Chapter 29, Problem 18P

(a)

To determine

To Find:

The resistance and inductive reactance of the plant’s total load.

Answer to Problem 18P

Resistance $R=571\Omega$ and inductive impedance $X_L$ = $266.4\Omega$ .

Explanation of Solution

Given:

Operated power of plant is $P=2.3\text{MW}$

Voltage supply to the plant is $\epsilon =40\text{kV}$

Frequency supply to the plant is $F=60\text{Hz}$

Resistance of the transmission line in $R=5.2\Omega$

Phase $\delta$ = $25°$

Formula used:

The relation between $Z$ and $\delta$ are given as

  $R=Z\cos \delta$ and $X_L=Z\sin \delta$.......... (1)

Impedance in terms of $I_{rms}$ and ${\epsilon }_{rms}$ ,

  $Z=\frac{{\epsilon }_{rms}}{I_{rms}}$.......... (2)

Average power $P_{av}={\epsilon }_{rms}{I}_{rms}\cos \delta$.......... (3)

Calculation:

From equation (3) rms current can be rewritten as

  $I_{rms}=\frac{P_{av}}{{\epsilon }_{rms}\cos \delta }$.......... (4)

Substitute the value $I_{rms}$ from equation (4) in equation (2)

  $Z=\frac{{\epsilon }^2{}_{rms}\cos \delta }{P_{av}}$.......... (5)

Substitute the values:

  $Z=\frac{{\left(40\text{kV}\right)}^2\cos 25°}{2.3\text{MW}}$

= $=630.3\Omega$

  $\begin{array}{c}R=630.3\cos 25°=571\Omega \\ X_L=630.3\Omega \sin 25°=266.4\Omega \end{array}$

Conclusion:

Hence, theResistance $R=571\Omega$ and inductive impedance $X_L$ = $266.4\Omega$ .

(b)

To determine

To Find:

RMS current and RMS voltage of the power line.

Answer to Problem 18P

  $63.4$ A; $40.3\text{kV}$

Explanation of Solution

Given:

Operated power of plant is $P=2.3\text{MW}$

Voltage supply to the plant is $\epsilon =40\text{kV}$

Frequency supply to the plant is $F=60\text{Hz}$

Resistance of the transmission line in $R=5.2\Omega$

Phase $\delta$ = $25°$

Formula use:

RMS current

  $I_{rms}=\frac{P_{av}}{{\epsilon }_{rms}\cos \delta }$.......... (1)

Calculation:

Substituting the values of $P_{av}$ , ${\epsilon }_{rms}$ and $\delta$ in equation (1)

  $\begin{array}{c}{I}_{rms}\text{=}\frac{\text{2}\text{.3 MW}}{\text{40 kV cos 25}°}\\ =63.4\text{A}\end{array}$

Apply Kirchhoff’s loop principle to the circuit:

  ${\epsilon }_{rms}-I_{rms}R-I_{rms}Z=0$

  ${\epsilon }_{rms}=I_{rms}\left(R+Z\right)$.......... (2)

Substituting the value of $I_{rms}$ , $R$ and $Z$ in equation (2)

  $\begin{array}{c}{\epsilon }_{rms}=63.4\times \left(5.2+630.3\right)\\ =40.3\text{kV}\end{array}$

Conclusion:

Hence,RMS current and RMS voltage are $63.4$ A and $40.3\text{kV}$ .

(c)

To determine

To Find:

Amount of power lost in transmission.

Answer to Problem 18P

  $20.9\text{kW}$

Explanation of Solution

Given:

Resistance of the transmission line in $R=5.2\Omega$

Transmission current $I_{rms}$ = 63.4 A

Formula use:

Transmission power

  $P_{trans}=I^2{}_{rms}R$ .......... (1)

Calculation:

Substituting the value of $I_{rms}$ , $R$ in equation (1)

  $\begin{array}{c}{P}_{trans}=I^2{}_{rms}R\\ $ ={(63.4\text{A})}^2\times 5.2\Omega $=20.9\text{kW}\end{array}$

Conclusion:

Hence,the power loss in the transmission line is $20.9\text{kW}$ .

(d)

To determine

To Find:

The amount of money that would be saved by the electric utility during one month of operation.

Answer to Problem 18P

$128

Explanation of Solution

Given:

Phase angle between voltage and current $\delta$ = $18°$

Operated power of plant is $P=2.3\text{MW}$

Voltage supply to the plant is $\epsilon =40\text{kV}$

Frequency supply to the plant is $F=60\text{Hz}$

Resistance of the transmission line in $R=5.2\Omega$

Formula used:

Cost saving equation $\Delta C=(P_{25°}-P_{18°})\Delta t$.......... (1)

Transmission power $P_{trans}=I^2{}_{rms}R$.......... (2)

RMS current $I_{rms}=\frac{P_{av}}{{\epsilon }_{rms}\cos \delta }$.......... (3)

Calculation:

Substituting the value of $P_{av}$ , ${\epsilon }_{rms}$ and $\delta$ in equation (3)

  $\begin{array}{c}{I}_{rms}=\frac{\text{2}\text{.3 MW}}{\text{40 kV cos18}°}\\ =60.5\text{A}\end{array}$

Substituting the value of $I_{rms}$ , $R$ in equation (2)

  $\begin{array}{c}{P}_{trans}=I^2{}_{rms}R\\ $ ={(60.5\text{A})}^2\times 5.2\Omega $=19\text{kW}\end{array}$

Cost saving is $\Delta C=\left(20.9\text{kW}-19\text{kW}\right)\left(16\text{h/d}\right)\left(30\text{d/month}\right)\left(\frac{\$0.14}{\text{kWh}}\right)$

  $=\$128$

Conclusion:

Hence, the cost is $\$128$ .

(e)

To determine

To Find:

The amount ofcapacitance required to achieve the charge.

Answer to Problem 18P

  $33\text{μF}$

Explanation of Solution

Given:

Phase angle between voltage and current $\delta$ = $18°$

Operated power of plant is $P=2.3\text{MW}$

Voltage supply to the plant is $\epsilon =40\text{kV}$

Frequency supply to the plant is $F=60\text{Hz}$

Resistance of the transmission line in $R=5.2\Omega$

Saved cost per kilowatt-hour is $128.

Formula used:

Capacitance $C=\frac{1}{2\pi f{X}_C}$ .......... (1)

Relation between $\delta$ , $X_C$ , $X_L$ and $R$ is

  $\tan \delta =\frac{X_L-X_C}{R}$.......... (2)

Conclusion:

From equation (2) $X_C$ can be rewritten as

  $X_C=X_L-R\tan \delta$

Substituting the value of $X_L$ , $R$ and $\delta$ in equation (1)

  $\begin{array}{l}C=\frac{1}{6.28\times 60{\text{s}}^{-1}\times \left(264.7-571\tan 18°\right)}\\ =33\text{μF}\end{array}$

Conclusion:

Hence, the required capacitance is $33\text{μF}$ .