Question

For Problem 8.25, determine the complex power delivered to the load in terms of symmetrical components. Verify the answer by adding up the complex power of each of the three phases.

Short answer

The power deliver to load with using symmetrical component$2922.77\angle 74.40°\text{\hspace{0.33em}kVA}$ and by adding up complex power of three phase is $2925.25\angle 74.40°\text{\hspace{0.33em}kVA}$which almost same.

Step-by-step solution

Write the given data from the question

Write the line to neutral voltages.

$\begin{array}{l}{V}_{an}=200\angle 25°\text{\hspace{0.33em}V}\\ {\text{V}}_{bn}=100\angle -155°\text{\hspace{0.33em}V}\\ {\text{V}}_{cn}=80\angle 100°\text{\hspace{0.33em}V}\end{array}$

The series impedance is $Z=6+j24\Omega$.

The mutual impedance is $Z_m=j3\Omega$.

The data from problem 8.25.

The sequence voltages$V_0,V_1$ and$V_2$ are$47.78\angle 57.63°\text{\hspace{0.33em}V,\hspace{0.33em}112.70}\angle \text{359.97}°\text{\hspace{0.33em}V}$ and$61.62\angle 45.88°\text{\hspace{0.33em}V}$ respectively.

The sequence currents$I_0,I_1$ and$I_2$ are$1.56\angle -21.06°\text{\hspace{0.33em}A,\hspace{0.33em}}5.16\angle 285.92°\text{\hspace{0.33em}A}$ and$5.16\angle -28.16°\text{\hspace{0.33em}A}$ respectively.

The phase currents$I_a,I_b$ and$I_c$ are$10.89\angle 313.03°\text{\hspace{0.33em}A,\hspace{0.33em}}6.93\angle 122.27°\text{A}$ and$0.8\angle 32.69°\text{\hspace{0.33em}A}$ respectively.

Determine the equation to calculate the complex power delivered to the load.

The equation to calculate the complex power delivered to the load in terms of symmetrical component is given as follows.

$S=3\left[V_0{\left(I_0\right)}^{*}+V_1{\left(I_1\right)}^{*}+V_2{\left(I_2\right)}^{*}\right]$ …… (1)

The equation to calculate the complex power without symmetrical component is given as follows.

$S=V_a{\left(I_a\right)}^{*}+V_b{\left(I_b\right)}^{*}+V_c{\left(I_c\right)}^{*}$ …… (2)

Calculate the complex power delivered to load with using symmetrical component and by adding up complex power of three phase.

Calculate the complex power delivered to the load in terms of symmetrical component.

Substitute $47.78\angle 57.63°\text{\hspace{0.33em}V,}$for $V_0$, $\text{112.70}\angle \text{359.97}°\text{\hspace{0.33em}V}$ for $V_1$, $61.62\angle 45.88°\text{\hspace{0.33em}V}$for $V_2$, $1.56\angle -21.06°\text{\hspace{0.33em}A,}$for $I_0$, $5.16\angle 285.92°\text{\hspace{0.33em}A}$ for $I_1$ and $5.16\angle -28.16°\text{\hspace{0.33em}A}$for $I_2$into equation (1).

$\begin{array}{l}S=3\left[\begin{array}{l}\left(47.78\angle 57.63°\right){\left(1.56\angle -21.06°\right)}^{*}+\left(\text{112.70}\angle \text{359.97}°\right){\left(5.16\angle 285.92°\right)}^{*}\\ +\left(61.62\angle 45.88°\text{\hspace{0.33em}V}\right){\left(5.16\angle -28.16°\right)}^{*}\end{array}\right]\\ S=3\left[\begin{array}{l}\left(47.78\angle 57.63°\right)\left(1.56\angle 21.06°\right)+\left(\text{112.70}\angle \text{359.97}°\right)\left(5.16\angle -285.92°\right)\\ +\left(61.62\angle 45.88°\text{\hspace{0.33em}V}\right)\left(5.16\angle 28.16°\right)\end{array}\right]\\ S=3\left[74.54\angle 78.69°+581.99\angle 74.05°+317.96\angle 74.05°\right]\\ S=2922.77\angle 74.40°\text{\hspace{0.33em}kVA}\end{array}$

Calculate the complex power by adding up complex power of three phase.

Substitute $200\angle 25°\text{\hspace{0.33em}V}$for $V_a$, $100\angle -155°\text{\hspace{0.33em}V}$for $V_1$, $80\angle 100°\text{\hspace{0.33em}V}$for $V_2$, $10.89\angle 313.03°\text{\hspace{0.33em}A}$for $I_0$, $6.93\angle 122.27°\text{A}$for $I_1$and $0.8\angle 32.69°\text{\hspace{0.33em}A}$for $I_2$into equation(2).

$\begin{array}{l}S=\left[\begin{array}{l}\left(200\angle 25°\right){\left(10.89\angle 313.03°\right)}^{*}+\left(100\angle -155°\right){\left(6.93\angle 122.27°\right)}^{*}\\ +\left(80\angle 100°\text{\hspace{0.33em}V}\right){\left(0.8\angle 32.69°\right)}^{*}\end{array}\right]\\ S=\left[\begin{array}{l}\left(200\angle 25°\right)\left(10.89\angle -313.03°\right)+\left(100\angle -155°\right)\left(6.93\angle -122.27°\right)\\ +\left(80\angle 100°\text{\hspace{0.33em}V}\right)\left(0.8\angle -32.69°\right)\end{array}\right]\\ S=2178\angle -288.03+693\angle 277.27°+\mathrm{64.67.31}°\\ S=2925.25\angle 74.40°\text{\hspace{0.33em}kVA}\end{array}$

Hence, the power deliver to load with using symmetrical component $2922.77\angle 74.40°\text{\hspace{0.33em}kVA}$and by adding up complex power of three phase is$2925.25\angle 74.40°\text{\hspace{0.33em}kVA}$ which almost same.