<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>

<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>

<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>

<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>

<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>

<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>

<p style="text-align:justify"><span style="font-size:14px">Base MVA = 150 MVA.</span></p>

<p style="text-align:justify"><span style="font-size:14px">Base kV = 132 kV.</span></p>

<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>

<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>

<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>

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

<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>

<p style="text-align:justify"><span style="font-size:14px">And for transmission lines, Z<sub>0</sub> = 2Z<sub>1</sub>.</span></p>

<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>

<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>

<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>

<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>

<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>

<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>

<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>

<p style="text-align:justify"><span style="font-size:14px">&nbsp;<span class="math-tex">\(I_Y=-j\sqrt{3I_1}\)</span></span></p>

<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>

<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>

<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>

<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>

<p style="text-align:justify"><span style="font-size:14px">The fault currents flowing in the generator windings will be:</span></p>

<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>

<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>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(I_R=I_1+I_2\)</span></span></p>

<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>

<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>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(I_Y=a^2I_1+aI_2\)</span></span></p>

<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>

<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>

<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>

<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>

<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>

<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>

<p style="text-align:justify"><span style="font-size:14px">&nbsp;&nbsp;&nbsp;&nbsp; I<sub>Y</sub> = 6692 amp.</span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp;&nbsp;&nbsp;&nbsp; I<sub>B</sub> = 13384 amp.</span></p>

<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>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(V_1=E-I_1Z_1\)</span></span></p>

<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>

<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>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; V<sub>2</sub> = V<sub>1</sub> = 0.5 pu.</span></p>

<p style="text-align:justify"><span style="font-size:14px">&nbsp; V<sub>0</sub> = 0</span></p>

<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>

<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>

<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>

<p style="text-align:justify">&nbsp; &nbsp; &nbsp;&nbsp;<span class="math-tex">\(=-V_1=-0.5\hspace{0.1cm}pu\)</span></p>

<p style="text-align:justify">&nbsp; &nbsp; &nbsp;&nbsp;<span class="math-tex">\(=-38.1kV\)</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(V_B=-38.1kV\)</span></span></p>

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

<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>

<p style="text-align:justify"><span style="font-size:14px">I<sub>LL</sub> at bus C = 1088 amp</span></p>

<p style="text-align:justify"><span style="font-size:14px">I<sub>LL</sub> at bus B = 1364 amp</span></p>

<p style="text-align:justify"><span style="font-size:14px">I<sub>LL</sub> at bus A = 1818 amp</span></p>

<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>

<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>

<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>

<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>

<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>

<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>

<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>

<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>

<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>

<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>

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

<p style="text-align:justify"><span style="font-size:14px">&nbsp;I<sub>Y</sub> = I<sub>B</sub> = 0</span></p>

<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>

<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>

<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>

<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>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(I_R=I_1+I_2\)</span></span></p>

<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>

<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>

<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>

<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>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(I_Y=a^2I_1+aI_2\)</span></span></p>

<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>

<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>

<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>

<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>

<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>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(I_B=aI_1+a^2I_2\)</span></span></p>

<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>

<p style="text-align:justify"><span style="font-size:14px">=0</span></p>

<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>

<p style="text-align:justify"><span style="font-size:14px">&nbsp;L-L fault.</span></p>

<p style="text-align:justify"><span style="font-size:14px"><span class="math-tex">\(V_1=E-I_1Z_1\)</span></span></p>

<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>

<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>

<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>

<p style="text-align:justify">&nbsp; &nbsp; &nbsp;&nbsp;<span class="math-tex">\(=0.3196\angle181.9^0\hspace{0.1cm}pu\)</span></p>

<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>

<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>

<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>

<p style="text-align:justify">&nbsp; &nbsp; &nbsp;&nbsp;<span class="math-tex">\(=\frac{132}{\sqrt3}\times0.994=78.75kV\)</span></p>

<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>

<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>

<p style="text-align:justify"><span style="font-size:14px">Similarly, Fault current for L-g fault at bus C = 1259 amp.</span></p>

<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>

<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>