<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>

<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>

<p><span style="font-size:14px">Therefore,</span></p>

<p><span style="font-size:14px">pu&nbsp;<span class="math-tex">\(??MVA(kVA)=\frac{MVA_{Actual}}{MVA_{Base}} \\\)</span></span></p>

<p><span style="font-size:14px">&nbsp;Where MVA<sub>Actual</sub> = Actual MVA</span></p>

<p><span style="font-size:14px">And MVA<sub>Base</sub> = Base MVA.</span></p>

<p><span style="font-size:14px">pu&nbsp;<span class="math-tex">\(Impedance=\frac{Z_{Actual}}{Z_{Base}} \\\)</span></span></p>

<p><span style="font-size:14px">&nbsp;&nbsp;&nbsp;Where Z<sub>Actual</sub> = Actual impedance in ohms</span></p>

<p><span style="font-size:14px">&nbsp;&nbsp;&nbsp;And Z<sub>Base</sub> = Base impedance in ohms.</span></p>

<p><span style="font-size:14px">Similar is the relation of current and voltage.</span></p>

<p><span style="font-size:14px"><strong><u>How to select base values?</u></strong></span></p>

<p>&nbsp;</p>

<p><span style="font-size:14px">Out of four system quantities,</span></p>

<p>&nbsp;</p>

<ul>
	<li>
	<p><span style="font-size:14px">MVA (kVA)</span></p>
	</li>
	<li>
	<p><span style="font-size:14px">kV (Line to line voltage)</span></p>
	</li>
	<li>
	<p><span style="font-size:14px">I (Line current)</span></p>
	</li>
	<li>
	<p><span style="font-size:14px">Z (Impedance)</span></p>
	</li>
</ul>

<p>&nbsp;</p>

<p><span style="font-size:14px">Base MVA (kVA) and base kV are to be selected. </span></p>

<p><span style="font-size:14px">Then,</span></p>

<p><span style="font-size:14px"><span class="math-tex">\(Base Current=\frac{Base KVA}{\sqrt{3}\times BasekV} \)</span></span></p>

<p><span style="font-size:14px"><span class="math-tex">\(Base Impedance=\frac{V_{ph}}{I} \)</span></span></p>

<p><span style="font-size:14px">Where V<sub>ph</sub> = ph-n value of base voltage</span></p>

<p><span style="font-size:14px">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;And I = Base current&nbsp;&nbsp;</span></p>

<p><span style="font-size:14px">Therefore,&nbsp; &nbsp;</span></p>

<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>

<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>

<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>

<p><span style="font-size:14px">where V = line to line voltage.</span></p>

<p><span style="font-size:14px">Actual Impedance in ohms</span></p>

<p><span style="font-size:14px">= pu Impedance&nbsp;<span class="math-tex">\(\times\frac{V^2}{VA}\)</span></span></p>

<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>

<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>

<p><span style="font-size:14px">Where kV= Line to line voltage in kV</span></p>

<p><span style="font-size:14px">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp;MVA= Three phase MVA</span></p>

<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>

<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>

<p>&nbsp;</p>

<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>

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

<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>

<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>

<p><span style="font-size:14px">Equating these two equations, we get pu impedance referred to new base,</span></p>

<p>&nbsp;</p>

<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>

<p>&nbsp;</p>