Chapter 12, Problem 81CP

A professional center is supplied by a balanced three-phase source. The center has four balance three-phase loads as follows:

Load 1: 150 kVA at 0.8 pf leading

Load 2: 100 kW at unity pf

Load 3: 200 kVA at 0.6 pf lagging

Load 4: 80 kW and 95 kVAR (inductive)

If the line impedance is 0.02 + j0.05 Ω per phase and the line voltage at the loads is 480 V, find the magnitude of the line voltage at the source.

To determine

Find the magnitude of the line voltage at the source.

Answer to Problem 81CP

The magnitude of the line voltage at the source is $\underset{\_}{516\text{V}}$.

Explanation of Solution

Given data:

A balanced three-phase source connected to four balanced three-phase loads, Those are,

Load 1:

The apparent power of the Load 1 $\left(S_1\right)$ is $150\text{kVA}$.

The power factor of the Load 1 is $0.8\left(\text{leading}\right)$.

Load 2:

The real power of the Load 2 $\left(P_2\right)$ is $100\text{kW}$.

The power factor of the Load 2 is unity.

Load 3:

The apparent power of the Load 3 $\left(S_3\right)$ is $200\text{kVA}$.

The power factor of the Load 3 is 0.6 (lagging).

Load 4:

The reactive power of the Load 4 $\left(S_4\right)$ is $95\text{kVAR}$ (inductive).

The real power of the Load 4 $\left(P_4\right)$ is $80\text{kW}$.

The line impedance $\left(Z_l\right)$ is $0.02+j0.05\text{Ω}$.

The line voltage at the loads $\left(V_L\right)$ is $480\text{V}$.

Formula used:

Write the expression to find the complex power of the Load 1.

$S_1=P_1+j{Q}_1$        (1)

Here,

$P_1$ is the real power of the Load 1.

$Q_1$ is the reactive power of the Load 1.

Write the expression to find the real power of the Load 1.

$P_1=S_1\cos {\theta }_1$        (2)

Here,

$S_1$ is the apparent power of the Load 1.

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

Write the expression to find the reactive power of the Load 1.

$Q_1=S_1\sin {\theta }_1$        (3)

Here,

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

Write the expression to find the complex power of the Load 2.

$S_2=P_2+j{Q}_2$        (4)

Here,

$P_2$ is the real power of the load 2.

$Q_2$ is the reactive power of the Load 2.

Write the expression to find the real power $\left(P_2\right)$.

$P_2=S_2\cos {\theta }_2$        (5)

Here,

$S_2$ is the apparent power of the Load 2.

$\cos {\theta }_2$ is the power factor of the Load 2.

Rearrange the equation (5) to find the apparent power $\left(S_2\right)$.

$S_2=\frac{P_2}{\cos {\theta }_2}$        (6)

Write the expression to find the reactive power $\left(Q_2\right)$.

$Q_2=S_2\sin {\theta }_2$        (7)

Here,

${\theta }_2$ is the power factor angle of the Load 2.

Write the expression to find the complex power of the Load 3.

$S_3=P_3+j{Q}_3$        (8)

Here,

$P_3$ is the real power of the Load 3.

$Q_3$ is the reactive power of the Load 3.

Write the expression to find the real power of the Load 3.

$P_3=S_3\cos {\theta }_3$        (9)

Here,

$S_3$ is the apparent power of the Load 3.

$\cos {\theta }_3$ is the power factor of the Load 3.

Write the expression to find the reactive power of the Load 3.

$Q_3=S_3\sin {\theta }_3$        (10)

Here,

${\theta }_3$ is the power factor angle of the Load 3.

Write the expression to find the complex power of the Load 4.

$S_4=P_4+j{Q}_4$        (11)

Here,

$P_4$ is the real power of the Load 4.

$Q_4$ is the real power of the Load 4.

Write the expression to find the total complex power $\left(S\right)$ absorbed by the load.

$S=S_1+S_2+S_3+S_4$        (12)

Here,

$S_1$ is the complex power of the Load 1.

$S_2$ is the complex power of the Load 2.

$S_3$ is the complex power of the Load 3.

$S_4$ is the complex power of the Load 4.

Write the expression to find the apparent power (S).

$S=\sqrt{3}{V}_L{I}_L$        (13)

Here,

$V_L$ is the line voltage.

$I_L$ is the line current.

Write the expression to find the complex power absorbed by the line.

$S_l=3I_L^2{Z}_L$        (14)

Here,

$Z_L$ is the line impedance.

Write the expression to find the total complex power at the source.

${\left(S_T\right)}_{\text{source}}=S_T+S_l$        (15)

Here,

$S_T$ is the total complex power absorbed by the load.

$S_l$ is the complex power absorbed by the line.

Write the expression for the apparent power $\left(S_T\right)$ of the source.

$S_T=\sqrt{3}{V}_T{I}_L$        (16)

Here

$V_T$ is the line voltage at the source.

$I_L$ is the line current.

Calculation:

Load 1:

The given leading power factor of the Load 1 is ,

$\cos {\theta }_1=0.8$

Re-write the equation to find the angle ${\theta }_1$.

$\begin{array}{c}{\theta }_1={\cos }^{-1}\left(0.8\right)\\ {\theta }_1=-36.86°\end{array}$

Substitute $-36.86°$ for ${\theta }_1$, and $150\text{kVA}$ for $S_1$ in equation (3) to find the reactive power of the Load 1.

$\begin{array}{c}{Q}_1=\left(150\text{kVA}\right)\sin \left(-36.86°\right)\left\{\because 1\text{k}={10}^3\right\}\\ =\left(150\times {10}^3\text{VA}\right)\left(-0.6\right)\\ =-90\text{kVAR}\end{array}$

Substitute $150\text{kVA}$ for $S_1$, and $0.8$ for $\cos {\theta }_1$ in equation (2) to find the real power $\left(P_1\right)$.

$\begin{array}{c}{P}_1=\left(150\text{kVA}\right)\left(0.8\right)\left\{\because 1\text{k}={10}^3\right\}\\ =\left(150\times {10}^3\text{VA}\right)\left(0.8\right)\left\{\because 1\text{W=1}\text{VA}\right\}\\ =120\text{kW}\end{array}$

Substitute $120\text{kW}$ for $P_1$, and $-90\text{kVAR}$ for $Q_1$ in equation (1) to find the complex power $S_1$.

$\begin{array}{c}{S}_1=120\text{kW}-j90\text{kVAR}\\ \text{=120}\text{kVA}-j90\text{kVAR}\left\{\because 1\text{W=1}\text{VA}\right\}\\ =120-j90\text{kVA}\end{array}$

Load 2:

Substitute $100\text{kW}$ for $P_2$, and $1$ for $\cos {\theta }_2$ in equation (6) to find the apparent power $\left(S_2\right)$.

$\begin{array}{c}{S}_2=\frac{100\text{kW}}{1}\\ =100\text{kVA}\left\{\because 1\text{W=1}\text{VA}\right\}\end{array}$

The given unity power factor of the Load 2,

$\cos {\theta }_2=1$

Rewrite the equation to find the angle ${\theta }_2$.

$\begin{array}{c}{\theta }_2={\cos }^{-1}\left(1\right)\\ {\theta }_2=0\end{array}$

Substitute $0$ for ${\theta }_2$, and $100\text{kVA}$ for $S_2$ in equation (7) to find the reactive power $\left(Q_2\right)$.

$\begin{array}{c}{Q}_2=100\text{kVA}\sin \left(0\right)\\ =0\text{kVAR}\end{array}$

Substitute $100\text{kW}$ for $P_2$, and $0\text{kVAR}$ for $Q_2$ in equation (4) to find the complex power $\left(S_2\right)$.

$\begin{array}{c}{S}_2=100\text{kW}+j0\text{kVAR}\\ \text{=100}\text{kVA}\left\{\because 1\text{W=1}\text{VA}\right\}\end{array}$

Load 3:

The given lagging power factor of the Load 3 is,

$\cos {\theta }_3=0.6$

Rewrite the equation to find the angle ${\theta }_3$.

$\begin{array}{c}{\theta }_3={\cos }^{-1}\left(0.6\right)\\ {\theta }_3=53.13°\end{array}$

Substitute $200\text{kVA}$ for $S_3$, and $0.6$ for $\cos {\theta }_3$ in equation (9) to find the real power $\left(P_3\right)$.

$\begin{array}{c}{P}_3=\left(200\text{kVA}\right)\left(0.6\right)\\ =120\text{kW}\left\{\because 1\text{W=1}\text{VA}\right\}\end{array}$

Substitute $200\text{kVA}$ for $S_3$, and $53.13°$ for ${\theta }_3$ in equation (10) to find the reactive power $\left(Q_3\right)$.

$\begin{array}{c}{Q}_3=\left(200\text{kVA}\right)\sin \left(53.13°\right)\\ =160\text{kVAR}\end{array}$

Substitute $120\text{kW}$ for $P_3$, and $160\text{kVAR}$ for $Q_3$ in equation (8) to find the complex power $\left(S_3\right)$ of the Load 3.

$\begin{array}{c}{S}_3=120\text{kW}+j160\text{kVAR}\\ =120\text{kVA}+j160\text{kVAR}\left\{\because 1\text{W=1}\text{VA}\right\}\\ =120+j160\text{kVA}\end{array}$

Load 4:

Substitute $80\text{kW}$ for $P_4$, and $95\text{kVAR}$ for $Q_4$ in equation (11) to find the complex power $\left(S_4\right)$ of the Load 4.

$\begin{array}{c}{S}_4=80\text{kW}+j95\text{kVAR}\\ =80\text{kVA}+j95\text{kVAR}\left\{\because 1\text{W=1}\text{VA}\right\}\\ =80+j95\text{kVA}\end{array}$

Substitute $120-j90\text{kVA}$ for $S_1$, $\text{100}\text{kVA}$ for $S_2$, $120+j160\text{kVA}$ for $S_3$, and $80+j95\text{kVA}$ for $S_4$ in equation (12) to find the total complex power $\left(S\right)$.

$\begin{array}{c}S=\left(120-j90\text{kVA}\right)+\left(\text{100}\text{kVA}\right)+\left(120+j160\text{kVA}\right)+\left(80+j95\text{kVA}\right)\\ =420+j165\text{kVA}\\ \text{=451}\text{.2}\angle \text{21}\text{.45}°\text{kVA}\end{array}$

Here, the apparent power is $S=\text{451}\text{.2}\text{kVA}$.

Substitute $\text{451}\text{.2}\text{kVA}$ for $S$, and $480\text{V}$ for $V_L$ in equation (13) to find the apparent power (S).

$\text{451}\text{.2}\text{kVA}=\sqrt{3}\left(480\text{V}\right)I_L$

Re-write the above equation to find the line current $\left(I_L\right)$.

$\begin{array}{c}{I}_L=\frac{\text{451}\text{.2}\text{kVA}}{\sqrt{3}\left(480\text{V}\right)}\\ =\frac{451.2\times {10}^3\text{VA}}{831.38\text{V}}\left\{\because 1\text{k}={10}^3\right\}\\ =542.7\text{A}\end{array}$

Substitute $0.02+j0.05\text{Ω}$ for $Z_l$, and $542.7\text{A}$ for $I_L$ in equation (14) to find the complex power absorbed by the line $\left(S_l\right)$.

$\begin{array}{c}{S}_l=3{\left(542.7\text{A}\right)}^2\left(0.02+j0.05\text{Ω}\right)\\ =17671+j44178{\text{A}}^2\cdot \text{Ω}\\ =17.671+j44.178\text{VA }\left\{\because \text{V}=\text{A}\cdot \Omega \right\}\\ =17.67+j44.18\text{kVA}\end{array}$

Substitute $17.67+j44.18\text{kVA}$ for $S_l$, and $\text{451}\text{.2}\angle \text{21}\text{.45}°\text{kVA}$ for $S_T$ in equation (15) to find ${\left(S_T\right)}_{\text{source}}$.

$\begin{array}{c}{\left(S_T\right)}_{\text{source}}=\left(17.67+j44.18\text{kVA}\right)+\left(\text{451}\text{.2}\angle \text{21}\text{.45}°\text{kVA}\right)\\ =\left(17.67+j44.18\text{kVA}\right)+\left(420+j165\text{kVA}\right)\\ =437.67+j209.18\text{kVA}\\ {\left(S_T\right)}_{\text{source}}\simeq 437.7+j209.2\text{kVA}\end{array}$

Re-write the above value as below,

${\left(S_T\right)}_{\text{source}}=485.1\angle 25.55°\text{ kVA}$

Substitute  $485.1\text{ kVA}$ for $S_T$, $542.7\text{A}$ for $I_L$ in equation (16) to find $\left(V_T\right)$.

$485.1\text{ kVA}=\sqrt{3}\left(542.7\text{A}\right)V_T$

Rewrite the above equation to find $\left(V_T\right)$.

$\begin{array}{c}{V}_T=\frac{485.1\text{ kVA}}{\sqrt{3}\left(542.7\text{A}\right)}\\ =\frac{485.1\times {10}^3\text{VA}}{940\text{A}}\left\{\because 1\text{k}={10}^3\right\}\\ =516.07\text{V}\\ V_T\cong 516\text{V}\end{array}$

Conclusion:

Thus, the magnitude of the line voltage at the source is $516\text{V}$.