Chapter 12, Problem 71P

In Fig. 12.73, two wattmeters are properly connected to the unbalanced load supplied by a balanced source such that with positive phase sequence.

1. (a) Determine the reading of each wattmeter.
2. (b) Calculate the total apparent power absorbed by the load.

To determine

Find the readings of wattmeter $W_1$ and wattmeter $W_2$.

Answer to Problem 71P

The readings of wattmeter $W_1$ and wattmeter $W_2$ are $\underset{\_}{2,590\text{W}}$ and $\underset{\_}{4,808\text{W}}$ respectively.

Explanation of Solution

Given data:

Refer to the Figure 12.73 in the textbook, which shows the unbalanced load circuit with two-wattmeters.

The balanced source connected the circuit has the line to line voltage $V_{ab}=208\angle 0°\text{ V}$.

Formula used:

Consider the positive phase sequence.

Write the expression to find the line to line voltage $V_{bc}$.

$V_{bc}=V_p\angle -120°$ (1)

Here,

$V_p$ is the magnitude of the phase voltage.

Write the expression to find the line to line voltage $V_{ca}$.

$V_{ca}=V_p\angle 120°$ (2)

Write the expression to find the line to line current $\left(I_{AB}\right)$.

$I_{AB}=\frac{V_{ab}}{Z_{AB}}$ (3)

Here,

$V_{ab}$ is the line to line voltage of $a\text{-}b$ source terminal.

$Z_{AB}$ is the line to line impedance of $A\text{-}B$ load terminal.

Write the expression to find the line to line to current $I_{BC}$.

$I_{BC}=\frac{V_{bc}}{Z_{BC}}$ (4)

Here,

$V_{bc}$ is the line to line voltage of $b\text{-}c$ source terminal.

$Z_{BC}$ is the line to line impedance of $B\text{-}C$ load terminal.

Write the expression to find the line to line current $I_{CA}$.

$I_{CA}=\frac{V_{ca}}{Z_{CA}}$                                                                                                    (5)

Here,

$V_{ca}$ is the line to line voltage of $c\text{-}a$ source terminal.

$Z_{CA}$ is the line to line impedance of $C\text{-}A$ load terminal.

Write the expression to find the line current $\left(I_{aA}\right)$.

$I_{aA}=I_{AB}-I_{CA}$ (6)

Write the expression to find the line current $\left(I_{cC}\right)$.

$I_{cC}=I_{CA}-I_{BC}$ (7)

Write the expression to find the power $\left(P_1\right)$.

$P_1=\left|V_{ab}\right|\left|I_{aA}\right|\cos \left({\theta }_{V_{ab}}-{\theta }_{I_{aA}}\right)$ (8)

Here,

${\theta }_{V_{ab}}$ is the phase angle of the source voltage $V_{ab}$.

${\theta }_{I_{aA}}$ is the phase angle of the line current $I_{aA}$.

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

$P_2=\left|V_{cb}\right|\left|I_{cC}\right|\cos \left({\theta }_{V_{cb}}-{\theta }_{I_{cC}}\right)$ (9)

Here,

${\theta }_{V_{cb}}$ is the phase angle of the source voltage $V_{cb}$.

${\theta }_{I_{cC}}$ is the phase angle of the line current $I_{cC}$.

Calculation:

The magnitude of the phase voltage $V_p$ is $208\text{V}$. Re-draw the circuit as shown in Figure 1 with the indication of phase and line currents.

Substitute $208\text{V}$ for $V_p$ in equation (1) to find the voltage $V_{bc}$.

$V_{bc}=208\angle -120°\text{V}$

Substitute $208\text{V}$ for $V_p$ in equation (2) to find the voltage $V_{ca}$.

$V_{ca}=208\angle 120°\text{V}$

Substitute $208\angle 0°\text{V}$ for $V_{ab}$ and $20\text{Ω}$ for $Z_{AB}$ in equation (3) to find the current $I_{AB}$.

$\begin{array}{c}{I}_{AB}=\frac{208\angle 0°\text{V}}{20\text{Ω}}\\ =10.4\angle 0°\frac{\text{V}}{\text{Ω}}\\ =10.4\angle 0°\text{A }\left\{\because \frac{\text{V}}{\text{Ω}}=\text{A}\right\}\end{array}$

Substitute $208\angle -120°\text{V}$ for $V_{bc}$ and $10-j10\text{Ω}$ for $Z_{BC}$ in equation (4) to find the $I_{BC}$$\begin{array}{c}{I}_{BC}=\frac{208\angle -120°\text{V}}{10-j10\text{Ω}}\\ =\frac{208\angle -120°\text{V}}{14.1421\angle -45°\text{Ω}}\\ =\frac{208}{14.1421}\angle \left(-120°-\left(-45°\right)\right)\text{ A}\left\{\because \frac{\text{V}}{\text{Ω}}=\text{A}\right\}\\ =14.708\angle -75°\text{ A}\end{array}$

Substitute $208\angle 120°\text{V}$ for $V_{ca}$ and $12+j5\text{Ω}$ for $Z_{CA}$ in equation (5) to find the current $I_{CA}$.

$\begin{array}{c}{I}_{CA}=\frac{208\angle 120°\text{V}}{12+j5\text{Ω}}\\ =\frac{208\angle 120°\text{V}}{13\angle 22.62°\text{Ω}}\\ =\frac{208}{13}\angle \left(120°-22.62°\right)\text{A}\left\{\because \frac{\text{V}}{\text{Ω}}=\text{A}\right\}\\ =16\angle 97.38°\text{A}\end{array}$

Substitute $10.4\angle 0°\text{A}$ for $I_{AB}$ and $16\angle 97.38°\text{A}$ for $I_{CA}$ in equation (6) to find the current $I_{aA}$.

$\begin{array}{c}{I}_{aA}=\left(10.4\angle 0°\text{A}\right)-\left(16\angle 97.38°\text{A}\right)\\ =\left(10.4+0j\text{A}\right)-\left(-2.055+j15.867\text{A}\right)\\ =12.455-j15.867\text{A}\\ =20.171\angle -51.869°\text{A}\\ I_{aA}\cong 20.171\angle -51.87°\text{A}\end{array}$

Substitute $16\angle 97.38°\text{A}$ for $I_{CA}$ and $14.708\angle -75°\text{A}$ for $I_{BC}$ in equation (7) to find $\left(I_{cC}\right)$.

$\begin{array}{c}{I}_{cC}=\left(16\angle 97.38°\text{A}\right)-\left(14.708\angle -75°\text{A}\right)\\ =\left(-2.0551+j15.867\text{A}\right)-\left(3.8067-j14.206\right)\text{A}\\ =-5.8617+j30.073\text{A}\\ =30.64\angle 101.03°\text{A}\end{array}$

Substitute $208\text{V}$ for $V_{ab}$, $20.171\text{A}$ for $I_{aA}$, $0°$ for ${\theta }_{V_{ab}}$ and $-51.87°$ for ${\theta }_{I_{aA}}$ in equation (8) to find $P_1$.

$\begin{array}{c}{P}_1=\left|208\text{V}\right|\left|20.171\text{A}\right|\cos \left(0°-\left(-51.87°\right)\right)\\ =\left(208\text{V}\right)\left(20.171\text{A}\right)\cos \left(51.87°\right)\\ =\left(4,195.568\text{VA}\right)\left(0.61744\right)\\ =2,590.51\text{W}\\ P_1\cong 2,590\text{kW}\end{array}$

Write the expression to find the voltage $V_{cb}$ is,

$V_{cb}=V_{bc}\angle 180°$ (10)

Substitute $208\angle -120°\text{V}$ for $V_{bc}$ in above equation to find the line to line source voltage $V_{cb}$.

$\begin{array}{c}{V}_{cb}=\left(208\angle -120°\text{V}\right)\left(\angle 180°\right)\\ =208\angle \left(-120°+180°\right)\text{V}\\ \text{=208}\angle \text{60}°\text{V}\end{array}$

Substitute $\text{208}\text{V}$ for $V_{cb}$, $30.64\text{A}$ for $I_{cC}$, $\text{60}°$ for ${\theta }_{V_{cb}}$ and $101.03°$ for ${\theta }_{I_{cC}}$ in equation (9) to find $P_2$.

$\begin{array}{c}{P}_2=\left|\text{208}\text{V}\right|\left|30.64\text{A}\right|\cos \left(60°-101.03°\right)\\ =\left(208\text{V}\right)\left(30.64\text{A}\right)\cos \left(-41.03\right)\\ =\left(6,373.12\text{VA}\right)\left(0.7543\right)\\ =4,807.244\text{W}\\ P_2\cong 4,808\text{W}\end{array}$

Conclusion:

Thus, the readings of wattmeter $W_1$ and wattmeter $W_2$ are $\underset{\_}{2,590\text{W}}$ and $\underset{\_}{4,808\text{W}}$ respectively.

To determine

Find the total apparent power absorbed by the unbalanced load.

Answer to Problem 71P

The total apparent power absorbed by the unbalanced load is $\underset{\_}{8,335\text{VA}}$.

Explanation of Solution

Formula used:

Write the expression to find the total real power is,

$P_T=P_1+P_2$ (11)

Here,

$P_1$ and $P_2$ are the readings of the wattmeters.

Write the expression to find the total reactive power is,

$Q_T=\sqrt{3}\left(P_2-P_1\right)$ (12)

Write the expression to find the total complex power is,

$S_T=P_T+j{Q}_T$ (13)

Write the expression to find the magnitude of the total apparent power is,

$S_T=\left|S_T\right|$ (14)

Here,

$S_T$ is the total complex power.

Calculation:

Substitute $2,590.51\text{W}$ for $P_1$ and $4,807.244\text{ W}$ for $P_2$ in equation (11) to find $P_T$.

$\begin{array}{c}{P}_T=\left(2,590.51\text{W}\right)+\left(4,807.244\text{ W}\right)\\ =7,397.754\text{W}\end{array}$

Substitute $2,590.51\text{W}$ for $P_1$ and $4,807.244\text{ W}$ for $P_2$ in equation (12) to find $Q_T$.

$\begin{array}{c}{Q}_T=\sqrt{3}\left(4,807.244\text{ W}-2,590.51\text{W}\right)\\ =3,840.4959\text{W}\\ =3,840.4959\text{VAR}\left\{\because 1\text{W=1}\text{VA}\right\}\end{array}$

Substitute $3,840.4959\text{VAR}$ for $Q_T$, and $7,397.754\text{W}$ for $P_T$ in equation (13) to find $S_T$.

$\begin{array}{c}{S}_T=7,397.754\text{W}+j3,840.4959\text{VAR}\\ =7,397.754\text{VA}+j3,840.4959\text{VAR}\left\{\because 1\text{W=1}\text{VA}\right\}\\ =8,335.236\angle 27.435°\text{VA}\\ S_T\cong 8,335\angle 27.44°\text{VA}\end{array}$

Substitute $8,335\angle 27.44°\text{VA}$ for $S_T$ in equation (14) to find $S_T$.

$\begin{array}{c}{S}_T=\left|8,335\angle 27.44°\text{VA}\right|\\ =8,335\text{VA}\end{array}$

Conclusion:

Thus, the total apparent power absorbed by the unbalanced load is $\underset{\_}{8,335\text{VA}}$.