<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>

<p>&nbsp;</p>

<ul>
	<li>
	<p><span style="font-size:14px">Generators</span></p>
	</li>
	<li>
	<p><span style="font-size:14px">Power Transformers</span></p>
	</li>
	<li>
	<p><span style="font-size:14px">Distribution Transformers</span></p>
	</li>
	<li>
	<p><span style="font-size:14px">Circuit Breakers</span></p>
	</li>
	<li>
	<p><span style="font-size:14px">Current Transformers (C. T.s)</span></p>
	</li>
	<li>
	<p><span style="font-size:14px">Transmission Lines</span></p>
	</li>
	<li>
	<p><span style="font-size:14px">Isolators</span></p>
	</li>
	<li>
	<p><span style="font-size:14px">Busbars</span></p>
	</li>
	<li>
	<p><span style="font-size:14px">Protective Relays</span></p>
	</li>
</ul>

<p>&nbsp;</p>

<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>

<p>&nbsp;</p>

<ol>
	<li>
	<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>
	</li>
</ol>

<p>&nbsp;</p>

<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>

<p>&nbsp;</p>

<p>&nbsp;</p>

<ol start="2">
	<li>
	<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>
	</li>
</ol>

<p>&nbsp;</p>

<ol start="3">
	<li>
	<p><span style="font-size:14px">Protective relay settings are calculated based on magnitude and power factor of the fault current.</span></p>
	</li>
</ol>

<p>&nbsp;</p>

<p>&nbsp;</p>

<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>