Chapter 12, Problem 48P

A balanced, positive-sequence wye-connected source has V_an = 240 $\angle 0°$ V rms and supplies an unbalanced delta-connected load via a transmission line with impedance 2 + j3 Ω per phase.

1. (a) Calculate the line currents if Z_AB = 40 + j15 Ω, Z_BC = 60 Ω, Z_CA = 18 − jl2 Ω.
2. (b) Find the complex power supplied by the source.

To determine

Calculate the line currents for the described circuit using PSpice.

Answer to Problem 48P

The value for the line currents $I_{aA}$, $I_{bB}$, and $I_{cC}$ are $24.92\angle -6.124\text{A}$, $9.723\angle -144.2\text{A}$, and $20.94\angle 136.5\text{A}$ respectively.

Explanation of Solution

Given data:

The phase voltage is $240\angle 0°\text{V}\text{rms}$.

The transmission line impedance is $2+j3\Omega$.

The value of the impedances $Z_{AB}=40+j15\Omega$, $Z_{BC}=60\Omega$, and $Z_{CA}=18-j12\Omega$.

Formula used:

Write the formulae for the conversion of delta connected impedances to star connected impedances.

$Z_A=\frac{Z_{AB}{Z}_{CA}}{Z_{AB}+Z_{BC}+Z_{CA}}$        (1)

$Z_A=\frac{Z_{AB}{Z}_{CA}}{Z_{AB}+Z_{BC}+Z_{CA}}$        (2)

$Z_A=\frac{Z_{AB}{Z}_{CA}}{Z_{AB}+Z_{BC}+Z_{CA}}$        (3)

Here,

$Z_{AB}$, $Z_{BC}$, and $Z_{CA}$ are the delta connected impedances, and

$Z_A$, $Z_B$, and $Z_C$ are the star connected impedances.

Write the expression for reactance of the inductor.

$X_L=j\omega L$        (4)

Here,

$\omega$ is the angular frequency, and

$L$ is the value of inductor.

Write the expression for reactance of the capacitor.

$X_C=\frac{1}{j\omega C}$        (5)

Here,

$C$ is the value of capacitor.

Calculation:

The given unbalanced delta connected load is shown in Figure 1.

Substitute $40+j15\Omega$ for $Z_{AB}$, $60\Omega$ for $Z_{BC}$, and $18-j12\Omega$ for $Z_{CA}$ in equation (1) to find $Z_A$.

$\begin{array}{c}{Z}_A=\frac{\left(40+j15\Omega \right)\left(18-j12\Omega \right)}{\left(40+j15\Omega \right)+\left(60\Omega \right)+\left(18-j12\Omega \right)}\\ =\frac{\left(900-j210\right){\Omega }^2}{\left(118+j3\right)\Omega }\\ =7.577-j1.972\Omega \end{array}$

Substitute $40+j15\Omega$ for $Z_{AB}$, $60\Omega$ for $Z_{BC}$, and $18-j12\Omega$ for $Z_{CA}$ in equation (2) to find $Z_B$.

$\begin{array}{c}{Z}_B=\frac{\left(40+j15\Omega \right)\left(60\Omega \right)}{\left(40+j15\Omega \right)+\left(60\Omega \right)+\left(18-j12\Omega \right)}\\ =\frac{\left(2400+j900\right){\Omega }^2}{\left(118+j3\right)\Omega }\\ =20.52+j7.105\Omega \end{array}$

Substitute $40+j15\Omega$ for $Z_{AB}$, $60\Omega$ for $Z_{BC}$, and $18-j12\Omega$ for $Z_{CA}$ in equation (3) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{\left(60\Omega \right)\left(18-j12\Omega \right)}{\left(40+j15\Omega \right)+\left(60\Omega \right)+\left(18-j12\Omega \right)}\\ =\frac{\left(1080-j720\right){\Omega }^2}{\left(118+j3\right)\Omega }\\ =8.992-j6.3303\Omega \end{array}$

The transformed circuit is shown in Figure 2.

The given balanced wye-connected source supplies the unbalanced delta connected load is shown in Figure 3.

Let us assume that the value of the angular frequency, $\omega =1\frac{\text{rad}}{\text{s}}$.

Calculate the frequency as follows.

$\begin{array}{c}2\pi f=1\frac{\text{rad}}{\text{s}}\\ f=\frac{1\frac{\text{rad}}{\text{s}}}{2\pi }\\ f=0.15915\text{Hz}\end{array}$

Substitute $1\frac{\text{rad}}{\text{s}}$ for $\omega$ and $j3\Omega$ for $X_L$ in equation (4) to find $L$.

$\begin{array}{c}j3\Omega =j\left(1\frac{\text{rad}}{\text{s}}\right)\left(L\right)\\ L=\frac{j3\Omega }{j\left(1\frac{\text{rad}}{\text{s}}\right)}\left\{\because 1\text{H}=1\Omega \cdot \text{s}\right\}\\ L=3\text{H}\end{array}$

Substitute $1\frac{\text{rad}}{\text{s}}$ for $\omega$ and $j7.105\Omega$ for $X_L$ in equation (4) to find $L$.

$\begin{array}{c}j7.105\Omega =j\left(1\frac{\text{rad}}{\text{s}}\right)\left(L\right)\\ L=\frac{j7.105\Omega }{j\left(1\frac{\text{rad}}{\text{s}}\right)}\left\{\because 1\text{H}=1\Omega \cdot \text{s}\right\}\\ L=7.105\text{H}\end{array}$

Substitute $1\frac{\text{rad}}{\text{s}}$ for $\omega$ and $-j1.972\Omega$ for $C$ in equation (5) to find $C$.

$\begin{array}{c}-j1.972\Omega =\frac{1}{j\left(1\frac{\text{rad}}{\text{s}}\right)C}\\ C=\frac{1}{j\left(1\frac{\text{rad}}{\text{s}}\right)\left(-j1.972\Omega \right)}\left\{1\text{F}=1\frac{\text{s}}{\Omega }\right\}\\ =0.507\text{F}\end{array}$

Substitute $1\frac{\text{rad}}{\text{s}}$ for $\omega$ and $-j6.3303\Omega$ for $C$ in equation (5) to find $C$.

$\begin{array}{c}-j6.3303\Omega =\frac{1}{j\left(1\frac{\text{rad}}{\text{s}}\right)C}\\ C=\frac{1}{j\left(1\frac{\text{rad}}{\text{s}}\right)\left(-j6.3303\Omega \right)}\left\{1\text{F}=1\frac{\text{s}}{\Omega }\right\}\\ =0.158\text{F}\end{array}$

The time domain representation of Figure 3 is shown in Figure 4.

PSpice Simulation:

Draw Figure 4 in PSpice as shown in Figure 5.

Provide the simulation setting as shown in Figure 6.

The obtained results are given below.

  FREQ        IM(V_PRINT1)  IP(V_PRINT1)

  1.592E-01   2.492E+01    -6.124E+00

  FREQ        IM(V_PRINT2)  IP(V_PRINT2)

  1.592E-01   9.723E+00    -1.442E+02

  FREQ       IM(V_PRINT3)   IP(V_PRINT3)

  1.592E-01   2.094E+01    1.365E+02

The obtained line currents are given below.

$\begin{array}{l}{I}_{aA}=24.92\angle -6.124\text{A}\\ I_{bB}=9.723\angle -144.2\text{A}\\ I_{cC}=20.94\angle 136.5\text{A}\end{array}$

Conclusion:

Thus, the value for the line currents $I_{aA}$, $I_{bB}$, and $I_{cC}$ are $24.92\angle -6.124\text{A}$, $9.723\angle -144.2\text{A}$, and $20.94\angle 136.5\text{A}$ respectively.

To determine

Calculate the total complex power supplied by the source.

Answer to Problem 48P

The total complex power supplied by the source is $\text{12}\text{.894}\angle 0.743\text{kVA}$.

Explanation of Solution

Calculation:

Write the expression for complex power delivered by source $a$.

$S_a=\left|I_{aA}\right|V_{an}$

Substitute $24.92\angle 6.124\text{A}$ for $\left|I_{aA}\right|$ and $240\angle 0°\text{V}$ for $V_{an}$.

$\begin{array}{c}{S}_a=\left|24.92\angle 6.124\text{A}\right|\left(240\angle 0°\text{V}\right)\\ =5980.8\angle 6.124\text{VA}\end{array}$

Write the expression for complex power delivered by source $b$.

$S_b=\left|I_{bB}\right|V_{bn}$

Substitute $9.723\angle 144.2\text{A}$ for $\left|I_{bB}\right|$ and $240\angle -120°\text{V}$ for $V_{bn}$.

$\begin{array}{c}{S}_b=\left|9.723\angle 144.2\text{A}\right|\left(240\angle -120°\text{V}\right)\\ =2333.52\angle 24.2\text{VA}\end{array}$

Write the expression for complex power delivered by source $c$.

$S_c=\left|I_{cC}\right|V_{cn}$

Substitute $20.94\angle -136.5\text{A}$ for $\left|I_{cC}\right|$ and $240\angle 120°\text{V}$ for $V_{cn}$.

$\begin{array}{c}{S}_c=\left|20.94\angle -136.5\text{A}\right|\left(240\angle 120°\text{V}\right)\\ =5025.6\angle -16.5\text{VA}\end{array}$

Write the expression for total complex power supplied by the source.

$S=S_a+S_b+S_c$

Substitute $5980.8\angle 6.124\text{VA}$ for $S_a$, $2333.52\angle 24.2\text{VA}$ for $S_b$ and $5025.6\angle -16.5\text{VA}$ for $S_c$.

$\begin{array}{c}S=\left(5980.8\angle 6.124\text{VA}\right)+\left(2333.52\angle 24.2\text{VA}\right)+\left(5025.6\angle -16.5\text{VA}\right)\\ =12894\angle 0.743\text{VA}\\ \text{=12}\text{.894}\angle 0.743\text{kVA}\end{array}$

Conclusion:

Thus, the total complex power supplied by the source is $\text{12}\text{.894}\angle 0.743\text{kVA}$.