<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>

<ol>
	<li style="text-align:justify">
	<p><span style="font-size:14px">pu value of reactor.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">MVAR capacity of reactor.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">Ohmic value of reactor.</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">The inductance of reactor</span></p>
	</li>
	<li style="text-align:justify">
	<p><span style="font-size:14px">The breaking capacity required for breakers B.</span></p>
	</li>
</ol>

<p>&nbsp;</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/16160639021616063902.jpg" style="height:491px; width:800px" /></span></p>

<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>

<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>

<p style="text-align:justify"><span style="font-size:14px">Source impedance is zero as transformer is connected to an infinite bus.</span></p>

<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>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify"><span style="font-size:14px">Transformer reactance =0.1 pu.</span></p>

<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>

<p style="text-align:justify"><span style="font-size:14px">Referring to figure 4.24,</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/16160639511616063951.jpg" style="height:491px; width:800px" /></span></p>

<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>

<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>

<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>

<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>

<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>

<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>

<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>

<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>

<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>

<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>

<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>

<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>

<p style="text-align:justify"><span style="font-size:14px">Henry = 69.3 mH</span></p>

<p style="text-align:justify"><span style="font-size:14px">For a three-phase fault at F<sub>2</sub>,</span></p>

<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>

<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>

<p style="text-align:justify">&nbsp;</p>

<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>

<p style="text-align:justify">&nbsp;</p>

<p style="text-align:justify">&nbsp;</p>