<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>

<ol>
	<li style="text-align:justify">
	<p><span style="font-size:14px">Short time withstand capacity of all 11 kV current transformers.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">Symmetrical breaking capacity (in MVA) of all 11 kV breakers.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">Short time withstand capacity of both 66/11 kV power transformers.</span></p>
	</li>
</ol>

<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>

<p style="text-align:justify"><span style="font-size:14px"><strong><u>Solution: </u></strong></span></p>

<p style="text-align:justify"><span style="font-size:14px">Using Equation 4.17,</span></p>

<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>

<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>

<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>

<p style="text-align:justify"><span style="font-size:14px">pu reactance of each power transformer on 500 MVA base,</span></p>

<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>

<p style="text-align:justify"><span style="font-size:14px">Equivalent impedance = j 1.5 pu.</span></p>

<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>

<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>

<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>

<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>

<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>