&nbsp;</p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">In these industries, cleanliness and hygiene standards are of top priority and have to be served. Along with cleanliness, other requirements are corrosion resistance, water and humidity resistance (IP rating), purged to prevent water entry and damage, covers with protections from condensation, etc.&nbsp;</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">The detailed standards for use of industrial robots in food and pharma can be obtained from &quot;3-A Sanitary Standard 3-A 103-00, Robot-based Automation Systems&quot;</span></span></li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><span style=\"color:#000000\"><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300528521630052852.gif\">Click here to visualize the topic better</a></span></strong></span></p>','','1','4.2.1',NULL,'1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4857,180,'4.2.2 Defence','','single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300529101630052910.png\" style=\"height:240px; width:237px\" /></p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">In the defence sector, robots are used for multiple tasks, from mine detection to transportation. The likes of UAVs and drones are the best example. The standards defined for the defence sector is termed as&nbsp; &ldquo;military-grade&#39;&#39; and differs from country to country and contains requirements for explosion resistance, maintenance requirements, hours and distance covered, etc.&nbsp;</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><span style=\"color:#000000\"><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300529591630052959.gif\">Click here to visualize the topic better</a></span></strong></span></p>\n\n<p>&nbsp;</p>','','1','4.2.2',NULL,'1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4858,180,'4.2.3 Warehouse and Packaging:','','single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300530341630053034.png\" style=\"height:336px; width:300px\" /></p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">In this specific sector, more mobile robots are used than static ones; the likes of Automated Guided Vehicles (AGVs), Industrial Mobile Robots (IMR), and task-specific robots.&nbsp; There are existing standards for these types of robots: </span><span style=\"color:#333333\">for AGVs, the U.S. standard B56.5 and the International Organization for Standardization (ISO) 3691-4; and for industrial robots and robot systems, R15.06 in the U.S., the national adoption of ISO 10218-1,2. The new and updated standard, the R15.08 standard covers such robots and their upgraded versions in detail.</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><span style=\"color:#333333\"><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300531181630053117.gif\">Click here to visualize the topic better</a></span></strong></span></p>','','1','4.2.3',NULL,'1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56');
INSERT INTO `tbl_course_question` VALUES (4859,180,'<p>4.3 IP protection classes</p>','','single','html','<p><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300532651630053265.png\" style=\"height:86px; width:100px\" /></span></p>\n\n<p>&nbsp;</p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Depending upon the type of task being performed, be it a delicate assembly process of PLC components or a MIG welding process; robots are classified and characterized based on indoor OR outdoor units. These units again are sub-classified based on the standards, types, requirement classes, etc. One such class is the IP Protection classes based on the process being automated.</span></span></li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p><span style=\"font-size:14px\"><span style=\"color:#000000\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300533001630053300.jpeg\" style=\"height:338px; width:600px\" /></span></span></p>\n\n<p>&nbsp;</p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Ingress Protection (IP) is a classification scale used to specify the degree of protection offered by an industrial robot against the intrusion of objects, dust or water. </span><span style=\"color:#333333\">Robot IP ratings are composed of two numbers. The first number shows the level of a robot&rsquo;s protection against solid particles, while the second number signifies a robot&rsquo;s level of protection from water. For example, the rating IP67 is common amongst </span><a href=\"https://robotsdoneright.com/applications/foundry-robots.html\" style=\"text-decoration:none\"><span style=\"color:#000306\">foundry robots</span></a><span style=\"color:#333333\"> and cleanroom robots. This means a robot is dust resistant and can withstand temporary immersion in water. These protection levels are important for both types of robots since foundries often involve operation in dirty, grimy, dusty environments. While </span><a href=\"https://robotsdoneright.com/applications/cleanroom-robots.html\" style=\"text-decoration:none\"><span style=\"color:#000306\">cleanroom robots</span></a><span style=\"color:#333333\"> need to be dust-tight and somewhat waterproof to prevent any contamination from spreading from the robot to the production line and to meet the stringent standards of cleanroom facilities. If there is no data available to determine a protection rating for one of the elements, then that digit is replaced with an &lsquo;X&rsquo; and if the robot has no protection, then a 0 is used. Below is a breakdown of solid level ratings followed by a breakdown of water level ratings.</span></span></li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#333333\">Solid Protection Level:</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<div>\n<table align=\"left\" border=\"1\" cellpadding=\"1\" cellspacing=\"1\" style=\"height:500px; width:800px\">\n	<thead>\n		<tr>\n			<th scope=\"col\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#000000\">Protection Level</span></span></div>\n			</th>\n			<th scope=\"col\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Meaning</span></span></div>\n			</th>\n		</tr>\n	</thead>\n	<tbody>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">0</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">No protection from solids.</span></span></div>\n			</td>\n		</tr>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">1</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Protection from large solid objects up to 50 mm in diameter.</span></span></div>\n			</td>\n		</tr>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">2</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Protection from medium objects up to 12.5 mm in diameter.</span></span></div>\n			</td>\n		</tr>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">3</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Protection from small objects up to 2.5 mm in diameter.</span></span></div>\n			</td>\n		</tr>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">4</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Protection from small objects up to 1 mm in diameter.</span></span></div>\n			</td>\n		</tr>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">5</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Dust Protected- The intrusion of dust is not entirely prevented, but it will not enter in sufficient enough quantities to interfere with the robot operation.</span></span></div>\n			</td>\n		</tr>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">6</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Dust Tight- The contact and intrusion of dust is completely protected against.</span></span></div>\n			</td>\n		</tr>\n	</tbody>\n</table>\n\n<p><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</span></p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\"><span style=\"color:#333333\">Water Protection Level:</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<table align=\"left\" border=\"1\" cellpadding=\"1\" cellspacing=\"1\" style=\"height:500px; width:800px\">\n	<thead>\n		<tr>\n			<th scope=\"col\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#000000\">Protection Level</span></span></div>\n			</th>\n			<th scope=\"col\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Meaning</span></span></div>\n			</th>\n		</tr>\n	</thead>\n	<tbody>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">0</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">No protection from any amount of water.</span></span></div>\n			</td>\n		</tr>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">1</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Protection from vertically dripping water.</span></span></div>\n			</td>\n		</tr>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">2</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Protection from inclined water drops from up to 15 degrees from vertical.</span></span></div>\n			</td>\n		</tr>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">3</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Protection from spraying water at any angle up to 60 degrees from vertical.</span></span></div>\n			</td>\n		</tr>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">4</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Protection from splashing water against the robot enclosure from all directions or angles.</span></span></div>\n			</td>\n		</tr>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">5</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Protection from water jets at any angle.</span></span></div>\n			</td>\n		</tr>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">6</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Protection from heavy jet spray from any angle.</span></span></div>\n			</td>\n		</tr>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">7</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Protection from temporary water immersion of up to 1 meter for up to 30 minutes.</span></span></div>\n			</td>\n		</tr>\n		<tr>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">8</span></div>\n			</td>\n			<td>\n			<div style=\"text-align:center\"><span style=\"font-size:28px\"><span style=\"color:#333333\">Complete protection of water submersion for an indefinite amount of time.</span></span></div>\n			</td>\n		</tr>\n	</tbody>\n</table>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\"><span style=\"color:#333333\">It is also important to understand when selecting a robot that not all parts of a robot will have the same IP rating. For example, some robots are IP54 rated for their body while their J3 arm and wrist are IP67. This means their body has limited dust intrusion and water spray protection, while their arm is dust-proof and can withstand some water immersion. </span></span></p>\n</div>','','1','4.3','IP protection classes','1','2','1','0','1','2022-08-01 17:02:56','2022-08-01 17:02:56'),(4860,180,'<p>4.4 Return On investment</p>','','single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300550021630055002.png\" style=\"height:251px; width:250px\" /></p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">When in any industry a process is to be automated, the entire planning starts with budget finalization and ends with the calculation of return on investment based on manpower reduction, cycle time reduction leading to electricity saved, products manufactured in a given amount of time and overheads saved. As, automation economy is of utmost importance for any industry, this factor is always given a lot of weightage.</span></span></p>','','1','4.4','Return On investment','1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4861,180,'4.4.1 Calculating the cost of installing a robot','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">In general, the cost of installing a robot isn&#39;t small, furthermore, it is never the case where only a robot is installed at a given location but also the peripherals that come along with it. These peripherals in themselves can vary from power units to large SPMs, conveyors and vision systems. </span><span style=\"color:#313131\">Next, one needs to consider variable costs, which include labour, energy, materials, ongoing maintenance and production supplies required to deploy a robot successfully. These costs vary depending on the type of manufacturing facility. The actual cost and ROI can only be calculated post the installation, as many factors like sudden breakdown/damage or unplanned downtimes are not considered during the initial calculations. Therefore, the cost of the robot, peripherals, SPMs, electricity, vision systems, manpower and inventory all add up to give the cost of installation and operation of a robotic cell.</span></span></p>','','1','4.4.1',NULL,'1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4862,180,'4.4.2 Additional Benefits of installing a Robot','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#313131\">The benefits of a robotic cell are not limited to the known ones like cycle time reduction, manpower reduction and cost reduction but at the same time, it reaps other benefits like saving space because a robotic cell can automate a task in a space which is generally less by 50% of the manual process because of its agile nature and compact design, increasing process efficiency and process flow in terms of material movement, increasing product consistency and quality like for example in a MIG welding cell where the weld pool remains consistent and the quality increases by removing the human error. Also, a batch process can be converted into a continuous process if required as a robotic cell can function continually without supervision 24/7. </span></span></p>','','1','4.4.2',NULL,'1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4863,180,'4.4.3 Calculating ROI','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300552371630055237.mp4\">Click here to visualize the topic better</a></strong></span></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\">\n	<p><span style=\"font-size:14px\">In many areas like chemical manufacturing and food-pharma where human intervention is not allowed, adding robots helps to save capital and in a way safeguard manpower from hazardous activities. In layman terms,ROI = Gains from Investment - Cost of Investment.</span></p>\n	</li>\n	<li style=\"list-style-type:disc; text-align:justify\">\n	<p><span style=\"font-size:14px\">The gains from investment are the reduced labour costs and additional earnings achieved&nbsp;due to an increase in&nbsp;productivity after the installation of a robotic cell.&nbsp;The cost of investment includes the one-time capital cost of purchasing and installing the robot, as well as the operating cost of the robotic system.</span></p>\n	</li>\n	<li style=\"list-style-type:disc; text-align:justify\">\n	<p><span style=\"font-size:14px\">The operational cost of the manual process is calculated by considering the following factors: number of workers per shift, number of shifts per day, total operators and annual compensation of the operators which gives us the total labour cost.</span></p>\n	</li>\n	<li style=\"list-style-type:disc; text-align:justify\">\n	<p><span style=\"font-size:14px\">The cost of investment comprises the following factors: total proposed robotic system price, number of robots in the system, number of robot operators per shift, number of shifts per day, total operators, the annual compensation of the operators and estimated electrical cost of the system which gives us the system cost.&nbsp;</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Taking a real-world robotic automation example that assumes the robot will work three&nbsp; shifts a day, five days a week, 50 weeks a year and neglecting electricity for ease of calculation:</span></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\">\n	<p><span style=\"font-size:14px\">Current labor costs: three operators per shift x 2 shifts = 6 operators x USD100,000/operator = USD 600,000/year</span></p>\n	</li>\n	<li style=\"list-style-type:disc; text-align:justify\">\n	<p><span style=\"font-size:14px\">Total system cost: USD 255,000 (robot cost of&nbsp; USD 85,000 x 3)</span></p>\n	</li>\n	<li style=\"list-style-type:disc; text-align:justify\">\n	<p><span style=\"font-size:14px\">Labor with robotic system: Estimate 25 percent of current labor costs, i.e., USD 600,000 x 25%&nbsp; = USD 150,000/year</span></p>\n	</li>\n	<li style=\"list-style-type:disc; text-align:justify\">\n	<p><span style=\"font-size:14px\">Labor savings: Current labor costs (USD 600,000) - labor for robotic system (USD 150,000) = USD 450,000</span></p>\n	</li>\n	<li style=\"list-style-type:disc; text-align:justify\">\n	<p><span style=\"font-size:14px\"><strong>ROI:</strong> USD 195,000 in the first year (USD 450,000 - USD 255,000)</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Example in INR (comparatively lesser&nbsp;labour costs and more number of labor per task):</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Current labor costs: Four operators per shift x 3 shifts = 12 operators x Rs.2,40,000/operator = Rs.28,80,000/year</span></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\">\n	<p><span style=\"font-size:14px\">Total system cost: Rs.18,00,000 (robot cost of Rs.6,00,000 x 3)</span></p>\n	</li>\n	<li style=\"list-style-type:disc; text-align:justify\">\n	<p><span style=\"font-size:14px\">Labor with robotic system: Estimate 25 percent of current labor costs, i.e., Rs.28,80,000 x 25%&nbsp; = Rs.7,20,000/year</span></p>\n	</li>\n	<li style=\"list-style-type:disc; text-align:justify\">\n	<p><span style=\"font-size:14px\">Labor savings: Current labor costs (Rs.28,80,000) - labor for robotic system (Rs.7,20,000) = Rs.21,60,000</span></p>\n	</li>\n	<li style=\"list-style-type:disc; text-align:justify\">\n	<p><span style=\"font-size:14px\"><strong>ROI:</strong> Rs.3,60,000 in the first year (Rs.21,60,000 - Rs.18,00,000)</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Hence, the cost of the robot system is paid for in just over a year. Those numbers look pretty good. In fact, the robot will last for many years (approximately 5-7 years). Therefore, once the initial robot cost is covered within the first year of installation, the payback dramatically increases. A good rule of thumb is to target no more than a two-year ROI, which will still drive five years of solid returns in a typical robot application. </span></p>\n\n<p>&nbsp;</p>','','1','4.4.3',NULL,'1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4864,180,'<p>4.5 Safety</p>','','single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300555351630055535.png\" style=\"height:247px; width:250px\" /></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Safety plays a crucial role while selecting an industrial robot. There are various safety standards applicable when installing industrial robots. In general, there are two broad classes of safety, namely, the safety of people operating the robots and the safety of robots. This section discusses some of them for awareness.</span></span></p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>','','1','4.5','Safety','1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4865,180,'4.5.1 Safety Standards and classes','','single','html','<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Mentioned below are some safety standards which are followed by all Original Equipment Manufacturers (OEM) as norms for robot manufacturing and operations.</span></span></li>\n</ul>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>R15.06</em>, Industrial Robots and Robot Systems - Safety Requirements. Provides safety requirements for industrial robot manufacture, remanufacture, and rebuild (Part 1); and robot system integration/installation (Part 2). R15.06 (ANSI/RIA R15.06-2012) is the U.S. National Adoption of ISO 10218-1,2:2011.</span></span>\n\n	<ul>\n		<li style=\"list-style-type:disc\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>RIA TR R15.306 (&quot;TR 306&quot;), Task-Based Risk Assessment Methodology</em>. Describes one method of risk assessment that would comply with the R15.06 requirements.</span></span></li>\n		<li style=\"list-style-type:disc\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>RIA TR R15.406 (&quot;TR 406&quot;), Safeguarding</em>. Explains how to design a system of safeguards to protect human workers in an industrial environment that also contains robot system(s). TR 406 contains greater detail than the 2012 R15.06 standard.</span></span></li>\n		<li style=\"list-style-type:disc\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>RIA TR R15.506 (&quot;TR 506&quot;), Applicability of R15.06-2012 for Existing Industrial Robot Applications</em>. The R15.06 standard is forward-looking; that is, its primary topic is the installation of an all-new robot system. TR 506 explains how to take the 2012 R15.06 standard into account for existing robot systems.</span></span></li>\n		<li style=\"list-style-type:disc\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>RIA TR R15.606 (&quot;TR 606&quot;), Collaborative Robot Safety</em>. Explains safety requirements specific to collaborative robots and robot systems, and is supplemental to the guidance in ANSI/RIA R15.06. TR 606 is the U.S. National Adoption of the ISO/TS 15066:2016.</span></span></li>\n		<li style=\"list-style-type:disc\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>RIA TR R15.806 (&quot;TR 806&quot;), Testing Methods for Power &amp; Force Limited Collaborative Applications</em>. Describes methods to test and verify that the forces exerted by a collaborative robot system remain within the allowable limits described in TR 606 (ISO/TS 15066).</span></span></li>\n	</ul>\n	</li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>B11.0, Safety of Machinery &ndash; General Requirements and Risk Assessment</em>. Applies to new, existing, modified, or rebuilt power-driven machines, not portable by hand while working, used to shape and/or form metal or other materials by cutting, dust-proof, Defence, pressure, electrical, or other processing techniques, or a combination of these processes. This can be a single machine, a machine tool, or a machine tool system(s), and may include industrial robot(s) and robot system(s). Also contains guidance on performing risk assessments.</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>B11.20, Safety Requirements for Integrated Manufacturing Systems</em>. Specifies the safety requirements for the design, construction, set-up, operation, and maintenance (including installation, dismantling and transport) of integrated manufacturing systems, which may include industrial robot(s) and robot system(s).</span></span></li>\n</ul>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>International Organization for Standardization (ISO)</em></span></span></p>\n\n<ul>\n	<li style=\"list-style-type:square; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>TC 299, Robotics</em>. Develops high quality standards for the safety of industrial robots and service robots to enable innovative robotic products to be brought onto the market. In addition, develops standards in fields like terminology, performance measurement and modularity.</span></span></li>\n	<li style=\"list-style-type:square; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>ISO 10218-1, Robots for industrial environments - Safety requirements - Part 1: Robots</em>. Specifies requirements and guidelines for the inherent safe design, protective measures, and information for use of industrial robots. It describes basic hazards associated with robots, and provides requirements to eliminate or adequately reduce the risks associated with these hazards.</span></span></li>\n	<li style=\"list-style-type:square; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>ISO 10218-2, Robots for industrial environments &ndash; Safety requirements &ndash; Part 2: Robot systems and system integration</em>. Specifies requirements and guidelines for the safe integration of an industrial robot into a complete robot system, which includes end-effectors and other related equipment. This document describes basic hazards associated with robot systems, and provides requirements to eliminate or adequately reduce the risks associated with these hazards.</span></span></li>\n	<li style=\"list-style-type:square; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">In the U.S., ISO 10218-1, has been Nationally Adopted as the single U.S. standard ANSI/RIA R15.06-2012 (see above for details).</span></span>\n	<ul>\n		<li style=\"list-style-type:circle\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Note: ISO 10218 does not apply to non-industrial robots although the safety principles established in ISO 10218 may be utilized for these other robots. Examples of non-industrial robot applications include, but are not limited to: undersea, military and space robots; tele-operated manipulators; prosthetics and other aids for the physically impaired; micro-robots (displacement &lt;1 mm); surgery or healthcare; and service or consumer products.</span></span></li>\n	</ul>\n	</li>\n	<li style=\"list-style-type:square; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>ISO/TS 15066, Collaborative Robot Safety</em>. Provides important information about how to implement a collaborative robot system in a manner that maintains safety for the human operator. In the U.S., this ISO TS has been Nationally Adopted as TR 606.</span></span></li>\n	<li style=\"list-style-type:square; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>ISO/TR 20218-1, Safety Design for End-effectors</em>. Describes how an industrial robot system should handle and manage end-effectors (end-of-arm tooling or EOAT) to maintain human safety, in either a collaborative or non-collaborative industrial environment.</span></span></li>\n	<li style=\"list-style-type:square; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\"><em>ISO/TR 20218-2, Safety Design for Manual Load/ Unload Stations</em>. Describes how to design a Manual Load/ Unload Station (MLUS) that will be safe and effective for the human worker to use.</span></span></li>\n</ul>','','1','4.5.1',NULL,'1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4866,180,'4.5.2 Earthing considerations','','single','html','<ul>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#212529\">Earthing is the first step towards electrical safety. The body of electrical equipment, appliances etc is to be connected to the earth for safe discharge of electric current, in the event of any leakage, fault etc. The earth is considered to be zero potential. Earthing is done to provide safety to users from electric shock.</span></span></li>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#333333\">There are various types of earthing like plate earthing, pipe earthing and strip earthing. Nowadays in robots, <strong>chemical earthing</strong> is preferred.</span></span></li>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#333333\">Chemical earthing is considered to be the most effective solution for consistent and permanent earthing. They are maintenance-free and have minimum fluctuations and are eco friendly. They are highly reliable for the safety of human life and have more surface area of conduction.</span></span></li>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#333333\">international standard IEC 60364 distinguishes three families of earthing arrangements, using the two-letter codes TN, TT and IT The first letter indicates the connection between the earth and the power supply equipment. ( transformer or generator)</span></span></li>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#333333\">T - Indicates one or more points of the supply are directly earthed ( For example, earthed neutral of the transformer ) I - Indicates either that the supply system is not earthed at all, or that the earthing includes a deliberately inserted impedance, the purpose of which is to limit the fault current. The second letter indicates the connection between the earth and the electrical device being supplied:</span></span></li>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#333333\">T - All exposed conductive metal-work is directly connected to the earth, independent of any other earth connection in the supply system.</span></span></li>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#333333\">N - All exposed conductive metal-work is connected directly to an earthed supply conductor provided by an Electric supply.</span></span></li>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#333333\"><strong>The third and fourth letters</strong> indicate the arrangement of the earthed supply conductor system.</span></span></li>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#333333\">S - Neutral and Earth conductor systems are quite separate</span></span></li>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#333333\">C - Neutral and Earth are combined into a single conductor</span></span></li>\n</ul>','','1','4.5.2',NULL,'1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4867,180,'<p>4.6 Payload Conditions</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">For any robot taken into consideration, the end of arm tooling in most cases does bear the load of the product, for this purpose, the type of robot and the end tool are selected post considering the payload. It is defined as the<span style=\"color:#666666\"> </span>weight the robot can lift and includes the weight of the End of Arm Tooling (EOAT) and the weight of the product being picked. To analyse this factor, we look at it from two perspectives:</span></span></p>\n\n<p>&nbsp;</p>','','1','4.6','Payload Conditions','1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4868,180,'4.6.1 Required load capacity based on the application','','single','html','<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">As a result of the influence of inertia, just selecting a robot with a payload rating higher than your End of Arm Tooling and the product weight is not enough as it is possible for a robot to have an acceptable payload but the inertia of the End of Arm Tooling and product is too high. This can slow the robot or cause the robot to be unable to run. <strong>For example:</strong></span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Assume a&nbsp; 6-DoF 6-joint robot has a payload rating of 7 kg and the EOAT will be a 0.1 m cube that weighs 5 kg.</span></span></li>\n	<li style=\"list-style-type:disc\">\n	<h3 style=\"margin-left:96px; text-align:justify\"><span style=\"font-size:48px\"><span style=\"color:#000000\">Acceptable Load: When the weight is mounted directly to the robot face plate the centre of gravity is 0.05 m and the inertia on the robot is 10-15% of the maximum allowed on robot axis 4 through 6 which is acceptable load</span>.</span></h3>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<h3 style=\"margin-left:96px; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">.</span><img src=\"https://lh5.googleusercontent.com/pWHlg7n6GDjPfXJRkxMlBTu2pytaxob3BQ9QrYiEJSapgiI-PIJ3-S9KGpEkteiWHmnAesbkCO52GSXmOcsFXlXObLiNeq0JMLAQt9ObYTdDw1q_i8y4TUJnQ3xQ9N5ZTO9Z2BA8=s0\" style=\"height:46px; width:100px\" /></span></h3>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<ul>\n	<li style=\"list-style-type:disc\">\n	<h3 style=\"margin-left:96px; text-align:justify\"><span style=\"font-size:48px\"><span style=\"color:#000000\">Borderline Load: When the weight is mounted .125 m away from the robot face plate the centre of gravity is 0.175 m and the inertia on the robot is 40-110% of the maximum allowed on robot axis 4 through 6. This load is borderline and would require the system Integrators and OEM to decide on a case by case basis.</span></span></h3>\n	</li>\n</ul>\n\n<p style=\"margin-left:144px; text-align:justify\">&nbsp;</p>\n\n<p style=\"margin-left:144px; text-align:justify\">&nbsp;</p>\n\n<h3 style=\"margin-left:96px; text-align:justify\"><span style=\"font-size:14px\"><img src=\"https://lh3.googleusercontent.com/SgJjFTtTEq8vb3MsSLSRMKJwRYQose3UI18I-38eLJs4Qz-3vqwqKfxEUeanOCo1-NOU8ekra6KdacOZhzSisgdu2YdPYvRgUaztesjSKoyiqnkqS2l5xDjYTv0MCI4NZjfsUdai=s0\" style=\"height:47px; width:100px\" /></span></h3>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<ul>\n	<li style=\"list-style-type:disc\">\n	<h3 style=\"margin-left:96px; text-align:justify\"><span style=\"font-size:48px\"><span style=\"color:#000000\">Unacceptable Load: When the weight is mounted .25 m away from the robot face plate the centre of gravity is 0.3 m and the inertia on the robot is 110-310% of the maximum allowed on robot axis 4 through 6 which is unacceptable.</span></span></h3>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><br />\n<span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<img src=\"https://lh4.googleusercontent.com/zvHSXDP2L1kDEpSW21m-GHO2z7oKnkWGlWokc4lqdebA6z_zkUt16a1F41H3DMSHlKunYfqYRFZMXiW-1FwOylPqkgi7fFl0a7nV8-gx1DKb7nYEgSfQlO1Ni1aU_S1jub3-D1OS=s0\" style=\"height:38px; width:100px\" /></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>','','1','4.6.1',NULL,'1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4869,180,'4.6.2 Load and Torque Specifications based on payload','','single','html','<p style=\"margin-left:48px; text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300644151630064415.jpeg\" style=\"height:703px; width:656px\" /></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">In its most b</span></span><span style=\"font-size:14px\"><span style=\"color:#000000\">asic form, the torque for a link (in the case of a robot) is calculated as a product of (mass x gravity x distance between the load and joint) i.e m*g*l. This basic case is only for horizontal systems which is seldom a case in industrial robots.</span></span></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Therefore, to calculate the torque based on the payload and link weight of the robot, the motor torque of each joint is considered. To do this, the total torque is divided into two parts, (1) holding torque and (2) torque required for motion.&nbsp;</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">The holding torque comes out to be individual terms of product of mass*distance from actuator (joint) being added to give the total torque. This is applicable for multiple actuators as well.</span></span></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>','','1','4.6.2',NULL,'1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4870,180,'<p>4.7 End of Arm Tooling</p>','','single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300567501630056750.png\" style=\"height:206px; width:150px\" /></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">They are the mounts at the end of the robot arm and are called end effectors. These end effectors vary based on the type of application like a weld-gun to pick and place grippers to suction cups. Listed below are the subtypes of the most commonly used grippers.</span></span></p>','','1','4.7','End of Arm Tooling','1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4871,180,'4.7.1 Gripper Types','','single','html','<p style=\"margin-left:48px; text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Pneumatic:</span></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Pneumatic grippers work by using a piston to open and close their fingers/jaws. When enough pressure is created in the piston chamber, the air will move the piston depending on its size</span></span></li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Electric Grippers:</span></span></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">For applications in which environmental/atmospheric conditions are controlled, pneumatic grippers are usually unsuitable. The main benefit here is that a compressor is not necessary. Compressors are inefficient, noisy, and expensive to run. The major advantage of electric grippers, such as those powered by stepper or servo motors, is that they can be controlled much more precisely.&nbsp; With the ability to adjust the jaw stroke to pick up many different types of parts, a single gripper may be used for delicate parts and strong parts.</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">It is possible to detect any quality outliers by using electric grippers, as they provide more precise positional feedback. Electric grippers usually include built-in position sensors, which eliminates the need for additional proximity or magnetic sensors.</span></span></li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Vacuum Grippers:&nbsp;</span></span></p>\n\n<ul>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">In addition to handling a wide range of applications, vacuum grippers are ideal for pick and place tasks in which uneven and even workpieces are present, such as cardboard, glass, sheet metal, and plastic. The vacuum cups (suction cups) are generally round in shape. Rubber or other elastic materials and soft plastics will be used to develop these cups. In some places, vacuum cups made of hard materials are used to handle soft material objects.</span></span></li>\n	<li style=\"list-style-type:disc; text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Two different devices are used in the suction cups for creating the vacuum, namely:</span></span></li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 1)Venturi</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2)Vacuum pump</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><span style=\"color:#000000\"><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300885811630088581.mp4\">Click here to visualize the topic better</a></span></strong></span></p>','','1','4.7.1',NULL,'1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4872,180,'<p>4.8 Mounting Conditions</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Based on the amount of space available for the robot and the type of application; a robot can be mounted in three different conditions:</span></span></p>\n\n<p>&nbsp;</p>','','1','4.8','Mounting Conditions','1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4873,180,'4.8.1 Floor','','single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300572361630057236.png\" style=\"height:279px; width:350px\" /></p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">This basic mounting is done for linked/ articulated robots in the presence of ample space as well as specific tasks like welding where the spark and weldpool depend on the torch angle. This mounting is also used in automobile industries for the assembly lines.</span></span></p>\n\n<p>&nbsp;</p>','','1','4.8.1',NULL,'1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4874,180,'4.8.2 Ceiling','','single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300573131630057313.png\" style=\"height:282px; width:170px\" /></p>\n\n<p>&nbsp;</p>\n\n<ul>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#000000\">This specific mount is used mainly for delta robots as the workplace like conveyors&nbsp; do not have space on the sides. Moreover, the ceiling mount helps access products in the middle of large conveyors.&nbsp;</span></span></li>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Care should be taken to insert special gravity dynamics in the robot program to handle the effects of gravity on the inverse system.</span></span></li>\n</ul>','','1','4.8.2',NULL,'1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4875,180,'4.8.3 Wall','','single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300573731630057373.png\" style=\"height:177px; width:300px\" /></p>\n\n<p>&nbsp;</p>\n\n<ul>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Whenever vertical space is limited, a wall-mounted Blastman robot is used. It is possible to have two robots working from opposite sides of the room or to have one robot by turning the workpiece accordingly.</span></span></li>\n	<li style=\"list-style-type: disc; text-align: justify;\"><span style=\"font-size:14px\"><span style=\"color:#000000\">In wall-mounted robots, the main movements are longitudinal and vertical.</span></span></li>\n</ul>','','1','4.8.3',NULL,'1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4876,180,'4.8.4 On a Mobile Robot','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Many times in warehousing and packaging, mobility becomes the governing factor without which the task becomes impossible. To facilitate such tasks, dark warehouses today are using mobile robots that have arms mounted on them to be able to both move and manage the packages. Moreover, such robots communicate with each other using various protocols to achieve higher efficiency.</span></span></p>','','1','4.8.4',NULL,'1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4877,180,'<p>5 Required Skills</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Once the type of robot has been narrowed down, it is essential to have basic skills to operate and understand the working of the robotic cell. To be thorough with this, a few basic skills required are mentioned below:</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><span style=\"color:#000000\"><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300575561630057556.mp4\">Click here to visualize the topic better</a></span></strong></span></p>','','1','5','Required Skills','1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4878,180,'<p>5.1 Basics of electronics and sensors</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300576851630057685.gif\">Click here to visualize the topic better</a></strong></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">A robotic cell is integrated with a lot of sensors which provide feedback as well as inputs to the robot to mark the start, stop, quality and flow of the process. To operate a robot with its peripherals it is important to know how your sensors work, what are the probable error causes, the concept they are running on, their specifications in terms of I/P, O/P and supply and how to integrate them with your robot.</span></span></p>','','1','5.1','Basics of electronics and sensors','1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4879,180,'<p>5.2 Basics of Programming</p>','','single','html','<p style=\"margin-left:48px; text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300577731630057772.gif\">Click here to visualize the topic better</a></strong></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">The reason behind opting for robots instead of special mechanisms and SPMs is to have high product variation on the line. For this, it is very important to know the robot programming language so as to understand the code written and make changes in case of a new product on the line. Also, an entire cell is integrated via PLC so it helps to know the basics of PLC programming as well.</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;</span></p>','','1','5.2','Basics of Programming','1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4880,180,'<p>5.3 Basics of motion control</p>','','single','html','<p style=\"margin-left:48px; text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300578621630057862.gif\">Click here to visualize the topic better</a></strong></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Robot is never in a static state while in operation; this makes it very important to understand how to control the motion of the robot and why it reacts to a certain input the way it does. For this, the basics of forward and inverse kinematics, the robot operating frame, the joint angles and speeds, the singularity of your robot and task specific motion controls like external axis synchronous with the robot. All of this helps to program and re-program the robot as and when required.</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;</span></p>','','1','5.3','Basics of motion control','1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4881,180,'<p>5.4 Basics of communication protocol</p>','','single','html','<p style=\"margin-left:48px; text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300579691630057969.gif\">Click here to visualize the topic better</a></strong></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Communication between robots and their peripherals, for example between the robot-HMI-sensors-PLC, occurs via standard communication protocols. All such protocols have their specific characteristics as well as restrictions. So firstly, the applicable characteristics have to be present, secondly all electronics should have the specific protocol required, also the latency and hardware requirements are all based on the communication protocol being selected. So it is important to have some basic knowledge of these protocols and their capabilities. Some famous protocols used in robotic cells are mentioned below:</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#444444\">EtherNet /IP (as standard), Modbus-TCP, DeviceNet, CANOpen, EtherCAT, Modbus-RTU and </span><span style=\"color:#000000\">RS-232.</span></span></p>','','1','5.4','Basics of communication protocol','1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4882,180,'<p>5.5 Basics of engineering drawing</p>','','single','html','<p style=\"margin-left:48px; text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><a href=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300580971630058097.gif\">Click here to visualize the topic better</a></strong></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span style=\"color:#000000\">Engineering drawing is a skill required to specifically cater to the manufacturing process that we are going to automate using the robot. For example, it is important to study and decipher a manufacturing drawing of a pipe which is welded on both ends to understand the weld radius, weld depth and weld pool diameter. Engineering drawings, then, become an essential skill when it comes to understanding the specifics of a task.</span></span></p>\n\n<p>&nbsp;</p>','','1','5.5','Basics of engineering drawing','1','2','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4883,180,'Which among these is not a characteristic taken into consideration when a robot is to be installed?','','multi','single','','','1','1',NULL,'4','1','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4884,180,'Which among these is a type of working environment that is used to understand the tasks done by manipulators:','','multi','single','','','1','1.1',NULL,'4','1','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4885,180,'What does the given diagram depict in terms of repeatability and accuracy?','','multi','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16300659011630065901.png','','1','1.2',NULL,'3','1','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4886,180,'For reliability as a factor of consideration, we use MTTF (Mean Time To Failure) to define and understand it. Given the theoretical method of calculating;  what does PHR stand for?','','multi','single','','','1','1.3',NULL,'1','1','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4887,180,'What is the limitation of using a custom mechanism like SPMs?','','multi','single','','','1','2.1',NULL,'3','3','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4888,180,'What is the full form of CNC?','','multi','single','','','1','2.2',NULL,'2','1','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4889,180,'CNC machines have a monopoly in which of the following:','','multi','single','','','1','2.2',NULL,'4','3','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4890,180,'What is the maximum number of axes available in CNC machines?','','multi','single','','','1','2.2',NULL,'2','1','1','1','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4891,180,'Which of the following is not a type of static robot:','','multi','single','','','1','3.1',NULL,'3','6','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4892,180,'“The two rotary joints linked to each other give a human arm type movement and the prismatic joint at the end effector is used for restricted motion in the Z-direction” is a defining characteristic of which static-type robot?','','multi','single','','','1','3.1',NULL,'4','1','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4893,180,'In a Delta robot, having all the arms connected to the end-effector results in:','','multi','single','','','1','3.1',NULL,'1','3','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4894,180,'There are three main environments for mobile robots that require significantly different design principles because they differ in the mechanism of motion, which of the following is not the correct answer:','','multi','single','','','1','3.2',NULL,'3','1','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4895,180,'Which of the following is a direct result of increased speed by automating a process:','','multi','single','','','1','4.1.1',NULL,'2','6','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4896,180,'“For any robot, there always exists a time difference between when the robot is commanded to move and when the movement is completed.” which part of time delay does this phrase point to?','','multi','single','','','1','4.1.2',NULL,'1','1','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4897,180,'Automated Guided Vehicles (AGVs) and Industrial Mobile Robots (IMRs) are generally used in which industry?','','multi','single','','','1','4.2.3',NULL,'3','3','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4898,180,'Robot IP ratings are composed of two numbers. The first number indicates a robot’s protection against _______, while the second number signifies a robot’s level of protection against _______. Fill in the blanks.','','multi','single','','','1','4.3',NULL,'4','1','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4899,180,'ROI = Gains from Investment - Cost of Investment. Where, the cost of investment includes: Select the odd one out','','multi','single','','','1','4.4.3',NULL,'4','3','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4900,180,'Nowadays in robots, _________ is preferred.','','multi','single','','','1','4.5.2',NULL,'4','1','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4901,180,'As a result of the influence of _______, just selecting a robot with a payload rating higher than your End of Arm Tooling and the product weight is not enough','','multi','single','','','1','4.6.1',NULL,'1','6','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4902,180,'This specific mounting is used mainly for delta robots as the workplace like conveyors do not have space on the sides. Moreover, this mount helps access products in the middle of large conveyors.','','multi','single','','','1','4.8.2',NULL,'2','4','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4903,180,'“The basics of forward and inverse kinematics, the robot operating frame, the joint angles and speeds, the singularity of your robot and task-specific controls like external axis synchronous with the robot” relates to which basic skill?','','multi','single','','','1','5.3',NULL,'3','3','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4904,180,'__________ is a skill required to specifically cater to the manufacturing process that we are going to automate using the robot.','','multi','single','','','1','5.5',NULL,'2','3','1','0','1','2022-08-01 17:02:57','2022-08-01 17:02:57'),(4905,181,'<p>1. Introduction</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660027126381660027126.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','1','1. Introduction','1','1','1','0','0','2022-08-09 12:09:23','2022-08-09 12:13:03'),(4906,181,'<p>2. Session Objectives &amp; Learning Outcomes</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660027438281660027438.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','2','2. Session Objectives & Learning Outcomes','1','3','1','0','0','2022-08-09 12:14:47','2022-08-09 12:14:47'),(4907,181,'<p>3. Session-1 Outline</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660027543961660027543.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','3','3. Session-1 Outline','1','1','1','0','0','2022-08-09 12:16:25','2022-08-09 12:16:25'),(4908,181,'<p>4. Session-2 Outline</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660027686701660027686.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','4','4. Session-2 Outline','1','1','1','0','0','2022-08-09 12:20:20','2022-08-09 12:29:00'),(4909,181,'<p>5. What is Avionics?</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660027933621660027933.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','5','5. What is Avionics?','1','3','1','0','0','2022-08-09 12:23:26','2022-08-09 12:23:26'),(4910,181,'<p>6. What is Avionics?</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660028110531660028110.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','6','5. What is Avionics?','1','3','1','0','0','2022-08-09 12:25:55','2022-08-09 12:25:55'),(4911,181,'<p>7. What is Avionics?</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660028220571660028220.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','7','7. What is Avionics?','1','3','1','0','0','2022-08-09 12:28:18','2022-08-09 12:28:18'),(4912,181,'<p>8. Avionics of a Rocket</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660028561211660028561.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','8','8. Avionics of a Rocket','1','3','1','0','0','2022-08-09 12:33:21','2022-08-09 12:33:21'),(4913,181,'<p>9. Avionics of a rocket</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660028715151660028715.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','9','9. Avionics of a rocket','1','3','1','0','0','2022-08-09 12:37:19','2022-08-09 12:37:19'),(4914,181,'<p>10. Avionics: Intuitive understanding</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660031712311660031712.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','10','10. Avionics: Intuitive understanding','1','3','1','0','0','2022-08-09 13:25:53','2022-08-09 13:26:18'),(4915,181,'<p>11. Avionics Subsystems</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660031830631660031830.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','11','11. Avionics Subsystems','1','1','1','0','0','2022-08-09 13:28:08','2022-08-09 13:28:08'),(4916,181,'<p>12. GNC: Example</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660031960171660031960.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','12','12. GNC: Example','1','3','1','0','0','2022-08-09 13:30:05','2022-08-09 13:30:05'),(4917,181,'<p>13. GNC: Definitions</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660032048761660032048.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','13','13. GNC: Definitions','1','2','1','0','0','2022-08-09 13:31:37','2022-08-09 13:31:37'),(4918,181,'<p>14. GNC: More Examples&nbsp;</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660032182251660032182.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','14','14. GNC: More Examples','1','3','1','0','0','2022-08-09 13:33:57','2022-08-09 13:33:57'),(4919,181,'<p>15. GNC: Practical Considerations</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660032282191660032282.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','15','15. GNC: Practical Considerations','1','3','1','0','0','2022-08-09 13:36:57','2022-08-09 13:36:57'),(4920,182,'<p>1. Recap</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660113524381660113524.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','1','1. Recap','1','1','1','0','0','2022-08-10 12:09:25','2022-08-10 12:09:25'),(4921,182,'<p>2. Session-2 Outline</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660113611551660113611.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','2','2. Session-2 Outline','1','3','1','0','0','2022-08-10 12:11:25','2022-08-10 12:11:25'),(4922,182,'<p>3. Sensors &amp; Computers</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660113819291660113819.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','3','3. Sensors & Computers','1','3','1','0','0','2022-08-10 12:15:32','2022-08-10 12:15:32'),(4923,182,'<p>4. Telemetry</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660114016941660114016.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','4','4. Telemetry','1','1','1','0','0','2022-08-10 12:26:58','2022-08-10 12:26:58'),(4924,182,'<p>5. Telemetry Requirements&nbsp;</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660114985791660114985.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','5','5. Telemetry Requirements','1','3','1','0','0','2022-08-10 12:33:50','2022-08-10 12:33:50'),(4925,182,'<p>6. Radio communication</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660116369981660116369.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','6','6. Radio communication','1','3','1','0','0','2022-08-10 13:05:20','2022-08-10 13:05:20'),(4926,182,'<p>7. INS: Gyroscope and Accelerometer</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660116995391660116995.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','7','7. INS: Gyroscope and Accelerometer','1','3','1','0','0','2022-08-10 13:07:35','2022-08-10 13:07:35'),(4927,182,'<p>8. INS: Drift</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660117097271660117097.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','8','8. INS: Drift','1','3','1','0','0','2022-08-10 13:08:52','2022-08-10 13:08:52'),(4928,182,'<p>9. INS: Sensor examples</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/16601171781001660117178.png\" style=\"height:450px; width:800px\" /></p>',NULL,'1','9','9. INS: Sensor examples','1','3','1','0','0','2022-08-10 13:10:08','2022-08-10 13:10:08'),(4929,182,'<p>10. GPS</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660117245381660117245.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','10','10. GPS','1','3','1','0','0','2022-08-10 13:11:16','2022-08-10 13:11:16'),(4930,182,'<p>11. Going Forward</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660117392871660117392.png\" style=\"height:450px; width:800px\" /></p>',NULL,'1','11','11. Going Forward','1','3','1','0','0','2022-08-10 13:13:50','2022-08-10 13:13:50'),(4932,183,'<p>Basics of Space Flight</p>',NULL,'single','html','<p><span style=\"font-size:14px\">Go through the Basics of Space Flight Course by NASA, available at the following link:</span></p>\r\n\r\n<p><span style=\"font-size:14px\"><a href=\"https://solarsystem.nasa.gov/basics/\" target=\"_blank\">Basics of Space Flight</a>&nbsp; &nbsp;</span></p>',NULL,'1','1','Basics of Space Flight','1','2','1','0','1','2022-08-11 11:29:56','2022-08-11 11:29:56'),(4933,183,'<p>Space Systems hardware</p>',NULL,'single','html','<p><span style=\"font-size:14px\">Please go through the section &quot;Space Hardware&quot; available at the following website:</span></p>\r\n\r\n<p><span style=\"font-size:14px\"><a href=\"http://www.braeunig.us/space/index.htm\" target=\"_blank\">Space Hardware</a>&nbsp;</span></p>',NULL,'1','2','Space System hardware','1','4','1','0','1','2022-08-11 11:33:24','2022-08-11 11:35:07'),(4934,183,'<p>Telemetry Game</p>',NULL,'single','html','<p>Play this Telemetry game available at the following link, and note the time taken to finish:</p>\r\n\r\n<p><a href=\"https://spaceplace.nasa.gov/dsn-game/en/\" target=\"_blank\">Uplink-Downlink: A Deep Space Network Game</a></p>',NULL,'1','3','Telemetry Game','1','3','1','0','1','2022-08-11 11:38:37','2022-08-11 11:38:37'),(4935,183,'<p>Guidance, Navigation and Control sensors</p>',NULL,'single','html','<p><span style=\"font-size:14px\">Please go through the GNC system. Note down the specifications of all sensors. Can you spot drift specification?&nbsp;</span></p>\r\n\r\n<p><span style=\"font-size:14px\"><a href=\"https://www.nasa.gov/smallsat-institute/sst-soa/guidance-navigation-and-control\" target=\"_blank\">Guidance, Navigation and Control subsystem</a></span></p>\r\n\r\n<p><span style=\"font-size:14px\">PDF of the same material is attached for offline reference. Please see the Resources section below.&nbsp;</span></p>',NULL,'1','4','GNC Sensors','1','3','1','0','1','2022-08-11 11:55:46','2022-08-11 11:55:46'),(4936,184,'<p>Avionics provide eyes and ears for successful flight of a rocket. Which of the following is not a sensor?&nbsp;&nbsp;</p>',NULL,'multi','single',NULL,NULL,'1','1','Sensor','3','1','1','0','1','2022-08-15 17:53:52','2022-08-15 17:53:52'),(4937,184,'<p>Please go through the help material attached.&nbsp;</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660566561601660566561.pdf',NULL,'1','1','Help resource','1','2','1','0','1','2022-08-15 17:59:21','2022-08-15 17:59:21'),(4938,184,'<p>Which of the follwoing is not a part of rocket telemetry?</p>',NULL,'multi','single',NULL,NULL,'1','1','Telemetry parts','4','1','1','0','1','2022-08-15 18:12:08','2022-08-15 18:12:08'),(4939,184,'<p>Please see the attached help material for GNC subsystem.</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660568295221660568295.pdf',NULL,'1','2','GNC Subsystem help','1','2','1','0','1','2022-08-15 18:28:15','2022-08-15 18:28:15'),(4940,184,'<p>Which of the following is not a driving performance parameters while selecting a gyroscope or an accelerometer?</p>',NULL,'multi','single',NULL,NULL,'1','2','Gyroscope parameters','3','1','1','0','1','2022-08-15 18:38:47','2022-08-15 18:38:47'),(4941,184,'<p>What is the mode of communication for telemetry?</p>',NULL,'multi','single',NULL,NULL,'1','2','Communication mode','1','1','1','0','1','2022-08-15 19:05:35','2022-08-15 19:05:35'),(4942,184,'<p>Which of the following relates to determination of position of a rocket along its flight path?</p>',NULL,'multi','single',NULL,NULL,'1','2','GNC terminology','4','3','1','0','1','2022-08-15 19:29:27','2022-08-15 19:29:27'),(4943,184,'<p>What does &quot;Guidance&quot; mean in terms of rocket GNC subsystem?&nbsp;</p>',NULL,'multi','single',NULL,NULL,'1','2','GNC Terminology','3','1','1','0','1','2022-08-15 19:32:41','2022-08-15 19:32:41'),(4944,184,'<p>Please go through the help material</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660572518741660572518.pdf',NULL,'3','3','Help material','1','2','1','0','1','2022-08-15 19:38:38','2022-08-15 19:38:38'),(4945,184,'<p>It is possible for a GPS satellite to give coverage beyong earth&#39;s atmosphere: True or false?&nbsp;</p>',NULL,'multi','single',NULL,NULL,'1','3','GPS coverage','1','1','1','0','1','2022-08-15 22:26:40','2022-08-15 22:26:40'),(4946,184,'<p>Which subsystem is repsonsible for making sure the ground station gets timely data from space?&nbsp;</p>',NULL,'multi','single','',NULL,'1','3','System ID','3','4','1','0','1','2022-08-15 22:36:50','2022-08-16 15:12:46'),(4947,184,'<p>Why are multiple raction wheels employed?&nbsp;</p>',NULL,'multi','single',NULL,NULL,'1','2','Reaction wheel','4','3','1','0','1','2022-08-15 22:45:23','2022-08-15 22:45:23'),(4948,184,'<p>Movable antennas are not needed for Geostationary satellites: True or False?</p>',NULL,'multi','single',NULL,NULL,'1','3','Antenna tracking','2','4','1','0','1','2022-08-15 22:54:09','2022-08-15 22:54:09'),(4949,184,'<p>What is the cause of drift in a gyroscope?</p>',NULL,'multi','single',NULL,NULL,'1','3','Drift','4','1','1','0','1','2022-08-15 22:56:25','2022-08-15 22:56:25'),(4950,184,'<p>What are the components required to make a tracking antenna?</p>',NULL,'multi','single',NULL,NULL,'1','3','Antenna tracking','4','3','1','0','1','2022-08-15 23:25:15','2022-08-15 23:25:15'),(4951,184,'<p>A star tracker sensor is used for</p>',NULL,'multi','single',NULL,NULL,'1','2','Star tracker','2','1','1','0','1','2022-08-15 23:28:48','2022-08-15 23:28:48'),(4952,184,'<p>In the telemetry game you played, how was the feedback control achieved?</p>',NULL,'multi','single',NULL,NULL,'1','3','Game','2','4','1','0','1','2022-08-15 23:37:50','2022-08-15 23:37:50'),(4953,184,'<p>Why are atomic clocks popular amongst the latest technological innvations?&nbsp;</p>',NULL,'multi','single',NULL,NULL,'1','3','Atomic Clock','3','3','1','0','1','2022-08-15 23:43:11','2022-08-15 23:43:11'),(4954,185,'<p>1. Recap</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660645056471660645056.PNG\" style=\"height:519px; width:800px\" /></p>',NULL,'1','1','1. Recap','1','3','1','0','0','2022-08-16 15:48:35','2022-08-16 15:48:35'),(4955,185,'<p>2. Concept of closed-loop control</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660645471981660645471.JPG\" style=\"height:155px; width:509px\" /></p>',NULL,'1','2','2. Concept of closed-loop control','1','3','1','0','1','2022-08-16 15:55:18','2022-08-16 15:55:18'),(4956,185,'<p>4. Rocket as a rigid body and it&#39;s attitude control</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1660645583661660645583.JPG\" style=\"height:599px; width:800px\" /></p>\r\n\r\n<p style=\"text-align:center\">Source: ISRO</p>',NULL,'1','4','4. Rocket as a rigid body and it\'s attitude control','1','3','1','0','0','2022-08-16 15:57:17','2022-08-16 16:40:25'),(4957,185,'<p>3. Tracking Control of Antenna&nbsp;</p>',NULL,'single','html','<p>Control system elements for telemetry:</p>\r\n\r\n<ol>\r\n	<li>Sensing the movement of the satellite</li>\r\n	<li>Sensing the power of the received radio signal&nbsp;</li>\r\n	<li>Create command signal&nbsp;</li>\r\n	<li>Compare with actual signal received with command</li>\r\n	<li>Move the actuator to reduce the error.</li>\r\n</ol>',NULL,'1','3','3. Tracking Control of Antenna','1','3','1','0','0','2022-08-16 16:43:04','2022-08-16 16:43:04'),(4958,186,'<p>1. Introduction</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661139283211661139283.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','1','1. Introduction','1','2','1','0','0','2022-08-22 09:06:18','2022-08-22 09:06:18'),(4959,186,'<p>2. Module Objectives</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661139495531661139495.png\" style=\"height:450px; width:800px\" /></p>',NULL,'1','2','2. Module Objectives','1','3','1','0','0','2022-08-22 09:09:08','2022-08-22 09:09:08'),(4960,186,'<p>3. Session-1 Outline</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661139623541661139623.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','3','3. Session-1 Outline','1','2','1','0','0','2022-08-22 09:11:18','2022-08-22 09:11:18'),(4961,186,'<p>4. Session-2 Outline</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661139715951661139715.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','4','4. Session-2 Outline','1','2','1','0','0','2022-08-22 09:12:51','2022-08-22 09:12:51'),(4962,186,'<p>5. What is CFD?</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661139849841661139849.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','5','5. What is CFD?','1','3','1','0','0','2022-08-22 09:14:48','2022-08-22 09:14:48'),(4963,186,'<p>6. CFD: A historical perspective</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661140019411661140019.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','6','6. CFD: A historical perspective','1','2','1','0','0','2022-08-22 09:18:12','2022-08-22 09:18:12'),(4964,186,'<p>7. Some Applications</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661140138661661140138.png\" style=\"height:450px; width:800px\" /></p>',NULL,'1','7','7. Some Applications','1','2','1','0','0','2022-08-22 09:19:49','2022-08-22 09:19:49'),(4965,186,'<p>8. When to use CFD?</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661140305391661140305.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','8','8. When to use CFD?','1','3','1','0','0','2022-08-22 09:22:50','2022-08-22 09:22:50'),(4966,186,'<p>9. Basic Principles of CFD</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661140419231661140419.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','9','9. Basic Principles of CFD','1','2','1','0','0','2022-08-22 09:24:28','2022-08-22 09:24:28'),(4967,186,'<p>10. Basic Stages of CFD</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661140512781661140512.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','10','10. Basic Stages of CFD','1','2','1','0','0','2022-08-22 09:26:05','2022-08-22 09:26:05'),(4968,186,'<p>11. Governing Equations</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661140629631661140629.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','11','11. Governing Equations','1','2','1','0','0','2022-08-22 09:29:04','2022-08-22 09:29:04'),(4969,186,'<p>12. A word on Discretization</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661142231911661142231.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','12','12. A word on Discretization','1','2','1','0','0','2022-08-22 09:55:19','2022-08-22 09:55:19'),(4970,186,'<p>13. Space Discretization</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661142375681661142375.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','13','13. Space Discretization','1','3','1','0','0','2022-08-22 09:57:27','2022-08-22 09:57:27'),(4971,186,'<p>14. Time discretization</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661142519111661142519.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','14','14. Time discretization','1','3','1','0','0','2022-08-22 09:59:33','2022-08-22 09:59:33'),(4972,186,'<p>15. Next session</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661142653581661142653.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','15','15. Next session','1','2','1','0','0','2022-08-22 10:02:14','2022-08-22 10:02:14'),(4973,187,'<p>Windows Subsystem for Linux&nbsp;</p>',NULL,'single','html','<p>OpenFOAM can be installed on Linux operating system. If your pc has linux installed, you can use the following link (choose Ubuntu 20.04 focal):</p>\r\n\r\n<p><a href=\"https://openfoam.org/download/8-ubuntu/\" target=\"_blank\">Installation of OpenFOAM on Linux</a></p>\r\n\r\n<p>Windows Subsystem for Linux (WSL) can be used on machines that do not have Linux distribution. Activate WSL and install openFOAM as per the following links:</p>\r\n\r\n<ul>\r\n	<li><a href=\"https://docs.microsoft.com/en-gb/windows/wsl/install\" target=\"_blank\">Activate WSL</a></li>\r\n	<li><a href=\"https://openfoam.org/download/windows/\" target=\"_blank\">Install OpenFOAM on Windows</a></li>\r\n</ul>\r\n\r\n<p>Please&nbsp;complete all the steps in the installation including enabling graphics application using VcXsrv and User Configuration, in case of Windows based installation.</p>',NULL,'1','1','WSL','1','2','1','0','1','2022-08-23 15:52:00','2022-08-23 15:52:00'),(4974,187,'<p>Confirming the installation:</p>',NULL,'single','html','<p>Use the following link (section Getting started) to make a new directory and run an in built tutorial:</p>\r\n\r\n<p><a href=\"https://openfoam.org/download/8-ubuntu/\" target=\"_blank\">Link for commands to run a turotial</a>&nbsp;&nbsp;</p>\r\n\r\n<p>FOr convenience, the commands are listed below:</p>\r\n\r\n<p>cd $HOME<br />\r\nmkdir OpenFOAM<br />\r\ncd OpenFOAM<br />\r\nmkdir MyDirectory<br />\r\ncd MyDirectory<br />\r\nmkdir -p $FOAM_RUN<br />\r\ncd $FOAM_RUN<br />\r\ncp -r $FOAM_TUTORIALS/incompressible/simpleFoam/pitzDaily .<br />\r\ncd pitzDaily<br />\r\nblockMesh<br />\r\nsimpleFoam<br />\r\nparaFoam</p>',NULL,'1','2','Verify installation','1','4','1','0','1','2022-08-23 17:11:20','2022-08-23 17:11:20'),(4975,187,'<p>Read all the blogs and watch videos available at the links&nbsp;included below:</p>',NULL,'single','html','<p>What are Wind tunnels?<br />\r\nhttps://www.nasa.gov/audience/forstudents/k-4/stories/nasa-knows/what-are-wind-tunnels-k4.html</p>\r\n\r\n<p>Some Siemens blogs on CFD for Aerospace:<br />\r\nhttps://blogs.sw.siemens.com/simcenter/10-insights-from-rocket-cfd-for-a-mind-blowing-moonshot/<br />\r\nhttps://www.youtube.com/watch?v=ec_cbhUtqvA&amp;t=219s<br />\r\nhttps://blogs.sw.siemens.com/simcenter/aerospace-aerodynamics-cfd-simulation/<br />\r\nhttps://www.youtube.com/watch?v=rVcIm5ZW8kk</p>\r\n\r\n<p>Multiphase and combustion applications:<br />\r\nhttps://blogs.sw.siemens.com/simcenter/reacting-flows-and-combustion-computational-fluid-dynamics-cfd-with-simcenter-star-ccm/<br />\r\nhttps://blogs.sw.siemens.com/simcenter/aerospace-aerodynamics-cfd-simulation/</p>',NULL,'1','3','1','1','2','1','0','1','2022-08-23 17:23:25','2022-08-23 17:23:25'),(4976,188,'<p>1. Recap of session 1</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661506213671661506213.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','1','1. Recap of session 1','1','1','1','0','0','2022-08-26 15:00:58','2022-08-26 15:00:58'),(4977,188,'<p>2. Session Outline</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661506672781661506672.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','2','2. Session Outline','1','2','1','0','0','2022-08-26 15:08:32','2022-08-26 15:08:32'),(4978,188,'<p>3. Steps of CFD</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661506835641661506835.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','3','3. Steps of CFD','1','2','1','0','0','2022-08-26 15:11:59','2022-08-26 15:11:59'),(4979,188,'<p>4. Conservation of Momentum</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661507153761661507153.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','4','4. Conservation of Momentum','1','3','1','0','0','2022-08-26 15:16:20','2022-08-26 15:16:20'),(4980,188,'<p>5. Conservation of Momentum Equations</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661507290851661507290.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','5','5. Conservation of Momentum Equations','1','2','1','0','0','2022-08-26 15:18:44','2022-08-26 15:18:44'),(4981,188,'<p>6.&nbsp;Conservation of Energy Equations</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661507382971661507382.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','6','6. Conservation of Energy Equations','1','3','1','0','0','2022-08-26 15:20:23','2022-08-26 15:20:23'),(4982,188,'<p>7. Solving equations</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661507478961661507478.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','7','7. Solving equations','1','3','1','0','0','2022-08-26 15:21:52','2022-08-26 15:21:52'),(4983,188,'<p>8. Working with OpenFOAM</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661507565641661507565.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','8','8. Working with OpenFOAM','1','2','1','0','0','2022-08-26 15:23:46','2022-08-26 15:23:46'),(4984,188,'<p>9. Sample output for elbow exercise</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661507694601661507694.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','9','9. Sample output for elbow exercise','1','4','1','0','0','2022-08-26 15:25:37','2022-08-26 15:25:37'),(4985,188,'<p>10. Next session...</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661507845921661507845.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','10','10. Next session...','1','2','1','0','0','2022-08-26 15:27:49','2022-08-26 15:27:49'),(4986,189,'<p>1. Basics of a typical C Development Environment&nbsp;</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661744571461661744571.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','1','1. Basics of a typical C Development Environment ','1','3','1','0','0','2022-08-29 09:14:27','2022-08-29 09:14:27'),(4987,189,'<p>2. Watch a video</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661744926381661744926.mp4',NULL,'1','2','2. Autonomy of a simple C program','1','2','1','0','0','2022-08-29 09:18:46','2022-08-29 09:18:46'),(4988,189,'<p>3. Anatomy of a C program</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661745118441661745118.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','3','3. Anatomy of a C program','1','3','1','0','1','2022-08-29 09:22:25','2022-08-29 09:22:25'),(4989,189,'<p>4. Which phase comes after edit phase?</p>',NULL,'user_input','single','',NULL,'1','4','4. Which phase comes after edit phase?','1','1','1','0','1','2022-08-29 09:25:28','2022-08-29 10:30:32'),(4990,189,'<p>5. Statements in C</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661745449321661745449.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','5','5. Statements in C','1','3','1','0','0','2022-08-29 09:28:00','2022-08-29 09:28:00'),(4991,189,'<p>6. In the statement &quot;<span style=\"font-size:12px\">#define KMS_PER_MILE 1.609</span>&quot;, what is&nbsp;<span style=\"font-size:12px\"><strong>KMS_PER_MILE</strong>?</span></p>',NULL,'multi','single','',NULL,'1','6','6. Exercise 1','2','3','1','0','1','2022-08-29 09:31:16','2022-08-29 10:30:03'),(4992,189,'<p>7. Embedding Third party tools</p>',NULL,'single','html','<p>Engage students with&nbsp;third party websites. For example, click on the following and ask students to code something and ask its output later on.</p>\r\n\r\n<p><a href=\"https://www.w3schools.com/c/tryc.php?filename=demo_helloworld\" target=\"_blank\">Try coding yourself - Click here</a>&nbsp;&nbsp;&nbsp;</p>',NULL,'1','7','7. Embedding Third party tools','1','2','1','0','0','2022-08-29 09:38:13','2022-08-29 09:38:13'),(4993,189,'<p>8. Confirm if they coded themseleves</p>',NULL,'multi','html','<p>What is the output of the following code?</p>\r\n\r\n<p>#include &lt;stdio.h&gt;</p>\r\n\r\n<p>int main() {<br />\r\n&nbsp; printf(&quot;Hello there!&quot;);<br />\r\n&nbsp; return 0;<br />\r\n}</p>',NULL,'1','8','8. Confirm if they coded themseleves','4','4','1','0','1','2022-08-29 09:40:32','2022-08-29 10:29:06'),(4994,190,'<p>How many tokens does this code contain?&nbsp;&nbsp;<span style=\"font-size:12pt\"><span style=\"font-size:11.5pt\">printf(&quot;Hello, World! \\n&quot;);</span></span></p>',NULL,'multi','single','',NULL,'1','1','How many tokens?','4','4','1','0','1','2022-08-29 10:00:33','2022-08-29 10:44:39'),(4995,190,'<p>Which character is used for ending each statement of C program?</p>',NULL,'multi','single','',NULL,'1','1','End character','3','1','1','0','1','2022-08-29 10:02:28','2022-08-29 10:43:24'),(4997,189,'<p>9. The whole ppt</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661748197431661748197.pptx',NULL,'1','9','9. The whole ppt','1','2','1','0','0','2022-08-29 10:13:18','2022-08-29 10:13:18'),(4998,190,'<p>Please read the atatched document.&nbsp;&nbsp;</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1661749963321661749963.docx',NULL,'1','1','Introduction','1','2','1','0','1','2022-08-29 10:42:43','2022-08-29 10:42:43'),(5001,190,'<p>What does the keyword &quot;int&quot; stand for?</p>',NULL,'user_input','single',NULL,NULL,'1','1','Int keyword','1','2','1','0','1','2022-08-30 18:28:00','2022-08-30 18:28:00'),(5002,191,'<p>1. Basics of a typical C program</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662273403301662273403.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','1','1. Basics of a typical C program','1','3','1','0','0','2022-09-04 12:07:38','2022-09-04 12:07:38'),(5003,191,'<p>2. Anatomy of a C program</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662273505921662273505.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','2','2. Anatomy of a C program','1','3','1','0','0','2022-09-04 12:08:55','2022-09-04 12:08:55'),(5004,191,'<p>3. Function syntax</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662273590581662273590.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','3','3. Function syntax','1','3','1','0','0','2022-09-04 12:10:26','2022-09-04 12:10:26'),(5005,191,'<p>4. Print &quot;Hello World&quot;</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662273684391662273684.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','4','4. Print \"Hello World\"','1','3','1','0','0','2022-09-04 12:11:50','2022-09-04 12:11:50'),(5006,191,'<p>5. return (0)</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662273843321662273843.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','5','5. return (0)','1','3','1','0','0','2022-09-04 12:14:34','2022-09-04 12:14:34'),(5007,191,'<p>6. Type of Statements in C</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/16622739211001662273921.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','6','6. Type of Statements in C','1','3','1','0','0','2022-09-04 12:18:08','2022-09-04 12:18:08'),(5008,191,'<p>7. Preprocessor in action</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662274227251662274227.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','7','7. Preprocessor in action','1','3','1','0','0','2022-09-04 12:21:00','2022-09-04 12:21:00'),(5009,191,'<p>8. Include standard libraries</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662274313541662274313.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','8','8. Include standard libraries','1','3','1','0','0','2022-09-04 12:22:40','2022-09-04 12:22:40'),(5010,191,'<p>9. Function main</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662274403321662274403.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','9','9. Function main','1','3','1','0','0','2022-09-04 12:24:14','2022-09-04 12:24:14'),(5011,191,'<p>10. main Function Definition</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662274659571662274659.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','10','10. main Function Definition','1','3','1','0','0','2022-09-04 12:28:17','2022-09-04 12:28:17'),(5012,191,'<p>11. Comments</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662275870191662275870.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','11','11. Comments','1','3','1','0','0','2022-09-04 12:48:57','2022-09-04 12:48:57'),(5013,191,'<p>12. Comment&nbsp;types</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662275984551662275984.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','12','12. Comment types','1','3','1','0','0','2022-09-04 12:50:15','2022-09-04 12:50:15'),(5014,191,'<p>13. CPU and Memory</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662276104711662276104.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','13','13. CPU and Memory','1','3','1','0','0','2022-09-04 12:52:34','2022-09-04 12:52:34'),(5015,191,'<p>14. Everything is Binary</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662276208651662276208.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','14','14. Everything is Binary','1','3','1','0','0','2022-09-04 12:53:51','2022-09-04 12:53:51'),(5016,191,'<p>15. Variables &amp; Data Types</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662276316471662276316.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','15','15. Variables & Data Types','1','3','1','0','0','2022-09-04 12:56:25','2022-09-04 12:56:25'),(5017,191,'<p>16. Declaration of Variables</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662276438601662276438.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','16','16. Declaration of Variables','1','3','1','0','0','2022-09-04 12:57:52','2022-09-04 12:57:52'),(5018,191,'<p>17. Examples of declaration</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662276535931662276535.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','17','17. Examples of declaration','1','3','1','0','0','2022-09-04 12:59:33','2022-09-04 12:59:33'),(5019,191,'<p>18. Syntax display for declarations&nbsp;</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662276655991662276655.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','18','18. Syntax display for declarations','1','3','1','0','0','2022-09-04 13:01:30','2022-09-04 13:01:30'),(5020,191,'<p>19. Identifiers</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662276747681662276747.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','19','19. Identifiers','1','3','1','0','0','2022-09-04 13:02:54','2022-09-04 13:02:54'),(5021,191,'<p>20. Reserved words</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662276854371662276854.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','20','20. Reserved words','1','3','1','0','0','2022-09-04 13:04:54','2022-09-04 13:04:54'),(5022,191,'<p>21. User defined identifiers</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662278035831662278035.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','21','21. User defined identifiers','1','3','1','0','0','2022-09-04 13:24:58','2022-09-04 13:24:58'),(5023,191,'<p>22. C is case-sensitive</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662278650281662278650.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','22','22. C is case-sensitive','1','3','1','0','0','2022-09-04 13:34:40','2022-09-04 13:34:40'),(5024,191,'<p>23. Common practice for definition</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662278742501662278742.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','23','23. Common practice for definition','1','3','1','0','0','2022-09-04 13:36:41','2022-09-04 13:36:41'),(5025,191,'<p>24. Programme Style hints</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662292116541662292116.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','24','24. Programme Style hints','1','3','1','0','0','2022-09-04 17:19:32','2022-09-04 17:19:32'),(5026,191,'<p>25. Style hints continued</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662292252211662292252.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','25','25. Style hints continued','1','3','1','0','0','2022-09-04 17:21:20','2022-09-04 17:21:20'),(5027,191,'<p>26. Summary of variables &amp; identifiers</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662292385241662292385.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','26','26. Summary of variables & identifiers','1','3','1','0','0','2022-09-04 17:23:28','2022-09-04 17:23:28'),(5028,191,'<p>27. Variable declarations and data types</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662292551351662292551.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','27','27. Variable declarations and data types','1','3','1','0','0','2022-09-04 17:26:25','2022-09-04 17:26:25'),(5029,191,'<p>28. Basic Data Types</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662292676501662292676.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','29','28. Basic Data Types','1','3','1','0','0','2022-09-04 17:28:26','2022-09-04 17:28:26'),(5030,191,'<p>29. Declaration of Variables</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662292795141662292795.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','29','29. Declaration of Variables','1','3','1','0','0','2022-09-04 17:30:34','2022-09-04 17:30:34'),(5031,191,'<p>30. Executable statements</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662292892881662292892.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','30','30. Executable statements','1','3','1','0','0','2022-09-04 17:32:10','2022-09-04 17:32:10'),(5032,191,'<p>31. Assignment Statements</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662292981371662292981.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','31','31. Assignment Statements','1','3','1','0','0','2022-09-04 17:33:36','2022-09-04 17:33:36'),(5033,191,'<p>32. Example of assignment statement</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662293114181662293114.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','32','32. Example of assignment statement','1','3','1','0','0','2022-09-04 17:35:49','2022-09-04 17:35:49'),(5034,191,'<p>33. The &quot;=&quot; sign</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662293204791662293204.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','33','33. The \"=\" sign','1','3','1','0','0','2022-09-04 17:37:07','2022-09-04 17:37:07'),(5035,191,'<p>34. Assignment statement: sum and sqrt&nbsp;</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662293404201662293404.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','34','34. Assignment statement: sum and sqrt ','1','3','1','0','0','2022-09-04 17:40:53','2022-09-04 17:40:53'),(5036,191,'<p>35. x+y = sum is invalid</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662293503771662293503.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','35','35. x+y = sum is invalid','1','3','1','0','0','2022-09-04 17:42:14','2022-09-04 17:42:14'),(5037,191,'<p>36. Assignment statement coding example</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662293588101662293588.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','36','36. Assignment statement coding example','1','3','1','0','0','2022-09-04 17:44:10','2022-09-04 17:44:10'),(5038,191,'<p>37. Initializing variables</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662293706481662293706.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','37','37. Initializing variables','1','3','1','0','0','2022-09-04 17:45:52','2022-09-04 17:45:52'),(5039,191,'<p>38. Syntax display for return statement&nbsp;</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662293821291662293821.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','38','38. Syntax display for return statement','1','3','1','0','0','2022-09-04 17:47:24','2022-09-04 17:47:24'),(5040,191,'<p>39. Summary: Day 1&nbsp;</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662293946901662293946.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','39','39. Summary: Day 1 ','1','3','1','0','0','2022-09-04 17:49:23','2022-09-04 17:49:23'),(5041,191,'<p>40. Review quiz 1</p>',NULL,'multi','html','<p>What is the data type of &quot;income&quot; in the following code?</p>\r\n\r\n<p style=\"text-align:left\"><span style=\"font-size:12px\"><span style=\"color:black\"><strong>int main (void) {</strong></span></span></p>\r\n\r\n<p style=\"text-align:left\"><span style=\"font-size:12px\"><span style=\"color:black\"><strong>&nbsp; &nbsp;&nbsp; double income; /* ??? */</strong></span></span></p>\r\n\r\n<p style=\"text-align:left\"><span style=\"font-size:12px\"><span style=\"color:black\"><strong>&nbsp; &nbsp;&nbsp; income = 35500.00; /* ??? */</strong></span></span></p>\r\n\r\n<p style=\"text-align:left\"><span style=\"font-size:12px\">&nbsp; &nbsp;&nbsp; <span style=\"color:black\"><strong>printf(&ldquo;Old income is %f&rdquo;, income);</strong></span></span></p>\r\n\r\n<p style=\"text-align:left\"><span style=\"font-size:12px\"><span style=\"color:black\"><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; income = 39000.00; /* ??? */</strong></span></span></p>\r\n\r\n<p style=\"text-align:left\"><span style=\"font-size:12px\">&nbsp; &nbsp;&nbsp; <span style=\"color:black\"><strong>printf(&ldquo;After raise: %f&rdquo;, income);</strong></span></span></p>\r\n\r\n<p style=\"text-align:left\"><span style=\"font-size:12px\"><span style=\"color:black\"><strong>&nbsp; }</strong></span></span></p>\r\n\r\n<p>&nbsp;</p>',NULL,'1','40','40. Review quiz','2','3','1','0','0','2022-09-04 17:53:29','2022-09-04 17:56:15'),(5042,191,'<p>41. Review quiz 2</p>',NULL,'multi','html','<p>How many assignment statemetns appear in the code below?</p>\r\n\r\n<p style=\"text-align:left\"><span style=\"font-size:12px\"><span style=\"color:black\"><strong>int main (void) {</strong></span></span></p>\r\n\r\n<p style=\"text-align:left\"><span style=\"font-size:12px\"><span style=\"color:black\"><strong>&nbsp; &nbsp;&nbsp; double income; /* ??? */</strong></span></span></p>\r\n\r\n<p style=\"text-align:left\"><span style=\"font-size:12px\"><span style=\"color:black\"><strong>&nbsp; &nbsp;&nbsp; income = 35500.00; /* ??? */</strong></span></span></p>\r\n\r\n<p style=\"text-align:left\"><span style=\"font-size:12px\">&nbsp; &nbsp;&nbsp; <span style=\"color:black\"><strong>printf(&ldquo;Old income is %f&rdquo;, income);</strong></span></span></p>\r\n\r\n<p style=\"text-align:left\"><span style=\"font-size:12px\"><span style=\"color:black\"><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; income = 39000.00; /* ??? */</strong></span></span></p>\r\n\r\n<p style=\"text-align:left\"><span style=\"font-size:12px\">&nbsp; &nbsp;&nbsp; <span style=\"color:black\"><strong>printf(&ldquo;After raise: %f&rdquo;, income);</strong></span></span></p>\r\n\r\n<p style=\"text-align:left\"><span style=\"font-size:12px\"><span style=\"color:black\"><strong>&nbsp; }</strong></span></span></p>',NULL,'1','41','41. Review quiz 2','3','3','1','0','0','2022-09-04 17:57:47','2022-09-04 17:57:47'),(5043,194,'<p>1. Arithmetic Expressions</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665337473361665337473.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','1','1. Arithmetic Expressions','1','3','1','0','0','2022-10-09 23:15:58','2022-10-09 23:15:58'),(5044,194,'<p>2. Arithmetic Operators</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665337618931665337618.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','2','2. Arithmetic Operators','1','3','1','0','0','2022-10-09 23:20:04','2022-10-09 23:20:04'),(5045,194,'<p>3. Datatype of an expression</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665337861331665337861.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','3','3. Datatype of an expression','1','3','1','0','0','2022-10-09 23:21:41','2022-10-09 23:21:41'),(5046,194,'<p>4. Mixed-Type Assignment Statement</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665337970351665337970.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','4','4. Mixed-Type Assignment Statement','1','3','1','0','0','2022-10-09 23:29:28','2022-10-09 23:29:28'),(5047,194,'<p>5. Rules for evaluating multiple operators</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665338424431665338424.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','5','5. Rules for evaluating multiple operators','1','3','1','0','0','2022-10-09 23:32:12','2022-10-09 23:32:12'),(5048,194,'<p>6. Writing mathematical formulas</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665338580561665338580.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','6','6. Writing mathematical formulas','1','3','1','0','0','2022-10-09 23:33:40','2022-10-09 23:33:40'),(5049,194,'<p>7. I/O Operations and the library functions</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665338667761665338667.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','7','7. I/O Operations and the library functions','1','3','1','0','0','2022-10-09 23:35:37','2022-10-09 23:35:37'),(5050,194,'<p>8. Printf and scanf: Format string</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665339389371665339389.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','8','8. Printf and scanf: Format string','1','3','1','0','0','2022-10-09 23:47:02','2022-10-09 23:47:02'),(5051,194,'<p>9. Format string: need of &#39;&amp;&#39;</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665339482961665339482.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','9','9. Format string: need of \'&\'','1','3','1','0','0','2022-10-09 23:48:22','2022-10-09 23:48:22'),(5052,194,'<p>10. Multiple placeholders</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665339555101665339555.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','10','10. Multiple placeholders','1','3','1','0','0','2022-10-09 23:50:06','2022-10-09 23:50:06'),(5053,194,'<p>11. Example of placeholder</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665339670281665339670.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','11','11. Example of placeholder','1','3','1','0','0','2022-10-09 23:51:31','2022-10-09 23:51:31'),(5054,194,'<p>12. Syntax display for printf function block</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665339740931665339740.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','12','12. Syntax display for printf function block','1','3','1','0','0','2022-10-09 23:52:40','2022-10-09 23:52:40'),(5055,194,'<p>13. Syntax display for scanf</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665339803801665339803.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','13','13. Syntax display for scanf','1','3','1','0','0','2022-10-09 23:53:54','2022-10-09 23:53:54'),(5056,194,'<p>14. Example: tell the user to enter a number</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665339913201665339913.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','14','14. Example: tell the user to enter a number','1','3','1','0','0','2022-10-09 23:55:37','2022-10-09 23:55:37'),(5057,194,'<p>15. Displaying blank lines</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665339977241665339977.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','15','15. Displaying blank lines','1','3','1','0','0','2022-10-09 23:57:00','2022-10-09 23:57:00'),(5058,194,'<p>16. Printing two or more variables</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665340074281665340074.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','16','16. Printing two or more variables','1','3','1','0','0','2022-10-09 23:59:23','2022-10-09 23:59:23'),(5059,194,'<p>17. Basic rule</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665340206751665340206.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','17','17. Basic rule','1','1','1','0','0','2022-10-10 00:00:30','2022-10-10 00:00:30'),(5060,194,'<p>18. Inputting a value into a variable</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665340342121665340342.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','18','18. Inputting a value into a variable','1','3','1','0','0','2022-10-10 00:12:02','2022-10-10 00:12:02'),(5061,194,'<p>19. Reading two or more data items</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665341040401665341040.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','19','19. Reading two or more data items','1','3','1','0','0','2022-10-10 00:14:34','2022-10-10 00:14:34'),(5062,194,'<p>20. Basic rule of placeholders</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665341119771665341119.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','20','20. Basic rule of placeholders','1','1','1','0','0','2022-10-10 00:15:49','2022-10-10 00:15:49'),(5063,194,'<p>21. Formatting numbers in program output</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665341244461665341244.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','21','21. Formatting numbers in program output','1','3','1','0','0','2022-10-10 00:18:04','2022-10-10 00:18:04'),(5064,194,'<p>22. Formatting values of type double</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665341329651665341329.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','22','22. Formatting values of type double','1','3','1','0','0','2022-10-10 00:19:10','2022-10-10 00:19:10'),(5065,194,'<p>23. Interactive mode, batch mode and data type</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665341388971665341388.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','23','23. Interactive mode, batch mode and data type','1','3','1','0','0','2022-10-10 00:31:09','2022-10-10 00:31:09'),(5066,194,'<p>24. Batch version of miles to KM code</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665342316211665342316.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','24','24. Batch version of miles to KM code','1','3','1','0','0','2022-10-10 00:39:19','2022-10-10 00:39:19'),(5067,194,'<p>25. Input and Output controlled by a program</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665342613531665342613.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','25','25. Input and Output controlled by a program','1','3','1','0','0','2022-10-10 00:41:22','2022-10-10 00:41:22'),(5068,194,'<p>26. Miles to KM program&nbsp;with named files</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665342747661665342747.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','26','26. Miles to KM program with named files','1','3','1','0','0','2022-10-10 00:43:23','2022-10-10 00:43:23'),(5069,194,'<p>27. Common Programming errors</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665342846871665342846.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','27','27. Common Programming errors','1','3','1','0','0','2022-10-10 00:48:03','2022-10-10 00:48:03'),(5070,194,'<p>28. Compiler listing of syntax errors</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665343177981665343177.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','28','28. Compiler listing of syntax errors','1','3','1','0','0','2022-10-10 00:50:16','2022-10-10 00:50:16'),(5071,194,'<p>29. A program with a run time error</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665343460861665343460.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','29','29. A program with a run time error','1','3','1','0','0','2022-10-10 00:54:58','2022-10-10 00:54:58'),(5072,194,'<p>30. Incorrect result due to &amp; omission</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665343575671665343575.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','30','30. Incorrect result due to & omission','1','3','1','0','1','2022-10-10 00:56:33','2022-10-10 00:57:37'),(5073,194,'<p>31. Summary</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1665343802481665343802.PNG\" style=\"height:600px; width:800px\" /></p>',NULL,'1','31','31. Summary','1','3','1','0','0','2022-10-10 01:00:21','2022-10-10 01:00:21'),(5074,195,'<p>1. A need for Digital Training</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1669096187631669096187.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','1','1. A need for Digital Training','1','3','1','0','0','2022-11-22 11:29:19','2022-11-22 12:05:33'),(5075,195,'<p>2. Our Approach</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1669096794511669096794.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','2','2. Our Approach','1','3','1','0','0','2022-11-22 11:36:16','2022-11-22 12:05:12'),(5076,195,'<p>3. Awareness Program</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1669097216751669097216.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','3','3. Awareness Program','1','1','1','0','0','2022-11-22 11:38:06','2022-11-22 12:04:17'),(5077,195,'<p>4. Digital Champion</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1669097328951669097328.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','4','4. Digital Champion','1','3','1','0','0','2022-11-22 11:39:42','2022-11-22 12:03:54'),(5078,195,'<p>5. Subject Experts and Mentors</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1669097420621669097420.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','5','5. Subject Experts and Mentors','1','2','1','0','0','2022-11-22 11:41:50','2022-11-22 12:03:35'),(5079,195,'<p>6. Evaluation Process</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1669097544191669097544.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','6','6. Evaluation Process','1','3','1','0','0','2022-11-22 11:43:30','2022-11-22 11:58:16'),(5080,195,'<p>7. Thank You</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1669097653861669097653.PNG\" style=\"height:450px; width:800px\" /></p>',NULL,'1','7','7. Thank You','1','2','1','0','0','2022-11-22 11:44:39','2022-11-22 11:57:53'),(5081,195,'<p>8. The whole Presentation and Resources</p>',NULL,'single','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1669097861141669097861.pptx',NULL,'1','8','8. The whole Presentation and Resources','1','2','1','0','0','2022-11-22 11:47:42','2022-11-22 11:57:34'),(5082,195,'<p>9. In session Assessment: Please answer the following: Which category does C4I4 belong to?</p>',NULL,'multi','single','',NULL,'1','9','9. In session Assessment','3','1','1','0','1','2022-11-22 11:53:08','2022-11-22 11:57:14'),(5083,196,'<p>1. Outline</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662120379441662120379.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','1','1. Outline','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5084,196,'<p>2. How electricity is generated?</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662121901231662121901.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','2','2. How electricity is generated?','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5085,196,'<p>3. Types of Power Generation</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662122291921662122291.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','3','3. Types of Power Generation','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5086,196,'<p>3.1 Types of Power Generation</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662122425711662122425.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','3.1',NULL,'1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5087,196,'<p>4. Installed capacity</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662122533781662122533.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','4','4. Installed capacity','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5088,196,'<p>4.1 Installed capacity</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662122679311662122679.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','4.1',NULL,'1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5089,196,'<p>4.2 Calculating consumption</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662123509911662123509.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','4.2',NULL,'1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5090,196,'<p>5. Thermal Power Plant</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662123628691662123628.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','5','5. Thermal Power Plant','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5091,196,'<p>5.1 Thermal Power Plant Schematic</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662123936811662123936.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','5.1','5.1 Thermal Power Plant Schematic','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5092,196,'<p>5.2 Thermal Power Plant: Boiler</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662124046611662124046.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','5.2','5.2 Thermal Power Plant: Boiler','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5093,196,'<p>5.3 View of 800 MW unit</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662124207421662124207.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','5.3','5.3 View of 800 MW unit','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5094,196,'<p>5.4 View of 250 MW unit</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662124503321662124503.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','5.4','5.4 View of 250 MW unit','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5095,196,'<p>5.5 Cross Section of Double Extraction IST</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662125118231662125118.png\" style=\"height:337px; width:680px\" /></p>',NULL,'1','5.5','5.5 Cross Section of Double Extraction IST','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5096,196,'<p>5.6 Assembly of 107 MW Double Extraction IST</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662125210551662125210.png\" style=\"height:523px; width:652px\" /></p>',NULL,'1','5.6','5.6 Assembly of 107 MW Double Extraction IST','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5097,196,'<p>5.7 Gas Turbine working principle</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662127149731662127149.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','5.7','5.7 Gas Turbine working principle','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5098,196,'<p>5.8 Gas Turbine View</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662127255701662127255.jpg\" style=\"height:384px; width:680px\" /></p>',NULL,'1','5.8','5.8 Gas Turbine View','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5099,196,'<p>5.9 Gas Turbine Combined Cycle</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662127357321662127357.jpg\" style=\"height:370px; width:680px\" /></p>',NULL,'1','5.9','5.9 Gas Turbine Combined Cycle','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5100,196,'<p>5.11 Gas Turbine generator</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662127605831662127605.png\" style=\"height:342px; width:680px\" /></p>',NULL,'1','5.11','5.11 Gas Turbine generator','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5101,196,'<p>5.12 Gas Power Plant</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662377406601662377406.jpg\" style=\"height:442px; width:624px\" /></p>',NULL,'1','5.12','5.12 Gas Power Plant','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5102,196,'<p>6. Nuclear Power Plant</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662377604321662377604.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','6','6. Nuclear Power Plant','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5103,196,'<p>6.1 Boiling Water Nuclear Power Plant</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662377779831662377779.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','6.1','6.1 Boiling Water Nuclear Power Plant','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5104,196,'<p>6.2 Pressurized Water Nuclear Power Plant</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662378300561662378300.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','6.12','6.2 Pressurized Water Nuclear Power Plant','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5105,196,'<p>6.3 Heavy Water Nuclear Power Plant</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662378508371662378508.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','6.3','6.3 Heavy Water Nuclear Power Plant','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5106,196,'<p>6.4 Sodium Cooled Fast Breeder Nuclear Power Plant</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662378732451662378732.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','6.4','6.4 Sodium Cooled Fast Breeder Nuclear Power Plant','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5107,196,'<p>7. Hydro-Electric Power Plant Dam</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662378949201662378949.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','7','7. Hydro-Electric Power Plant Dam','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5108,196,'<p>7.1 View of Underground Hydro Electric Power Station&nbsp;</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662379172911662379172.jpg\" style=\"height:498px; width:680px\" /></p>',NULL,'1','7.1','7.1 View of Underground Hydro Electric Power Station','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5109,196,'<p>7.2 Working Principle of Hydro Electric Power Station</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662379331471662379331.png\" style=\"height:372px; width:680px\" /></p>',NULL,'1','7.2','7.2 Working Principle of Hydro Electric Power Station','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5110,196,'<p>7.3 Hydro Electric Turbine-Generator</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662379464731662379464.jpg\" style=\"height:534px; width:507px\" /></p>',NULL,'1','7.3','7.3 Hydro Electric Turbine-Generator','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5111,196,'<p>7.4 Hydro-Electric Power Plant working principle</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662379670851662379670.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','7.4','7.4 Hydro-Electric Power Plant working principle','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5112,196,'<p>7.5 View of Mini/Micro Hydro Power Plant</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662379854301662379854.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','7.5','7.5 View of Mini/Micro Hydro Power Plant','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5113,196,'<p>8. Tidal Power Plant</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662380053791662380053.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','8','8. Tidal Power Plant','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5114,196,'<p>8.1 View of Tidal Power Plant</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662380212301662380212.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','8.1','8.1 View of Tidal Power Plant','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5115,196,'<p>8.2 Statistics of Tidal Power</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662380320591662380320.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','8.2','8.2 Statistics of Tidal Power','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5116,196,'<p>9. Solar Power Plant</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662381487111662381487.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','9','9. Solar Power Plant','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5117,196,'<p>9.1 Solar Power Plant Working Principle</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662381577641662381577.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','9.1','9.1 Solar Power Plant Working Principle','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5118,196,'<p>10. Wind Power Plant</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662381826261662381826.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','10','10. Wind Power Plant','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5119,196,'<p>10.1 View of Wind Power Plant</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662381945991662381945.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','10.1','10.1 View of Wind Power Plant','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5120,196,'<p>11. Geo&nbsp;Thermal Power Plant</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662382199481662382199.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','11','11. Geo Thermal Power Plant','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5121,196,'<p>12. Thank&nbsp;You</p>',NULL,'single','html','<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1662382269191662382269.PNG\" style=\"height:382px; width:680px\" /></p>',NULL,'1','12','Thank You','1','2','1','0','1','2024-08-02 13:28:27','2024-08-02 13:28:27'),(5122,197,'<p>4 Fault Current Calculations</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">Faults are inevitable. They do occur. The large fault currents will pass through the equipment of power system. Hence, equipments are required to be short-time rated for these fault currents. The circuit breakers are to be designed to interrupt this high value of fault current. The relay setting is decided based on this fault current. Thus, fault current calculations and fault analysis are very essential not only for existing power system but also while designing and installing the equipments of power system. We will discuss different types of faults, their causes and their consequences. We will also learn the procedure of calculating these fault currents.</span></p>','','1','4','Fault Current Calculations','1','2','1','0','1','2024-08-02 13:33:03','2024-08-02 13:33:03'),(5123,197,'<p>4.1 Types Of Faults</p>','','single','html','<ol>\n	<li>\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\"><u>Symmetrical (Balanced) Faults:</u></span></p>\n	</li>\n</ol>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Symmetrical faults are faults involving all the three phases. Triple line or L-L-L (R-Y-B) and triple line to ground or L-L-L-g (R-Y-B-g) faults are symmetrical faults.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<ol start=\"2\">\n	<li>\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\"><u>Asymmetrical Unbalanced) Faults:</u></span></p>\n	</li>\n</ol>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The faults involving only one or two phases are categorized as asymmetrical faults. Single line to ground (L-g), double line to ground (L-L-g) and double line (L-L) faults are asymmetrical faults. For a three-phase power system, there can be 10 different faults. These are R-g, Y-g,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;B-g, R-Y, Y-B, B-R, R-Y-g, Y-B-g, B-R-g and R-Y-B. R-Y-B and R-Y-B-g are the same faults as far as magnitude of fault currents is concerned. When an R-Y-B-g fault occurs, the three phase currents (I<sub>Rf</sub>, I<sub>Yf</sub> and I<sub>Bf</sub>) may be ten times (or even more) the rated full load current (referred to as CMR or Continuous Maximum Rating) for which the electrical equipment is designed. However, the vector sum of these three currents is zero,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">i.e., I<sub>Rf</sub> + I<sub>Yf</sub> + I<sub>Bf</sub>= 0.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Hence no current passes through the ground (Figure 4.1).</span></p>\n\n<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16159626641615962664.jpg\" style=\"height:551px; width:800px\" /></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Therefore, with reference to fault current values, R-Y-B and R-Y-B-g faults are same. Hence their fault current analysis is identical. These faults, however, differ in case of problem of overvoltage as will be discussed later.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">If one conductor of transmission line gets broken, it is a kind of abnormality which is known as single phasing. This abnormality is also called &ldquo;open conductor&rdquo; or &ldquo;broken conductor&rdquo;.</span></p>','','1','4.1','Types Of Faults','1','2','1','0','1','2024-08-02 13:33:03','2024-08-02 13:33:03'),(5124,197,'<p>4.2 Causes Of Fault</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">A symmetrical triple line to ground fault can occur in case of switching ON a circuit breaker when the earthing switch is inadvertently kept ON. There is a kind of mechanical or electrical interlock between a circuit breaker of a transmission line and its earthing switch such that if the earthing switch is ON, circuit breaker cannot be switched ON. However, a protection engineer always has to consider a probability of failure of that interlocking arrangement and design a protection scheme to look after this fault.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Two phases can be bridged together either in generators or motors or in power transformers/ distribution transformers because of failure of insulation between phases, particularly when conductors of different phases are in the same slot of a stator of a generator or motor (this is very unusual as machines are rarely designed with two conductors of different phases in the same slot, hence probability of L-L fault in machines is only in overhung portion).</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">In the transmission lines, two phase wires can be shorted by kite strings or tree limbs. Moreover, in monsoon, the two phase conductors may swing (as they are suspended using suspension insulators which are free to move) due to winds and storms. While swinging two conductors (in horizontal line configuration) may come nearer, the dielectric strength of air might have dropped to as bad as </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">5 kV (peak)/ cm leading to flashover between two phase wires. This is the cause of L-L fault in lines.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Line to ground fault (L-g) is most common fault and can occur&nbsp; because of flashover across the line insulator due to lightning or switching overvoltage. Line to ground fault can occur in machines and transformers as well because of failure of insulation between a live conductor and earthed core.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">L-L-g fault can occur when L-L (say R-Y) and L-g (say Y-g) faults occur simultaneously.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">These are common and known causes. It is difficult to enlist the causes of faults as the faults occur in the field practices because of innumerable number of causes. Some of them can be listed as follows:</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Terminal block in an indoor control panel burnt due to overheating (there can be some cause of this overheating) and terminals shorted. The current juice from C. T. secondary to relay via terminal block did not reach relay from one pair of C. T.s in case of differential protection scheme due to this shorting of terminals. The current juice from another pair of C. T.s passed through protective relay, mal-operating the differential protection of generator. Class A trip occurred, there was a complete collapse of generator, GT, boiler, turbine and auxiliaries. There was no internal fault in the generator, but differential protection operated just because of mere terminal block.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:11.0pt\">There was a ground fault (L-g) in the cable connected between generator and GT. This L-g (say R-g) fault resulted in increase of V<sub>YB</sub> voltage by approximately </span><span class=\"math-tex\">\\(\\sqrt3\\)</span><span style=\"font-size:14px\">&nbsp;times rated line to line voltage. Because of this overvoltage, there was flashover between Y and B phases in generator bus P. T. panel which was part of differential protection zone. This resulted in class A trip of generator protection scheme.</span></p>\n	</li>\n</ul>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">In case of differential protection of transformer, measuring C. T. connections were wrongly wired to a differential relay from one side (say, secondary) of transformer C. T.s. (current transformers). This error was caught while primary injection testing of differential protection scheme, as differential current of certain milliamperes (as against zero) was measured under normal condition. If this problem would not have been caught, the differential protection would certainly have mal-operated due to heavy through fault (for which, it must be stable).</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">While making a circuit breaker OFF at an industry (66 kV consumer) for shut-down and scheduled maintenance, the biased differential protection of 132/66 kV power transformer&nbsp;&nbsp;&nbsp; (of state electricity transmission corporation)&nbsp; mal-operated leading to unwanted tripping of 132/66 kV transformer which was feeding many other consumers obviously. This problem was corrected by changing the setting of second harmonic stabilization.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">At one generating station, rotor earth-fault protection operated because of uneven d. c. current flow in parallel paths of carbon brushes due to small variation in brush dimensions.</span></p>\n	</li>\n</ul>','','1','4.2','Causes Of Fault','1','2','1','0','1','2024-08-02 13:33:03','2024-08-02 13:33:03'),(5125,197,'<p>4.3 Consequence Of Faults</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">The electrical equipment will be damaged due to faults, they will be required to be repaired and till then there will be an outage of the portion of the power system. The duty of protection and maintenance engineers is to reduce down time (time elapsed from clearance of the fault to resumption of the portion of the power system back to consumers) to minimum.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong>Damage due to faults are of two kinds:</strong></span></p>\n\n<p>&nbsp;</p>\n\n<ol>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\"><strong><u>Thermal Damage:</u> </strong>Fault current magnitude of 2 to 5 times (or even up to 8 times) the CMR (Continuous Maximum Rating) of the electrical equipment will heat the conductor (heat produced is proportional to I<sup>2</sup>Rt, I= Current passing through the conductor, R= Resistance of the conductor and t=time for which current is passed) and hence the insulation around it. When the current equal to CMR is passing, the conductor and insulation around it are also heated because of which the temperature rises. But at the same time heat is dissipated which is proportional to (T-T<sub>R</sub>)<sup>4</sup>. The heat is dissipated either naturally or using forced cooling means. As heat is dissipated at a higher rate than heat generated, temperature will stabilize at certain value. This equilibrium temperature reached for current equal to CMR will be withstood by the insulation of the equipment because the equipment is properly so designed. But when the fault current, as mentioned, passes through the equipment, the heat will generate at a faster rate and the equilibrium temperature reached in this case will be higher. This temperature will be certainly more than withstand limit of the insulation of the equipment, insulation will thermally break down, resulting into another fault if the remedial actions of the first fault are not taken. As thermal break down process is a function of time, the protective relaying has to act just before the time (leaving 10 to 20 % time margin) at which the insulation is going to be damaged. The time of operation of protective relay should, obviously, not more than this time at which insulation is damaged. But it should not be very earlier than this time. This is because the fault may be of transient nature and it may vanish before the withstand temperature limit of insulation is reached. In this case unnecessary operation of a relay and consequent tripping of a breaker will lead to unwanted outage and discomfort to consumers. Therefore, the function of protective gear is very critical. The relays used are time-delayed overcurrent relays in this case.</span></p>\n	</li>\n</ol>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<ol start=\"2\">\n	<li>\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><u>Electrodynamic Damage:</u></strong></span>&nbsp;<span style=\"font-size:14px\">When the fault current exceeds 8 to 10 times the full load rating of the equipment, this kind of damage can occur. This fault current flows through parallel conductors placed in transformers or in slots of generators, the repellent force is produced which would destruct the whole equipment structurally within few (8 to 10 or in certain cases little more) cycles (one cycle in 50 Hz a. c. system equals 20 milliseconds) and the equipment may burst. The structural or mechanical design of equipment is made such that these electrodynamic repellent forces produced by normal operating current of the equipment and those produced by smaller (2 to 8 times CMR) magnitudes of fault currents can be withstood by the equipment and it will not be electrodynamically damaged (the equipment can be damaged thermally as discussed, but this happens after delay in time, for which protective scheme is there). But for large short-circuit currents the devastating damage envisaged is to be avoided by using instantaneous overcurrent relay which operates within 1 to 3 cycles.</span></p>\n	</li>\n</ol>\n\n<p style=\"margin-left:72px; text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">It would be worthwhile to have an idea of probability of incidence of faults on the different elements of a power system (Table 4.1).</span><span style=\"font-size:11pt\">&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span></p>\n\n<table align=\"center\" border=\"1\" cellspacing=\"0\" dir=\"ltr\" style=\"width:800px\">\n	<caption>\n	<div><span style=\"font-size:28px\">Table 4.1: Fault Statistics</span></div>\n	</caption>\n	<thead>\n		<tr>\n			<th scope=\"col\" style=\"border-color:black; vertical-align:top; width:235px\">\n			<div><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Element</span></div>\n			</th>\n			<th scope=\"col\" style=\"vertical-align:top; width:191px\">\n			<div><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp; Percentage of Faults</span></div>\n			</th>\n		</tr>\n	</thead>\n	<tbody>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:235px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp; Overhead Transmission Lines</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:191px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 45-55</span></div>\n			</td>\n		</tr>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:235px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp; Underground Cables</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:191px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 10-12</span></div>\n			</td>\n		</tr>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:235px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp; Switchgears</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:191px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 10-15</span></div>\n			</td>\n		</tr>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:235px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp; Power Transformers</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:191px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 10-15</span></div>\n			</td>\n		</tr>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:235px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp; C. T. and P. T.</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:191px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;1-3</span></div>\n			</td>\n		</tr>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:235px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp; Control Circuits</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:191px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2-4</span></div>\n			</td>\n		</tr>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:235px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp; Miscellaneous</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:191px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 7-10</span></div>\n			</td>\n		</tr>\n	</tbody>\n</table>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Faults occurring in overhead transmission lines are sharing the highest percentage. Therefore, the statistics of different kinds of faults occurring in an overhead transmission line will be, really, interesting (Table 4.2).</span><span style=\"font-size:11pt\">&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span></p>\n\n<table align=\"center\" border=\"1\" cellspacing=\"0\" dir=\"ltr\" style=\"width:800px\">\n	<caption>\n	<div><span style=\"font-size:28px\">Table 4.2: Statistics of Faults occurring in Overhead Lines</span></div>\n	</caption>\n	<thead>\n		<tr>\n			<th scope=\"col\" style=\"border-color:black; vertical-align:top; width:169px\">\n			<div><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;Type of Fault</span></div>\n			</th>\n			<th scope=\"col\" style=\"vertical-align:top; width:219px\">\n			<div><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Percentage Occurrence</span></div>\n			</th>\n		</tr>\n	</thead>\n	<tbody>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:169px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; L-g Faults</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:219px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;80-90</span></div>\n			</td>\n		</tr>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:169px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; L-L Faults</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:219px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 10-12</span></div>\n			</td>\n		</tr>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:169px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; L-L-g Faults</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:219px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 3-5</span></div>\n			</td>\n		</tr>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:169px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;L-L-L-g Faults</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:219px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 0.1-0.2</span></div>\n			</td>\n		</tr>\n	</tbody>\n</table>\n\n<p style=\"text-align:justify\"><span style=\"font-size:11pt\"><span style=\"color:white\">It is very clear tha<span style=\"font-size:14px\">t </span></span><span style=\"font-size:14px\">Line to Ground faults are the most commonly occurring faults.</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">It is very necessary to learn that 85% of faults listed in Table 4.2 are transient in nature. For example, the flashover occurs across the line insulator due to lightning or switching overvoltage. The overvoltage due to any of these two causes decays to normal voltage within milliseconds (maximum 5 milliseconds). However, the arc due to flashover maintains up to about 0.3 second. At the end of this time, the air around insulator regains its dielectric strength and fault is automatically rectified or self-healed. This is one kind of transient fault. In this case, either the relay should not operate, and breaker should not trip up to 0.3 second or if tripping occurs, circuit breaker is automatically reclosed using auto-recloser relays.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">For the remaining 15% sustained faults, maintenance staff have to rectify the fault. (Obviously first locate the fault in a long transmission line which was time consuming as one has to patrol on the route of line to locate the fault. With the advent of numerical relays, this task has been made very easy as the relay indicates the location very accurately). </span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">To summarize, it can be stated that the critical clearing time of the faults by protective relays is very important. Later operation damages the electrical equipment and too earlier operation unnecessarily trip the breaker which may result in power system instability. Hence, tripping a breaker unnecessarily is many a times as equally harmful as not tripping a breaker in case of fault. Therefore, the role of protection engineer is very crucial. </span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Other important consequences of occurrence of faults are:</span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Because of damage or destruction of electrical equipment, they will be required to be repaired which results in long down time before portion of power system is resumed back to service.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Voltage drops.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Loss of revenue to electricity company and loss of production (and hence revenue) to industries.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Inconvenience caused to consumers.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Power system instability: The synchronism between machines working in power system could be lost because of fault occurrence and if this asynchronous operation is beyond transient stability limit of the generator, it will be isolated from bus because of operation of pole-slipping protection. If this is repeated for all generators of the system, there will be cascaded tripping (as it is called) and complete dark-out in the state grid.</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Therefore, proper protective system has to be designed to avoid these consequences or (if avoiding is not possible) minimize the effects of these consequences. For this purpose, minimum faulted portion of the power system has to be selectively isolated from the remnant of the healthy power system as quickly as it is possible. This is so often quite complicated, and we will enter into these complicacies little later and will also appreciate how modern Numerical protective relays make this task possible.</span></p>','','1','4.3','Consequence Of Faults','1','2','1','0','1','2024-08-02 13:33:03','2024-08-02 13:33:03'),(5126,197,'<p>4.4 Why Fault Analysis And Fault Current Calculation</p>','','single','html','<p><span style=\"font-size:14px\">When the process of designing&nbsp; power system is initiated or there is an addition of portion of the power system to cope with the new load demands, it is very essential to prepare the specifications of all electrical equipment of the power system to initiate the procurements of these equipment. A small list (not an exhaustive) of the equipment is:</span></p>\n\n<p>&nbsp;</p>\n\n<ul>\n	<li>\n	<p><span style=\"font-size:14px\">Generators</span></p>\n	</li>\n	<li>\n	<p><span style=\"font-size:14px\">Power Transformers</span></p>\n	</li>\n	<li>\n	<p><span style=\"font-size:14px\">Distribution Transformers</span></p>\n	</li>\n	<li>\n	<p><span style=\"font-size:14px\">Circuit Breakers</span></p>\n	</li>\n	<li>\n	<p><span style=\"font-size:14px\">Current Transformers (C. T.s)</span></p>\n	</li>\n	<li>\n	<p><span style=\"font-size:14px\">Transmission Lines</span></p>\n	</li>\n	<li>\n	<p><span style=\"font-size:14px\">Isolators</span></p>\n	</li>\n	<li>\n	<p><span style=\"font-size:14px\">Busbars</span></p>\n	</li>\n	<li>\n	<p><span style=\"font-size:14px\">Protective Relays</span></p>\n	</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\"><strong>Fault current analysis and calculations for knowing the magnitude of fault currents is a must for following reasons:</strong></span></p>\n\n<p>&nbsp;</p>\n\n<ol>\n	<li>\n	<p><span style=\"font-size:14px\">One important specification of all equipment listed above is the short time withstand current capacity. This means they should be able to withstand fault current for short time without being damaged. It is a practice to specify this short time as 1 second, 3 seconds or 5 seconds. Highest short time is a very conservative specification. Most common is 3 seconds. It is obvious that higher this short time, costlier will be the equipment. This is because more heat will develop and therefore larger conductor dimensions are required to contain this heat. Also, as temperature reached, because of fault current passage for more time, will be higher, the insulation used must withstand the higher temperature. Both these, the bigger conductor and better insulating material will cost more.</span></p>\n	</li>\n</ol>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">This requirement (to specify the fault current) necessitates the calculation of magnitude of fault current well in advance when the power system is being designed.</span></p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<ol start=\"2\">\n	<li>\n	<p><span style=\"font-size:14px\">The symmetrical rms breaking capacity and making capacity of a circuit breaker are also the important specifications while asking for quotations of a circuit breaker. This can be known only if we know the magnitude of the fault current and power factor of the fault current in advance.</span></p>\n	</li>\n</ol>\n\n<p>&nbsp;</p>\n\n<ol start=\"3\">\n	<li>\n	<p><span style=\"font-size:14px\">Protective relay settings are calculated based on magnitude and power factor of the fault current.</span></p>\n	</li>\n</ol>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">Therefore, the exercise we will be doing in following paragraphs is very important in designing and project engineering of power system.</span></p>','','1','4.4','Fault Analysis And Fault Current Calculation','1','2','1','0','1','2024-08-02 13:33:03','2024-08-02 13:33:03'),(5127,197,'<p>4.5 Analysis Of Symmetrical Faults</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">As discussed, triple line or triple line to ground fault is a symmetrical fault. In this section, we will carry out the analysis of this kind of fault and calculate the magnitude of the fault current in case of such a fault.</span></p>','','1','4.5','Analysis Of Symmetrical Faults','1','2','1','0','1','2024-08-02 13:33:03','2024-08-02 13:33:03'),(5128,197,'4.5.1 Per Unite Method','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;Fault current calculations can be made by taking either the actual values of different quantities (such as voltage, current, impedance etc.) or the per unit (pu) values. Using per unit values simplifies the calculations to a great extent.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The per unit value of the quantity is the ratio of the actual value of that quantity to an arbitrarily selected base value. Therefore, per unit (shortly termed as pu) value has no units.</span></p>\n\n<p><span style=\"font-size:14px\">Therefore,</span></p>\n\n<p><span style=\"font-size:14px\">pu&nbsp;<span class=\"math-tex\">\\(??MVA(kVA)=\\frac{MVA_{Actual}}{MVA_{Base}} \\\\\\)</span></span></p>\n\n<p><span style=\"font-size:14px\">&nbsp;Where MVA<sub>Actual</sub> = Actual MVA</span></p>\n\n<p><span style=\"font-size:14px\">And MVA<sub>Base</sub> = Base MVA.</span></p>\n\n<p><span style=\"font-size:14px\">pu&nbsp;<span class=\"math-tex\">\\(Impedance=\\frac{Z_{Actual}}{Z_{Base}} \\\\\\)</span></span></p>\n\n<p><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;Where Z<sub>Actual</sub> = Actual impedance in ohms</span></p>\n\n<p><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;And Z<sub>Base</sub> = Base impedance in ohms.</span></p>\n\n<p><span style=\"font-size:14px\">Similar is the relation of current and voltage.</span></p>\n\n<p><span style=\"font-size:14px\"><strong><u>How to select base values?</u></strong></span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">Out of four system quantities,</span></p>\n\n<p>&nbsp;</p>\n\n<ul>\n	<li>\n	<p><span style=\"font-size:14px\">MVA (kVA)</span></p>\n	</li>\n	<li>\n	<p><span style=\"font-size:14px\">kV (Line to line voltage)</span></p>\n	</li>\n	<li>\n	<p><span style=\"font-size:14px\">I (Line current)</span></p>\n	</li>\n	<li>\n	<p><span style=\"font-size:14px\">Z (Impedance)</span></p>\n	</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">Base MVA (kVA) and base kV are to be selected. </span></p>\n\n<p><span style=\"font-size:14px\">Then,</span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Base Current=\\frac{Base KVA}{\\sqrt{3}\\times BasekV} \\)</span></span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Base Impedance=\\frac{V_{ph}}{I} \\)</span></span></p>\n\n<p><span style=\"font-size:14px\">Where V<sub>ph</sub> = ph-n value of base voltage</span></p>\n\n<p><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;And I = Base current&nbsp;&nbsp;</span></p>\n\n<p><span style=\"font-size:14px\">Therefore,&nbsp; &nbsp;</span></p>\n\n<p><span style=\"font-size:14px\">&nbsp;<span class=\"math-tex\">\\(Base\\hspace{0.1cm} Impedance=\\frac{\\sqrt{3}\\times V_{ph}}{\\sqrt{3}\\times I} \\)</span>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span></p>\n\n<p><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp; &nbsp;<span class=\"math-tex\">\\(=\\frac{\\sqrt{3}V_{ph}\\times\\sqrt{3}V_{ph}} {3V_ph\\times I} \\\\ \\)</span></span></p>\n\n<p><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=\\frac{V^2}{\\mathrm{Three \\hspace{0.1cm} Phase} \\hspace{0.1cm} VA} \\)</span></span></p>\n\n<p><span style=\"font-size:14px\">where V = line to line voltage.</span></p>\n\n<p><span style=\"font-size:14px\">Actual Impedance in ohms</span></p>\n\n<p><span style=\"font-size:14px\">= pu Impedance&nbsp;<span class=\"math-tex\">\\(\\times\\frac{V^2}{VA}\\)</span></span></p>\n\n<p><span style=\"font-size:14px\">&nbsp;<span class=\"math-tex\">\\(\\frac{Z_{ohm}}{Z_{pu}}=\\frac{\\left(\\frac{V}{1000}\\right)^2}{\\frac{VA}{10^6}} \\\\\\)</span></span></p>\n\n<p><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=\\frac{\\left(KV\\right)^2}{MVA} \\\\\\)</span>&nbsp;</span></p>\n\n<p><span style=\"font-size:14px\">Where kV= Line to line voltage in kV</span></p>\n\n<p><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp;MVA= Three phase MVA</span></p>\n\n<p><span style=\"font-size:14px\">or, &nbsp;<span class=\"math-tex\">\\(Z_{ohm}=\\frac{\\left(KV\\right)^2}{MVA}\\times Z_{pu} \\\\\\)</span>&nbsp;&hellip;&hellip;&hellip;.(Eq 4.1)</span></p>\n\n<p><span style=\"font-size:14px\">or,&nbsp;<span class=\"math-tex\">\\(Z_{pu}=\\frac{MVA}{\\left(KV\\right)^2}\\times Z_{ohm} \\\\\\)</span>&hellip;&hellip;&hellip;&hellip;(Eq 4.2)</span></p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">So often, during fault calculations, it is required to deal with two different base values. However, calculations are done on common base. Hence, conversion from one base value to another is required. The following expressions hold good for such conversion:</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Z_{ohm}=\\frac{\\left(KV\\right)^2_{old}}{MVA_{old}}\\times Z_{puold} \\\\ \\)</span></span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Z_{ohm}=\\frac{\\left(KV\\right)^2_{new}}{MVA_{new}}\\times Z_{punew} \\)</span></span></p>\n\n<p><span style=\"font-size:14px\">Equating these two equations, we get pu impedance referred to new base,</span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Z_{punew}=\\frac{MVA_{new}}{MVA_{old}}\\frac{\\left(KV\\right)^2_{old}}{\\left(KV\\right)^2_{new}}\\times Z_{puold} \\\\\\)</span></span></p>\n\n<p>&nbsp;</p>','','1','4.5.1',NULL,'1','2','1','0','1','2024-08-02 13:33:03','2024-08-02 13:33:03'),(5129,197,'4.5.2 Advantages Of  pu System','','single','html','<ol start=\"1\">\n	<li>\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\">Manufacturers usually specify the impedance values of the equipment in pu of the rating of the Equipment.</span></p>\n	</li>\n	<li>\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\">Per unit data representation yields important information about relative magnitudes.</span></p>\n	</li>\n	<li>\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\">In power system there are large number of transformers. The ohmic value of impedance as referred to secondary is different from the value as referred to primary. But the pu values on both the sides are same. This simplifies the calculation of fault currents.</span></p>\n	</li>\n</ol>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Ohmic value of the equivalent impedance referred to primary of transformer = Z<sub>1</sub></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Ohmic value of the equivalent impedance referred to secondary of transformer = Z<sub>2</sub></span></p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">We know that,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Z<sub>2</sub> = Z<sub>1</sub>K<sup>2</sup></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Where&nbsp;&nbsp;<span class=\"math-tex\">\\(K=\\frac{V_2}{V_1}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;V<sub>1</sub> = Primary Voltage</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;V<sub>2</sub> = Secondary Voltage</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">pu Z <span class=\"math-tex\">\\(=\\frac{IZ}{V}\\)</span>&nbsp;</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Where&nbsp; I = Full load rated current of the transformer.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Z = Equivalent Impedance in ohms.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; V = Rated Voltage.</span></p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">pu Z referred to secondary,</span></p>\n\n<p style=\"text-align:justify\"><span class=\"math-tex\">\\(=\\frac{I_2Z_2}{V_2}=\\frac{I_2Z_1K^2}{V_2}=\\frac{Z_1\\frac{V_2^2}{V_1^2}}{\\frac{V_2}{I_2}}\\)</span></p>\n\n<p style=\"text-align:justify\"><span class=\"math-tex\">\\(=\\frac{Z_1V_2^2I_2}{V_2V_1^2}=\\frac{Z_1V_2^2I_1V_1}{V_2V_1^2V_2} \\\\\\)</span>&hellip;&hellip;&nbsp;As&nbsp;&nbsp;<span class=\"math-tex\">\\(\\frac{V_2}{V_1}=\\frac{I_1}{I_2}\\)</span></p>\n\n<p style=\"text-align:justify\"><span class=\"math-tex\">\\(=\\frac{I_1Z_1}{V_1}=\\)</span>&nbsp;<span style=\"font-size:14px\">pu Z referred to primary.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Hence pu values are same on both sides of transformer.</span></p>','','1','4.5.2',NULL,'1','2','1','0','1','2024-08-02 13:33:03','2024-08-02 13:33:03'),(5130,197,'4.5.3 Calculation of Fault Current for Triple Line to Ground Fault (Symmetrical Fault: L-L-L-g)','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">Let us plot the vector diagrams for three phase voltages and currents as shown in figure 4.2.</span></p>\n\n<p><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160588211616058821.jpg\" style=\"float:left; height:524px; width:800px\" /></span></p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">These are known as balanced voltages and currents. The electrical vector quantities are said to be balanced if:</span></p>\n\n<ul>\n	<li>\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\">The magnitudes of all three phase quantities are equal.</span></p>\n	</li>\n	<li>\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\">The vectors are at an angle of 120<sup>0</sup>&nbsp;with each other.</span></p>\n	</li>\n	<li>\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\">The phase sequence is R-Y-B. This means that, when the vectors are rotated anticlockwise, vector Y will replace vector R and vector B will replace vector Y.</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">However, when fault occurs, the voltages and currents are no more balanced. The unbalanced voltages and currents can be represented as shown in figure 4.3.</span></p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160588711616058871.jpg\" style=\"float:left; height:524px; width:800px\" /></span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span></p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; I<sub>R</sub>&nbsp;= I<sub>R1</sub>&nbsp;+ I<sub>R2</sub>&nbsp;+ I<sub>R0</sub></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;I<sub>Y</sub>&nbsp;= I<sub>Y1</sub>&nbsp;+ I<sub>Y2</sub>&nbsp;+ I<sub>Y0</sub></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;I<sub>B</sub>&nbsp;= I<sub>B1</sub>&nbsp;+ I<sub>B2</sub>&nbsp;+ I<sub>B0</sub></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Positive sequence currents mean the balanced currents with phase sequence R-Y-B. Negative sequence currents are also balanced (vector sum I<sub>R2</sub>&nbsp;+ I<sub>Y2</sub>&nbsp;+ I<sub>B2</sub>&nbsp;= 0) but with phase sequence R-B-Y. Zero sequence currents are in phase with equal magnitudes as shown.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p><span style=\"font-size:14px\">&nbsp;Similarly,</span></p>\n\n<p><span style=\"font-size:14px\">&nbsp;V<sub>R</sub>&nbsp;= V<sub>R1</sub>&nbsp;+ V<sub>R2</sub>&nbsp;+ V<sub>R0</sub></span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">V<sub>Y</sub>&nbsp;= V<sub>Y1</sub>&nbsp;+ V<sub>Y2</sub>&nbsp;+ V<sub>Y0</sub></span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">V<sub>B</sub>&nbsp;= V<sub>B1</sub>&nbsp;+ V<sub>B2</sub>&nbsp;+ V<sub>B0</sub></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now, let us introduce vector operator &ldquo;a&rdquo; (figure 4.4).</span></p>\n\n<p><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160589031616058903.jpg\" style=\"height:524px; width:800px\" /></span></p>\n\n<p><span style=\"font-size:14px\">Vector operator a is a unit vector when operated on any other vector, rotates that vector through 120<sup>0</sup>&nbsp;in positive (anticlockwise) direction (obviously without changing the magnitude of the vector).</span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">Summarizing,</span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(a=1\\angle120^0\\)</span></span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(=-0.5+j\\hspace{0.1cm}0.866\\)</span></span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(a^2=1\\angle240^0\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp;= - 0.5- j 0.866</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(a^3=1\\angle0^0\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(a^4=a=1\\angle120^0\\)</span></span></p>\n\n<p><span style=\"font-size:14px\">&nbsp;And so on.</span></p>\n\n<p><span style=\"font-size:14px\">&nbsp;1 + a + a<sup>2</sup>&nbsp;= 0</span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">Hence w. r. t. figure 4.3,</span></p>\n\n<p><span style=\"font-size:14px\">&nbsp;I<sub>R1</sub>&nbsp;= I<sub>1</sub>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;I<sub>R2</sub>&nbsp;= I<sub>2</sub></span></p>\n\n<p><span style=\"font-size:14px\">I<sub>Y1</sub>&nbsp;= a<sup>2</sup>&nbsp;I<sub>R1</sub>&nbsp;= a<sup>2</sup>&nbsp;I<sub>1&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;</sub>I<sub>Y2</sub>&nbsp;= a I<sub>2&nbsp;</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span></p>\n\n<p><span style=\"font-size:14px\">I<sub>R0</sub>&nbsp;= I<sub>Y0</sub>&nbsp;= I<sub>B0</sub>&nbsp;= I<sub>0&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</sub></span></p>\n\n<p><span style=\"font-size:14px\"><sub>&nbsp;</sub>I<sub>B1</sub>&nbsp;= a I<sub>1</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;I<sub>B2</sub>&nbsp;= a<sup>2</sup>&nbsp;I<sub>2</sub></span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">Therefore,</span></p>\n\n<p><span style=\"font-size:14px\">&nbsp;I<sub>R</sub>&nbsp;= I<sub>1</sub>&nbsp;+ I<sub>2</sub>&nbsp;+ I<sub>0</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(Eq 4.3)</span></p>\n\n<p><span style=\"font-size:14px\">&nbsp; I<sub>Y</sub>&nbsp;= a<sup>2</sup>&nbsp;I<sub>1</sub>&nbsp;+ a I<sub>2&nbsp;</sub>+ I<sub>0</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (Eq 4.4)</span></p>\n\n<p><span style=\"font-size:14px\">&nbsp;&nbsp;I<sub>B</sub>&nbsp;= a I<sub>1</sub>&nbsp;+ a<sup>2</sup>&nbsp;I<sub>2</sub>&nbsp;+ I<sub>0</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (Eq 4.5)</span></p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Similar expressions for voltages can be written.</span></p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">If we multiply equation 4.4 by a and equation 4.5 by a<sup>2</sup>,</span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">&nbsp;I<sub>R</sub>&nbsp;= I<sub>1</sub>&nbsp;+ I<sub>2</sub>&nbsp;+ I<sub>0</sub>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; (Eq 4.6)</span></p>\n\n<p><span style=\"font-size:14px\">a I<sub>Y</sub>&nbsp;= I<sub>1</sub>&nbsp;+ a<sup>2</sup>&nbsp;I<sub>2</sub>&nbsp;+ a I<sub>0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</sub>(Eq 4.7)</span></p>\n\n<p><span style=\"font-size:14px\">a<sup>2</sup>&nbsp;I<sub>B</sub>&nbsp;= I<sub>1</sub>&nbsp;+ a I<sub>2</sub>&nbsp;+ a<sup>2</sup>&nbsp;I<sub>0</sub>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;(Eq 4.8)</span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">Therefore,</span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_1=\\frac{1}{3}\\left(I_R+aI_Y+a^2I_B\\right)\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;(Eq 4.9)</span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">Similarly,</span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_2=\\frac{1}{3}\\left(I_R+a^2I_Y+aI_B\\right)\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (Eq 4.10)</span></p>\n\n<p><span style=\"font-size:14px\">And&nbsp;&nbsp;<span class=\"math-tex\">\\(I_0=\\frac{1}{3}\\left(I_R+I_Y+I_B\\right)\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (Eq 4.11)</span></p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">All these expressions can be represented in matrix form also as follows:</span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\left[\\matrix{I_R&amp;\\cr I_Y&amp;\\cr I_B}\\right]=\\left[\\matrix{1 &amp;\\cr 1&amp; \\cr 1 }\\matrix{1 &amp;\\cr a^2 &amp;\\cr a}\\matrix{1&amp; \\cr a &amp;\\cr a^2 }\\right]\\left[\\matrix{I_0 &amp;\\cr I_1 &amp;\\cr I_2 }\\right]\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp; Or,&nbsp;&nbsp;I<sub>RYB</sub>&nbsp;= [A] I<sub>012</sub></span></p>\n\n<p><span style=\"font-size:14px\">Where,&nbsp;<span class=\"math-tex\">\\(\\left[A\\right]=\\left[\\matrix{1 &amp;\\cr 1 &amp;\\cr 1 }\\matrix{1 &amp;\\cr a^2 &amp;\\cr a}\\matrix{1 &amp;\\cr a &amp;\\cr a^2 }\\right]\\)</span></span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\left[\\matrix{I_0 &amp;\\cr I_1 &amp;\\cr I_2 }\\right]=\\left[A\\right]^{-1}\\left[\\matrix{I_R &amp;\\cr I_Y &amp;\\cr I_B }\\right]\\)</span>&nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(Where\\hspace{0.1cm}\\left[A\\right]^{-1}=\\frac{1}{3}\\left[\\matrix{1 &amp;\\cr 1&amp; \\cr 1 }\\matrix{1 &amp;\\cr a &amp;\\cr a^2}\\matrix{1 &amp;\\cr a^2 &amp;\\cr a }\\right]\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp; Or, I<sub>012</sub>&nbsp;= [A]<sup>-1</sup>&nbsp;I<sub>RYB</sub></span></p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Similar expressions for voltages can be written.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Generator at the power plant generates only positive sequence emf E. When fault occurs the sequence voltages can be found out as follows:</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Figure 4.5 shows a star connected generator with neutral impedance Z<sub>n</sub>&nbsp;connected in neutral circuit. When fault occurs, there will be three sequence networks as shown in figures 4.6, 4.7 and 4.8. The positive sequence emf E is seen only in positive sequence network of figure 4.6 and not in figures 4.7 and 4.8. Currents I<sub>R1</sub>, I<sub>Y1</sub>&nbsp;and I<sub>B1</sub>&nbsp;pass only through phase windings (figure 4.5) and not through neutral as I<sub>R1</sub>&nbsp;+ I<sub>Y1</sub>&nbsp;+ I<sub>B1</sub>&nbsp;= 0.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160589991616058999.jpg\" style=\"float:left; height:524px; width:800px\" /></span></p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160590491616059049.jpg\" style=\"float:left; height:524px; width:800px\" /><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160590761616059076.jpg\" style=\"height:524px; width:800px\" /><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160590951616059095.jpg\" style=\"height:524px; width:800px\" /></span></p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Hence, w. r. t. positive sequence network shown in figure 4.6,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">V<sub>1</sub>&nbsp;= E &ndash; I<sub>1</sub>&nbsp;Z<sub>1</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(Eq 4.12)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Where Z<sub>1</sub>&nbsp;= Positive Sequence Impedance of the Generator.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">As discussed earlier the value of Z<sub>1</sub>&nbsp;will be X<sub>d</sub>? (sub-transient direct axis reactance) if analysis of fault events is being done during one cycle following the inception of fault. The magnitude of fault current will be very large as X<sub>d</sub>? is usually very small. If analysis is being made for transient period (15 to 30 cycles following the instant of fault inception, time depending on X/R ratio) X<sub>d</sub>? has to be considered. The magnitude of the fault current, in this case will be less than that during sub-transient period, but yet quite large as&nbsp; transient reactance ( X<sub>d</sub>? ) of the large generator is of the order of 30 % (sub-transient reactance is of the order of 20 %). If one is calculating the fault current in a steady-state period, X<sub>d</sub>&nbsp;(or X<sub>s</sub>) is to be considered. This steady-state reactance, X<sub>s</sub>, is of the order of 200 %. Hence, steady-state fault current is less than the full load rated current of the generator.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Refer negative sequence network shown in figure 4.7. It can be appreciated that the sum of the three negative sequence currents I<sub>R2</sub>, I<sub>Y2</sub>&nbsp;and I<sub>B2</sub>&nbsp;is zero, hence there is no current through neutral circuit (refer figure 4.5).</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Therefore, V<sub>2</sub>&nbsp;= - I<sub>2</sub>&nbsp;Z<sub>2</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(Eq 4.13)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Where Z<sub>2</sub>&nbsp;= Negative Sequence Impedance of Generator.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">But, I<sub>R0</sub>&nbsp;+ I<sub>Y0</sub>&nbsp;+ I<sub>B0</sub>&nbsp;= 3 I<sub>0</sub>. Hence, referring to figure 4.5, the zero sequence currents passing through generator windings are I<sub>R0</sub>, I<sub>Y0</sub>&nbsp;and I<sub>B0</sub>&nbsp;respectively, but current will pass through neutral circuit and its magnitude will be 3I<sub>0</sub>&nbsp;causing a voltage drop I<sub>0</sub>&nbsp;(Z<sub>g0</sub>&nbsp;+ 3 Z<sub>n</sub>)</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Therefore,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; Z<sub>0</sub>&nbsp;= Z<sub>g0</sub>&nbsp;+ 3 Z<sub>n</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (Eq 4.14)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Referring to zero sequence network shown in figure 4.8,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">V<sub>0</sub>&nbsp;= - I<sub>0</sub>&nbsp;Z<sub>0</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (Eq 4.15)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now resuming back to calculation of fault current for a three-phase fault (L-L-L-g or L-L-L), let us write the conditions [figure 4.1(b)]:</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160591671616059167.jpg\" style=\"height:564px; width:800px\" /></span></p>\n\n<p><span style=\"font-size:14px\">V<sub>R</sub>&nbsp;= V<sub>Y</sub>&nbsp;= V<sub>B</sub>&nbsp;= 0, at the point of fault.</span></p>\n\n<p><span style=\"font-size:14px\">I<sub>R</sub>&nbsp;+ I<sub>Y</sub>&nbsp;+ I<sub>B</sub>&nbsp;= 0</span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{.045in}.}\\hspace{.3in}I_0=\\frac{1}{3}\\left(I_R+I_Y+I_B\\right)=0\\)</span></span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(V_1=\\frac{1}{3}\\left(V_R+aV_Y+a^2V_B\\right)=0\\)</span></span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(V_2=\\frac{1}{3}\\left(V_R+a^2V_Y+aV_B\\right)=0\\)</span></span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_0=\\frac{1}{3}\\left(I_R+I_Y+I_B\\right)=0\\)</span></span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{.045in}.}\\hspace{.3in}V_0=-I_0Z_0=0\\)</span></span></p>\n\n<p><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; V<sub>2</sub>&nbsp;= - I<sub>2</sub>&nbsp;Z<sub>2</sub></span></p>\n\n<p><span style=\"font-size:14px\"><sub><span class=\"math-tex\">\\(\\dot{.\\hspace{.045in}.}\\hspace{.3in}0=-I_2\\hspace{0,1cm}Z_2\\)</span></sub></span></p>\n\n<p><span style=\"font-size:14px\"><sub><span class=\"math-tex\">\\(\\dot{.\\hspace{.045in}.}\\hspace{.3in}I_2=0\\)</span></sub></span></p>\n\n<p>&nbsp;</p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Hence negative and zero sequence networks do not exist, and only positive sequence network is there. Hence, only positive sequence currents are existing. Hence, this fault (L-L-L-g) is said to be symmetrical fault.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Figure 4.9 is the positive sequence network (or reactance diagram) for L-L-L-g fault.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160591971616059197.jpg\" style=\"height:524px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">V<sub>1</sub> = E &ndash; I<sub>1</sub> Z<sub>1</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (refer figure 4.6)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><sub>&nbsp;<sub><span class=\"math-tex\">\\(\\dot{.\\hspace{.025in}.}\\hspace{.2in}0=E-I_1Z_1\\)</span></sub></sub></span></p>\n\n<p><span style=\"font-size:14px\"><sub><sub><span class=\"math-tex\">\\(\\dot{.\\hspace{.025in}.}\\hspace{.2in}I_1=\\frac{E}{Z_1}\\)</span></sub></sub>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;(Eq 4.16)</span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_R=I_1+I_2+I_0=\\frac{E}{Z_1}\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_Y=a^2I_1+aI_2+I_0=a^2\\frac{E}{Z_1}\\)</span></span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_B=aI_1+a^2I_2+I_0=a\\frac{E}{Z_1}\\)</span></span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">These are the three phase fault currents.</span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Fault\\hspace{0.1cm} Current=\\frac{V_{Base}}{Z_1}\\)</span></span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">If base voltage is considered as 1 pu&nbsp;<span class=\"math-tex\">\\((1\\angle0^0)\\)</span>&sbquo;</span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(pu\\hspace{0.1cm}Fault\\hspace{0.1cm} Current=\\frac{1}{pu Z_1}\\)</span></span></p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\sqrt3\\times Base\\hspace{0.1cm}Current\\times pu\\hspace{0.1cm} Fault\\hspace{0.1cm}Current\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</span></p>\n\n<p><span class=\"math-tex\">\\(=\\frac{\\sqrt3\\times V_{Base}\\times I_{Base}}{puZ_1}\\)</span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">As Base Voltage = 1 pu</span></p>\n\n<p><span style=\"font-size:14px\">And Base Current = 1 pu</span></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{.025in}.}\\hspace{.2in}Fault\\hspace{0.1cm} Level \\hspace{0.1cm}in \\hspace{0.1cm}MVA=\\frac{Base\\hspace{0.1cm} MVA}{pu\\hspace{0.1cm} Z_1}\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (Eq 4.17) &nbsp;</span></p>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Fault current for three-phase fault can be calculated using Equation 4.16 or Equation 4.17. &nbsp; &nbsp; &nbsp;&nbsp;</span></p>','','1','4.5.3',NULL,'1','2','1','0','1','2024-08-02 13:33:03','2024-08-02 13:33:03'),(5131,197,'4.5.4  Application Example','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><u>Example 4.1:</u></strong> Refer figure 4.10. Calculate the fault currents for three-phase faults at buses A, B, C, D and E. Ignore resistance of transmission lines.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160599181616059918.jpg\" style=\"float:left; height:515px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><u>Solution:</u></strong>&nbsp;&nbsp;Base MVA = 150 MVA.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Base kV = 132 kV line to line.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">pu reactance of Generator = 0.2 pu on its own base</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Generator\\hspace{0.1cm}MVA=\\frac{120}{0.85}=141.17\\hspace{0.1cm}MVA\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">pu reactance of Generator referred to</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;<span class=\"math-tex\">\\(150\\hspace{0.1cm}MVA=\\frac{MVA_{new}}{MVA_{old}}\\times Z_{puold}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=\\frac{150}{141.17}\\times 0.2=0.2125\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">pu reactance of transformer = 0.1 pu.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">For transmission lines,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(pu\\hspace{0.1cm}X=\\frac{MVA}{\\left(KV\\right)^2}\\times Z_{ohm}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(pu\\hspace{0.1cm}X_{AB}=\\frac{150}{132^2}\\times 6=0.0516\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Similarly,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><em>pu X</em><sub><em>BC</em></sub>&nbsp;= 0.0344 pu</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><em>pu X</em><sub><em>CD</em></sub>&nbsp;= 0.0344 pu</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><em>pu X</em><sub><em>DE</em></sub>&nbsp;= 0.0344 pu</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Referring to reactance diagram shown in figure 4.11,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160599561616059956.jpg\" style=\"height:515px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong>For a three-phase fault at bus A,</strong></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Pu X = 0.15625 pu</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Fault\\hspace{0.1cm}MVA=\\frac{Base\\hspace{0.1cm}MVA}{pu\\hspace{0.1cm}X}\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=\\frac{150}{0.15625}\\)</span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=960\\hspace{0.1cm}MVA\\)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Fault\\hspace{0.1cm}Current=\\frac{960\\times 10^6}{\\sqrt3\\times132\\times 10^3}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=4198.9\\hspace{0.1cm}Amp\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><strong><span style=\"font-size:14px\">Similarly, for fault at bus B,</span></strong></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Pu X = 0.20785 pu.&nbsp;&nbsp;&nbsp;&nbsp; </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Fault\\hspace{0.1cm}MVA=\\frac{150}{0.20785}\\)</span>&nbsp;</span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=721.67\\hspace{0.1cm}MVA\\)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Fault Current = 3156.5 Amp.</span></p>\n\n<h1 style=\"text-align:justify\">&nbsp;</h1>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong>For fault at bus C,</strong></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Fault MVA = 619.195 MVA.&nbsp;&nbsp;&nbsp; Fault Current, I<sub>f</sub> = 2708.27 Amp.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">For fault at bus D,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Fault MVA = 542.20 MVA.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">I<sub>f</sub> = 2371.5 Amp.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">For fault at bus E,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Fault MVA = 482.24 MVA.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">I<sub>f</sub> = 2109.24 Amp</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The reader shall note that while carrying out this fault analysis, pre-fault currents are not considered as usually the fault currents are much larger than the load currents. Hence, the load currents are ignored in comparison to the fault currents. The fault analysis considering the pre-fault currents, however, can be carried out, but it is little complicated. It is a usual practice to ignore the pre-fault currents.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>','','1','4.5.4',NULL,'1','2','1','0','1','2024-08-02 13:33:03','2024-08-02 13:33:03'),(5132,197,'4.5.5 Application Example','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><u>Example 4.2:</u></strong> With reference to figure 4.12, calculate the fault currents for three-phase faults at buses A, B, C and D respectively. Also suggest the symmetrical breaking capacities of breakers 1, 2, 3 and 4. For a fault at bus C, what will be the fault current in 132 kV transmission line and fault current shared by each generator?</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160610981616061098.jpg\" style=\"height:524px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><u>Solution:</u></strong> Base MVA =100 MVA</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">pu reactance of generators,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(X_g=\\frac{100}{82.35}\\times 0.15\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;<span class=\"math-tex\">\\(Generator\\hspace{0.1cm}MVA=\\frac{70}{0.85}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=82.35\\hspace{0.1cm}MVA\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 0.182 pu.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">pu reactance of 100 MVA generator transformers,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">X<sub>T1</sub> = 0.12 pu.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Resistance and reactance of 132 kV transmission line,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(pu\\hspace{0.1cm}R_{T1}=\\frac{100}{132^2}\\times 3=0.017\\hspace{0.1cm}pu \\\\ \\)</span></span></p>\n\n<p style=\"text-align:justify\"><span class=\"math-tex\">\\(pu\\hspace{0.1cm}X_{T1}=\\frac{100}{132^2}\\times 12=0.069\\hspace{0.1cm}pu\\)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">pu reactance of 50 MVA transformer on 100 MVA base,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(X_{T2}=\\frac{100}{50}\\times 0.1=0.2\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Resistance and reactance of 66 kV transmission line on 100 MVA base,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(pu\\hspace{0.1cm}R_{T2}=\\frac{100}{66^2}\\times 2=0.046\\hspace{0.1cm}pu \\\\ \\)</span></span></p>\n\n<p style=\"text-align:justify\"><span class=\"math-tex\">\\(pu\\hspace{0.1cm}X_{T2}=\\frac{100}{66^2}\\times 6=0.1377\\hspace{0.1cm}pu\\)</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Referring to reactance diagram shown in figure 4.13,</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160611241616061124.jpg\" style=\"height:524px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong>For a three-phase fault at bus A,</strong></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Fault\\hspace{0.1cm}MVA=\\frac{Base\\hspace{0.1cm}MVA}{pu\\hspace{0.1cm}Z}=\\frac{100}{0.151}\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span style=\"font-size:14px\"><span class=\"math-tex\">\\(=662.25\\hspace{0.1cm}MVA\\)</span></span></p>\n\n<p><span style=\"font-size:14px\">Fault current, <span class=\"math-tex\">\\(I_f=\\frac{662.25\\times 10^3}{\\sqrt3\\times 132}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <span class=\"math-tex\">\\(=2896.6\\hspace{0.1cm}Amp\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong>For fault at bus B,</strong></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Fault\\hspace{0.1cm}MVA=\\frac{100}{0.017+j\\hspace{0.1cm}0.22}\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=\\frac{100}{0.22\\angle85.58}\\)</span> &nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=454.55\\angle-85.58^0\\hspace{0.1cm}MVA\\)</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_f=1988\\angle-85.58^0\\hspace{0.1cm}Amp\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong>For a fault at bus C,</strong></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Fault\\hspace{0.1cm}MVA=\\frac{100}{0.017+j\\hspace{0.1cm}0.42}\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <span class=\"math-tex\">\\(=\\frac{100}{0.42\\angle87.68}\\)</span>&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=238.09\\angle-87.68^0\\hspace{0.1cm}MVA\\)</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_f=\\frac{238.09\\angle-87.68\\times10^3}{\\sqrt3\\times66}\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=2082.79\\angle-87.68^0\\hspace{0.1cm}Amp\\)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">For the same fault, fault current in primary of 50 MVA transformer (or in 132 kV line),</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_{f132}=\\frac{2082.79}{2}=1041.4\\hspace{0.1cm}Amp\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">This fault current when shared by each generator can be calculated as follows:</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">I<sub>f</sub> shared by each generator <span class=\"math-tex\">\\(=1041.4\\times\\frac{132}{11}\\times\\frac{1}{2}\\)</span>&nbsp;</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 6248.4 Amp.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong>For a three-phase fault at bus D,</strong></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Fault\\hspace{0.1cm}MVA=\\frac{100}{0.063+j\\hspace{0.1cm}0.5577}\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=\\frac{100}{0.561\\angle83.55}\\)</span> &nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=178.25\\angle-83.55^0\\hspace{0.1cm}MVA\\)</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">I<sub>f </sub>=1559.31 Amp.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Symmetrical breaking capacities will be decided based on the fault MVA at the bus near the breaker.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Therefore, symmetrical breaking capacity required for</span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Breaker 1: 750 MVA</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Breaker 2: 500 MVA</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Breaker 3: 300 MVA</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Breaker 4: 200 MVA</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Rounded off values are selected.</span></p>','','1','4.5.5',NULL,'1','2','1','0','1','2024-08-02 13:33:03','2024-08-02 13:33:03'),(5133,197,'4.5.6  Application Example','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><u>Example 4.3:</u></strong> Figure 4.14 shows a single line diagram of a 66/11 kV substation. Calculate</span></p>\n\n<ol>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Short time withstand capacity of all 11 kV current transformers.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Symmetrical breaking capacity (in MVA) of all 11 kV breakers.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Short time withstand capacity of both 66/11 kV power transformers.</span></p>\n	</li>\n</ol>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160612091616061209.jpg\" style=\"height:627px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><u>Solution: </u></strong></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Using Equation 4.17,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Fault level in&nbsp;&nbsp;<span class=\"math-tex\">\\(MVA=\\frac{Base\\hspace{0.1cm}MVA}{pu\\hspace{0.1cm}Z}\\)</span> </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Fault level in MVA at 66 kV bus is obviously 500 MVA as breaker selected is of breaking capacity 500 MVA. If base MVA is selected as 500 MVA, pu Z from generation (source) to 66 kV bus is 1 pu. This is known as source impedance. Reactance diagram, accordingly, is drawn as shown in figure 4.15.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160616601616061660.jpg\" style=\"height:524px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">pu reactance of each power transformer on 500 MVA base,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:11pt\"><span class=\"math-tex\">\\(=\\frac{500}{50}\\times 0.1=1\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Equivalent impedance = j 1.5 pu.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Fault MVA on 11 kV bus&nbsp;<span class=\"math-tex\">\\(=\\frac{500}{1.5}=333.33\\hspace{0.1cm}MVA\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Hence all 11 kV breakers shall be required of symmetrical breaking capacity 350 MVA (rounded off).</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Fault current,&nbsp;<span class=\"math-tex\">\\(I_f=\\frac{333.33\\times 10^3}{\\sqrt3\\times 11}=17495\\hspace{0.1cm}MVA\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Hence, short time current withstand capacity for all 11 kV C. T.s is to be demanded as 20 kA (rounded off) for 3 seconds. The same value is the short time withstand capacity of power transformers in secondary side. For primary it will be</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(20\\times \\frac{11}{66}=3.33\\hspace{0.1cm}Amp\\)</span></span></p>','','1','4.5.6',NULL,'1','2','1','0','1','2024-08-02 13:33:03','2024-08-02 13:33:03'),(5134,197,'4.5.7 Application Example','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><u>Example 4.4:</u></strong> Consider a three-bus system of figure 4.16. Both the generators are 120 MW, 0.85 pf each with transient reactance 15 %. Both the transformers are of 150 MVA capacity with equivalent leakage reactance 10 % each. Calculate the fault current for a three-phase fault as shown in figure. Assume all line reactances are given on 150 MVA base.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16194186601619418660.jpg\" style=\"height:675px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><u>Solution:</u> </strong>The reactance diagram of figure 4.16 is as shown in figure 4.17. </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160621251616062125.jpg\" style=\"height:675px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Base MVA = 150.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Generator reactance at 150 MVA base,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(=\\frac{150\\times 0.85}{120}\\times 0.15=0.1594\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now, for further analysis, delta to star transformation is required. The reactances in star formation are each equal with values</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\frac{0,1\\times 0.1}{0.1+0.1+0.1}=0.03333\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The reactance diagram after delta to star conversion is given in figure 4.18.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160621551616062155.jpg\" style=\"height:675px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Equivalent reactance,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">X<sub>eq</sub> = 0.18 pu.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Fault MVA &nbsp;<span class=\"math-tex\">\\(=\\frac{150}{0.18}\\)</span>&nbsp;=834.75 MVA.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Fault current,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_f=\\frac{834.75\\times 10^3}{\\sqrt3\\times 132}=3651\\hspace{0.1cm}Amp\\)</span></span></p>','','1','4.5.7',NULL,'1','2','1','0','1','2024-08-02 13:33:03','2024-08-02 13:33:03'),(5135,197,'4.5.8  Current Limiting Reactors','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">If the country or a state is very small with reference to its geographical area, the transmission lines will be too short, resulting in very large fault currents. These fault currents are required to be limited or reduced. This can be done by using fault current limiting reactors. The advantage of using reactors is that there will be no power loss, but the disadvantage is that power factor worsens. Yet a very useful application of fault limiting reactors is discussed as follows:</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">There is a power system network as shown in figure 4.19. All 66 kV breakers used are of symmetrical breaking capacity 750 MVA. After 5 years (or may be 10 years), 50 MVA capacity is required to be added as load demand has increased. This necessitated to install one more 66 kV incomer and one more 66/11 kV, 50 MVA power transformer. But as third parallel incomer has been added feeding more power to 66 kV bus, the fault current for a fault at 66 kV bus will increase. This will necessitate all the breakers (the new ones to be installed and the old ones which are already existing) to be of higher breaking capacity. The new ones can be purchased with higher breaking capacity, but the old breakers will have to be replaced with breakers having higher breaking capacity. This is not economically viable. In this case, a bus reactor can be installed as shown in figure 4.20. If need be, a similar fault limiting reactor can be installed at 11 kV bus too. This reactor will limit the fault current at 11 kV bus as required.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160622531616062253.jpg\" style=\"height:631px; width:800px\" /><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160622761616062276.jpg\" style=\"height:569px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160623181616062318.jpg\" style=\"height:491px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The same motive can be achieved in yet different way. Refer figure 4.21. The old installations are shown with full lines and the new installations are shown with dotted lines. If old breakers are not to be replaced with breakers having higher breaking capacity, the only requirement is an interlock such that:</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<ol>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">When both the incomer circuit breakers CB<sub>1</sub> and CB<sub>2</sub> are closed (ON), bus sectionalizing breaker cannot be made ON.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Bus sectionalizing breaker can be made ON only if either of the incomer circuit breaker is OFF.</span></p>\n	</li>\n</ol>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">This interlock serves the purpose of limiting fault current as two incomers are never in parallel. Other advantages of using sectionalizing breaker are:</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">The disadvantage of low power factor with use of bus reactor is not there when we use bus sectionalizing breaker.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">In case of bus fault, whole of the bus has to be evacuated when bus reactor is used. Whereas half of the bus can be used in case of bus fault when sectionalizing breaker is used, as the healthy part of the bus can be used as the concerned incomer breaker and sectionalizing breaker will be tripped by bus protection system.</span></p>\n	</li>\n</ul>','','1','4.5.8',NULL,'1','2','1','0','1','2024-08-02 13:33:03','2024-08-02 13:33:03'),(5136,197,'4.5.10  Transients in R-L series circuits','','single','html','<p><span style=\"font-size:14px\">?We have seen that a power system is rich of inductive reactance. Resistive component is negligibly small. Refer to single phase equivalent circuit of a typical power system network as shown in figure 4.25.</span></p>\n\n<p><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16164831591616483159.jpg\" style=\"height:491px; width:800px\" /></span></p>\n\n<p><span style=\"font-size:14px\">When fault occurs, the load and system capacitance are shorted. Using Kirchoff&rsquo;s voltage law, we can write,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(L\\frac{di}{dt}+Ri=e=E_m\\sin(\\omega t+\\theta)\\)</span>&nbsp; &nbsp; &nbsp;&nbsp;(Eq 4.18)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The voltage waveshape is shown in figure 4.26.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16161466691616146669.jpg\" style=\"height:491px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Equation 4.18 is an equation written for fault current i when there is a sudden short circuit. This equation is a differential equation, and the solution is the sum of complementary function (transient component) and the particular integral (steady-state component). The transient solution can be obtained by writing,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(L\\frac{di}{dt}+R\\hspace{0.1cm}i=0\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}\\frac{di}{i}=\\frac{-R}{L}\\hspace{0.1cm}dt\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Integrating,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(ln\\hspace{0.1cm}i=\\frac{-Rt}{L}+K\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Where K = Constant of integration. </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{.045cm}.}\\hspace{0.1cm}ln\\hspace{0.1cm}i=ln\\hspace{0.1cm}e^{\\frac{-Rt}{L}}+ln\\hspace{0.1cm}A\\)</span>&nbsp; &nbsp;&nbsp;</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Where A = Constant,&nbsp;<span class=\"math-tex\">\\(ln\\hspace{0.1cm}A=K\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{.045cm}.}\\hspace{0.1cm}\\hspace{0.1cm}i=A\\hspace{0.1cm}e^{\\frac{-Rt}{L}}\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (Eq 4.19)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">It is seen that the transient component is the exponentially decaying component or d. c. offset as it is called. It decays to a very small value after about 15 to 30 cycles depending upon time constant L/R. The transient component is not d. c., it decays after about 30 cycles as stated. But for the operation of an instantaneous overcurrent relay which operates within 1 to 3 cycles, this transient component can be assumed to remain constant. That is why the transient component is named d. c. offset.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">For obtaining steady-state solution,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(i=C\\cos(\\omega t+\\theta)+D\\sin(\\omega t+\\theta)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\frac{\\mathrm{d}i}{\\mathrm{d}t}=-C\\hspace{0.1cm}\\omega \\hspace{0.1cm}\\sin(\\omega t+\\theta)+D\\hspace{0.1cm}\\omega \\hspace{0.1cm}\\cos(\\omega t+\\theta)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Substituting these values of i and&nbsp;<span class=\"math-tex\">\\(\\frac{\\mathrm{d}i}{\\mathrm{d}t}\\)</span>&nbsp;in Equation 4.18,</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(-C\\hspace{0.1cm}\\omega \\hspace{0.1cm}L\\sin(\\omega t+\\theta)+D\\hspace{0.1cm}\\omega \\hspace{0.1cm}L\\cos(\\omega t+\\theta)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(+R \\hspace{0.1cm}C\\hspace{0.1cm} \\cos(\\omega t+\\theta)+ R \\hspace{0.1cm}D\\hspace{0.1cm} \\sin(\\omega t+\\theta)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(=E_m \\hspace{0.1cm} \\sin(\\omega t+\\theta)\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Equating coefficients of sin and cos terms,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(R\\hspace{0.1cm}C+D\\hspace{0.1cm}\\omega\\hspace{0.1cm}L=0,\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;<span class=\"math-tex\">\\(R\\hspace{0.1cm}D-C\\hspace{0.1cm}\\omega\\hspace{0.1cm}L=E_m\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{.045cm}.}\\hspace{0.3cm}C= \\frac{-D\\hspace{0.1cm}\\omega \\hspace{0.1cm}L}{R}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{.045cm}.}\\hspace{0.3cm}R\\hspace{0.1cm}D+ \\frac{D\\hspace{0.1cm}\\omega^2 \\hspace{0.1cm}L^2}{R}=E_m\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{.045cm}.}\\hspace{0.3cm}D[\\frac{R^2+\\omega^2\\hspace{0.1cm}L^2}{R}]=E_m\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{.045cm}.}\\hspace{0.3cm}D=\\frac{R\\hspace{0.1cm}E_m}{R^2+\\omega^2\\hspace{0.1cm}L^2}\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(C=\\frac{-\\omega\\hspace{0.1cm}L\\hspace{0.1cm}E_m}{R^2+\\omega^2\\hspace{0.1cm}L^2}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Referring to impedance triangle shown in figure 4.27,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16161467031616146703.jpg\" style=\"height:491px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\cos\\phi=\\)</span>&nbsp;Power factor,&nbsp;&nbsp; </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(Z=\\sqrt{(R^2+\\omega^2\\hspace{0.1cm}L^2)}\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(\\sin\\phi=\\frac{\\omega\\hspace{0.1cm}L}{Z}\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;<span class=\"math-tex\">\\(\\cos\\phi=\\frac{R}{Z}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">These relations give the final steady-state solution as follows:</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(i=\\frac{E_m}{Z}[\\sin(\\omega\\hspace{0.1cm}t+\\theta)\\cos\\phi\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(-\\cos(\\omega\\hspace{0.1cm}t+\\theta)\\sin\\phi]\\)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(=\\frac{E_m}{Z}\\hspace{0.1cm}\\sin(\\omega\\hspace{0.1cm}t+\\theta-\\phi)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(=I_m\\hspace{0.1cm}\\sin(\\omega\\hspace{0.1cm}t+\\theta-\\phi)\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Where&nbsp;<span class=\"math-tex\">\\(\\frac{E_m}{Z}=I_m\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Total solution is </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(i=A\\hspace{0.1cm}e^{-\\frac{RT}{L}}+I_m\\hspace{0.1cm}\\sin(\\omega\\hspace{0.1cm}t+\\theta-\\phi)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now at t = 0 (just before fault inception), i = 0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (ignoring pre-fault current)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(0=A\\hspace{0.1cm}+I_m\\hspace{0.1cm}\\sin(\\theta-\\phi)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.1cm}A\\hspace{0.1cm}=I_m\\hspace{0.1cm}\\sin(\\phi-\\theta)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Therefore,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(i\\hspace{0.1cm}=I_m\\hspace{0.1cm}[\\sin(\\omega t+\\theta-\\phi)+sin(\\phi -\\theta)e^{\\frac{-Rt}{L}}]\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">It is seen that the fault current is a sinusoidal wave superimposed on exponentially decaying d. c. offset.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now, the fault can occur at any instant of voltage wave. In figure 4.26, fault is shown to have occurred at <span class=\"math-tex\">\\(\\theta=45^0\\)</span> approximately. If fault occurs at e = 0, <span class=\"math-tex\">\\(\\theta=0^0\\)</span>and if we assume &omega;L &gt;&gt; R (the assumption which is not far from reality), <span class=\"math-tex\">\\(\\phi=90^0\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(i=I_m\\hspace{0.1cm}[\\sin(\\omega t-90)+e^{\\frac{-RT}{L}}]\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The waveshape will be as shown in figure 4.28. The waveshape shows that there will be maximum asymmetry. If fault occurs at e = E<sub>m</sub>, &theta; = 90<sup>0</sup>.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16161467471616146747.jpg\" style=\"height:491px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(i=I_m\\hspace{0.1cm}\\sin\\hspace{0.1cm}\\omega t\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The waveshape of fault current will now be symmetrical w. r. t. time axis. But one never knows the instant at which fault will occur. Hence, a switchgear engineer (or a protection engineer) has to design the power system for worst asymmetry as shown in figure 4.28 wherein the peak of the current will be 2 I<sub>m</sub> (or 2.8 &times; I<sub>rms</sub>). This is known as dynamic current.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong>The significance of this asymmetry in waveshape is many-fold: </strong></span></p>\n\n<ol>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">When fault occurs, all the equipment of the power system (generators, transformers, isolators, circuit breakers, busbars etc.) have to withstand this dynamic current as this current can be a reason for electrodynamic failure of the equipment and, in turn, bursting of the equipment.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">In the era of very fast operating relays (one cycle) and breakers having very low breaker time (one or two cycles), the circuit breaker will be interrupting the asymmetric fault current. Hence, while specifying, a switchgear engineer shall have to ask for higher breaking capacity than the symmetrical breaking capacity calculated in foregoing section. The multiplying factor for specifying asymmetric breaking capacity for breakers is tabulated in Table 4.3 below:</span></p>\n	</li>\n</ol>\n\n<table align=\"center\" border=\"1\" dir=\"ltr\" style=\"width:800px\">\n	<caption>\n	<div style=\"text-align:center\"><span style=\"font-size:28px\">Table 4.3: Multiplying Factor for Asymmetric Breaking Current</span></div>\n	</caption>\n	<thead>\n		<tr>\n			<th scope=\"col\" style=\"border-color:black; vertical-align:top; width:197px\">\n			<div><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Circuit Breaker Time</span></div>\n			</th>\n			<th scope=\"col\" style=\"vertical-align:top; width:209px\">\n			<div><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Multiplying Factor</span></div>\n			</th>\n		</tr>\n	</thead>\n	<tbody>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:197px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; More than 8 cycles</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:209px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">1.0</span></div>\n			</td>\n		</tr>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; text-align:center; vertical-align:middle; width:197px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">5 cycles</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:209px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">1.1</span></div>\n			</td>\n		</tr>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:197px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">3 cycles</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:209px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">1.2</span></div>\n			</td>\n		</tr>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:197px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">&nbsp;2 cycles</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:209px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">1.4</span></div>\n			</td>\n		</tr>\n		<tr>\n			<td style=\"border-bottom:1px solid black; border-left:1px solid black; border-right:1px solid black; border-top:none; vertical-align:top; width:197px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">1 cycle</span></div>\n			</td>\n			<td style=\"border-bottom:1px solid black; border-left:none; border-right:1px solid black; border-top:none; vertical-align:top; width:209px\">\n			<div style=\"text-align:center\"><span style=\"font-size:28px\">1.6</span></div>\n			</td>\n		</tr>\n	</tbody>\n</table>\n\n<ol start=\"3\">\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">The instantaneous overcurrent relays and distance relays (in first zone) will overreach due to this d. c. offset. This will be discussed later.?</span></p>\n	</li>\n</ol>\n\n<p><span style=\"font-size:14px\">?</span></p>','','1','4.5.10',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5137,197,'4.5.9 Application Example','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><u><strong>Example 4.5</strong>:</u> Figure 4.22 shows a single line diagram of power system network. The breaking capacity of breakers A is 1000 MVA. Find out:</span></p>\n\n<ol>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">pu value of reactor.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">MVAR capacity of reactor.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Ohmic value of reactor.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">The inductance of reactor</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">The breaking capacity required for breakers B.</span></p>\n	</li>\n</ol>\n\n<p>&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160639021616063902.jpg\" style=\"height:491px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><u>Solution:</u></strong> Base MVA is selected as 100 MVA. Let us first draw reactance diagram as shown in figure 4.23.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160639251616063925.jpg\" style=\"height:491px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Source impedance is zero as transformer is connected to an infinite bus.</span></p>\n\n<p><span style=\"font-size:14px\">Generator reactance,&nbsp;<span class=\"math-tex\">\\(X_g=\\frac{100}{50}\\times 0.12=0.24\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Transformer reactance =0.1 pu.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">For a three-phase fault at F<sub>1</sub>, fault MVA = 1000 MVA, as circuit breakers A have breaking capacity 1000 MVA.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Referring to figure 4.24,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16160639511616063951.jpg\" style=\"height:491px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(X_{eq}=\\frac{0.12(a+0.1)}{a+0.12+0.1}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; Fault MVA &nbsp;<span class=\"math-tex\">\\(=\\frac{100}{X_{eq}}=1000\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{.3in}\\frac{100(a+0.22)}{0.12(a+0.1)}=1000\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{.3in}\\)</span>100 a + 22 = 120 (a+0.1)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{.3in}\\)</span>20 a = 10</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{.3in}\\)</span>a = 0.5 pu.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The reactor of 0.5 pu (at 100 MVA base) is to be installed to limit the fault current for three-phase fault at F<sub>1</sub>.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now, <span class=\"math-tex\">\\(\\frac{(kV)^2}{MVA}\\times pu\\hspace{0.1cm}X=X_{ohm}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{.3in}\\)</span>&nbsp;Ohmic value of reactor <span class=\"math-tex\">\\(=\\frac{66^2}{100}\\times 0.5=21.78\\hspace{0.1cm}\\Omega\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">MVAR capacity of the reactor&nbsp;<span class=\"math-tex\">\\(=\\sqrt3\\times kV\\times\\frac{kV}{\\sqrt3\\times 21.78}=\\frac{66^2}{21.78}=200\\hspace{0.1cm}MVAR\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">pu value of the reactor on its own base&nbsp;<span class=\"math-tex\">\\(=\\frac{200}{100}\\times 0.5=1.0\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Inductance of the reactor &nbsp;<span class=\"math-tex\">\\(=\\frac{X}{2\\pi f}=\\frac{21.78}{2\\pi \\times 50}=0.0693\\)</span>&nbsp;</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Henry = 69.3 mH</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">For a three-phase fault at F<sub>2</sub>,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(X_{eq}=\\frac{(0.12+0.5)\\times 0.1}{0.12+0.5+0.1}=0.08611\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp; Fault MVA&nbsp;<span class=\"math-tex\">\\(=\\frac{100}{0.08611}\\)</span>= 1161.29 MVA.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">This means, the symmetrical breaking capacity to be specified while floating the tender should be 1250 MVA (rounded off). These are not just rounded off values, these are standard values also. In the text later the standard values will be listed.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>','','1','4.5.9',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5138,197,'<p>4.6 Fault Current Calculations for Unsymmetrical (Unbalanced) Faults</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">L-L, L-g and L-L-g faults are unsymmetrical faults. When such faults occur in power system network, negative and zero sequence voltages and currents are produced. When negative sequence currents flow in the network, rotating magnetic field (RMF) is produced in all rotating machines (synchronous generators, synchronous motors and induction motors) which rotate at synchronous speed in reverse direction w. r. t. RMF produced by positive sequence currents. The physical rotor is always rotating in the same direction as RMF due to positive sequence currents. Slip due to positive RMF is zero in synchronous machines and about 4 % (0.04 pu) in induction motors. Hence, slip frequency eddy currents are nil in synchronous machines and negligible in induction machines. Therefore, there is no eddy current loss (or negligible loss) in rotor iron in all rotating machines under normal operating condition as there is only positive sequence current present. But slip frequency due to negative RMF (due to presence of negative sequence currents in case of unbalanced faults) is 200 % (196 % in case of induction motors) and hence large slip frequency eddy currents flow in rotor windings as well as rotor iron. Rotor of synchronous generators, as discussed, is not laminated and hence large eddy currents and large eddy current loss (eddy current loss is proportional to f<sup>2 </sup>and sf being 200 % or 100 Hz which gives rise to very large eddy current loss). This loss manifests in terms of heat and development of large temperature due to heat. This heat can make the rotor red hot if protective actions are not taken early. Also, if the temperature developed is more than withstand limit of the insulation used, the insulation of the rotor winding will be thermally damaged resulting in rotor earth fault.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">As a thumb rule, if heat produced by positive sequence current I<sub>1</sub> is H [proportional to (I<sub>1</sub>)<sup>2</sup>Rt], heat produced by equal magnitude of negative sequence current I<sub>2</sub> will be 6H. Thus, even if the fault current due to unsymmetrical fault is less than CMR of the equipment, it will be harmful to all rotating machines.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Zero sequence current is third harmonic current. Refer figure 4.33. It can be appreciated that the third harmonic or zero sequence currents (or voltages) are in phase. The hysteresis loss is proportional to frequency, f; but the eddy current loss is proportional to square of the frequency, f<sup>2</sup>. Hence large iron losses are produced because of zero sequence currents. The magnitude of zero sequence current may be less than that of fundamental (positive sequence) current, but as eddy current loss is proportional to f<sup>2</sup>, the heat produced will be quite high. Hence zero sequence currents are harmful to rotating machines and transformers as well.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:11pt\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16161527141616152714.jpg\" style=\"height:451px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Thus, when we will enter in power system protection, we will see that fault currents are usually larger than CMR of the equipment, but in certain cases, the fault currents are smaller than CMR of the equipment and yet they are harmful.</span></p>','','1','4.6','Fault Current Calculations for Unsymmetrical (Unbalanced) Faults','1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5139,197,'4.6.1 Line to Line Fault (L-L Fault)','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">Let us consider Y-B fault. Referring to figure 4.34, </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16161538841616153884.jpg\" style=\"height:545px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">the conditions for Y-B fault are:</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">I<sub>R</sub> = 0, I<sub>Y</sub> = -<sub> </sub>I<sub>B</sub>, V<sub>Y</sub> = V<sub>B</sub> + I<sub>Y</sub> Z<sub>f</sub>.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_0=\\frac{1}{3}(I_R+I_Y+I_B)=0\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">V<sub>0</sub> = - I<sub>0</sub> Z<sub>0</sub> = 0</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Thus, zero sequence current and voltage are zero.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp; I<sub>Y</sub> = - I<sub>B</sub> &nbsp;&nbsp;</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(a^2I_1+aI_2=-(aI_1+a^2I_2)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}I_1(a^2+a)=-I_2(a^2+a)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}I_1=-I_2\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (EQ 4.21)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp; V<sub>Y</sub> = V<sub>B</sub> + I<sub>Y</sub> Z<sub>f</sub> </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(a^2\\hspace{0.1cm}V_1+a\\hspace{0.1cm}V_2=a\\hspace{0.1cm}V_1+a^2\\hspace{0.1cm}V_2+(a^2\\hspace{0.1cm}I_1+a\\hspace{0.1cm}I_2)Z_f\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(V_1(a^2-a)=V_2(a^2-a)+(a^2I_1-aI_1)Z_f\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">V<sub>1</sub> = V<sub>2</sub> + I<sub>1</sub> Z<sub>f</sub>&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (Eq 4.22)</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">For satisfying equations 4.21 and 4.22, sequence networks are to be connected as shown in figure 4.35.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16161539111616153911.jpg\" style=\"height:545px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">From figure 4.35,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_1=\\frac{E}{Z_1+Z_2+Z_f}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Z<sub>f</sub> is the arc resistance due to flashover between two conductors of different phases. Z<sub>f</sub> is resistive, hence we can write R<sub>f</sub> also.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_Y=-I_B=a^2I_1+aI_2=I_1(a^2-a)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=\\frac{-j\\sqrt{3E}}{Z_1+Z_2+Z_f}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Z<sub>f</sub> (or R<sub>f</sub>) is quite small as only arc (flashover) resistance is involved. It can be 2 to 4 ohms.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>','','1','4.6.1',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5140,197,'4.6.2 delta-Wye Transformer Between Source and a Fault','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">If delta-wye transformer (figure 4.36) is present between source and a fault (L-L-L-g, L-L-L, L-L, L-L-g or L-g), the relation between positive sequence currents on primary and secondary sides of a power transformer can be seen in figure 4.37.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16161542591616154259.jpg\" style=\"height:545px; width:800px\" /><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16161542781616154278.jpg\" style=\"height:545px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">By kirchoff&rsquo;s current law, it can be written as,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_{RL}=I_R-I_B\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(I_{YL}=I_Y-I_R\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(I_{BL}=I_B-I_Y\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Comparing primary and secondary line currents, secondary line current lead respective (say, R-phase) primary line current by 30<sup>0</sup>. This delta-wye connection is also known as DY-11 or 11 o&rsquo;clock connection. This 30<sup>0</sup> phase-shift is there in any kind of fault (or for that matter normal load currents also) i.e.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;L-L-L-g, L-L-g, L-L or L-g. While carrying out fault analysis, this phase-shift has to be considered. It was not mentioned in analysis of three phase fault because the three line currents after phase shift were yet balanced. The calculation shown was of magnitude of fault currents only.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">There is another kind of delta-wye connection also which is known as DY-1 connection. In this connection I<sub>r1</sub> will lag I<sub>RL1</sub>, hence there is -30<sup>0</sup> phase shift. Unless specifically mentioned, we will assume delta-wye connection to be DY-11.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The relation between negative sequence currents on both sides of a power transformer is different. This is shown in figure 4.38. Secondary line current, in this case, lag respective primary line current by 30<sup>0</sup>. </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16164835781616483578.jpg\" style=\"height:545px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">For zero sequence values,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_{RL0}=I_{R0}-I_{B0}=0\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Similarly, I<sub>YL0</sub> and I<sub>&shy;BL0</sub> are zero. This means that zero sequence currents which are present in star side are not reflected in delta side. Repeating the same sentence in different words, third harmonic currents present in wye side are not transformed to delta (source) side. Thus, third (and hence all triplen) harmonics are suppressed by delta-wye connection and damages to electrical equipment due to third harmonic currents (or voltages) is avoided. This is one important reason why delta-wye connected transformers are used in power systems.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;</span></p>','','1','4.6.2',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5141,197,'4.6.3  Single line to Ground Fault (L-g Fault)','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">Let us consider R-g fault (figure 4.39).</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16162247931616224793.jpg\" style=\"height:545px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The conditions for R-g fault are:</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">I<sub>Y</sub> = 0, I<sub>B</sub> = 0, V<sub>R</sub> = I<sub>R</sub> Z<sub>f</sub> </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><em>a<sup>2</sup>I<sub>1</sub>+aI<sub>2</sub>+I<sub>0</sub>=aI<sub>1</sub>+a<sup>2</sup>I<sub>2</sub>+I<sub>0</sub>&nbsp; &nbsp;&nbsp;</em>&nbsp;(as I<sub>Y</sub> =I<sub>B</sub>)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}I_1(a^2-a)=I_2(a^2-a)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}I_1=I_2\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now I<sub>Y</sub> = 0</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(a^2I_1+aI_2+I_0=0 \\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}a^2I_1+aI_1+I_0=0 \\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}I_1(-1)+I_0=0 \\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}I_1=I_0 \\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}I_1=I_2=I_0 \\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}I_R=I_1+I_2+I_0=3I_1 \\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}I_1=\\frac{I_R}{3}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(V_R=V_1+V_2+V_0=I_R\\hspace{0.1cm}Z_f=3I_1Z_f\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}I_1=\\frac{V_R}{3Z_f}=\\frac{V_1+V_2+V_0}{3Z_f}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">V<sub>1</sub> + V<sub>2</sub> + V<sub>0</sub> = I<sub>1 </sub>&times; 3Z<sub>f</sub></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">These relations between sequence quantities can be satisfied if sequence networks are drawn as shown in figure 4.40.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16162248221616224822.jpg\" style=\"height:545px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">From figure 4.40,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_1=I_2=I_0=\\frac{E}{Z_1+Z_2+Z_0+3Z_f}\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">If deliberate impedance Z<sub>n</sub> is connected in the neutral circuit of power system network,</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_1=I_2=I_0=\\frac{E}{Z_1+Z_2+Z_0+3Z_f+3Z_n}\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">This fault current can be quite small in magnitude. This is because, Z<sub>1</sub> =Z<sub>2</sub> (for transmission line), Z<sub>0</sub> is approximately four times Z<sub>1</sub> for long E. H. V. lines and Z<sub>f</sub> = Flashover resistance + Tower resistance + Tower footing resistance (soil resistance which is variable and very high in case of asphalt soils). Z<sub>f</sub> can be as high as 20-50 ohms.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_R=3I_1=\\frac{3E}{Z_1+Z_2+Z_0+3Z_f+3Z_n}\\)</span></span></p>','','1','4.6.3',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5142,197,'4.6.4 Comparison Between Single Line to Ground Fault and Three Phase Fault Currents','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">Usually, power system reactance is much larger than its resistance. Therefore, we can write:</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Z<sub>1</sub> = X<sub>1</sub>, Z<sub>2</sub> = X<sub>2,</sub> Z<sub>0</sub> = X<sub>0</sub>. And assuming Z<sub>f</sub> = R<sub>f</sub> = 0,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Magnitude of fault current for a three-phase fault,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_{LLL}=\\frac{E}{X_1} \\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">And that for a single line to ground fault</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_{Lg}=\\frac{3E}{X_1+X_2+X_0+3X_n}\\)</span>&nbsp;</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">We have introduced pure reactor in the neutral circuit, which is usually the case.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now there are three possibilities:</span></p>\n\n<ol>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Solidly grounded or effectively grounded system: X<sub>n</sub> = 0, X<sub>1</sub> = X<sub>2</sub> for all static electrical equipment (i.e., transformers and transmission lines). If we assume X<sub>1</sub> = X<sub>2</sub> for generators also, which is not much far from reality. X<sub>0</sub> &lt;&lt; X<sub>1</sub> for generators. Substituting these conditions,</span></p>\n\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;<span class=\"math-tex\">\\(I_{Lg}=\\frac{3E}{2X_1}\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(I_{LLL}=\\frac{E}{X_1}\\)</span></span></p>\n\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}I_{Lg}&gt;I_{LLL}\\)</span></span></p>\n\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\">This means that a single line to ground fault at the terminals of a generator with solidly grounded neutral is more severe than a three-phase fault at the same location.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\">Non-effective grounding of neutral of the generator:</span></p>\n\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\">In this case,</span></p>\n\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_{Lg}=\\frac{3E}{2X_1+3X_n}\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;I<sub>LLL</sub> does not change as it does not depend upon X<sub>n</sub>.</span></p>\n\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\">For I<sub>LLL</sub> and I<sub>Lg</sub> to be equal,</span></p>\n\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\frac{3E}{2X_1+3X_n}=\\frac{E}{X_1}\\)</span></span></p>\n\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}X_n=\\frac{X_1}{3}\\)</span></span></p>\n	</li>\n</ol>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">If <span class=\"math-tex\">\\(X_n&lt;\\frac{X_1}{3}\\)</span>, L-g fault is more severe than L-L-L-g fault and if <span class=\"math-tex\">\\(X_n&gt;\\frac{X_1}{3}\\)</span>, L-L-L-g fault is more severe than L-g fault. The probability of L-g fault is much more than that of L-L-L-g fault. Therefore, generators are usually grounded through very high impedance (reactance). The consequence of this high value of neutral reactance is high overvoltage as shown in figure 4.41. Neutral point (star point) is at zero potential under normal operating conditions. But when single line to ground fault (or double line to ground fault) occurs as shown in figure 4.41, neutral voltage is equal to I<sub>f</sub> X<sub>n</sub> w. r. t. ground. Therefore, voltage of R phase (Y phase and B phase also) w. r. t. ground,<img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16162247391616224739.jpg\" style=\"height:545px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">V<sub>Rg</sub> = V<sub>ph-n</sub> + I<sub>f</sub> X<sub>n</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(obviously, vectorially) </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Similarly, there will be overvoltage in Y and B phase also. If the neutral impedance is very large (Z<sub>n</sub> in case of 210 MW generators is of the order of 4000 ohms), the voltage of R, Y and B phases w. r. t. ground will be approximately line to line voltage of the generator. Therefore, line to line voltages V<sub>RY</sub>, V<sub>YB</sub> and V<sub>BR</sub> will be <span class=\"math-tex\">\\(\\sqrt3\\)</span>&nbsp;times normal line to line voltage of the generator.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">This means that the insulation of the generator windings and generator connections to the generator transformer (GT) has to be designed to withstand high voltage. This adds a lot to the cost of generator. But as GT is 10-20 meters away from the generator and hence the increased insulation level is only for generator, connecting leads (bus duct) up to GT and GT primary, the high impedance grounding is preferred. In case of generator, limiting the magnitude of earth-fault currents assumes priority. </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">In general, we can conclude that all low voltage and medium voltage (6.6 kV, 11 kV) systems are non-effectively grounded. Whereas, all high voltage (66 kV, 132 kV), extra high voltage (220 kV, 400 kV) and ultra high voltage (765 kV and beyond) systems are solidly grounded. If we use non-effective grounding for such systems, the insulation cost of line insulators for line lengths ranging from 50 km to 500 km will be exorbitantly high. The fault current for earth-faults in the vicinity of transformers will be high in this case, but sensitive and fast-acting earth-fault protection can relieve the transformer from damages.</span></p>\n\n<ol start=\"3\">\n	<li style=\"text-align:justify\">\n	<p style=\"text-align:justify\"><span style=\"font-size:14px\">For faults on transmission line away from source, as X<sub>0</sub> &gt;&gt;X<sub>1</sub>, I<sub>LLL</sub> &gt;&gt; I<sub>Lg</sub></span></p>\n	</li>\n</ol>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Therefore, the electrical equipments are relieved of high earth-fault currents (when earth-faults frequently occur) and damages due to that.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\">&nbsp;</p>','','1','4.6.4',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5143,197,'4.6.5 Double Line to Ground Fault','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">Consider Y-B-g fault (figure 4.42). </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16164840251616484025.jpg\" style=\"height:675px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">conditions are:</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;&nbsp; I<sub>R</sub> =0, V<sub>Y</sub> = V<sub>B</sub> = (I<sub>Y</sub> + I<sub>B</sub>) Z<sub>f</sub>,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}\\)</span>I<sub>1</sub> + I<sub>2</sub> + I<sub>0</sub> = 0</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp; V<sub>Y</sub> = V<sub>B</sub></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}a^2V_1+aV_2+V_0=aV_1+a^2V_2+V_0\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}\\)</span>&nbsp;V<sub>1</sub> = V<sub>2</sub></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;V<sub>Y</sub> = (I<sub>Y</sub> + I<sub>B</sub>) Z<sub>f</sub> </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:12px\"><span class=\"math-tex\">\\(a^2V_1+aV_2+V_0\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span class=\"math-tex\">\\(=(a^2I_1+aI_2+I_0+aI_1+a^2I_2+I_0)Z_f\\)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}V_0-V_1=3I_0Z_f\\)</span>&nbsp;&nbsp;&hellip;&hellip;.as V<sub>1</sub> = V<sub>2</sub> and</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp;I<sub>1</sub> + I<sub>2</sub> + I<sub>0</sub> = 0</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">These relations between sequence quantities are satisfied in figure 4.43.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16164840771616484077.jpg\" style=\"height:545px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">From figure 4.43,</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_1=\\frac{E}{Z_1+\\frac{Z_2(Z_0+3Z_f)}{Z_2+Z_0+3Z_f}}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">I<sub>2</sub> and I<sub>0</sub> can be found from figure 4.43 or from figure 4.44 (circuit of figure 4.43 redrawn in a different way to understand). </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16162258811616225881.jpg\" style=\"height:460px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Then V<sub>1</sub>, V<sub>2</sub> and V<sub>0</sub> can be found and finally V<sub>R</sub>, V<sub>Y</sub>, V<sub>B</sub>, I<sub>R</sub>, I<sub>Y</sub> and I<sub>B</sub> can be calculated.</span></p>','','1','4.6.5',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5144,197,'4.6.6 Application Example','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong><u>Example 4.6:</u></strong> Refer figure 4.45. We want to calculate fault currents for L-g and L-L faults at different buses. Also fault current fed by generator windings will be calculated. Moreover, voltages at fault point will be found out.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16162399151616239915.jpg\" style=\"height:665px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Sequence networks can be drawn as follows (figure 4.46, figure 4.47 and figure 4.48):</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16162399441616239944.jpg\" style=\"height:406px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16162399701616239970.jpg\" style=\"height:406px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16162400021616240002.jpg\" style=\"height:406px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Base MVA = 150 MVA.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Base kV = 132 kV.</span></p>\n\n<p><span style=\"font-size:14px\">Base current&nbsp;<span class=\"math-tex\">\\(=\\frac{150\\times10^3}{\\sqrt3\\times132}=656\\hspace{0.1cm}amp \\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">We assume, Z<sub>f</sub> = 0, X<sub>g1</sub> = X<sub>g2</sub></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">For transformer and transmission lines, X<sub>1</sub> = X<sub>2</sub></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">For transformer, we assume X<sub>T1</sub> = X<sub>T2</sub> = X<sub>T0</sub>&hellip;..(Not far from reality)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">And for transmission lines, Z<sub>0</sub> = 2Z<sub>1</sub>.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Zero sequence current is not reflected in primary side of delta-wye transformer. That is the reason why there is a discontinuity in figure 4.48.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Let us consider L-L fault at bus D. The pu values are calculated as follows:</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_1=\\frac{1\\angle0^0}{2(0.069+j\\hspace{0.1cm}0.5875)}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=\\frac{1\\angle0^0}{2(0.59\\angle83.30^0)}\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp;<span style=\"font-size:14px\"><span class=\"math-tex\">\\(=0.847\\angle-83.30^0\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_2=-I_1=0.847\\angle96.7^0\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_R=0\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;<span class=\"math-tex\">\\(I_Y=-j\\sqrt{3I_1}\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=1\\angle-90^0\\times 1.467\\angle-83.3^0\\)</span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=1.467\\angle-173.3^0\\hspace{0.1cm}pu\\)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_B=-I_Y=1.467\\angle6.7^0\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Fault current, &nbsp;<span class=\"math-tex\">\\(\\mid I_Y\\mid =\\mid I_B \\mid=1.467\\times656=962\\hspace{0.1cm}amp \\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The fault currents flowing in the generator windings will be:</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_1=0.847\\angle -113.3^0\\hspace{0.1cm}pu\\)</span>&nbsp;as there will be -30<sup>0</sup> phase-shift.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Similarly,&nbsp;<span class=\"math-tex\">\\(I_2=0.847\\angle126.7^0\\hspace{0.1cm}pu\\)</span>, as there will be +30<sup>0</sup> phase-shift.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_R=I_1+I_2\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=0.847\\angle-113.3^0+0.847\\angle126.7^0\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=0.85\\angle186.74^0\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_Y=a^2I_1+aI_2\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;<span class=\"math-tex\">\\(=(1\\angle240^0\\times0.847\\angle-113.3^0)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(+ (1\\angle120^0\\times0.847\\angle126.7^0)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(=0.85\\angle186.74^0\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_B=aI_1+a^2I_2=1.7\\angle6.7^0\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Base current <span class=\"math-tex\">\\(=\\frac{150\\times10^3}{\\sqrt3\\times11}=7873\\hspace{0.1cm}amp\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\)</span>&nbsp; &nbsp;&nbsp;I<sub>R</sub> = 6692 amp.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;&nbsp; I<sub>Y</sub> = 6692 amp.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;&nbsp; I<sub>B</sub> = 13384 amp.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now for L-L fault at bus D, the voltage at fault point can be calculated.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(V_1=E-I_1Z_1\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=1\\angle0^0-0.847\\angle-83.3^0\\\\\\times0.59\\angle83.3^0\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=0.5\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; V<sub>2</sub> = V<sub>1</sub> = 0.5 pu.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; V<sub>0</sub> = 0</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; V<sub>R</sub> = V<sub>1</sub> + V<sub>2</sub> + V<sub>0</sub> =1.0 pu</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp;&nbsp; <span class=\"math-tex\">\\(=\\frac{132KV}{\\sqrt3}\\)</span>=76.21 kV</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(V_Y=a^2V_1+aV_2=V_1(a^2+a)\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=-V_1=-0.5\\hspace{0.1cm}pu\\)</span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=-38.1kV\\)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(V_B=-38.1kV\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now, we will write only final calculation for L-L fault on buses A, B, and C.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">I<sub>LL</sub> at bus C = 1088 amp</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">I<sub>LL</sub> at bus B = 1364 amp</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">I<sub>LL</sub> at bus A = 1818 amp</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Voltages at any point or at generator terminals for L-L fault at bus A, B, C or D can also be calculated. This exercise is left to the reader.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now let us calculate the fault current for line to ground fault (L-g) at bus D.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_1=I_2=I_0=\\frac{E}{Z_1+Z_2+Z_0}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Where&nbsp; Z<sub>1</sub> = 0.069 + j0.5875, Z<sub>2</sub> = 0.069 + j0.5875 and Z<sub>0</sub> = 0.138 + j0.65</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;Hence I<sub>1</sub> = I<sub>2</sub> = I<sub>0</sub> <span class=\"math-tex\">\\(=\\frac{1\\angle0^0}{0.276+j\\hspace{0.1cm}1.825}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=\\frac{1\\angle0^0}{1.846\\angle 81.4^0}\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=0.5417\\angle-81.4^0\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_R=I_1+I_2+I_0\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=1.625\\angle-81.4^0\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=1.625\\times656=1066\\hspace{0.1cm}amp\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;I<sub>Y</sub> = I<sub>B</sub> = 0</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The currents flowing in the windings of the generator for L-g fault at bus D can be calculated as follows:</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_1=0.5417\\angle-111.4^0\\)</span>&nbsp;pu&hellip;&hellip;..(-30<sup>0</sup> phase-shift from wye to delta)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_2=0.5417\\angle-51.4^0\\)</span>&nbsp;pu&hellip;&hellip;&hellip;..(+30<sup>0</sup> phase-shift from wye to delta)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_0=0\\)</span>&nbsp;(as zero sequence currents are not reflected in delta side)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_R=I_1+I_2\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; <span class=\"math-tex\">\\(=(0.5417\\angle-111.4^0)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <span class=\"math-tex\">\\(+ (0.5417\\angle-51.4^0)\\)</span>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=0.9375\\angle-81.4^0\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=7380.94\\angle-81.4^0\\hspace{0.1cm}amp\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_Y=a^2I_1+aI_2\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(= (1\\angle240^0\\times0.5417\\angle-111.4^0)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(+(1\\angle120^0\\times0.5417\\angle-51.4^0)\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(=0.9375\\angle98.6^0\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(=7380.94\\angle98.6^0\\hspace{0.1cm}amp\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}I_R-I_Y\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(I_B=aI_1+a^2I_2\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(=0.5417\\angle8.6^0+0.5417\\angle188.6^0\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">=0</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Thus, L-g fault on star side of power transformer (delta-wye connected) is reflected on delta side as</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;L-L fault.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(V_1=E-I_1Z_1\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp; &nbsp;<span class=\"math-tex\">\\(=1\\angle0^0-0.5417\\angle-81.4^0\\times0.59\\angle83.3^0\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=0.6807\\angle-0.926^0\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(V_2=-I_2Z_2=0.5417\\angle98.6^0\\times0.59\\angle83.3^0\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=0.3196\\angle181.9^0\\hspace{0.1cm}pu\\)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(V_0=-I_0Z_0=0.36\\angle176.6^0\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(\\dot{.\\hspace{0.045in}.}\\hspace{0.3cm}V_R=0\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(V_Y=a^2V_1+aV_2+V_0=0.994\\angle237^0\\hspace{0.1cm}pu\\)</span></span></p>\n\n<p style=\"text-align:justify\">&nbsp; &nbsp; &nbsp;&nbsp;<span class=\"math-tex\">\\(=\\frac{132}{\\sqrt3}\\times0.994=78.75kV\\)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><span class=\"math-tex\">\\(V_B=1.05\\angle120^0\\hspace{0.1cm}pu=80.02\\hspace{0.1cm}kV\\)</span></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;It can be checked that V<sub>YB</sub> = 132 kV. (V<sub>YB</sub> = V<sub>Y</sub> - V<sub>B</sub>)</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Similarly, Fault current for L-g fault at bus C = 1259 amp.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;Fault current for L-g fault at bus B = 1732 amp.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Fault current for L-g fault at bus A = 2714 amp.</span></p>','','1','4.6.6',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5145,197,'Refer to the media and select an answer.','','multi','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">If fault current due to L-L-L-g fault is I<sub>LLLg</sub> and that due to triple line fault is I<sub>LLL</sub>,</span></p>','','1','4.1',NULL,'3','3','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5146,197,'Thermal damage to the electrical equipment occurs due to...','','multi','single','','','1','4.3',NULL,'2','3','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5147,197,'Double line fault can occur because of flashover across the line insulator.','','multi','single','','','1','4.2',NULL,'2','3','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5148,197,'The most frequently occurring faults are....','','multi','single','','','1','4.3',NULL,'1','1','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5149,197,'Maximum number of faults occurring in the power system are in...','','multi','single','','','1','4.3',NULL,'4','1','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5150,197,'Fault current calculations are required','','multi','single','','','1','4.4',NULL,'4','3,6','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5151,197,'Refer to the media and select an answer.','','multi','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">If Z<sub>1</sub> = equivalent impedance of transformer referred to primary in pu, Z<sub>2</sub> = equivalent impedance of transformer referred to secondary in pu and V<sub>2</sub>/V<sub>1</sub> = K (voltage ratio of transformer),</span></p>','','1','4.5.2',NULL,'3','6','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5152,197,'In case of transmission line, when fault occurs,','','multi','single','','','1','4.6.1',NULL,'1','3,6','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5153,197,'Refer figure 4.29. Fault current for a three-phase fault at bus E is...','','multi','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163047221616304722.jpg','','1','4.5.4',NULL,'3','4,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5154,197,'Refer figure 4.29. Fault current for a triple line fault at bus D is','','multi','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163047791616304779.jpg','','1','4.5.4',NULL,'2','4,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5155,197,'Refer figure 4.29. Fault current for a triple line fault at bus A is','','multi','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163048171616304817.jpg','','1','4.5.4',NULL,'1','4,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5156,197,'Refer figure 4.29. At night, the load is less. One gas turbine power plant unit 4 is shut down and disconnected from bus. Three phase fault current, now, for a bus fault at bus B is','','multi','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163048751616304875.jpg','','1','4.5.4',NULL,'4','4,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5157,197,'Refer figure 4.30. Fault current for a triple line fault at 6.6 kV bus is...','','multi','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163049211616304921.jpg','','1','4.5.5',NULL,'2','4,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5158,197,'Refer figure 4.30. For a three-phase fault at 415 V bus, the fault current will','','multi','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163049681616304968.jpg','','1','4.5.5',NULL,'3','4,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5159,197,'Refer to the media and select an answer.','','multi','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">Refer figure 4.30. </span></p>\n\n<p style=\"text-align:justify\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163051171616305117.jpg\" style=\"height:397px; width:500px\" /></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The dynamic current for CT<sub>2</sub> to be specified is...</span></p>','','1','4.5.10',NULL,'4','4,6,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5160,197,'Refer figure 4.31. The symmetrical breaking capacity of breakers B has to be equal to','','multi','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163051851616305185.jpg','','1','4.5.6',NULL,'4','4,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5161,197,'Refer to the media and select an answer.MVAR rating required for fault limiting reactor is....','','multi','html','<ol>\n	<li style=\"text-align:justify\"><span style=\"font-size:11pt\">Refer figure 4.32. </span><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163054931616305493.jpg\" style=\"height:422px; width:500px\" /><span style=\"font-size:11pt\">Data for figure 4.32:</span></li>\n</ol>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">Generators 1, 2, 3: 11 kV, 120 MW, 0.85 pf,X&#39;<sub>d&nbsp;</sub>=0.2 pu</span></p>\n\n<p><span style=\"font-size:14px\">Generator Transformers: 11/132 kV, 150 MVA, %X =14%.</span></p>\n\n<p><span style=\"font-size:14px\">Symmetrical breaking capacity of breakers A = 1000 MVA.</span></p>\n\n<p><span style=\"font-size:14px\">Generators 1 and 2 with their GTs were connected to 132 kV bus. After 10 years, generator 3 with its GT is added. </span></p>','','1','4.5.9',NULL,'1','4,6,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5162,197,'Referring to figure 4.32, if the provision of reactor is not made, the breaking capacity required for all breakers A and B (with generating plants 1, 2 and 3) would have been...','','multi','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163055751616305575.jpg','','1','4.5.9',NULL,'3','4,6,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5163,197,'There are no zero sequence fault currents when the fault is...','','multi','single','','','1','4.6.1',NULL,'2','6,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5164,197,'The method of neutral grounding affects','','multi','single','','','1','4.6.4',NULL,'3','6,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5165,197,'The positive sequence voltage at fault point is zero in case of...','','multi','single','','','1','4.5.3',NULL,'1','6,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5166,197,'If the transmission line is transposed properly,','','multi','html','','','1','4.6.4',NULL,'3','6,7','1','1','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5167,197,'Refer to the media and select an answer.','','multi','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">For which fault, out of the faults given below, I<sub>1</sub> = I<sub>2</sub> = I<sub>0</sub>?</span></p>','','1','4.6.3',NULL,'4','6,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5168,197,'The zero-sequence current will flow through a transmission line when there is...','','multi','single','','','1','4.6.5',NULL,'4','6,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5169,197,'Refer to the media and select an answer.','','multi','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">For a single line to ground fault if I<sub>0</sub> = 2.2 pu, the current flowing in the neutral circuit of power system network will be</span></p>','','1','4.6.3',NULL,'1','6,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5170,197,'If zero sequence current is zero for a fault in power system network, the kind of fault is...','','multi','single','','','1','4.6.1',NULL,'3','6,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5171,197,'If the asymmetrical fault occurs in power system network, the fault current less than the CMR of any equipment of the system can be dangerous.','','multi','single','','','1','4.6',NULL,'1','3,6','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5172,197,'Slip frequency of the eddy currents generated in the rotor iron of a synchronous generator due to NPS currents flowing in the stator of the generator (system frequency is 50 Hz) is...','','multi','single','','','1','4.6',NULL,'4','6,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5173,197,'If there is a single line to ground fault in secondary of YY0 power transformer, zero sequence currents are not transformed/reflected in primary of the transformer.','','multi','single','','','1','4.6.2',NULL,'2','3,6','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5174,197,'Refer to the media and select an answer.The fault current for L-L fault at bus D is...','','multi','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">Refer figure 4.12. </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:11pt\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163038581616303858.jpg\" style=\"height:327px; width:500px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">100 MVA and 50 MVA transformers are DY11 and star point solidly grounded for both. For both transformers, Z<sub>1</sub> = Z<sub>2</sub> = Z<sub>0</sub>. For 132 kV line, Z<sub>0</sub> = 2Z<sub>1</sub> and for 66 kV line Z<sub>0</sub> = 1.5 Z<sub>1</sub>. Z<sub>f</sub> is assumed to be zero.</span></p>','','1','4.6.6',NULL,'1','4,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5175,197,'Refer to the media andselect an answer. The fault current for L-L fault at bus C is...','','multi','html','<p><span style=\"font-size:14px\">Refer figure 4.12.</span></p>\n\n<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163038581616303858.jpg\" style=\"height:327px; width:500px\" /></p>\n\n<p><span style=\"font-size:14px\">100 MVA and 50 MVA transformers are DY11 and star point solidly grounded for both. For both transformers, Z<sub>1</sub> = Z<sub>2</sub> = Z<sub>0</sub>. For 132 kV line, Z<sub>0</sub> = 2Z<sub>1</sub> and for 66 kV line Z<sub>0</sub> = 1.5 Z<sub>1</sub>. Z<sub>f</sub> is assumed to be zero.</span></p>','','1','4.6.6',NULL,'3','4,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5176,197,'Refer to the media and select an answer. The fault current for L-L fault at bus B is','','multi','html','<p><span style=\"font-size:14px\">Refer figure 4.12.</span></p>\n\n<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163038581616303858.jpg\" style=\"height:327px; width:500px\" /></p>\n\n<p><span style=\"font-size:14px\">100 MVA and 50 MVA transformers are DY11 and star point solidly grounded for both. For both transformers, Z<sub>1</sub> = Z<sub>2</sub> = Z<sub>0</sub>. For 132 kV line, Z<sub>0</sub> = 2Z<sub>1</sub> and for 66 kV line Z<sub>0</sub> = 1.5 Z<sub>1</sub>. Z<sub>f</sub> is assumed to be zero.</span></p>','','1','4.6.6',NULL,'4','4,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5177,197,'Refer to the media and select an answer. The fault current for L-L fault at bus A is...','','multi','html','<p>Refer figure 4.12.</p>\n\n<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163038581616303858.jpg\" style=\"height:327px; width:500px\" /></p>\n\n<p>100 MVA and 50 MVA transformers are DY11 and star point solidly grounded for both. For both transformers, Z<sub>1</sub> = Z<sub>2</sub> = Z<sub>0</sub>. For 132 kV line, Z<sub>0</sub> = 2Z<sub>1</sub> and for 66 kV line Z<sub>0</sub> = 1.5 Z<sub>1</sub>. Z<sub>f</sub> is assumed to be zero.</p>','','1','4.6.6',NULL,'2','4,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5178,197,'Refer to the media and select an answer. The fault current in B phase lines of 132 kV line for Y-B fault at bus D is...','','multi','html','<p><span style=\"font-size:14px\">Refer figure 4.12.</span></p>\n\n<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163038581616303858.jpg\" style=\"height:327px; width:500px\" /></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">100 MVA and 50 MVA transformers are DY11 and star point solidly grounded for both. For both transformers, Z<sub>1</sub> = Z<sub>2</sub> = Z<sub>0</sub>. For 132 kV line, Z<sub>0</sub> = 2Z<sub>1</sub> and for 66 kV line Z<sub>0</sub> = 1.5 Z<sub>1</sub>. Z<sub>f</sub> is assumed to be zero.</span></p>','','1','4.6.6',NULL,'1','4,6,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5179,197,'Refer to the media and select an answer. The fault current for L-g fault at bus D is ...','','multi','html','<p><span style=\"font-size:14px\">Refer figure 4.12.</span></p>\n\n<p><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163038581616303858.jpg\" style=\"height:327px; width:500px\" /></p>\n\n<p>&nbsp;</p>\n\n<p><span style=\"font-size:14px\">100 MVA and 50 MVA transformers are DY11 and star point solidly grounded for both. For both transformers, Z<sub>1</sub> = Z<sub>2</sub> = Z<sub>0</sub>. For 132 kV line, Z<sub>0</sub> = 2Z<sub>1</sub> and for 66 kV line Z<sub>0</sub> = 1.5 Z<sub>1</sub>. Z<sub>f</sub> is assumed to be zero.</span></p>','','1','4.6.6',NULL,'4','4,7','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5180,197,'11 kV, 500 MVA (breaking capacity) breaker is damaged in a substation and hence sent to manufacturer for repairs. One 66 kV, 500 MVA breaker is available in store department as spare. Will you recommend to use this breaker to replace the damaged 11 kV breaker?','','multi','single','','','1','4.5.3',NULL,'2','4,6,8','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5181,197,'<p>5 Substations</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">As already discussed, voltage of transmission is a function of the amount of power to be transmitted and the distance to which this power is required to be transmitted. This voltage is &nbsp;optimum and is an economical choice. Hence, there will be different voltages of transmission and distribution used from source to final consumer. Hence at source, voltage is stepped up and further towards distribution end, stepped down. This requires power transformers to be installed at many locations from source to load. This requires substations. Also, a power system is an interconnected network in which there are many nodes. These nodes are known as substations. </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">In this substation, many (two or more than two) transmission lines are incoming and many lines are outgoing. Voltage is stepped up or stepped down. Measurement of current, voltage, power and other electrical quantities is required. The equipment protections require protective gear. Hence there are many equipments in the substation. Substation design and project engineering is a subject by itself and shall be offered as a separate module, but we will have a bird&#39;s eye view of a substation. </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The equipment in a substation and their functions in brief are enlisted below. The relative locations of these equipment can be well appreciated in the single line diagram of figure 5.1.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; </span><span style=\"font-size:16px\">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163216051616321605.jpg\" style=\"float:left; height:1102px; width:600px\" /></span></p>','','1','5','Substations','1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5182,197,'<p>5.1 Outdoor Equipment</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">We will give a very short treatise to different outdoor equipment of the substation. Detailed treatment can be given in a separate module of Substation Design and Practices.</span></p>','','1','5.1','Outdoor Equipment','1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5183,197,'5.1.1 Lightning Arresters (L.A.)','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163217831616321783.jpg\" style=\"height:513px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">In monsoon season, there is lightning. The high frequency overvoltages are developed due to lightning. A typical wave shape of this overvoltage is shown in figure 5.2. The peak of this overvoltage can be as high as 1000 kV in the worst case. This wave of lightning overvoltage travels over transmission Line (from wherever it is incident) as a travelling wave and will reach a substation or a power station. The insulation of the electrical equipment will be intrinsically damaged by this overvoltage which is rising to its peak value within a very short time (typically 1.2 microsecond).</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">High frequency overvoltages are developed by switching actions also. Switching off or ON is inevitable in any power system either manually or automatically by operation of protective gear. These overvoltages also travel over a line, reach the substation and can damage the insulation of the equipment. </span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Lightning arresters are provided to protect the electrical equipment of a substation against overvoltages mentioned above. Lightning arrester (or Surge arrester) diverts (hence the name Surge diverter also) the energy of the overvoltage to ground, thereby reducing the peak of overvoltage. The action is so fast (5 to 10 milliseconds) that, however grounding the energy is a ground fault, protective relay does not operate so fast. LA is the first equipment at the entry of the transmission line in a substation.</span></p>','','1','5.1.1',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5184,197,'5.1.2 Earthing Switches','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">Whenever any sustained fault occurs in a transmission line, so often the disc of suspension insulator is required to be replaced or some repair is required to be carried out. Also, the scheduled maintenance requires the line to be made off. But we do not know the instant of the voltage waveshape at which the breaker was switched off. The transmission line will be charged by the voltage at the instant of switching off. This charge will, however, exponentially decay, but it takes long time as the d. c. resistance of line insulators is in terms of Megaohms.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Hence it is necessary to discharge the trapped charges in the capacitance of the transmission line. This is done by an earthing switch. After making a breaker off (or after the breaker opens due to signal received by a protective relay), an associated earthing switch is to be manually switched ON. Earthing switches on both sides of a transmission line are to be switched ON as a usual practice to discharge the trapped charges in the line.</span></p>','','1','5.1.2',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5185,197,'5.1.3 Isolators','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">Isolators are off load operating switches. They are provided on both sides of an associated breaker. When any maintenance of the breaker (cleaning of contacts, adjusting spring pressure etc.) is required to be carried out, the breaker will be made off. Referring to figure 5.1, it can be appreciated that the fixed contacts of breaker are yet live even after opening it. Hence maintenance of the breaker is possible only after opening the isolators. </span></p>\n\n<p style=\"text-align:justify\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163219011616321901.jpg\" style=\"height:918px; width:500px\" /></p>','','1','5.1.3',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5186,197,'5.1.4  Circuit Breakers','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">Circuit breakers are on-load switches. The functions of a circuit breaker are:</span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">To carry full load rated current with minimal millivolt drop across contacts of the breaker.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">To safely interrupt the rated current and the fault current at very low power factor without being damaged.</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">This is to introduce a circuit breaker. There are many more functions and details of a circuit breaker, which will be dealt with little later.</span></p>','','1','5.1.4',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5187,197,'5.1.5 Current Transformers','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">Current flowing through transmission lines ranges from100 amp to 1000 amp under normal operating conditions. This current is required to be measured. Also, it is required to measure power, energy, kVAR (MVAR) etc. No measuring instrument can handle this large current. Hence this current is stepped down to appropriate value (1 amp or 5 amp) which can be conveniently and safely measured by measuring instruments. </span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Moreover, when fault occurs in the power system, it is sensed by the protective relays, the relay will operate, give signal to the breaker which will interrupt the fault current to protect the power system. This fault current is required to be stepped down [as it is 3 to 10 times (or even more) the CMR (Continuous Maximum Rating)] to a value which a protective relay can handle.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">This stepping down of the current is carried out by the current transformer. These current transformers have single primary winding but 2, 3, 4, 5 or more (according to requirement) number of cores and secondary windings to feed to measuring instruments and protective relays. The core requirement for measurement and different protections are different. Hence, multicore C. T.s as shown in figure 5.1 are used.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16194205811619420581.jpg\" style=\"height:1469px; width:800px\" /></span></p>','','1','5.1.5',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5188,197,'5.1.6  Potential Transformers','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">As currents in power system are large, voltages are also high (11 kV, 66 kV, 132 kV, 220 kV etc.). These voltages are to be stepped down to a value (110 V) which can be fed to measuring instruments and protective relays. There are as many C.T.s required as are the number of transmission lines emanating from the substation. But in case of potential transformers, one P. T. per bus is required.</span></p>','','1','5.1.6',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5189,197,'5.1.7 Busbars','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">Busbar is a node to which feeders or transmission lines are connected. Busbar is an ACSR conductor of suitable cross-section to meet the total load of all feeders connected to it.</span></p>\n\n<p>&nbsp;</p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p style=\"text-align:left\"><span style=\"font-size:14px\"><strong>Busbar Arrangements: </strong></span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Different busbar arrangements used in practice are as follows:</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<ol>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\"><strong><u>Single Busbar Arrangement:</u> </strong>Refer figure 5.3.</span></p>\n	</li>\n</ol>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16164847331616484733.jpg\" style=\"height:694px; width:800px\" /></span></p>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\"><strong><u>Advantages: </u></strong></span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Most economical arrangement.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">The arrangement occupies less floor space.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Simple arrangement</span></p>\n	</li>\n</ul>\n\n<h1 style=\"margin-left:48px; text-align:justify\">&nbsp;</h1>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\"><strong><u>Disadvantages: </u></strong></span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">There is no flexibility offered. If there is a bus fault, there will be complete dark out for the substation. No power can be fed to any consumer from that substation where there is a bus fault.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">In case of bus fault, fault current will be fed to the fault by all transmission lines connected to that bus. This situation may lead to wide-spread instability of the power system if bus fault is not cleared fast. In the worst case, there can be cascaded tripping resulting in complete dark out of the grid.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Long down time (interruption time) in case of bus fault and related repairs.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">In case of maintenance of a circuit breaker, the concerned transmission line is required to be interrupted.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">In case of busbar maintenance, complete shut-down of substation is required.</span></p>\n	</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<ol start=\"2\">\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\"><strong><u>Single Bus Arrangement with Sectionalizing Isolator:</u></strong></span></p>\n	</li>\n</ol>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\">Such an arrangement is used for 11 kV busbars. Sectionalizing isolator is normally closed. Refer figure 5.4.</span></p>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163221321616322132.jpg\" style=\"height:695px; width:800px\" /></span></p>\n\n<p style=\"margin-left:44px; text-align:justify\"><span style=\"font-size:14px\"><strong>&nbsp;<u>&nbsp;Advantages: </u></strong></span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">The arrangement occupies less floor space.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Yet a simple scheme.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Some flexibility added w. r. t. the preceding arrangement. If there is a bus fault on the left side of the sectionalizing isolator, initially all the breakers of the bus will trip; but immediately afterwards, the sectionalizing isolator is opened and power feed from right side of the bus can be resumed by manually closing the breakers on that side.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">In case of busbar maintenance, all the breakers are required to be opened; but following this operation, sectionalizing isolator can be opened. Then, power feed can be resumed for half of the bus. Bus maintenance can be carried out on remnant half. If it is properly planned, the bus maintenance of the whole bus can be done.</span></p>\n	</li>\n</ul>\n\n<p style=\"margin-left:44px; text-align:justify\"><span style=\"font-size:14px\">&nbsp;</span></p>\n\n<p style=\"margin-left:44px; text-align:justify\"><span style=\"font-size:14px\"><strong>&nbsp;<u>Disadvantages:</u></strong> </span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Cost of three sectionalizing isolators (for three phases) is added w. r. t. preceding arrangement. But isolators are not very costly.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">There is no guarantee of power system stability as in preceding case as sectionalizing isolators cannot interrupt fault current. Hence all breakers will trip threatening system stability.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Long down time in case of bus fault (obviously for half the bus).</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">In case of maintenance of a circuit breaker, the concerned transmission line is required to be interrupted.</span></p>\n	</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<ol start=\"3\">\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\"><strong><u>Single Busbar Arrangement with Sectionalizing Breaker:</u></strong></span></p>\n	</li>\n</ol>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\">This is used for 66 kV buses and sometimes (in the old installations) for 132 kV buses too. The sectionalizing breaker is normally closed. The arrangement is shown in single line diagram of figure 5.5.<u> </u></span></p>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163230751616323075.jpg\" style=\"height:685px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<strong> <u>Features of the scheme:</u></strong> </span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">More flexibility than arrangement number 2. In case of bus fault (say, left half), breakers 1, 3, 4 and sectionalizing breaker will trip. Breakers 2, 5 and 6 will not trip. Hence, there will be no interruption of power feed from other half of the bus even for a moment. Hence power system stability is also partially guaranteed.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Busbar maintenance can be done in a scheduled mode as in case of earlier arrangement. For maintenance, sectionalizing breaker and other breakers in the portion to be maintained can be made off. The remnant half will remain live without any interruption of power.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Cost of six isolating switches and one h. v. circuit breaker results in more costly arrangement than preceding scheme.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">One bay (that for sectionalizing breaker as shown in figure 5.5) is added, hence the arrangement occupies little more space than single bus scheme.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">The arrangement is not so simple as it looks because the bus protection scheme is little complicated.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Long down time in case of bus fault (for partial bus).</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">In case of maintenance of a circuit breaker, the concerned transmission line is required to be interrupted.</span></p>\n	</li>\n</ul>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\">However, it seems that there are more demerits than merits, the arrangement is superior to the earlier ones as the improved flexibility and partial guarantee of power system stability are much more important. The chief demerit is only little more cost of equipment, land and protection.</span></p>\n\n<p style=\"margin-left:48px; text-align:justify\">&nbsp;</p>\n\n<ol start=\"4\">\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\"><strong><u>Double Bus Arrangement: </u></strong></span></p>\n	</li>\n</ol>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\">In this scheme, two sets of busbars are used (main bus I and main bus II). Both are usually identical. Also, it is a general practice that approximately half the power is transferred through bus I and the other half through bus II. New installations of 132 kV bus and exclusively all 220 kV bus installations use double bus arrangement. The single line diagram of this kind of bus arrangement is shown in figure 5.6.</span></p>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163222131616322213.jpg\" style=\"height:938px; width:800px\" /></span></p>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\"><strong><u>The Status of Breakers and Isolating Switches under Normal Operating Conditions:</u></strong></span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">B<sub>1</sub>, B<sub>2, </sub>B<sub>3, </sub>B<sub>4</sub> are closed.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">I<sub>1</sub> to I<sub>8</sub> are closed.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">BPI<sub>1</sub> to BPI<sub>4</sub> are open.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">BC is closed.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">BCI<sub>1</sub>, BCI<sub>2</sub> are closed.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">BSI<sub>1</sub>, BSI<sub>3</sub> are closed.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">BSI<sub>6</sub>, BSI<sub>8</sub> are closed.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">BSI<sub>2</sub>, BSI<sub>4</sub> are open.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">BSI<sub>5</sub>, BSI<sub>7</sub> are open.</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">The direction of power(current) flow shown in figure 5.6 is typical. Such substations are a part of interconnected network. Hence direction of power flow depends upon generating conditions (how many generators are operating) and load conditions. On both sides (incomer as shown in figure 5.6 and outgoing feeders as shown) there are generators (may be far from the substation or just at the station) and load.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now, we will discuss the sequence of operation of the breakers and isolators in case of different conditions depicted below:</span></p>\n\n<p style=\"margin-left:48px; text-align:justify\">&nbsp;</p>\n\n<ol start=\"1\" style=\"list-style-type:upper-alpha\">\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\"><strong><u>Maintenance of main breaker I:</u></strong></span></p>\n	</li>\n</ol>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Close BSI<sub>2</sub>.</span></p>\n	</li>\n	<li>\n	<p><span style=\"font-size:14px\">Close BSI<sub>4</sub>.</span></p>\n	</li>\n	<li>\n	<p><span style=\"font-size:14px\">Open BSI<sub>1</sub> and BSI<sub>3</sub>.</span></p>\n	</li>\n</ul>\n\n<p><span style=\"font-size:14px\">These isolator operations could be carried out on-load because when closing (before closing) BSI<sub>2</sub> (and hence BSI<sub>4</sub> too), the potential difference across its poles was zero. If this potential difference is high (system voltage), only off-load operation is possible. This is because there is no arc-quenching facility in the isolators. Similarly, after closing BSI<sub>2</sub>, while opening (after opening) BSI<sub>1</sub> (and for that matter BSI<sub>3</sub> also), the potential difference across poles of BSI<sub>1</sub> is zero.</span></p>\n\n<p>&nbsp;</p>\n\n<ul>\n	<li>\n	<p><span style=\"font-size:14px\">Make BC off.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Open isolators BCI<sub>1</sub> and BCI<sub>2</sub></span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now bus I is evacuated and can be maintained. There will be no interruption or discontinuity of power fed to consumers. The procedure depicted at A above is a sequence and cannot be altered.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<ol start=\"2\" style=\"list-style-type:upper-alpha\">\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\"><strong><u>There is a bus fault on main bus I:</u> </strong></span></p>\n	</li>\n</ol>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">All breakers B<sub>1</sub>, B<sub>2</sub>, B<sub>3</sub> and B<sub>4</sub> will trip due to operation of buszone protection. The power interruption will occur till the switching operation is done. However, it is not required to wait till the fault rectification is done on bus I if it is a sustained fault. Power flow can be shifted to bus II. But switching operations will take few (5 to 10) minutes and hence power system stability cannot be guaranteed.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Following switching operations (considering normal operation as original) are to be carried out.</span></p>\n\n<ul style=\"list-style-type:disc\">\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Close BSI<sub>2</sub> and BSI<sub>4</sub>.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Open BSI<sub>1</sub> and BSI<sub>3</sub>.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Make BC off.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Open BCI<sub>1</sub> and BCI<sub>2</sub>.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Now make B<sub>1</sub> to B<sub>4</sub> ON.</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now power is transferred to main bus II. Once the fault is rectified, original situation can be restored. If, during the time of rectification and repairs on bus I, bus II cannot handle total power, some load shedding is required.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<ol start=\"3\" style=\"list-style-type:upper-alpha\">\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\"><strong><u>Maintenance of Breaker (say, B<sub>1</sub>):</u></strong></span></p>\n	</li>\n</ol>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Following is the sequence of operation:</span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Close BSI<sub>2</sub>.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Open BSI<sub>1</sub>.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Close BPI<sub>1</sub>.</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">This could be done as potential difference across poles of BPI<sub>1</sub> was zero.</span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Close BSI<sub>5</sub>, Open BSI<sub>6</sub>.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Close BSI<sub>7</sub>, Open BSI<sub>8</sub>.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Make B<sub>1</sub> off.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Open I<sub>1</sub>, I<sub>2</sub>.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Change the selector switch to position &ldquo;b&ldquo; in d. c. control circuit (figure 5.7).</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now breaker B<sub>1</sub> can be maintained without interrupting power in incomer 1. Now if there is a fault in incomer 1, the fault current paths are as follows:</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<ul>\n	<li style=\"list-style-type:none\">\n	<ul>\n		<li style=\"list-style-type:none\">\n		<ul>\n			<li style=\"list-style-type:none\">\n			<ul>\n				<li style=\"list-style-type:none\">\n				<ul>\n					<li style=\"list-style-type:none\">\n					<ul>\n						<li style=\"list-style-type:none\">\n						<p>&nbsp;</p>\n						</li>\n					</ul>\n					</li>\n				</ul>\n				</li>\n			</ul>\n			</li>\n		</ul>\n		</li>\n	</ul>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">From line 3:</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">B<sub>3</sub>- BSI<sub>3</sub>- BC- BSI<sub>2</sub>- BPI<sub>1</sub>- Fault in incomer 1.</span></p>\n\n<ul>\n	<li style=\"list-style-type:none\">\n	<ul>\n		<li style=\"list-style-type:none\">\n		<ul>\n			<li style=\"list-style-type:none\">\n			<ul>\n				<li style=\"list-style-type:none\">\n				<ul>\n					<li style=\"list-style-type:none\">\n					<ul>\n						<li style=\"list-style-type:none\">\n						<p>&nbsp;</p>\n						</li>\n					</ul>\n					</li>\n				</ul>\n				</li>\n			</ul>\n			</li>\n		</ul>\n		</li>\n	</ul>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">From line 4:</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">B<sub>4</sub>- BSI<sub>7</sub>- BC- BSI<sub>2</sub> -BPI<sub>1</sub>- Fault in incomer 1.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">From incomer 2:</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">B<sub>2</sub>- BSI<sub>5</sub>- BC- BSI<sub>2</sub>- BPI<sub>1</sub>- Fault in incomer 1</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Thus, fault current is passing through bus coupler. Also, referring to figure 5.7, closing of relay contact of protective relay for incomer 1, auxiliary tripping relay will operate which will trip bus coupler and not breaker 1 (breaker 1 is already under maintenance) as selector switch (which is located in control room) is at position &ldquo;b&rdquo; (which normally was at position &ldquo;a&rdquo;). Thus, function of breaker is done by bus coupler and protection (isolation of faulty part from healthy system) is taken care of. However, bus coupler can replace only one breaker and therefore breaker maintenance of only one breaker at a time can be done.</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16194394241619439424.jpg\" style=\"height:1060px; width:800px\" /></span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">At many installations, failures of by-pass isolators were reported. Hence practice of providing such isolators is discontinued at many installations. In this case, maintenance of breaker cannot be done without power interruption in concerned line.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Now, let us summarize quickly, the features of this arrangement:</span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Very costly arrangement.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Large floor space required.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Sequence of operation of isolators and breakers and bus protection scheme are very complex.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Hundred percent flexibility. No discontinuity of power to consumers during bus maintenance. Power interruption only for few minutes in case of bus fault.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Power system stability is not guaranteed in case of bus fault.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Very short down time in case of bus fault.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Breaker can be maintained without interrupting power flowing in the transmission line.</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<ol start=\"5\">\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\"><strong><u>One and A Half Breaker Arrangement:</u></strong></span></p>\n	</li>\n</ol>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\">Figure 5.8 shows a single line diagram of the bus arrangement. There are two buses as in double bus scheme. There is no bus coupler. There are three breakers (say B<sub>1</sub>, B<sub>2</sub> and B<sub>3</sub>) for two lines (say line 1 and line 2). Thus, there are one and a half breaker for a line. That is the reason why the scheme is named as one and a half breaker scheme.</span></p>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\">Such an arrangement of bus is very costly and is used for all 400 kV busbar installations and bus installations beyond 400 kV (765 kV and 1200 kV).</span></p>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\">All breakers and isolators in figure 5.8 are ON under normal service conditions.</span></p>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163223351616322335.jpg\" style=\"height:1135px; width:800px\" /></span></p>\n\n<p style=\"margin-left:48px; text-align:justify\"><span style=\"font-size:14px\"><strong><u>Features of the scheme: </u></strong></span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Very costly scheme.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">The horizontal clearance (due to many isolators) in 400 kV (and beyond) will be very large. This would require large floor space. Hence in such a system, pantograph type isolators (figure 5.9) are used where there are vertical clearances and hence smaller floor space.<img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163231911616323191.jpg\" style=\"height:478px; width:800px\" /></span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">The arrangement is very complex and bus protection scheme is also very complex.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Highly flexible scheme. There are multiple routes of power flow. In case of bus maintenance of (say) bus I, just switch off breakers B<sub>1</sub>, B<sub>4</sub>, and B<sub>7</sub> and open isolators I<sub>1</sub>, I<sub>2</sub>, I<sub>7</sub>, I<sub>8</sub>, I<sub>13</sub>, I<sub>14</sub>. There is no discontinuity of power flow. There is neither any discontinuity of power due to fault on bus (say) I, as breakers B<sub>1</sub>, B<sub>4</sub> and B<sub>7</sub> will be tripped by buszone protection scheme.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Power system stability is fully guaranteed as the bus fault does not result in any power interruption. If a single bus cannot handle total power delivered, some load shedding may be required.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">There is no problem of down time due to fault rectification and repairs.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Breakers can be maintained without interrupting power flowing in the transmission line. More than one breaker can be maintained simultaneously. </span></p>\n	</li>\n</ul>','','1','5.1.7',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5190,197,'<p>5.2 Indoor Equipment</p>','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">We will, now, have a bird&rsquo;s eye-view on the indoor equipment required in the substation.</span></p>','','1','5.2','Indoor Equipment','1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5191,197,'5.2.1 Draw-out Type Switchgear','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">These are indoor circuit breakers which can be drawn out of the control panel after switching off a breaker. Hence no isolators are required.</span></p>','','1','5.2.1',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5192,197,'5.2.2 Measuring Instruments','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">These are ammeters, voltmeters, watt meters, VARh meters etc.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>','','1','5.2.2',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5193,197,'5.2.3 Protective Relays','','single','html','<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Directional Overcurrent and Earth-fault Relays (67, 67N)</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Non-directional Time-delayed Overcurrent Relays (51, 51N)</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Differential Relays (87)</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">Buszone relays are not shown in figure 5.1. usually, 11 kV and 66 kV buses are not provided with buszone protection because of economic reasons to be discussed later. But for 132 kV buses and beyond 132 kV, buszone protection will be provided.</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><img alt=\"\" src=\"https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16194207621619420762.jpg\" style=\"height:1469px; width:800px\" />???????</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>','','1','5.2.3',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5194,197,'5.2.4 Control Panels','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong>Following control panels are there in a typical control room:</strong></span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">One control panel for each incomer.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">One control panel for each outgoing feeder.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">One control panel for each power transformer.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">One control panel for OLTC (On-load Tap Changer) of each transformer.</span></p>\n	</li>\n</ul>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">If there are 2 incomers, 4 outgoing feeders and 2 power transformers in a substation,</span></p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\">Number of panels for these purposes will be 10.</span></p>\n\n<p style=\"text-align:justify\">&nbsp;</p>\n\n<p style=\"text-align:justify\"><span style=\"font-size:14px\"><strong>Other panels required are:</strong></span></p>\n\n<ul>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">One panel for sectionalizing breaker or bus coupler.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">One high voltage bus P. T. panel.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">One low voltage bus P. T. panel.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">One panel for battery charger.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">One rectifier panel.</span></p>\n	</li>\n	<li style=\"text-align:justify\">\n	<p><span style=\"font-size:14px\">One panel for shunt capacitors.</span></p>\n	</li>\n</ul>','','1','5.2.4',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5195,197,'5.2.5 Rectifier','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">Control supply for controlling (making ON and OFF) breakers and isolators and auto control of breakers through protective gear is always d. c., as in case of station supply failure, control is not lost. This d. c. supply (usually 220 V d. c.) is obtained using rectifier. In all substations there is 11 kV/ 415 V station transformer. This 415 V (230 V phase to neutral) supply is rectified to 220 V d. c. to feed to control circuits.</span></p>','','1','5.2.5',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5196,197,'5.2.6 Battery and Battery Charger','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">In case of station supply failure, rectifier cannot feed control circuits. Hence chemical batteries are always installed in all substations and power stations to feed the control circuits. These batteries are regularly maintained. For charging the batteries, battery charger is also installed.</span></p>','','1','5.2.6',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5197,197,'5.2.7 Shunt Capacitors','','single','html','<p style=\"text-align:justify\"><span style=\"font-size:14px\">These are installed for power factor improvement.</span></p>','','1','5.2.7',NULL,'1','2','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5198,197,'The peak of the overvoltage due to lightning can be as high as','','multi','single','','','1','5.1.1',NULL,'4','1','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5199,197,'Lightning arresters protect the electrical equipment against...','','multi','single','','','1','5.1.1',NULL,'2','1,3','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5200,197,'Isolators can be operated...','','multi','single','','','1','5.1.7',NULL,'3','6','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5201,197,'Which of the following statements is correct?','','multi','single','','','1','5.1.4',NULL,'4','1,3','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5202,197,'Refer figure 5.10. For a bus fault at F as shown, breakers','','multi','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/hint/uploads/images/16163229061616322906.jpg','','1','5.1.7',NULL,'3','3,6','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5203,197,'With reference to power system stability,','','multi','single','','','1','5.1.7',NULL,'2','1,3','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5204,197,'With reference to continuity of power in case of bus maintenance,','','multi','single','','','1','5.1.7',NULL,'4','1,3','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5205,197,'With reference to floor space required and cost of arrangement,','','multi','single','','','1','5.1.7',NULL,'1','1,3','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5206,197,'Lightning and switching surges decay to normal power frequency voltage within...','','multi','single','','','1','5.1.1',NULL,'1','1,3','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5207,197,'Earthing switches are provided in substations for the safety of operation and maintenance staff of a substation.','','multi','single','','','1','5.1.2',NULL,'1','1','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5208,197,'Isolators are provided in a substation for the breaker maintenance.','','multi','single','','','1','5.1.3',NULL,'1','1','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5209,197,'The CT core which is used for measurement purpose can also be used for feeding protective relays.','','multi','single','','','1','5.1.5',NULL,'2','1,3','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5210,197,'If transmission line is to be maintained, the sequence of operation is...','','multi','single','','','1','5.1.7',NULL,'3','6','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5211,197,'Pantograph type isolators are used in...','','multi','single','','','1','5.1.7',NULL,'4','1,3','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5212,197,'Isolators are not provided in draw-out type switchgears.','','multi','single','','','1','5.2.1',NULL,'1','1','1','0','1','2024-08-02 13:33:04','2024-08-02 13:33:04'),(5213,197,'If the transmission line is transposed properly,','','multi','single','','','1','4.6.4',NULL,'3','6,7','1','0','1','2024-08-02 13:33:05','2024-08-02 13:33:05'),(5215,179,'<p>test</p>',NULL,'single','single',NULL,NULL,'easy','1','test','3','2','1','0','1','2024-09-27 13:04:33','2024-09-27 13:04:33'),(5216,179,'<p>this is a test question</p>',NULL,'multi','single','https://edupmquestionhelp.s3.ap-south-1.amazonaws.com/assets/images/1727806122391727806122.png',NULL,'easy','1','test','1','2','1','0','1','2024-10-01 23:38:42','2024-10-01 23:38:42');

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