Chapter 12, Problem 82CP

A balanced three-phase system has a distribution wire with impedance 2 + j6 Ω per phase. The system supplies two three-phase loads that are connected in parallel. The first is a balanced wye-connected load that absorbs 400 kVA at a power factor of 0.8 lagging. The second load is a balanced delta-connected load with impedance of 10 + j8 Ω per phase. If the magnitude of the line voltage at the loads is 2400 V rms, calculate the magnitude of the line voltage at the source and the total complex power supplied to the two loads.

To determine

Find the magnitude of the line voltage at the source and the total complex power supplied to the two loads.

Answer to Problem 82CP

The magnitude of the line voltage is $4.959\text{ kV}$ at the source and the total complex power supplied to the loads are $\underset{\_}{1.37377-j1.0829\text{MVA}}$.

Explanation of Solution

Given data:

Three-phase distribution system has line impedance per phase is $2+j6\text{Ω}$. The two balanced loads are,

Load 1:

The apparent power of the three phase load is $400\text{kVA}$.

The power factor is $0.8$ (lagging).

Load 2:

The impedance per phase for delta connected load is $10+j8\text{Ω}$.

The magnitude of the line voltage at loads is $2400\text{V}\left(\text{rms}\right)$.

Formula used:

Load 1: Wye-connected

Write the expression to find the real power $\left(P_1\right)$ of the load 1.

$P_1=S_1\cos {\theta }_1$        (1)

Here,

$S_1$ is the apparent power of the star connected load 1.

$\cos {\theta }_1$ is the power factor of the load 1.

Write the expression to find the reactive power $\left(Q_1\right)$ of the load 1.

$Q_1=S_1\sin {\theta }_1$        (2)

Here,

${\theta }_1$ is the power factor angle of load 1.

Write the expression to find the complex power of the three-phase load.

$S_1=P_1+j{Q}_1$        (3)

Here,

$P_1$ is the real power, and

$Q_1$ is the reactive power.

Load 2: Delta-connected

Write the expression to find the apparent power $S_2$ of Load 2.

$S_2=\frac{3V_p^2}{Z_p^{*}}$        (4)

Here,

$V_p$ is the phase voltage at load 2 , and

$Z_p^{*}$ is the conjugate of the phase impedance $Z_p$.

The phase voltage is equal to line voltage for delta connection.

$V_p=V_L$

Write the expression to find the total apparent power $S$.

$S=S_1+S_2$        (5)

Here,

$S_1$, and $S_2$ are apparent powers of load 1 and load 2.

Write the expression to find the apparent power for the wye-connected load 1.

$S_1=3V_p{I}_1^{*}$        (6)

Here,

$I_1^{*}$ is the conjugate of the current $I_1$.

The phase voltage and line voltage relation for wye-connection is,

$V_p=\frac{V_L}{\sqrt{3}}$

Substitute $\frac{V_L}{\sqrt{3}}$ for $V_p$ in equation (6).

$S_1=3\left(\frac{V_L}{\sqrt{3}}\right)I_1^{*}$

$S_1=\sqrt{3}{V}_L{I}_1^{*}$

$I_1^{*}=\frac{S_1}{\sqrt{3}{V}_L}$        (7)

$I_1$ is a complex conjugate of the current $I_1^{*}$.

Write the expression to find the line current equivalent wye-load of the delta-connected load 2.

$I_2=I_p\sqrt{3}\angle -30°$        (8)

Here,

$I_p$ is the phase current.

Write the expression to find the phase current of load 2.

$I_p=\frac{V_p}{Z_{\Delta }}$

Here,

$Z_{\Delta }$ is the impedance for the delta connected load.

Substitute $\frac{V_p}{Z_{\Delta }}$ for $I_p$ in above equation

$I_2=\frac{V_p}{Z_{\Delta }}\sqrt{3}\angle -30°$        (9)

Write the expression to find the total current $\left(I\right)$.

$I=I_1+I_2$        (10)

Here,

$I_1$ and $I_2$ are the currents.

Write the expression to find the source voltage $V_s$.

$V_s=V_L+V_{\text{line}}$        (11)

Here,

$V_L$ is the line voltage, and

$V_{\text{line}}$ is the line voltage of the delta connected load.

Write the expression to find the line voltage of the delta connected load.

$V_{\text{Line}}=I_{\text{Line}}{Z}_{\text{Line}}$        (12)

Here,

$I_{\text{Line}}$ is the line current, and

$Z_{\text{Line}}$ is the line impedance.

Calculation:

Load1:

Substitute $400\text{kVA}$ for $S_1$, and $0.8$ for $\cos {\theta }_1$ in equation (1) to find $P_1$.

$\begin{array}{c}{P}_1=\left(400\text{kVA}\right)\left(0.8\right)\\ =320\text{kVA}\end{array}$

The given leading power factor is,

$\begin{array}{l}\cos {\theta }_1=0.8\\ {\theta }_1={\cos }^{-1}\left(0.8\right)\\ {\theta }_1=39.86°\end{array}$

Substitute $39.86°$ for ${\theta }_1$, and $400\text{kVA}$ for $S_1$ in equation (2) to find the reactive power $Q_1$.

$\begin{array}{c}{Q}_1=\left(400\text{kVA}\right)\sin \left(36.86°\right)\\ =\left(400\text{kVA}\right)\left(0.6\right)\\ =240\text{kVAR}\end{array}$

Substitute $320\text{kVA}$ for $P_1$, and $240\text{kVAR}$ for $Q_1$ in equation (3) to find $S_1$.

$\begin{array}{c}{S}_1=320\text{kVA}+j240\text{kVAR}\\ =320+j240\text{kVA}\end{array}$

Substitute $2400\text{V}$ for $V_L$, $320+j240\text{kVA}$ for $S_1$ in equation (7).

$\begin{array}{l}{I}_1^{*}=\frac{320+j240\text{kVA}}{\sqrt{3}\left(2400\text{V}\right)}\\ I_1^{*}\cong 76.98+j57.735\text{A}\end{array}$

Thus, the line current $I_1$ is $76.98-j57.735\text{A}$.

Load 2:

Substitute $2400\text{V}$ for $V_L$ and $10-j8\text{Ω}$ for $Z_p^{*}$ in equation (4) to find $S_2$.

$\begin{array}{c}{S}_2=3\frac{{\left(2400\text{V}\right)}^2}{10-j8\text{Ω}}\left\{\because Z_p=Z_p^{*}\right\}\\ =3\left(\frac{5760\times {10}^3{\text{V}}^2}{12.806\angle -38.66°\frac{\text{V}}{\text{A}}}\right)\left\{\because 1\text{Ω}=1\frac{\text{V}}{\text{A}}\right\}\\ =3\left(449789.16\angle 38.66°\right)\text{ VA}\\ S_2=1349367.483\angle 38.66°\text{VA}\end{array}$

Substitute $2400\text{V}$ for $V_p$, and $10+j8\text{Ω}$ for $Z_{\Delta }$ in equation (9) to find the line current $I_2$.

$\begin{array}{c}{I}_2=\left(\frac{2400}{10+j8}\text{A}\right)\sqrt{3}\angle -30°\\ =\left(\frac{2400}{12.81\angle 38.66°}\right)\sqrt{3}\angle -30°\\ =\left(187.353\angle -38.66°\right)\left(\sqrt{3}\angle -30°\right)\text{A}\\ I_2=324.505\angle -68.66°\text{A}\end{array}$

The current $I_2$ is,

$I_2=118.087-j302.256\text{A}$

Substitute $118.087-j302.256\text{A}$ for $I_2$ and $76.98-j57.735\text{A}$ for $I_1$ in equation (10) to find $I$.

$\begin{array}{c}I=\left(118.087-j302.256\text{A}\right)+\left(76.98-j57.735\text{A}\right)\\ =195.067-j359.991\text{A}\end{array}$

The line current is $I=I_{\text{line}}$.

Substitute $195.067-j359.991\text{A}$ for $I_{\text{line}}$ and $2+j6\text{Ω}$ for $Z_{\text{line}}$ in equation (12) to find $V_{\text{Line}}$.

$\begin{array}{c}{V}_{\text{Line}}=\left(195.067-j359.991\text{A}\right)\left(2+j6\text{Ω}\right)\\ =\left(409.444\angle -61.54°\text{A}\right)\left(6.324\angle 71.56°\text{Ω}\right)\\ V_{\text{Line}}=\text{2589}\text{.323}\angle \text{10}\text{.02}°\text{V}\end{array}$

Substitute $\text{2589}\text{.323}\angle \text{10}\text{.02}°\text{V}$ for $V_{\text{Line}}$ and $2400\angle 0°\text{V}$ for $V_L$ in equation (11) to find $V_s$.

$\begin{array}{c}{V}_s=\left(2400\text{V}\right)+\left(\text{2589}\text{.323}\angle \text{10}\text{.02}°\text{V}\right)\\ =\left(2400\text{V}\right)+\left(2549.828+j450.521\text{V}\right)\\ =4959+j450.521\text{V}\\ =4.959\angle 5.2°\text{ kV}\end{array}$

The magnitude of the line voltage at the source is $4.959\text{ kV}$.

Substitute $1349367.483\angle 38.66°\text{VA}$ for $S_2$ and $320+j240\text{kVA}$ for $S_1$ in equation (5) to find total complex power $S$.

$\begin{array}{c}S=\left(1349367.483\angle 38.66°\text{VA}\right)+\left(320+j240\text{kVA}\right)\\ =\left(1053676.158+j842946.71\text{VA}\right)+\left(320000+j240000\text{VA}\right)\\ =1373676.158-j1082946.71\text{VA}\\ S=1.37377-j1.0829\text{MVA}\end{array}$

Conclusion:

Thus, the magnitude of the line voltage is $4.959\text{ kV}$ at the source and the total complex power supplied to the loads are $\underset{\_}{1.37377-j1.0829\text{MVA}}$.