Chapter 12, Problem 69P

A certain store contains three balanced three-phase loads. The three loads are:

Load 1: 16kVA at 0.85 pf lagging

Load 2: 12 kVA at 0.6 pf lagging

Load 3: 8 kW at unity pf

The line voltage at the load is 208 V rms at 60 Hz, and the line impedance is $0.4+j0.8\Omega$. Determine the line current and the complex power delivered to the loads.

To determine

Find the line currents and the complex power delivered to the loads.

Answer to Problem 69P

The line currents $I_a$, $I_b$ and $I_c$ are $\underset{\_}{94.32\angle -62.05°\text{A}}$, $\underset{\_}{94.32\angle 177.95°\text{A}}$, and $\underset{\_}{\text{94}\text{.32}\angle \text{57}\text{.95}°\text{A}}$  respectively.

The complex power delivered to the loads is $\underset{\_}{28.8+j18.03\text{kVA}}$.

Explanation of Solution

Given data:

The given three balanced three-phase loads are,

The reactive power of the Load 1 is $16\text{kVA}$ and the power factor is 0.85 (lagging).

The reactive power of the Load 2 is 12 kVA and the power factor is 0.6 (lagging).

The real power of the Load 3 is $8\text{kW}$ and the power factor is unity.

The line voltage at the load is $208\text{V}\left(\text{rms}\right)$ at 60 Hz.

The line impedance is $0.4+j0.8\text{Ω}$.

Formula used:

Write the expression to find the complex power $S_1$ for 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 average power $P_1$.

$P_1=S_1\cos {\theta }_1$ (2)

Here,

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

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

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

Write the expression to find the reactive power $Q_1$.

$Q_1=S_1\sin {\theta }_1$ (3)

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 of the Load 2.

$P_2=S_2\cos {\theta }_2$ (5)

Here,

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

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

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

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

$Q_2=S_2\sin {\theta }_2$ (6)

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

$S_3=P_3+j{Q}_3$ (7)

Here,

$P_3$ is the real power of the Load 3, and

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

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

$P_3=S_3\cos {\theta }_3$ (8)

Here,

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

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

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

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

$Q_3=S_3\sin {\theta }_3$ (9)

Write the expression to find the total complex power.

$S=S_1+S_2+S_3$ (10)

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.

Write the expression to find the phase voltage.

$V_p=\frac{V_L}{\sqrt{3}}$ (11)

Here,

$V_L$ is the line voltage at the load.

Write the expression to find the line to neutral voltage $\left(V_{an}\right)$.

$V_{an}=V_p\angle -30°$ (12)

Here,

$V_p$ is the phase voltage.

Write the expression to find the complex power $S$.

$S=3V{I}^{*}$ (13)

Here,

$V$ is the voltage, and

$I^{*}$ is the conjugate of the current $I_a$.

Write the expression to find the line current $I_b$.

$I_b=I_a\angle +240°$ (14)

Here,

$I_a$ is the line current.

Write the expression to find the line current $I_c$.

$I_c=I_a\angle +120°$ (15)

Calculation:

The given lagging power factor of the Load 1 is,

$\cos {\theta }_1=0.85$

Rewrite the above equation to find the angle ${\theta }_1$.

$\begin{array}{c}{\theta }_1={\cos }^{-1}\left(0.85\right)\\ =31.788°\\ {\theta }_1\cong 31.79°\end{array}$

Substitute $0.85$ for $\cos {\theta }_1$, and $16\text{kVA}$ for $S_1$ in equation (2) to find $P_1$.

$\begin{array}{c}{P}_1=\left(16\text{kVA}\right)\left(0.85\right)\\ =13.6\text{kVA}\left\{\because 1\text{W=1}\text{VA}\right\}\\ P_1=13.6\text{kW}\end{array}$

Substitute $31.79°$ for ${\theta }_1$, and $16\text{kVA}$ for $S_1$ in equation (3) to find $Q_1$.

$\begin{array}{c}{Q}_1=\left(16\text{kVA}\right)\sin \left(31.79°\right)\\ =\left(16\text{kVA}\right)\left(0.5268\right)\\ =8.428\text{kVAR}\\ Q_1\cong 8.43\text{kVAR}\end{array}$

Substitute $13.6\text{kW}$ for $P_1$, and $8.43\text{kVAR}$ for $Q_1$ in equation (1) to find $S_1$.

$\begin{array}{c}{S}_1=13.6\text{kW}+j8.43\text{kVAR}\\ =\left(13.6+j8.43\right)\text{kVA}\left\{\because 1\text{W=1}\text{VA}\right\}\end{array}$

The given lagging power factor of the Load 2 is,

$\cos {\theta }_2=0.6$

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

$\begin{array}{c}{\theta }_2={\cos }^{-1}\left(0.6\right)\\ =53.130°\end{array}$

Substitute $0.6$ for $\cos {\theta }_2$, and $12\text{kVA}$ for $S_2$ in equation (5) to find $P_2$.

$\begin{array}{c}{P}_2=\left(12\text{kVA}\right)\left(0.6\right)\\ =7.2\text{kW}\end{array}$

Substitute $53.130°$ for ${\theta }_2$, and $12\text{kVA}$ for $S_2$ in equation (6) to find $Q_2$.

$\begin{array}{c}{Q}_2=\left(12\text{kVA}\right)\sin \left(53.130°\right)\\ =\left(12\text{kVA}\right)\left(0.8\right)\\ Q_2=9.6\text{kVAR}\end{array}$

Substitute $9.6\text{kVAR}$ for $Q_2$, and $7.2\text{kW}$ for $P_2$ in equation (4) to find $S_2$.

$\begin{array}{c}{S}_2=7.2\text{kW}+j9.6\text{kVAR}\\ =\left(7.2+j9.6\right)\text{kVA}\left\{\because 1\text{W=1}\text{VA}\right\}\end{array}$

The given unity power factor of the Load 3 is,

$\cos {\theta }_3=1$

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

$\begin{array}{c}{\theta }_3={\cos }^{-1}\left(1\right)\\ =0\end{array}$

Substitute $1$ for $\cos {\theta }_3$, and $8\text{kW}$ for $P_3$ in equation (8).

$8\text{kW}=S_3\left(1\right)$

Rewrite the above equation to find $S_3$.

$\begin{array}{c}{S}_3=8\text{kW}\\ =\text{8}\text{kVA}\left\{\because 1\text{W=1}\text{VA}\right\}\end{array}$

Substitute $\text{8}\text{kVA}$ for $S_3$, and 0 for ${\theta }_3$ in equation (9) to find $Q_3$.

$\begin{array}{c}{Q}_3=\left(8\text{kVA}\right)\sin \left(0\right)\\ =0\text{kVAR}\end{array}$

Substitute $8\text{kW}$ for $P_3$, and $0\text{kVAR}$ for $Q_3$ in equation (7) to find $S_3$.

$\begin{array}{c}{S}_3=8\text{kW}+j0\text{kVAR}\\ =8\text{kVA}\left\{\because 1\text{W=1}\text{VA}\right\}\end{array}$

Substitute $13.6+j8.43\text{kVA}$ for $S_1$, $7.2+j9.6\text{kVA}$ for $S_2$, and $8\text{kVA}$ for $S_3$ in equation (10) to find $S$.

$\begin{array}{c}S=\left[13.6+j8.43\text{kVA}\right]+\left[7.2+j9.6\text{kVA}\right]+\left[8\text{kVA}\right]\\ =28.8+j18.03\text{kVA}\end{array}$

Substitute $208\text{V}$ for $V_L$ in equation (11) to find $V_p$.

$\begin{array}{c}{V}_p=\frac{208\text{V}}{\sqrt{3}}\\ =120.09\text{V}\end{array}$

Substitute $120.09\text{V}$ for $V_p$ in equation (12) to find $V_{an}$.

$V_{an}=120.09\angle -30°\text{V}$

Substitute $120.09\angle -30°\text{V}$ for $V$, and $28.8+j18.03\text{kVA}$ for $S$ in equation (13).

$28.8+j18.03\text{kVA}=3\left(120.09\angle -30°\text{V}\right)I^{*}$

Re-write the above equation to find the current $I^{*}$.

$\begin{array}{c}{I}^{*}=\frac{28.8+j18.03\text{kVA}}{3\left(120.09\angle -30°\text{V}\right)}\\ =\frac{33.98\angle 32.05°\text{kVA}}{360.27\angle -30°\text{V}}\\ =\frac{33.98}{360.27}\angle \left(32.05°+30°\right)\text{kA}\\ I^{*}=0.094318\angle 62.05°\text{kA}\end{array}$

The complex current $I^{*}$ is approximated as follows.

$I^{*}=94.32\angle 62.05°\text{kA}$

The line current $I_a$ is the complex conjugate of the current $I^{*}$.

$I_a=94.32\angle -62.05°\text{A}$

Substitute $94.32\angle -62.05°\text{A}$ for $I_a$ in equation (14) to find the line current $I_b$.

$\begin{array}{c}{I}_b=\left(94.32\angle -62.05°\text{A}\right)\left(\angle +240°\right)\\ =94.32\angle \left(-62.05°+240°\right)\text{A}\\ I_b=94.32\angle 177.95°\text{A}\end{array}$

Substitute $94.32\angle -62.05°\text{A}$ for $I_a$ in equation (15) to find the line current $I_c$.

$\begin{array}{c}{I}_c=\left(94.32\angle -62.05°\text{A}\right)\left(\angle 120°\right)\\ =94.32\angle \left(-62.05°+120°\right)\text{A}\\ I_c=\text{94}\text{.32}\angle \text{57}\text{.95}°\text{A}\end{array}$

Conclusion:

Thus,

The line currents $I_a$, $I_b$ and $I_c$ are $\underset{\_}{94.32\angle -62.05°\text{A}}$, $\underset{\_}{94.32\angle 177.95°\text{A}}$, and $\underset{\_}{\text{94}\text{.32}\angle \text{57}\text{.95}°\text{A}}$ respectively.

The complex power delivered to the loads is $\underset{\_}{28.8+j18.03\text{kVA}}$.