Question

Using the voltages of Problem 8.6(a) and the currents of Problem 8.5, compute the complex power dissipated based on (a) phase components and (b) symmetrical components.

Short answer

(a) The complex power dissipated based on phase component is $2.05\angle 12.84°\text{\hspace{0.33em}kVA}$.

(b) The complex power based on sequence component is $1.97\angle 12.31°\text{\hspace{0.33em}kVA}$.

Step-by-step solution

Write the given data from the question.

Write the phase component of the voltages.

$\begin{array}{l}{V}_0=10\angle 0°\text{\hspace{0.33em}V}\\ V_2=80\angle 30°\text{\hspace{0.33em}V}\\ V_2=40\angle -30°\text{\hspace{0.33em}V}\end{array}$

Write the phase component of the current.

$\begin{array}{l}{I}_a=10\angle 0°\text{\hspace{0.33em}A}\\ I_b=8\angle -90°\text{\hspace{0.33em}A}\\ I_c=6\angle 150°\text{\hspace{0.33em}A}\end{array}$

Determine the equation to calculate the complex power dissipated baes on phase component and symmetrical component.

The equation to calculate the phase voltages is given as follows.$\left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{V}_0\\ V_1\\ V_2\end{array}\right]$ ...........(1)

The equation to calculate the complex power based on phase 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)

The equation to calculate the sequence currents is given as follows.

$\left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]$ .........(3)

The equation to calculate the complex power based on sequence component is given as follows.

$S=V_0{\left(I_0\right)}^{*}+V_1{\left(I_1\right)}^{*}+V_2{\left(I_2\right)}^{*}$ ..........(4)

Calculate the complex power based on phase sequence component.

(b)

Calculate the phase voltages.

Substitute $10\angle 0°\text{\hspace{0.33em}V}$for $V_0$, $80\angle 30°\text{\hspace{0.33em}V}$for $V_1$and $40\angle -30°\text{\hspace{0.33em}V}$for$V_2$into equation (1).

$\begin{array}{l}\left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle -120& 1\angle 120\\ 1& 1\angle 120& 1\angle -120\end{array}\right]\left[\begin{array}{c}10\angle 0°\\ 80\angle 30°\\ 40\angle -30°\end{array}\right]\\ \left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{c}10\angle 0°+80\angle 30°+40\angle -30°\\ 10\angle 0°+\left(80\angle 30°\right)\left(1\angle -120\right)+\left(40\angle -30°\right)\left(1\angle 120\right)\\ 10\angle 0°+\left(80\angle 30°\right)\left(1\angle 120\right)+\left(40\angle -30°\right)\left(1\angle -120\right)\end{array}\right]\\ \left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{c}10\angle 0°+80\angle 30°+40\angle -30°\\ 10\angle 0°+\left(80\angle -90°\right)+\left(40\angle 90°\right)\\ 10\angle 0°+\left(80\angle 150°\right)+\left(40\angle -150°\right)\end{array}\right]\\ \left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{c}115.66\angle 9.96°\\ 41.23\angle -75.96°\\ 96.02\angle 167.97°\end{array}\right]\text{\hspace{0.33em}V}\end{array}$

Calculate the complex power dissipated based on phase component.

Substitute $115.66\angle 9.96°\text{\hspace{0.33em}V}$for $V_a$,$41.23\angle -75.96°\text{\hspace{0.33em}V}$for$V_b$and$96.02\angle 167.97°\text{\hspace{0.33em}V}$for $V_c$,$10\angle 0°\text{\hspace{0.33em}A}$for $I_a$,$8\angle -90°\text{\hspace{0.33em}A}$for$I_b$and$6\angle 150°\text{\hspace{0.33em}A}$for$I_c$into equation (2).

$\begin{array}{l}S=\left(115.66\angle 9.96°\right){\left(10\angle 0°\right)}^{*}+\left(41.23\angle -75.96°\right){\left(8\angle -90°\right)}^{*}+\left(96.02\angle 167.97°\right){\left(6\angle 150°\right)}^{*}\\ S=\left(115.66\angle 9.96°\right){\left(10\angle -0°\right)}^{*}+\left(41.23\angle -75.96°\right)\left(8\angle 90°\right)+\left(96.02\angle 167.97°\right)\left(6\angle -150°\right)\\ S=1156.6\angle 9.96°+329.84\angle 14.04°+576.12\angle 17.97°\\ S=2.05\angle 12.84°\text{\hspace{0.33em}kVA}\end{array}$

Hence, the complex power dissipated based on phase component is $2.05\angle 12.84°\text{\hspace{0.33em}VA}$.

Calculate the complex power based on sequence component.

(b)

Calculate the sequence current.

Substitute $10\angle 0°\text{\hspace{0.33em}A}$for $I_a$,$8\angle -90°\text{\hspace{0.33em}A}$for$I_b$and$6\angle 150°\text{\hspace{0.33em}A}$for$I_c$into equation (3)

$\begin{array}{l}\left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle 120°& 1\angle -120°\\ 1& 1\angle -120°& 1\angle 120°\end{array}\right]\left[\begin{array}{c}10\angle 0°\\ 8\angle -90°\\ 6\angle 150°\end{array}\right]\\ \left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}10\angle 0°+8\angle -90°+6\angle 150°\\ 10\angle 0°+\left(8\angle -90°\right)\left(1\angle 120°\right)+\left(6\angle 150°\right)\left(1\angle -120°\right)\\ 10\angle 0°+\left(8\angle -90°\right)\left(1\angle -120°\right)+\left(6\angle 150°\right)\left(1\angle 120°\right)\end{array}\right]\\ \left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}10\angle 0°+8\angle -90°+6\angle 150°\\ 10\angle 0°+\left(8\angle 30°\right)+\left(6\angle 30°\right)\\ 10\angle 0°+\left(8\angle -210°\right)\left(1\angle -120°\right)+\left(6\angle 270°\right)\end{array}\right]\\ \left[\begin{array}{c}{I}_0\\ I_1\\ I_2\end{array}\right]=\left[\begin{array}{c}4\angle 66.408°\\ 7.734\angle 17.556°\\ 1.221\angle 33.073°\end{array}\right]\end{array}$

Calculate the complex power based on sequence component.

Substitute$10\angle 0°\text{\hspace{0.33em}V}$ for $V_0$, $80\angle 30°\text{\hspace{0.33em}V}$for$V_1$ and$40\angle -30°\text{\hspace{0.33em}V}$ for $V_2$,$4\angle 66.408°$ for $I_0$, $7.734\angle 17.556°$for$I_1$ and$1.221\angle 33.073°$ for$I_2$ into equation (4).

s

Hence, the complex power based on sequence component is$1.97\angle 12.31°\text{\hspace{0.33em}kVA}$

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