STEM Problem Detail

Problem #41

Electrical Engineering Hard

Problem Statement

Prompt

Provide step by step solution of the question number 2.

Solution Rendering...

## Problem Solution: Power Flow Calculation (General)

**(sol)**

\begin{align*}
S_{12} &= \frac{|V_1|^2}{|Z|} \angle \theta_z - \frac{|V_1||V_2|}{|Z|} \angle \theta_z \angle \theta_{12} = \frac{|V_1|^2}{|1|} \angle 85^\circ - \frac{|V_1||V_2|}{|1|} \angle 85^\circ \angle 10^\circ = |V_1|^2 \angle 85^\circ - |V_1||V_2| \angle 95^\circ \\
-S_{21} &= -\frac{|V_2|^2}{|Z|} \angle \theta_z + \frac{|V_1||V_2|}{|Z|} \angle \theta_z \angle -\theta_{12} = -\frac{|V_2|^2}{|1|} \angle 85^\circ + \frac{|V_1||V_2|}{|1|} \angle 85^\circ \angle -10^\circ = -|V_2|^2 \angle 85^\circ + |V_1||V_2| \angle 75^\circ
\end{align*}

**(a)** $$|V_1| = |V_2| = 1.0$$
\begin{align*}
S_{12} &= 1 \angle 85^\circ - 1 \angle 95^\circ = \underline{\underline{0.1743 + j0}} = 0.1743 \angle 0^\circ (5\%) \\
-S_{21} &= -1 \angle 85^\circ + 1 \angle 75^\circ = \underline{\underline{0.1717 - j0.0303}} = 0.1743 \angle -10^\circ (5\%)
\end{align*}

**(b)** $$|V_1| = 1.1, |V_2| = 0.9$$
\begin{align*}
S_{12} &= 1.1^2 \angle 85^\circ - (1.1 \times 0.9) \angle 95^\circ = 1.21 \angle 85^\circ - 0.99 \angle 95^\circ \\
&= \underline{\underline{0.1917 + j0.2192}} = 0.2912 \angle 48.8178^\circ (5\%) \\
-S_{21} &= -0.9^2 \angle 85^\circ + (1.1 \times 0.9) \angle 75^\circ = -0.81 \angle 85^\circ + 0.99 \angle 75^\circ \\
&= \underline{\underline{0.1856 + j0.1493}} = 0.2383 \angle 38.8178^\circ (5\%)
\end{align*}

## Problem Solution: $$Z_{\text{line}} = 0.1 \angle 85^\circ$$

**(Sol)**
$$Z_{\text{line}} = 0.1 \angle 85^\circ; |V_1| = 1.05, |V_2| = 0.95$$

\begin{align*}
S_{12} &= \frac{|V_1|^2}{|Z|} e^{j\theta_z} - \frac{|V_1||V_2|}{|Z|} e^{j\theta_z} e^{j\theta_{12}} = \frac{1.05^2}{0.1} \angle 85^\circ - \frac{1.05 \cdot 0.95}{0.1} \angle 85^\circ \angle \theta_{12} \\
&= 11.025 \angle 85^\circ - 9.975 \angle(85^\circ + \theta_{12}) \\
P_{12} &= 11.025 \cos 85^\circ - 9.975 \cos(85^\circ + \theta_{12})
\end{align*}

\begin{align*}
-S_{21} &= -\frac{|V_2|^2}{|Z|} \angle \theta_z + \frac{|V_1||V_2|}{|Z|} \angle (\theta_z - \theta_{12}) = -\frac{0.95^2}{0.1} \angle 85^\circ + \frac{1.05 \cdot 0.95}{0.1} \angle (85^\circ - \theta_{12}) \\
&= -9.025 \angle 85^\circ + 9.975 \angle (85^\circ - \theta_{12}) \\
-P_{21} &= -9.025 \cos 85^\circ + 9.975 \cos(85^\circ - \theta_{12})
\end{align*}

(b) $$\theta_{12} = 180 - \theta_z = 180 - 85^\circ = 95^\circ \Rightarrow P_{12\text{maximum}}$$
(a) $$P_{12\text{max}} = 11.025 \cos 85^\circ - 9.975 \cos(85^\circ + 95^\circ) = 10.936$$
(d) $$\theta_{12} = \theta_z = 85^\circ \Rightarrow -P_{21} \text{ maximum}$$
(c) $$-P_{12\text{max}} = -9.025 \cos 85^\circ + 9.975 \cos(85^\circ - 85^\circ) = 9.1884$$
(e) $$P_{\text{loss}} = P_{12} - (-P_{21}); \theta_{12} = 10^\circ$$

\begin{align*}
P_{12} &= 11.025 \cos 85^\circ - 9.975 \cos(85^\circ + 10^\circ) = 1.8303 \\
-P_{21} &= -9.025 \cos 85^\circ + 9.975 \cos(85^\circ - 10^\circ) = 1.7951 \\
P_{\text{loss}} &= P_{12} - (-P_{21}) = 1.8303 - 1.7951 = 0.0352
\end{align*}

(a) $$P_{12\text{max}} = 10.9 \sim 11.0 (4\%)$$; 10.936
(b) $$\theta_{12} = 95^\circ (\text{完全正確才給分}(4\%))$$
(c) $$-P_{12\text{max}} = 9.0 \sim 9.2 (4\%)$$; 9.1884
(d) $$\theta_{12} = 85^\circ (\text{完全正確才給分}(4\%))$$
(e) $$P_{\text{loss}} = 0.0350 \sim 0.0354 (4\%)$$; 0.0352

## Problem Solution: Prob 4.26 Modified ($$Z_{\text{line}}=0.1\angle 80^\circ$$)

**(Sol prob4.26 改 $$Z_{\text{line}}=0.1\angle 80^\circ$$)**
$$Z_{\text{line}} = 0.1 \angle 80^\circ; |V_1| = 1.05, |V_2| = 0.95$$

\begin{align*}
S_{12} &= \frac{|V_1|^2}{|Z|} e^{j\theta_z} - \frac{|V_1||V_2|}{|Z|} e^{j\theta_z} e^{j\theta_{12}} = \frac{1.05^2}{0.1} \angle 80^\circ - \frac{1.05 \cdot 0.95}{0.1} \angle 80^\circ \angle \theta_{12} \\
&= 11.03 \angle 80^\circ - 9.975 \angle(80^\circ + \theta_{12}) \\
P_{12} &= 11.03 \cos 80^\circ - 9.975 \cos(80^\circ + \theta_{12})
\end{align*}

\begin{align*}
-S_{21} &= -\frac{|V_2|^2}{|Z|} \angle \theta_z + \frac{|V_1||V_2|}{|Z|} \angle (\theta_z - \theta_{12}) = -\frac{0.95^2}{0.1} \angle 80^\circ + \frac{1.05 \cdot 0.95}{0.1} \angle (80^\circ - \theta_{12}) \\
&= -9.025 \angle 80^\circ + 9.975 \angle(80^\circ - \theta_{12}) \\
-P_{21} &= -9.025 \cos 80^\circ + 9.975 \cos(80^\circ - \theta_{12})
\end{align*}

(b) $$\theta_{12} = 180 - \theta_z = 180 - 80^\circ = 100^\circ \Rightarrow P_{12\text{maximum}}$$
(a) $$P_{12\text{max}} = \underbrace{11.03 \cos 80^\circ}_{1.915} - 9.975 \cos(80^\circ + 100^\circ) = 11.89 \doteq 11.9$$
(d) $$\theta_{12} = \theta_z = 80^\circ \Rightarrow -P_{21} \text{ maximum}$$
(c) $$-P_{21\text{max}} = \underbrace{-9.025 \cos 80^\circ}_{-1.567} + 9.975 \cos(80^\circ - 80^\circ) = 8.408 \doteq 8.4$$
(e) $$P_{\text{loss}} = P_{12} - (-P_{21}); \theta_{12} = 10^\circ$$

\begin{align*}
P_{12} &= \underbrace{11.03 \cos 80^\circ}_{1.915} - \underbrace{9.975 \cos(80^\circ + 10^\circ)}_{0} = 1.915 \\
-P_{21} &= \underbrace{-9.025 \cos 80^\circ}_{-1.567} + \underbrace{9.975 \cos(80^\circ - 10^\circ)}_{3.412} = 1.845
\end{align*}

$$P_{\text{loss}} = P_{12} - (-P_{21}) = 1.915 - 1.845 = 0.07$$

(a) $$P_{12\text{max}} = 11.89 \doteq 11.9 (3\%)$$
(b) $$\theta_{12} = 100^\circ (\text{完全正確才給分}(3\%))$$
(c) $$-P_{21\text{max}} = 8.408 \doteq 8.4 (3\%)$$
(d) $$\theta_{12} = 80^\circ (3\%)$$
(e) $$P_{\text{loss}} = 0.07 (3\%)$$

## Problem Solution: P4.27 (Case A)

**(Sol P4.27)**
$$|V_2| = 1, \quad Z = |Z|\angle \theta_z = 0.01 + j0.1 = 0.1005 \angle 84.2894^\circ$$

- $$S_{D2} = 0.5 + j0.5 \Rightarrow P_{D2} = 0.5P_{21} = -P_{D2} = -0.5$$

\[
P_{21} = \text{Re}\left[ -\frac{|V_2|^2}{|Z|} \angle \theta_z - \frac{|V_1||V_2|}{|Z|} \angle(\theta_z - \theta_{12}) \right] = -\frac{|V_2|^2}{|Z|} \cos \theta_z - \frac{|V_1||V_2|}{|Z|} \cos(\theta_z - \theta_{12})
\]

\[
-0.5 = -\underbrace{\frac{(1)^2}{0.1005} \cos(84.2894^\circ)}_{0.9901} - \underbrace{\frac{(1 \times 1)}{0.1005}}_{9.9504} \cos(84.2894^\circ - \theta_{12})
\]

\[
-0.5 = -0.9901 - 9.9504 \cos(\alpha)
\]

\[
\alpha = 84.2894^\circ - \theta_{12} = \cos^{-1}\left( \frac{0.5+0.9901}{9.9504} \right) = 81.3874^\circ
\]

\[
\theta_{12} = 84.2894^\circ - 81.3874^\circ = 2.902^\circ
\]

$$\therefore V_2 = 0.9987 - j0.0506 = 1 \angle -2.902^\circ (7\%)$$
- $$S_{12} = \frac{|V_1|^2}{|Z|} \angle \theta_z - \frac{|V_1||V_2|}{|Z|} \angle(\theta_z + \theta_{12}) = \frac{1^2}{0.1005} \angle 84.2894^\circ - \frac{1 \times 1}{0.1005} \angle(84.2894^\circ + 2.902^\circ)$$
\[
= 9.9504 \angle 84.2894^\circ - 9.9504 \angle 87.1914^\circ = 0.5027 - j0.0374
\]
$$S_{G1} = S_{D1} + S_{12} \Rightarrow S_{G1} = (0.5 + j0.5) + (0.503 - j0.037) = 1.0025 + j0.4626 = 1.104 \angle 24.77^\circ (7\%)$$
- $$S_{21} = -\frac{|V_2|^2}{|Z_{\text{line}}|} \angle \theta_z + \frac{|V_1||V_2|}{|Z_{\text{line}}|} \angle(\theta_z - \theta_{12}) = -\frac{1^2}{0.1005} \angle 84.2894^\circ + \frac{1 \times 1}{0.1005} \angle(84.2894^\circ - 2.902^\circ)$$
\[
= -9.9504 \angle 84.2894^\circ + 9.9504 \angle 81.3874^\circ = -0.500 + j0.0628
\]
$$S_{G2} = S_{D2} + S_{21} \Rightarrow S_{G2} = (0.5 + j0.5) + (-0.500 + j0.063) = j0.5628$$

$$Q_{G2} = 0.5628 (7\%)$$

## Problem Solution: P4.27 (Case B)

**(Sol P4.27) P21**
$$|V_2| = 1, \quad Z = |Z|\angle \theta_z = 0.01 + j0.1 = 0.1005 \angle 84.29^\circ$$

- $$S_{D2} = 0.4 + j0.3 \Rightarrow P_{D2} = 0.4P_{21} = -P_{D2} = -0.4$$

\[
-0.4 = -\underbrace{\frac{(1)^2}{0.1005} \cos(84.29^\circ)}_{0.9900} - \underbrace{\frac{(1 \times 1)}{0.1005}}_{9.950} \cos(84.29^\circ - \theta_{12})
\]

\[
-0.4 = -0.99 - 9.95 \cos(\alpha)
\]

\[
\alpha = 84.29^\circ - \theta_{12} = \cos^{-1}\left( \frac{0.4+0.99}{9.95} \right) = 81.97^\circ
\]

\[
\theta_{12} = 84.29^\circ - 81.97^\circ = 2.32^\circ
\]

$$\therefore V_2 = 1 \angle -2.32^\circ = 0.9992 - j0.0405 (7\%)$$
- $$S_{12} = \frac{|V_1|^2}{|Z|} \angle \theta_z - \frac{|V_1||V_2|}{|Z|} \angle(\theta_z + \theta_{12}) = \frac{1^2}{0.1005} \angle 84.29^\circ - \frac{1 \times 1}{0.1005} \angle(84.29^\circ + 2.32^\circ)$$
\[
= 9.95 \angle 84.29^\circ - 9.95 \angle(86.61^\circ) = 0.4029 \angle -4.55^\circ = 0.4016 - j0.03196
\]
$$S_{G1} = S_{D1} + S_{12} \Rightarrow S_{G1} = (0.5 + j0.5) + (0.4016 - j0.03196) = 0.9016 + j0.4680 = 1.016 \angle 27.43^\circ (7\%)$$
- $$S_{21} = \frac{|V_2|^2}{|Z_{\text{line}}|} \angle \theta_z - \frac{|V_1||V_2|}{|Z_{\text{line}}|} \angle(\theta_z - \theta_{12}) = \frac{1^2}{0.1005} \angle 84.29^\circ - \frac{1 \times 1}{0.1005} \angle(84.29^\circ - 2.32^\circ)$$
\[
= 9.95 \angle 84.29^\circ - 9.95 \angle 81.97^\circ = 0.4029 \angle 173.1^\circ = -0.4 + j0.0484
\]
$$S_{G2} = S_{D2} + S_{21} \Rightarrow S_{G2} = (0.4 + j0.3) + (-0.4 + j0.0484) = j 0.3484$$

$$Q_{G2} = 0.3484 \approx 0.348 (7\%)$$

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