Question

The two parallel lines in Example 3.13 supply a balanced load with a load current of $1\angle -30°$per unit. Determine the real and reactive power supplied to the load bus from each parallel line with (a) no regulating transformer, (b) the transformer in Example 3.13(a), and (c) the phase-angle-regulating transformer in Example Assume that the voltage at bus $abc$is adjusted so that the voltage at bus $a\text{'}b\text{'}c\text{'}$remains constant at $1\angle 0°$per unit. Also assume positive sequence. Comment on the effects of the regulating transformer.

Short answer

a)The real and reactive power supplied to the load bus by Line$L1$ and$L2$ are$0.480+j0.277\text{\hspace{0.17em}pu}$ and $0.386+j0.233\text{\hspace{0.17em}pu}$, respectively.

(b)The real and reactive power supplied to the load bus by Line$L2$ and $L1$are$0.4915+j0.1802\text{\hspace{0.17em}pu}$and $0.3745+j0.3198\text{\hspace{0.17em}pu}$,respectively. The voltage magnitude regulation transformer increases the reactive and real power supplied to load bus by the line$L1$ .

(c) The real and reactive power supplied to the load bus by Line $L2$and$L1$ are$0.3585+j0.2876\text{\hspace{0.17em}pu}$ and $0.5075+j0.2124\text{\hspace{0.17em}pu}$$0.5075+j0.2124\text{\hspace{0.17em}pu}$respectively. The phase angle regulating transformer increase the real power of the line$L1$ with the small change in the reactive power.

Step-by-step solution

Write the given data from the question. 

The load current$I_L=1\angle -30°\text{\hspace{0.17em}A}$.

The voltage at$a\text{'}b\text{'}c\text{'}$$V_1=1\angle 0°\text{\hspace{0.17em}pu}$.

The reactance of the line, L1.$X_{L1}=j0.25\text{\hspace{0.17em}pu}$

The reactance of the line$X_{L2}=j0.20\text{\hspace{0.17em}pu}$, .

 Step 2: Determine the formulas to calculate the real and reactive power supplied to the load bus by each parallel line.

The equation for complex power supplied by line$L1$and $L2$ is as follows:

$P_{Load}+j{Q}_{Load}=V_1{I}_L^{*}$…… (1)

Here,$P_{Load}$is the real power,and $Q_{Load}$is the reactive power supplied to the load bus.

Theequationto calculate the line current is$L2$as follows:

$I_{L2}=\left(\frac{X_{L1}}{X_{L1}+X_{L2}}\right)I_L$…… (2)

Theequationto calculate the real and reactive power supplied to load bus by line$L2$ is givenas follows:

$P_{L2}+j{Q}_{L2}=V_1{I}_{L2}^{*}$…… (3)

Here,$P_{L2}$is the real power,and $Q_{L2}$is the reactive power supplied to the load bus by line$L2$ .

Theequationto calculate the real and reactive power supplied to load bus by line $L1$ is givenas follows:

$P_{L1}+j{Q}_{L1}=\left(P_{Load}+j{Q}_{Load}\right)-\left(P_{L2}+j{Q}_{L2}\right)$…… (4)

Here$P_{L1}$,is the real power,and $Q_{L1}$is the reactive power supplied to the load bus by line$L2$.

The admittance matrix is given as follows:

$\left[\begin{array}{c}I\\ I_L\end{array}\right]=\left[\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right]\left[\begin{array}{c}V\\ V_1\end{array}\right]$…… (5)

Here, $Y_{11},Y_{12},Y_{21}\mathbf{and}{Y}_{22}$are the admittance of the lines.

Determine the real and reactive power supplied by each parallel line with no regulation transformer.

(a)

Draw the circuit of the system.

Calculate the real and reactive power supplied to the load bus.

Substitute$1\angle 0°\text{\hspace{0.17em}pu}$ for $V_1$and $1\angle -30°\text{\hspace{0.17em}A}$for$I_L$into equation (1).

$\begin{array}{l}{P}_{Load}+j{Q}_{Load}=\left(1\angle 0°\right){\left(1\angle -30°\right)}^{*}\\ P_{Load}+j{Q}_{Load}=1\angle 30°\\ P_{Load}+j{Q}_{Load}=0.866+j0.5\text{\hspace{0.17em}pu}\end{array}$

Calculate the current for line $L1$.

Substitute$j0.25\text{\hspace{0.17em}pu}$for$X_{L1}$,$j0.20\text{\hspace{0.17em}pu}$for$X_{L2}$,and $1\angle -30°\text{\hspace{0.17em}A}$for$I_L$into equation (2).

$\begin{array}{l}{I}_{L2}=\left(\frac{0.25}{0.25+0.20}\right)1\angle -30°\\ I_{L2}=\left(\frac{0.25}{0.45}\right)1\angle -30°\\ I_{L2}=0.555\angle -30°\text{\hspace{0.17em}pu}\end{array}$

Calculate the real and the reactive power supplied to load bus by line$L2$.

Substitute $1\angle 0°\text{\hspace{0.17em}pu}$for$V_1$and $0.555\angle -30°\text{\hspace{0.17em}pu}$for $I_{L1}$into equation (3).

$\begin{array}{l}{P}_{L2}+j{Q}_{L2}=\left(1\angle 0°\right){\left(0.555\angle -30°\right)}^{*}\\ P_{L2}+j{Q}_{L2}=0.555\angle 30°\\ P_{L2}+j{Q}_{L2}=0.480+j0.277\text{\hspace{0.17em}pu}\end{array}$

Calculate the real and reactive power supplied to the load bus by line$L1$.

Substitute $0.480+j0.277\text{\hspace{0.17em}pu}$for$P_{L2}+j{Q}_{L2}$and $0.866+j0.5\text{\hspace{0.17em}pu}$for $P_{Load}+j{Q}_{Load}$into equation (4).

$\begin{array}{l}{P}_{L1}+j{Q}_{L1}=\left(0.866+j0.5\right)-\left(0.480+j0.277\right)\\ P_{L1}+j{Q}_{L1}=0.386+j0.233\text{\hspace{0.17em}pu}\end{array}$

Hence, the real and reactive power supplied to the load bus by Line$L2$ and $L1$are $0.480+j0.277\text{\hspace{0.17em}pu}$and$0.386+j0.233\text{\hspace{0.17em}pu}$ , respectively.

Step 4:Determine the real and reactive power supplied by each parallel line with line no regulation transformer.

(b)

The$2\times 2$admittance matrix.

From example 3.13,

.$\left[\begin{array}{c}I\\ -1\angle -30°\end{array}\right]=\left[\begin{array}{cc}-j9& j8.810\\ j8.810& -j8.628\end{array}\right]\left[\begin{array}{c}V\\ 1\angle 0°\end{array}\right]$

Solve the second equation.

$\begin{array}{c}-1\angle -30°=j8.810V-j9\left(1\angle 30°\right)\\ V=\frac{8.628\angle 90\angle -1\angle -30°}{8.810\angle 90°}\\ V=\frac{0.866+j9.128}{8.810\angle 90}\\ V=1.04\angle 5.42°\text{\hspace{0.17em}pu}\end{array}$

Calculate the current$I_{L2}$from the figure shown in the step1.

$I_{L2}=\frac{V-V_1}{j{X}_{L1}}$

Substitute $1.04\angle 5.42°\text{\hspace{0.17em}pu}$for , for$V_1$, and $j0.20\text{\hspace{0.17em}pu}$for $X_{L1}$into the above equation.

$\begin{array}{l}{I}_{L2}=\frac{1.04\angle 5.42°°-1\angle 0°\text{\hspace{0.17em}pu}}{j0.20}\\ I_{L2}=\frac{0.1048\angle 69.71}{0.20\angle 90°}\\ I_{L2}=0.524\angle -20.29°\text{\hspace{0.17em}pu}\end{array}$

Calculate the real and reactive power supplied to the load by line$L2$.

Substitute$0.524\angle -20.29°\text{\hspace{0.17em}pu}$ for $I_{L2}$and $1\angle 0°\text{\hspace{0.17em}pu}$ for $V_1$into equation (3).

$\begin{array}{l}{P}_{L2}+j{Q}_{L2}=\left(1\angle 0°\right){\left(0.524\angle -20.29°\right)}^{*}\\ P_{L2}+j{Q}_{L2}=0.524\angle 20.29°\\ P_{L2}+j{Q}_{L2}=0.4915+j0.1802\text{\hspace{0.17em}pu}\end{array}$

Calculate the real and reactive power supplied to load by the line $L1$.

Substitute$0.474+j0.228\text{\hspace{0.17em}pu}$for $P_{L2}+j{Q}_{L2}$and $0.866+j0.5\text{\hspace{0.17em}pu}$for$P_{Load}+j{Q}_{Load}$ into equation (4).

$\begin{array}{l}{P}_{L1}+j{Q}_{L1}=\left(0.866+j0.5\right)-\left(0.4915+j0.1802\right)\\ P_{L1}+j{Q}_{L1}=0.3745+j0.3198\text{\hspace{0.17em}pu}\end{array}$

Hence, the real and reactive power supplied to the load bus by Line$L2$ and $L1$are $0.4915+j0.1802\text{\hspace{0.17em}pu}$and $0.3745+j0.3198\text{\hspace{0.17em}pu}$, respectively. The voltage magnitude regulation transformer increases the reactive and real power supplied to load bus by the line $L1$.

Determine the real and reactive power supplied to the load bus with a phase angle regulation transformer

(c)

The admittance the example 3.13(b)is calculated as follows:

$\begin{array}{l}{Y}_{21}=-0.2093+J8.8945\text{\hspace{0.17em}pu}\\ Y_{22}=-j9\text{\hspace{0.17em}pu}\end{array}$

The voltage $V$can be calculated asfollows:

$V=\frac{Y_{22}{V}_1-I_{Load}}{Y_{21}}$

Substitute$-0.2093+J8.8945\text{\hspace{0.17em}pu}$for $Y_{21}$,$-j9\text{\hspace{0.17em}pu}$for $Y_{22}$,$1\angle 0°\text{\hspace{0.17em}pu}$for$V_1$, and$1\angle -30°\text{\hspace{0.17em}A}$for $I_L$in the above equation.

$\begin{array}{l}V=\frac{\left(-j9\right)\left(1\angle 0°\text{\hspace{0.17em}pu}\right)-1\angle -30°\text{\hspace{0.17em}A}}{-0.2093+J8.8945}\\ V=\frac{9.53\angle 95.21°}{8.99\angle 91.33°}\\ V=1.06\angle 3.89°\text{\hspace{0.17em}pu}\end{array}$

Calculate the current$I_{L2}$from the figure shown in step1.

$I_{L2}=\frac{V-V_1}{j{X}_{L1}}$

Substitute$1.06\angle 3.89°\text{\hspace{0.17em}pu}$for$V$,$1\angle 0°\text{\hspace{0.17em}pu}$for$V_1$, and $j0.20\text{\hspace{0.17em}pu}$for$X_{L1}$into the above equation.

$\begin{array}{l}{I}_{L2}=\frac{1.06\angle 3.89°\text{\hspace{0.17em}pu}°-1\angle 0°\text{\hspace{0.17em}pu}}{j0.20}\\ I_{L2}=\frac{0.0919\angle 51.24}{0.20\angle 90°}\\ I_{L2}=0.4595\angle -38.76°\text{\hspace{0.17em}pu}\end{array}$

Calculate the real and reactive power supplied to the load by line $L2$.

Substitute$0.4595\angle -38.76°\text{\hspace{0.17em}pu}$for$I_{L2}$and$1\angle 0°\text{\hspace{0.17em}pu}$for$V_1$into equation (3).

$\begin{array}{l}{P}_{L2}+j{Q}_{L2}=\left(1\angle 0°\right){\left(0.4595\angle -38.76°\right)}^{*}\\ P_{L2}+j{Q}_{L2}=0.4595\angle 38.76°\\ P_{L2}+j{Q}_{L2}=0.3585+j0.2876\text{\hspace{0.17em}pu}\end{array}$

Calculate the real and reactive power supplied to load by the line $L1$.

Substitute$0.3585+j0.2876\text{\hspace{0.17em}pu}$for$P_{L2}+j{Q}_{L2}$and $0.866+j0.5\text{\hspace{0.17em}pu}$for $P_{Load}+j{Q}_{Load}$into equation (4).

$\begin{array}{l}{P}_{L1}+j{Q}_{L1}=\left(0.866+j0.5\right)-\left(0.3585+j0.2876\right)\\ P_{L1}+j{Q}_{L1}=0.5075+j0.2124\text{\hspace{0.17em}pu}\end{array}$

Hence, the real and reactive power supplied to the load bus by Line $L2$and $L1$are $0.3585+j0.2876\text{\hspace{0.17em}pu}$and$0.5075+j0.2124\text{\hspace{0.17em}pu}$ respectively. The phase angle regulating transformer increase the real power of the line$L1$ with the small change in the reactive power.