Chapter 11, Problem 73P

A 240-V rms 60-Hz supply serves a load that is 10 kW (resistive), 15 kVAR (capacitive), and 22 kVAR (inductive). Find:

1. (a) the apparent power
2. (b) the current drawn from the supply
3. (c) the kVAR rating and capacitance required to improve the power factor to 0.96 lagging
4. (d) the current drawn from the supply under the new power-factor conditions

To determine

Find the apparent power for the given loads.

Answer to Problem 73P

The apparent power for the given loads is $12.21\text{kVA}$.

Explanation of Solution

Given data:

The voltage $V_{\text{rms}}$ is $240\text{V}$.

The Frequency $f$ is $60\text{Hz}$.

The real power $P$ is $10\text{kW}\left(\text{resistive}\right)$.

The reactive power is,

$Q_1=15\text{kVAR}\left(\text{capacitive}\right)$ and $Q_2=22\text{kVAR}\left(\text{inductive}\right)$.

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.

Calculation:

Substitute $10\text{kW}\left(\text{resistive}\right)$ for P, $15\text{kVAR}\left(\text{capacitive}\right)$ for $Q_1$, and $22\text{kVAR}\left(\text{inductive}\right)$ for $Q_2$ in equation (1) to find the complex power $S$ in VA.

$S=\left(10+j15+j22\right)\text{kVA}$

As the reactive power $Q_1$ is capacitive, the equation becomes,

$\begin{array}{l}S=\left(10-j15+j22\right)\text{kVA}\\ S=\left(10+j7\right)\text{kVA}\end{array}$

On comparing the above equation with equation (1).

$P=10\text{kW}$ and $Q=7\text{kVAR}$

The apparent power $S$ is,

$\begin{array}{c}S=\left|S\right|\\ =\sqrt{{10}^2+7^2}\\ =12.21\text{kVA}\end{array}$

Conclusion:

Thus, the apparent power for the given loads is $12.21\text{kVA}$.

To determine

Find the current drawn from the supply.

Answer to Problem 73P

The current drawn from the supply is $50.86\angle -35°\text{A}$.

Explanation of Solution

Given data:

From Part (a), the complex power is,

$S=\left(10+j7\right)\text{kVA}$

The voltage $V_{\text{rms}}$ is $240\text{V}$.

Formula used:

Write the expression to find the rms current.

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

Calculation:

Substitute $\left(10+j7\right)\text{kVA}$ for $S$ and $240\text{V}$ for $V_{\text{rms}}$ in equation (2) to find the rms current in amperes.

$\begin{array}{l}{I}_{\text{rms}}^{\text{*}}=\frac{\left(10+j7\right)\times {\text{10}}^3\text{VA}}{240\text{V}}\left\{\because 1\text{k}={10}^3\right\}\\ I_{\text{rms}}^{\text{*}}=\frac{\left(10000+j7000\right)\text{A}}{240}\\ I_{\text{rms}}^{\text{*}}=\left(41.667+j29.167\right)\text{A}\\ I_{\text{rms}}=\left(41.667-j29.167\right)\text{A}\end{array}$

Convert the equation from rectangular to polar form.

$I_{\text{rms}}=50.86\angle -35°\text{A}$

Conclusion:

Thus, the current drawn from the supply is $50.86\angle -35°\text{A}$.

To determine

Find the value of the capacitance and reactive power to raise the power factor to 0.96 lagging.

Answer to Problem 73P

The value of capacitance C is $188.03\mu \text{F}$ and the reactive power $Q_C$ is $4.083\text{kVAR}$.

Explanation of Solution

Given data:

From Part (a),

$P=10\text{kW}$ and $Q=7\text{kVAR}$

The frequency $f=60\text{Hz}$.

Power factor $\text{pf}$ is 0.96 lagging.

The rms voltage is $240\text{V}$.

Formula used:

Write the expression to find the phase angle $\theta$.

$\theta ={\cos }^{-1}\left(\text{pf}\right)$ (3)

Here,

$\theta$ is the phase angle.

Write the expression for phase angle $\theta$.

$\theta ={\tan }^{-1}\left(\frac{Q}{P}\right)$ (4)

Write the expression to find the value of the capacitance.

$C=\frac{Q_C}{2\pi f{V}_{\text{rms}}^{\text{2}}}$ (5)

Here,

$Q_C$ is the reactive power, and

$f$ is the frequency.

Calculation:

Substitute $10\text{kW}$ for $P$ and $7\text{kVAR}$ for $Q$ in equation (4) to find the phase angle ${\theta }_1$.

$\begin{array}{c}{\theta }_1={\tan }^{-1}\left(\frac{7\text{kVAR}}{10\text{kW}}\right)\\ =35°\end{array}$

The power factor is raised to 0.96 lagging.

Substitute 0.96 for $\text{pf}$ in equation (3) to find the phase angle ${\theta }_2$.

$\begin{array}{c}{\theta }_2={\cos }^{-1}\left(0.96\right)\\ =16.26°\end{array}$

The reactive power is calculated as follows.

$Q_C=P\left[\tan {\theta }_1-\tan {\theta }_2\right]$ (6)

Substitute $10\text{kW}$ for P, $35°$ for ${\theta }_1$ and $16.26°$ for ${\theta }_2$ in the equation (6) to find the reactive power due to the shunt capacitor in VAR.

$\begin{array}{c}{Q}_C=10\text{k}\left[\tan \left(35°\right)-\tan \left(16.26°\right)\right]\\ =10\text{k}\times 0.4083\\ =4.083\text{kVAR}\end{array}$

Substitute $4.083\text{kVAR}$ for $Q_C$, $240\text{V}$ for $V_{\text{rms}}$, and $60\text{Hz}$ for $f$ in equation (5) to find the value of the capacitance in farads.

$\begin{array}{c}C=\frac{4.083\times {\text{10}}^3\text{VAR}}{2\pi \left(60\text{Hz}\right){\left(240\text{V}\right)}^2}\left\{\because 1\text{k}={10}^3\right\}\\ =\frac{4.083\times {\text{10}}^3\text{VA}}{2\pi \left(60\text{Hz}\right)\left(57600{\text{V}}^2\right)}\\ =\frac{4.083\times {\text{10}}^3\text{VA}}{2\pi \left(60\frac{\text{1}}{\text{s}}\right)\left(57600{\text{V}}^2\right)}\left\{\because 1\text{Hz}=\frac{\text{1}}{\text{s}}\right\}\\ =1.8803\times {\text{10}}^{-4}\times {\text{10}}^2\times {\text{10}}^{-2}\text{F}\left\{\because 1\text{F}=\frac{\text{A}\cdot \text{s}}{\text{V}}\right\}\end{array}$

Simplify the equation as follows,

$\begin{array}{c}C=\text{188}\text{.03}\times {\text{10}}^{-6}\text{F}\\ =188.03\mu \text{F}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Conclusion:

Thus, the value of capacitance C is $188.03\mu \text{F}$ and the reactive power $Q_C$ is $4.083\text{kVAR}$.

To determine

Find the current drawn from the supply under the new power factor condition.

Answer to Problem 73P

The current drawn from the supply is $43.4\angle -16.26°\text{A}$.

Explanation of Solution

Given data:

The voltage $V_{\text{rms}}$ is $240\text{V}$.

The real power is,

$P_2=P_1=1\text{0}\text{kW}$

From Part (a),

$Q=7\text{kVAR}$

From Part (c),

$Q_C=4.083\text{kVAR}$

Calculation:

The complex power $S_2$ under the new power factor condition is,

$S_2=P_2+j{Q}_2$ (7)

The reactive power $Q_2$ is,

$Q_2=Q-Q_C$

Substitute $7\text{kVAR}$ for $Q$ and $4.083\text{kVAR}$ for $Q_C$ to find the reactive power $Q_2$ in VAR.

$\begin{array}{c}{Q}_2=7\text{kVAR}-4.083\text{kVAR}\\ =2.917\text{kVAR}\end{array}$

Substitute $2.917\text{kVAR}$ for $Q_2$ and $1\text{0}\text{kW}$ for $P_2$ in equation (7) to find the complex power $S_2$ in VA.

$S_2=\left(1\text{0}+j2.917\right)\text{kVA}$

Substitute $\left(10+j2.917\right)\text{kVA}$ for $S_2$ and $240\text{V}$ for $V_{\text{rms}}$ in equation (2) to find the rms current in amperes.

$\begin{array}{l}{I}_{\text{rms}}^{\text{*}}=\frac{\left(10+j2.917\right)\times {\text{10}}^3\text{VA}}{240\text{V}}\left\{\because 1\text{k}={10}^3\right\}\\ I_{\text{rms}}^{\text{*}}=\frac{\left(10000+j2917\right)\text{A}}{240}\\ I_{\text{rms}}^{\text{*}}=\left(10000+j12.154\right)\text{A}\\ I_{\text{rms}}=\left(10000-j12.154\right)\text{A}\end{array}$

Convert the equation from rectangular to polar form.

$I_{\text{rms}}=43.4\angle -16.26°\text{A}$

Conclusion:

Thus, the current drawn from the supply is $43.4\angle -16.26°\text{A}$.