The 500 kVA, 69 kV/4160 V, 60 Hz transformer shown in Fig. 30 has a resistance $R_{ p } \text { of } 150 \Omega$ and a leakage reactance $X_{ p } \text { of } 632 \Omega$ . Using the per-unit method, calculate

a. the voltage regulation when the load varies between zero and 250 kVA at a lagging power factor of 80%
b. the actual voltage across the 250 kVA load
c. the actual line current $I_{1}$

Question

The 500 kVA, 69 kV/4160 V, 60 Hz transformer shown in Fig. 30 has a resistance [latex]R_{ p } ext { of } 150 \Omega[/latex] and a leakage reactance [latex]X_{ p } ext { of } 632 \Omega[/latex]. Using the per-unit method, calculate

a. the voltage regulation when the load varies between zero and 250 kVA at a lagging power factor of 80%
b. the actual voltage across the 250 kVA load
c. the actual line current [latex]I_{1}[/latex]

Accepted Answer

In examining Fig. 30, it is clear that the presence of the magnetizing branch does not affect the voltage drop across [latex]R_{ p } ext { and } X_{ p } ext {. }[/latex] Consequently, the magnetizing branch does not affect the voltage regulation.

To determine the voltage regulation, we will refer all voltages, impedances, and currents to the HV (69 kV) side. We assume the voltage between terminals 1, 2 is 69 kV, and that it remains fixed.

The base power [latex]P_{ B } ext { is } 500 kVA ext {. }[/latex]

The base voltage [latex]E_{ B } ext { is } 69 kV[/latex].

Consequently, the base current is

[latex]\begin{aligned}I_{ B } &=P_{ B } / E_{ B }=500000 / 69000 \&=7.25 A\end{aligned}[/latex]

and the base impedance is

[latex]Z_{ B }=E_{ B } / I_{ B }=69000 / 7.25=9517 \Omega[/latex]

The per-unit value of [latex]R_{ p } ext { is }[/latex]

[latex]R_{ p }( pu )=150 / 9517=0.0158[/latex]

The per-unit value of [latex]X_{ p } ext { is }[/latex]

[latex]X_{ p }( pu )=632 / 9517=0.0664[/latex]

The per-unit value of voltage [latex]E_{ 12 } ext { is }[/latex]

[latex]E_{12}( pu )=69000 / 69 kV =1.0[/latex]

The per-unit value of the apparent power absorbed by the load is

[latex]S( pu )=250 kVA / 500 kVA =0.5[/latex]

The per-unit value of the active power absorbed by the load is

[latex]P( pu )=S( pu ) \cos heta=0.5 imes 0.8=0.4[/latex]

The per-unit value of the reactive power absorbed by the load is

[latex]\begin{aligned}Q( pu ) &=\sqrt{S^{2}( pu )-P^{2}( pu )} \&=\sqrt{0.5^{2}-0.4^{2}} \&=0.3\end{aligned}[/latex]

The per-unit load resistance [latex]R_{ L }[/latex] corresponding to P is

[latex]R_{ L }( pu )=\frac{E^{2}( pu )}{P( pu )}=\frac{1.0^{2}}{0.4}=2.50[/latex]

The per-unit load reactance [latex]X_{ L }[/latex] corresponding to Q is

[latex]X_{ L }( pu )=\frac{E^{2}( pu )}{Q( pu )}=\frac{1.0^{2}}{0.3}=3.333[/latex]

We now draw the equivalent per-unit circuit shown in Fig. 33. The magnetizing branch is not shown because it does not enter into the calculations. Note that the load appears across the primary terminals 3, 4 of the circuit shown in Fig. 30. (These terminals are not accessible; they exist only in the equivalent circuit diagram.) The per-unit impedance between terminals 3, 4 is

[latex]\begin{aligned}Z_{34}( pu ) &=\frac{2.50 imes j 3.33}{250+ j 3.33} \&=2 \angle 36.87^{\circ} \&=1.6+ j 1.2\end{aligned}[/latex]

The per-unit impedance between terminals 1, 2 is

[latex]\begin{aligned}Z_{12}( pu )=& 0.0158+1.6 \&+ j (1.2+0.0664) \=& 1.616+ j 1.266 \=& 2.053 \angle 38.07^{\circ}\end{aligned}[/latex]

The per-unit current [latex]I_{1} ext { is }[/latex]

[latex]\begin{aligned}I_{1}( pu ) &=\frac{E_{12}( pu )}{Z_{12}( pu )}=\frac{1.0}{2.053 \angle 38.07^{\circ}} \&=0.4872 \angle-38.07^{\circ}\end{aligned}[/latex]

The per-unit voltage [latex]E_{34}[/latex] across the load is

[latex]\begin{aligned}E_{34}( pu ) &=I_{1}( pu ) imes Z_{34}( pu ) \&=\left(0.4872 \angle-38.07^{\circ} ight)\left(2 \angle 36.87^{\circ} ight) \&=0.9744 \angle-1.20^{\circ}\end{aligned}[/latex]

The per-unit voltage regulation is

[latex]\frac{E_{34}( pu ) ext { at no-load }-E_{34}( pu ) ext { at full-load }}{E_{34}( pu ) ext { at full-load }}[/latex]

[latex]=\frac{1.0-0.9744}{0.9744}=0.0263[/latex]

a. The voltage regulation is therefore 2.63%.

We can now calculate the actual values of the voltage and current as follows:
Voltage across terminals 3, 4 is

[latex]\begin{aligned}E_{34} &=E_{34}( pu ) imes E_{ B } \&=0.9744 imes 69000 \&=67.23 kV\end{aligned}[/latex]

b. Actual voltage across the load is

[latex]\begin{aligned}E_{56} &=E_{34} imes(4160 / 69000) \&=67.23 imes 10^{3} imes 0.0603 \&=4054 V\end{aligned}[/latex]

c. Actual line current is

[latex]\begin{aligned}I_{1} &=I_{1}( pu ) imes I_{ B } \&=0.4872 imes 7.246 \&=3.53 A\end{aligned}[/latex]

Question 10.10: The 500 kVA, 69 kV/4160 V, 60 Hz transformer shown in Fig. 3...

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