Chapter 12, Problem 56P

Using Fig. 12.63, design a problem to help other students to better understand unbalanced three-phase systems.

Figure 12.63

For Prob. 12.56.

To determine

Design a problem using Figure 12.63 for the better understand of the unbalanced three-phase system.

Explanation of Solution

Problem design:

For the circuit in Figure 12.63, the parameters are $X_L=\text{ 10}\Omega$, $X_C=5\Omega$  and $R=20\Omega$. The phase voltage $V_p=440\text{ V}$. Find the line currents, real power absorbed by the load and the total complex power supplied from the source.

Calculation:

Sketch the given circuit as shown in Figure 1.

Write the voltage equation for loop 1 using Kirchhoff’s voltage law.

$\begin{array}{l}-\left(440\angle 0°\text{ V}\right)+\left(440\angle -120°\text{V}\right)+\left(j10\Omega \right)\left(I_1-I_3\right)=0\\ \left(j10\Omega \right)\left(I_1-I_3\right)=\left(440\angle 0°\text{ V}\right)-\left(440\angle -120°\text{V}\right)\\ I_1-I_3=\frac{\left(440\angle 0°-440\angle -120°\right)\text{ V}}{j10\Omega }\\ I_1-I_3=\frac{440-\left(-220-j381.05\right)}{j10}\frac{\text{V}}{\Omega }\end{array}$

Reduce the equation as follows.

$\begin{array}{l}{I}_1-I_3=\frac{660+j381.05}{j10}\text{ A}\\ I_1-I_3=\frac{762.10\angle 30°}{10\angle 90°}\text{ A}\end{array}$

$I_1-I_3=76.21\angle -60°\text{ A}$        (1)

Write the voltage equation for loop 2 using Kirchhoff’s voltage law.

$\begin{array}{l}\left(440\angle 120°\text{ V}\right)-\left(440\angle -120°\text{V}\right)+\left(20\Omega \right)\left(I_2-I_3\right)=0\\ -\left(20\Omega \right)\left(I_2-I_3\right)=\left(440\angle 120°\text{ V}\right)-\left(440\angle -120°\text{V}\right)\\ I_3-I_2=\frac{\left(440\angle 120°\text{ V}\right)-\left(440\angle -120°\text{V}\right)\text{ V}}{20\Omega }\\ I_3-I_2=\frac{-220+j381.05-\left(-220-j381.05\right)}{20}\frac{\text{V}}{\Omega }\end{array}$

Reduce the equation as follows.

$\begin{array}{l}{I}_3-I_2=\frac{j762.1}{20}\text{ A}\\ I_3-I_2=\frac{762.1\angle 90°}{20}\text{ A}\\ I_3-I_2=38.105\angle 90°\text{ A}\end{array}$

$I_3-I_2=38.1\angle 90°\text{ A}$        (2)

Write the voltage equation for loop 3 using Kirchhoff’s voltage law.

$\begin{array}{l}\left(j10\Omega \right)\left(I_3-I_1\right)+\left(20\Omega \right)\left(I_3-I_2\right)-\left(j5\Omega \right)I_3=0\\ -\left(j10\Omega \right)\left(I_1-I_3\right)+\left(20\Omega \right)\left(I_3-I_2\right)=\left(j5\Omega \right)I_3\end{array}$

$I_3=\frac{-\left(j10\Omega \right)\left(I_1-I_3\right)+\left(20\Omega \right)\left(I_3-I_2\right)}{j5\Omega }$        (3)

Substitute $76.21\angle -60°\text{ A}$ for $I_1-I_3$ and $38.1\angle 90°\text{ A}$ for $I_3-I_2$ in equation (3).

$\begin{array}{l}{I}_3=\frac{-\left(j10\Omega \right)\left(76.21\angle -60°\text{ A}\right)+\left(20\Omega \right)\left(38.1\angle 90°\text{ A}\right)}{j5\Omega }\\ I_3=\frac{-\left(j10\right)\left(76.21\angle -60°\right)+\left(20\right)\left(38.1\angle 90°\right)}{j5}\frac{\Omega \cdot \text{A}}{\Omega }\\ I_3=\frac{-\left(10\angle 90°\right)\left(76.21\angle -60°\right)+\left(20\right)\left(38.1\angle 90°\right)}{5\angle 90°}\text{A}\\ I_3=\frac{-762.1\angle 30°+762\angle 90°}{5\angle 90°}\text{A}\end{array}$

Reduce the equation as follows.

$\begin{array}{l}{I}_3=\frac{-\left(659.91+j381\right)+\left(j762\right)}{5\angle 90°}\text{A}\\ I_3=\frac{-659.91+j381}{5\angle 90°}\text{A}\\ I_3=\frac{762\angle 150°}{5\angle 90°}\text{A}\end{array}$

$I_3=152.4\angle 60°\text{A}$

Substitute $152.4\angle 60°\text{A}$ for $I_3$ in equation (1) to find the current $I_1$.

$\begin{array}{l}{I}_1-152.4\angle 60°\text{A}=76.21\angle -60°\text{ A}\\ I_1=\left(76.21\angle -60°\text{ A}\right)+\left(152.4\angle 60°\text{A}\right)\\ I_1=38.11-j66+76.2+j131.98\text{ A}\\ I_1=114.31+j65.98\text{ A}\end{array}$

Rewrite the equation as,

$\begin{array}{l}{I}_1=131.99\angle 29.99°\text{A}\\ I_1\cong 132\angle 30°\text{A}\end{array}$

Substitute $152.4\angle 60°\text{A}$ for $I_3$ in equation (2) to find the current $I_2$.

$\begin{array}{l}152.4\angle 60°\text{A}-I_2=38.1\angle 90°\text{ A}\\ I_2=152.4\angle 60°\text{A}-38.1\angle 90°\text{ A}\\ I_2=\left(76.2+j131.98\text{A}\right)-\left(j38.1\text{ A}\right)\end{array}$

Reduce the equation as follows.

$\begin{array}{l}{I}_2=76.2+j93.9\text{A}\\ I_2=120.93\angle 50.94°\text{A}\end{array}$

From the figure 1, the current line currents can be defend as follows.

$I_a=I_1$

Substitute $132\angle 30°\text{ A}$ for $I_1$ in above equation to find  line current $I_a$.

$I_a=132\angle 30°\text{ A}$

And

$I_b=I_2-I_1$

Substitute $120.93\angle 50.94°\text{ A}$ for $I_2$ and $132\angle 30°\text{ A}$ for $I_1$ in above equation to find  line current $I_b$.

$\begin{array}{l}{I}_b=120.93\angle 50.94°-132\angle 30°\text{ A}\\ I_b=76.2+j93.90-\left(114.32+j66\right)\text{ A}\end{array}$

$I_b=-38.12+j27.9\text{ A}$

$I_b=47.23\angle 143.8°\text{ A}$

And

$I_c=-I_2$

Substitute $120.93\angle 50.94°\text{ A}$ for $I_2$ in above equation to find line current $I_c$.

$I_c=-120.93\angle 50.94°\text{ A}$

$\begin{array}{l}{I}_c=120.93\angle \left(50.94°+180°\right)\text{A}\\ I_c=120.93\angle 230.94°\text{A}\end{array}$

$I_c\cong 120.9\angle 230.9°\text{A}$

Therefore, the line currents from source to load are,

$I_a=132\angle 30°\text{ A}$

$I_b=47.23\angle 143.8°\text{ A}$

$I_c=120.9\angle 230.9°\text{A}$

Write the expression to find the total complex power absorbed by the load as,

$S={\left|I_1-I_3\right|}^2\left(j10\Omega \right)+{\left|I_2-I_3\right|}^2\left(20\Omega \right)+{\left|I_3\right|}^2\left(-j5\Omega \right)$

Substitute $76.21\angle -60°\text{ A}$ for $I_1-I_3$, $38.105\angle 90°\text{ A}$ for $I_3-I_2$ and $152.4\angle 60°\text{A}$ for $I_3$ to find $S$.

$\begin{array}{l}S={\left|76.21\angle -60°\text{ A}\right|}^2\left(j10\Omega \right)+{\left|38.105\angle 90°\text{ A}\right|}^2\left(20\Omega \right)+{\left|152.4\angle 60°\text{A}\right|}^2\left(-j5\Omega \right)\\ S={\left|76.21\text{ A}\right|}^2\left(j10\Omega \right)+{\left|38.105\text{ A}\right|}^2\left(20\Omega \right)+{\left|152.4\text{ A}\right|}^2\left(-j5\Omega \right)\\ S=j58,079.641+29,039.8205-j116,128.8{\text{ A}}^2\cdot \Omega \\ S\cong 29.04-j58.05\text{ kVA}\end{array}$

The real power absorbed by the load is $29.04\text{ kW}$.

The total complex power absorbed by the load is equal to the total complex power supplied by the source.

Thus, the line currents $I_a$, $I_b$ and $I_c$ are $\underset{\_}{132\angle 30°\text{ A}}$, $\underset{\_}{47.23\angle 143.8°\text{ A}}$ and $\underset{\_}{120.9\angle 230.9°\text{A}}$ respectively. The real power absorbed by the load is $\underset{\_}{29.04\text{ kW}}$. And total complex power supplied by the source is $\underset{\_}{29.04-j58.05\text{ kVA}}$.

Conclusion:

Thus, a problem designed, and solved for the better understand of the unbalanced three-phase system.