Chapter 11, Problem 75P

Consider the power system shown in Fig. 11.90. Calculate:

1. (a) the total complex power
2. (b) the power factor
3. (c) the parallel capacitance necessary to establish a unity power factor

To determine

Calculate the complex power of the circuit shown in Figure 11.90.

Answer to Problem 75P

The total complex power for the given circuit is $32.14\text{kW}+j7.357\text{kVAR}$.

Explanation of Solution

Given data:

Refer to Figure 11.90 in the textbook.

The voltage $V_{\text{rms}}$ is $440\text{V}$.

The capacitance C is $-j30\Omega$.

The inductance L is $j10\Omega$.

Formula used:

Write the expression to find the complex power.

$S=P+jQ$ (1)

Here,

$P$ is the real power, and

$Q$ is the reactive power.

Write the expression to find the complex power.

$S=\frac{V_{\text{rms}}^{\text{2}}}{Z^{*}}$ (2)

Here,

$V_{\text{rms}}$ is the rms voltage, and

$Z$ is the impedance.

Calculation:

Refer to figure 11.90 in the textbook.

Consider the impedance $Z_1$ in the circuit is,

$Z_1=\left(40-j30\right)\Omega$

Consider the impedance $Z_2$ in the circuit is,

$Z_2=\left(10-j10\right)\Omega$

Consider the impedance $Z_3$ in the circuit is,

$Z_3=10\Omega$

Substitute $440\text{V}$ for $V_{\text{rms}}$ and $\left(40-j30\right)\Omega$ for $Z_1$ in equation (2) to find the complex power $S_1$ in VA.

$\begin{array}{c}{S}_1=\frac{{\left(440\text{V}\right)}^2}{\left(40-j30\right){\Omega }^{*}}\\ =\frac{193600{\text{V}}^2}{\left(40+j30\right)\frac{\text{V}}{\text{A}}}\left\{\because 1\Omega =\frac{\text{V}}{\text{A}}\right\}\\ =3872\angle -36.87°\text{VA}\\ =\left(3097.6-j2323.2\right)\text{VA}\end{array}$

Substitute $440\text{V}$ for $V_{\text{rms}}$ and $\left(10-j10\right)\Omega$ for $Z_2$ in equation (2) to find the complex power $S_2$ in VA.

$\begin{array}{c}{S}_2=\frac{{\left(440\text{V}\right)}^2}{\left(10-j10\right){\Omega }^{*}}\\ =\frac{193600{\text{V}}^2}{\left(10+j10\right)\frac{\text{V}}{\text{A}}}\left\{\because 1\Omega =\frac{\text{V}}{\text{A}}\right\}\\ =13689.7\angle 45°\text{VA}\\ =\left(9680+j9680\right)\text{VA}\end{array}$

Substitute $440\text{V}$ for $V_{\text{rms}}$ and $10\Omega$ for $Z_3$ in equation (2) to find the complex power $S_3$ in VA.

$\begin{array}{c}{S}_3=\frac{{\left(440\text{V}\right)}^2}{10{\Omega }^{*}}\\ =\frac{193600{\text{V}}^2}{10\frac{\text{V}}{\text{A}}}\left\{\because 1\Omega =\frac{\text{V}}{\text{A}}\right\}\\ =19360\text{VA}\end{array}$

The total complex power is,

$S=S_1+S_2+S_3$

Substitute $\left(3097.6-j2323.2\right)\text{VA}$ for $S_1$, $\left(9680+j9680\right)\text{VA}$ for $S_2$, and $19360\text{VA}$ for $S_3$ in the equation to find the total complex power in VA.

$\begin{array}{c}S=\left(3097.6-j2323.2\right)\text{VA}+\left(9680+j9680\right)\text{VA}+19360\text{VA}\\ =\left(\left(\text{32137}\text{.6}\times {\text{10}}^3\times {\text{10}}^{-3}\right)+\left(j\text{7356}\text{.8}\times {\text{10}}^3\times {\text{10}}^{-3}\right)\right)\text{VA}\\ =\left(\text{32}\text{.14}+j7.357\right)\text{kVA}\left\{\because 1\text{k}={10}^3\right\}\end{array}$

Comparing the above equation with equation (1).

$P=32.14\text{kW}$ and $Q=7.357\text{kVAR}$

Hence, the total complex power is,

$S=32.14\text{kW}+j7.357\text{kVAR}$ (3)

Conclusion:

Thus, the total complex power for the given circuit is $32.14\text{kW}+j7.357\text{kVAR}$.

To determine

Find the power factor for the given circuit.

Answer to Problem 75P

The power factor for the given circuit is $0.9748$.

Explanation of Solution

Given data:

The voltage $V_{\text{rms}}$ is $440\text{V}$.

From Part (a),

The real power and the reactive power is,

$P=32.14\text{kW}$ and $Q=7.357\text{kVAR}$

Formula used:

Write the expression to find the power factor $\text{pf}$.

$\text{pf}=\cos \left(\theta \right)$ (4)

Here,

$\theta$ is the phase angle.

Write the expression for phase angle $\theta$.

$\theta ={\tan }^{-1}\left(\frac{Q}{P}\right)$ (5)

Calculation:

Substitute $32.14\text{kW}$ for $P$ and $7.357\text{kVAR}$ for $Q$ in equation (5) to find the phase angle $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{7.357\text{kVAR}}{32.14\text{kW}}\right)\\ =12.893°\end{array}$

Substitute $12.893°$ for $\theta$ in equation (4) to find the power factor.

$\begin{array}{c}\text{pf}=\cos \left(12.893°\right)\\ =0.9748\end{array}$

Conclusion:

Thus, the power factor for the given circuit is $0.9748$.

To determine

Find the parallel capacitance value required to establish a unity power factor.

Answer to Problem 75P

The value of capacitance is $100.8\mu \text{F}$.

Explanation of Solution

Given data:

The voltage $V_{\text{rms}}$ is $440\text{V}$.

Formula used:

Write the expression to find the value of the capacitance.

$C=\frac{Q_C}{2\pi f{V}_{\text{rms}}^{\text{2}}}$ (6)

Here,

$Q_C$ is the reactive power, and

$f$ is the frequency.

Calculation:

Consider the frequency is $f=60\text{Hz}$.

From equation (3), the reactive power is,

$Q=Q_C=7.357\text{kVAR}$

Substitute $7.357\text{kVAR}$ for $Q_C$, $440\text{V}$ for $V_{\text{rms}}$, and $60\text{Hz}$ for $f$ in equation (6) to find the value of the capacitance in farads.

$\begin{array}{c}C=\frac{7.357\times {\text{10}}^3\text{VAR}}{2\pi \left(60\text{Hz}\right){\left(440\text{V}\right)}^2}\left\{\because 1\text{k}={10}^3\right\}\\ =\frac{7.357\times {\text{10}}^3\text{VA}}{2\pi \left(60\text{Hz}\right)\left(193600{\text{V}}^2\right)}\\ =\frac{7.357\times {\text{10}}^3\text{VA}}{2\pi \left(60\frac{\text{1}}{\text{s}}\right)\left(193600{\text{V}}^2\right)}\left\{\because 1\text{Hz}=\frac{\text{1}}{\text{s}}\right\}\\ =1.008\times {\text{10}}^{-4}\times {\text{10}}^2\times {\text{10}}^{-2}\text{F}\left\{\because 1\text{F}=\frac{\text{A}\cdot \text{s}}{\text{V}}\right\}\end{array}$

Simplify the equation as follows,

$\begin{array}{c}C=\text{100}\text{.8}\times {\text{10}}^{-6}\text{F}\\ =100.8\mu \text{F}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Conclusion:

Thus, the value of capacitance is $100.8\mu \text{F}$.