Question

Repeat Problem 6.38 except use an initial voltage guess of $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\angle }{\mathbf{30}}^{\mathbf{°}}$

Short answer

The value of power mismatch is within the convergence limit of 0.1 MVA is

${\delta }_2=-11.4^{°}$

$V_2=0.902\mathrm{p}.\mathrm{u}$

Step-by-step solution

Write the given data from the question.

The generator bus supplies $180\mathrm{MW}$at a$1.0\mathrm{per}\mathrm{unit}$voltage.

The 60 Mvar of load supplies through a lossless transmission line.

A $100\mathrm{MVA}$base impedance of $j0.1$and no line charging.

Starting initial voltage guess is$1.0\angle {30}^{°}$.

Determine the formula of mismatch real and reactive power.

Write the formula of mismatch real power is

$∆P_{\mathit{2}}\left(0\right)=P_{\mathit{2}}\mathit{-}{P}_{\mathit{2}}\left(0\right)$ …… (1)

Here,$P_2$is generator bus power and $P_2\left(0\right)$is real power.

Write the formula of mismatch reactive power is

$∆Q_{\mathit{2}}\left(0\right)=Q_{\mathit{2}}\mathit{-}{Q}_{\mathit{2}}\left(0\right)$ …… (2)

Here, $Q_2$is load supply of lossless transmission line and $Q_2\left(0\right)$ is reactive power.

Write the formula of second iteration of real power is

$P_2\left(1\right)=V_2\left[Y_{21}{V}_1\cos \left({\delta }_2-{\delta }_1-{\varphi }_{21}\right)+V_2{Y}_{22}\cos \left(-{\varphi }_{22}\right)\right]$ …… (3)

Here,$V_{\mathit{2}}$is voltage $2$and $Y_{21}$is bus impedance, $V_1$is voltage $1$and $Y_{22}$is bus impedance.

Write the formula of second iteration of reactive power is

$Q_2\left(1\right)=V_2\left[Y_{21}{V}_1\cos \left({\delta }_2-{\delta }_1-{\varphi }_{21}\right)+V_2{Y}_{22}\cos \left(-{\varphi }_{22}\right)\right]$ …… (4)

Here,$V_2$is voltage $2$and$Y_{21}$is bus impedance,$V_1$is voltage $1$and $Y_{22}$ is bus impedance.

Determine the different values of power mismatch and iteration of real and reactive power.

Determine the required mismatch is,

$\left[\begin{array}{c}∆P_2\\ ∆Q_2\end{array}\right]=\left[\frac{{\displaystyle \frac{{\displaystyle \frac{\partial P_2}{\partial {\delta }_2}}}{\partial Q_2}}}{\partial {\delta }_2}\frac{{\displaystyle \frac{{\displaystyle \frac{\partial P_2}{\partial V_2}}}{\partial Q_2}}}{\partial V_2}\right]\left[\begin{array}{c}∆\partial 2\left(0\right)\\ ∆V_2\left(0\right)\end{array}\right]$

Determine the real power ${\mathrm{P}}_2\left(0\right)$

$$\begin{gathered} {\mathrm{P}}_2\left(0\right)={\mathrm{V}}_2\left[{\mathrm{Y}}_{21}{\mathrm{V}}_1\cos \left({\mathrm{\delta }}_2-{\mathrm{\delta }}_1-{\mathrm{\theta }}_{21}\right)+{\mathrm{V}}_2{\mathrm{Y}}_{22}\cos \left(-{\mathrm{\theta }}_{22}\right)\right] \\ =\left(1\right)\left[\left(\frac{1}{0.1}\right)\left(1\right)\cos \left(30°-90°\right)+\left(1\right)\left(\frac{1}{0.1}\right)\cos \left(90°\right)\right] \\ =\left(1\right)\left[\left(10\right)\left(5\right)+\left(10\right)\left(0\right)\right] \\ =5\mathrm{p}.\mathrm{u} \end{gathered}$$

Determine the reactive power$Q_2\left(0\right)$.

$$\begin{gathered} {\mathrm{Q}}_2\left(0\right)={\mathrm{V}}_2\left[{\mathrm{Y}}_{21}{\mathrm{V}}_1\sin \left({\mathrm{\delta }}_2-{\mathrm{\delta }}_1-{\mathrm{\theta }}_{21}\right)+{\mathrm{V}}_2{\mathrm{Y}}_{22}sin\left(-{\mathrm{\theta }}_{22}\right)\right] \\ =\left(1\right)\left[\left(\frac{1}{0.1}\right)\left(1\right)\sin \left(30°-90°\right)+\left(1\right)\left(\frac{1}{0.1}\right)\sin \left(90°\right)\right] \\ =\left[\left(10\right)\left(-0.866\right)+\left(10\right)\left(1\right)\right] \\ =1.34\mathrm{p}.\mathrm{u} \end{gathered}$$

The mismatch power is,

$$\begin{gathered} \mathrm{substitute}-\frac{180\times {10}^6}{100\times {10}^6}\mathrm{for}{\mathrm{P}}^2\mathrm{and}5\mathrm{for}{\mathrm{P}}_2\left(0\right)\mathrm{into}\mathrm{equation}\left(1\right). \\ ∆{\mathrm{P}}_2=-\frac{180\times {10}^6}{100\times {10}^6}-5 \\ =-6.8\mathrm{pu} \\ \mathrm{substitute}-\frac{60\times {10}^6}{100\times {10}^6}\mathrm{for}{\mathrm{Q}}^2\mathrm{and}1.34\mathrm{for}{\mathrm{Q}}_2\left(0\right)\mathrm{into}\mathrm{equation}\left(2\right). \\ ∆{\mathrm{Q}}_2\left(0\right)=-\frac{60\times {10}^6}{100\times {10}^6}-1.34 \\ =-1.94\mathrm{pu} \\ \mathrm{The}\mathrm{elements}\mathrm{of}\mathrm{the}\mathrm{Jacobian}\mathrm{matrix}\mathrm{are}, \end{gathered}$$

$$\begin{gathered} \frac{\partial {\mathrm{P}}_2}{\partial {\mathrm{\delta }}_2}=-{\mathrm{V}}_2\left[{\mathrm{Y}}_{21}{\mathrm{V}}_1\sin \left({\mathrm{\delta }}_2-{\mathrm{\delta }}_1-{\mathrm{\theta }}_{21}\right)\right] \\ =-\left(1\right)\left[\left(\frac{1}{0.1}\right)\left(1\right)\sin \left(30°-90°\right)\right] \\ =8.66 \end{gathered}$$

Solve further as,

$$\begin{gathered} \frac{\partial {\mathrm{P}}_2}{\partial {\mathrm{V}}_2}={\mathrm{Y}}_{21}{\mathrm{V}}_1\cos \left({\mathrm{\delta }}_2-{\mathrm{\delta }}_1-{\mathrm{\theta }}_{21}\right) \\ =\left[\left(\frac{1}{0.1}\right)\left(1\right)\cos \left(30°-90°\right)\right] \\ =5 \end{gathered}$$

Solve further as,

$$\begin{gathered} \frac{\partial {\mathrm{Q}}_2}{\partial {\mathrm{\delta }}_2}=-{\mathrm{V}}_2\left[{\mathrm{Y}}_{21}{\mathrm{V}}_1\cos \left({\mathrm{\delta }}_2-{\mathrm{\delta }}_1-{\mathrm{\theta }}_{21}\right)\right] \\ =-\left(1\right)\left[\left(\frac{1}{0.1}\right)\left(1\right)\cos \left(30°-90°\right)\right] \\ =5 \end{gathered}$$

Solve further as,

$$\begin{gathered} \frac{\partial {\mathrm{Q}}_2}{\partial {\mathrm{V}}_2}=-{\mathrm{V}}_2\left[{\mathrm{V}}_{22}{\mathrm{Y}}_{22}\sin \left(-{\mathrm{\theta }}_{22}\right)+[{\mathrm{Y}}_{21}{\mathrm{V}}_1\sin \left({\mathrm{\delta }}_2-{\mathrm{\delta }}_1-{\mathrm{\theta }}_{21}\right)\right] \\ =-\left(1\right)\left(\frac{1}{0.1}\right)\sin \left(90°\right)+\left[\left(\frac{1}{0.1}\right)\left(1\right)\sin \left(30°-90°\right)\right] \\ =-\left(10\right)\left(1\right)+\left[\left(10\right)\left(-0.866\right)\right] \\ =-18.66 \end{gathered}$$

Solve further as,

The power mismatch equation is

$\left[\begin{array}{c}∆P_2\\ ∆Q_2\end{array}\right]=\left[\frac{{\displaystyle \frac{{\displaystyle \frac{\partial P_2}{\partial {\delta }_2}}}{\partial Q_2}}}{\partial {\delta }_2}\frac{{\displaystyle \frac{{\displaystyle \frac{\partial P_2}{\partial V_2}}}{\partial Q_2}}}{\partial V_2}\right]\left[\begin{array}{c}∆{\partial }_2\left(0\right)\\ ∆V_2\left(0\right)\end{array}\right]$

The equation can be written as,

$\left[\begin{array}{c}∆{\delta }_2\left(1\right)\\ ∆V_2\left(1\right)\end{array}\right]=\left[\frac{{\displaystyle \frac{{\displaystyle \frac{\partial P_2}{\partial {\delta }_2}}}{\partial Q_2}}}{\partial {\delta }_2}\frac{{\displaystyle \frac{{\displaystyle \frac{\partial P_2}{\partial V_2}}}{\partial Q_2}}}{\partial V_2}\right]\left[\begin{array}{c}∆P_2\\ ∆Q_2\end{array}\right]$ …… (5)

localid="1655897554304" $$\begin{gathered} Substitute8.66for\frac{\partial P_2}{\partial {\delta }_2},5for\frac{\partial P_2}{\partial V_2}and5for\frac{\partial Q_2}{\partial {\delta }_2},-18.66for\frac{\partial Q_2}{\partial {\delta }_2},-6.8pufor \\ ∆P_2\left(0\right)and-1.92pufor∆Q_2\left(0\right)intoequation\left(5\right). \\ \left[\begin{array}{c}∆{\delta }_2\\ ∆V_2\end{array}\right]={\left[\begin{array}{cc}8.66& 5\\ 5& -18.66\end{array}\right]}^{-1}\left[\begin{array}{c}-6.8\\ -1.92\end{array}\right] \\ =\left[\begin{array}{cc}0.1& 0.0267\\ 0.0267& -0.046\end{array}\right]\left[\begin{array}{c}-6.8\\ -1.92\end{array}\right]\begin{array}{}\end{array} \\ =\left[\begin{array}{c}-0.68-0.0512\\ -0.182+0.088\end{array}\right] \\ =\left[\begin{array}{c}-0.7312\\ -0.094\end{array}\right] \end{gathered}$$

Therefore the values of voltage and $\delta$are

$$\begin{gathered} ∆{\mathrm{\delta }}_2=-0.7312\mathrm{rad} \\ =-41.9° \end{gathered}$$

Solve further as,

$∆{\mathrm{V}}_2\left(0\right)=-0.094\mathrm{p}.\mathrm{u}.$

The new values of voltage and $\delta$are,

$$\begin{gathered} {\delta }_2\left(1\right)=30°+\left(-41.9°\right) \\ =-11.9° \end{gathered}$$

Solve for the voltage.

$$\begin{gathered} {\mathrm{V}}_2\left(1\right)=1+\left(-0.094\right) \\ =0.906\mathrm{p}.\mathrm{u} \end{gathered}$$

The second iteration,

The value of real power ${\mathrm{P}}_2\left(1\right)\mathrm{is},$

$$\begin{gathered} P_2\left(1\right)=-{\mathrm{V}}_2\left[{\mathrm{Y}}_{21}{\mathrm{V}}_1\cos \left({\mathrm{\delta }}_2-{\mathrm{\delta }}_1-{\mathrm{\theta }}_{21}\right)+{\mathrm{V}}_2{\mathrm{Y}}_{22}\cos \left(-{\mathrm{\theta }}_{22}\right)\right] \\ =\left(0.906\right)\left[\left(\frac{1}{10}\right)\left(1\right)\cos \left(-11°-90°\right)+\left(0.906\right)\left(10\right)\cos \left(90°\right)\right] \\ =-1.878 \end{gathered}$$

The value of reactive power $Q_2\left(1\right)$is,

$$\begin{gathered} Q_2\left(1\right)=-{\mathrm{V}}_2\left[{\mathrm{Y}}_{21}{\mathrm{V}}_1\cos \left({\mathrm{\delta }}_2-{\mathrm{\delta }}_1-{\mathrm{\theta }}_{21}\right)+{\mathrm{V}}_2{\mathrm{Y}}_{22}\cos \left(-{\mathrm{\theta }}_{22}\right)\right] \\ =\left(0.906\right)\left[\left(10\right)\left(1\right)\sin \left(-11.9°-90°\right)+\left(0.906\right)\left(10\right)\sin \left(90°\right)\right] \\ =\left(0.906\right)\left[-9.785+9.06\right] \\ =-0.656 \end{gathered}$$

The mismatch power is,

$∆{\mathrm{P}}_2\left(1\right)={\mathrm{P}}_2-{\mathrm{P}}_2\left(1\right)$ …… (6)

Here, $P_2$is generator bus power and $P_2\left(1\right)$is real power.

Substitute $-1.8$for $P_2$and $-1.878for{P}_2\left(1\right)$into equation (6).

$$\begin{gathered} ∆{\mathrm{P}}_2\left(1\right)=-1.8-(-1.878) \\ =0.078\mathrm{p}.\mathrm{u}. \end{gathered}$$

Solve for reactive power as,

$∆Q_2\left(1\right)=Q_2-Q_1\left(1\right)$ …… (7)

Here, $Q_2$is load supply of lossless transmission line and $Q_2\left(1\right)$is reactive power.

$$\begin{gathered} \mathrm{Substitute}-0.6\mathrm{for}{\mathrm{Q}}_2\mathrm{and}-0.656\mathrm{for}{\mathrm{Q}}_2\left(1\right)\mathrm{into}\mathrm{equation}\left(7\right) \\ ∆{\mathrm{Q}}_2\left(1\right)=-0.6-\left(-0.656\right) \\ =0.056\mathrm{p}.\mathrm{u}. \end{gathered}$$

The elements of the Jacobian matrix are,

$$\begin{gathered} \frac{\partial {\mathrm{P}}_2}{\partial {\mathrm{\delta }}_2}=-{\mathrm{V}}_2\left[{\mathrm{Y}}_{21}{\mathrm{V}}_1\sin \left({\mathrm{\delta }}_2-{\mathrm{\delta }}_1-{\mathrm{\theta }}_{21}\right)\right] \\ =-\left(0.906\right)\left[\left(10\right)\left(1\right)\sin \left(-11.9°-90°\right)\right] \\ =8.865 \end{gathered}$$

$$\begin{gathered} \frac{\partial {\mathrm{P}}_2}{\partial {\mathrm{V}}_2}=-Y_{21}\left[{\mathrm{V}}_1\cos \left({\mathrm{\delta }}_2-{\mathrm{\delta }}_1-{\mathrm{\theta }}_{21}\right)\right] \\ =\left(10\right)\left(1\right)\cos \left(-11.9°-90°\right) \\ =-0.206 \end{gathered}$$

$$\begin{gathered} \frac{\partial {\mathrm{Q}}_2}{\partial {\mathrm{\delta }}_2}=-{\mathrm{V}}_2\left[{\mathrm{Y}}_{21}{\mathrm{V}}_1\cos \left({\mathrm{\delta }}_2-{\mathrm{\delta }}_1-{\mathrm{\theta }}_{21}\right)\right] \\ =\left(0.906\right)\left[\left(10\right)\left(1\right)\cos \left(-11.9°-90°\right)\right] \\ =-1.868 \end{gathered}$$

$$\begin{gathered} \frac{\partial {\mathrm{Q}}_2}{\partial {\mathrm{V}}_2}=-{\mathrm{V}}_2{\mathrm{Y}}_{22}\sin \left(-{\mathrm{\theta }}_{22}\right)+\left[{\mathrm{Y}}_{21}{\mathrm{V}}_1\sin \left({\mathrm{\delta }}_2-{\mathrm{\delta }}_1-{\mathrm{\theta }}_{21}\right)\right] \\ =-\left(0.906\right)\left(10\right)\sin \left(90°\right)+\left[\left(10\right)\left(1\right)\sin \left(-11.9°-90°\right)\right] \\ =-18.845 \end{gathered}$$

The power mismatch equation is

$\left[\begin{array}{c}∆P_2\\ ∆Q_2\end{array}\right]=\left[\frac{{\displaystyle \frac{{\displaystyle \frac{\partial P_2}{\partial {\delta }_2}}}{\partial Q_2}}}{\partial {\delta }_2}\frac{{\displaystyle \frac{{\displaystyle \frac{\partial P_2}{\partial V_2}}}{\partial Q_2}}}{\partial V_2}\right]\left[\begin{array}{c}∆{\partial }_2\left(2\right)\\ ∆V_2\left(2\right)\end{array}\right]$

The equation can be written as,

$\left[\begin{array}{c}∆{\delta }_2\left(2\right)\\ ∆V_2\left(2\right)\end{array}\right]=\left[\frac{{\displaystyle \frac{{\displaystyle \frac{\partial P_2}{\partial {\delta }_2}}}{\partial Q_2}}}{\partial {\delta }_2}\frac{{\displaystyle \frac{{\displaystyle \frac{\partial P_2}{\partial V_2}}}{\partial Q_2}}}{\partial V_2}\right]\left[\begin{array}{c}∆P_2\\ ∆Q_2\end{array}\right]$ …… (8)

$$\begin{gathered} Substitute8865for\frac{\partial P_2}{\partial {\delta }_2},-0.206for\frac{\partial P_2}{\partial V_2}and-1.868for\frac{\partial Q_2}{\partial {\delta }_2},-18.845for\frac{\partial Q_2}{\partial V_2},0.078pufor \\ ∆P_{2}and0.056pufor∆Q_{2}intoequation\left(8\right). \end{gathered}$$

$$\begin{gathered} \left[\begin{array}{c}∆{\delta }_2\\ ∆V_2\end{array}\right]={\left[\begin{array}{cc}8.66& -0.206\\ -1.868& -18.845\end{array}\right]}^{-1}\left[\begin{array}{c}0.078\\ 0.056\end{array}\right] \\ =\left[\begin{array}{cc}0.1125& -0.00123\\ -0.011& -0.053\end{array}\right]\left[\begin{array}{c}0.078\\ 0.056\end{array}\right]\begin{array}{}\end{array} \\ =\left[\begin{array}{c}0.0087\\ -0.0038\end{array}\right] \end{gathered}$$

Therefore the values of voltage and $\delta$are

$$\begin{gathered} ∆{\delta }_2=0.0087rad \\ =0.5° \\ And, \\ ∆V_2=-0.0038 \\ Thenewvaluesofvoltageand\delta are \\ {\delta }_2\left(2\right)=-11.9°+0.5° \\ =-11.4° \end{gathered}$$

And,

$$\begin{gathered} {\mathrm{V}}_2\left(2\right)=0.906-0.0038 \\ =0.902\mathrm{p}.\mathrm{u} \end{gathered}$$

$$\begin{gathered} \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{real}\mathrm{power}{\mathrm{P}}_2\left(2\right)\mathrm{is}, \\ {\mathrm{P}}_2\left(2\right)={\mathrm{V}}_2\left[{\mathrm{Y}}_{21}{\mathrm{V}}_1\cos \left({\mathrm{\delta }}_2-{\mathrm{\delta }}_1-{\mathrm{\theta }}_{21}\right)+{\mathrm{V}}_2{\mathrm{Y}}_{22}\cos \left(-{\mathrm{\theta }}_{22}\right)\right] \\ =\left(0.902\right)\left[\left(10\right)\left(1\right)\cos \left(-11.4°-90°\right)\right] \\ =-1.783 \end{gathered}$$

$$\begin{gathered} \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{reactive}\mathrm{power}{\mathrm{Q}}_2\left(2\right)\mathrm{is}, \\ {\mathrm{Q}}_2\left(2\right)={\mathrm{V}}_2\left[{\mathrm{Y}}_{21}{\mathrm{V}}_1\cos \left({\mathrm{\delta }}_2-{\mathrm{\delta }}_1-{\mathrm{\theta }}_{21}\right)+{\mathrm{V}}_2{\mathrm{Y}}_{22}\cos \left(-{\mathrm{\theta }}_{22}\right)\right] \\ =\left(0.902\right)\left[\left(10\right)\left(1\right)\sin \left(-11.4°-90°\right)+\left(0.902\right)\left(10\right)\sin \left(90°\right)\right] \\ =\left(0.902\right)\left[-9.8+9.02\right] \\ =-0.703 \end{gathered}$$

The mismatch values are,

$∆{\mathrm{P}}_2\left(2\right)={\mathrm{P}}_2-{\mathrm{P}}_2\left(2\right)$

$\mathrm{Here},{\mathrm{P}}_2\mathrm{is}\mathrm{generator}\mathrm{bus}\mathrm{power}\mathrm{and}{\mathrm{P}}_2\left(2\right)\mathrm{is}\mathrm{real}\mathrm{power}.$

$\mathrm{Substitute}-1.8\mathrm{for}{\mathrm{P}}_2\mathrm{and}-1.783\mathrm{for}{\mathrm{P}}_2\left(2\right)\mathrm{into}\mathrm{equation}\left(9\right).$

$$\begin{gathered} ∆{\mathrm{P}}_2\left(2\right)=-1.8-(-1.783) \\ =0.002\mathrm{p}.\mathrm{u}. \end{gathered}$$

Also,

$∆Q_2\left(2\right)=Q_2-Q_2\left(2\right)$ …… (10)

$$\begin{gathered} \mathrm{Here},{\mathrm{Q}}_2\mathrm{is}\mathrm{load}\mathrm{supply}\mathrm{of}\mathrm{lossless}\mathrm{transmission}\mathrm{line}\mathrm{and}{\mathrm{Q}}_2\left(2\right)\mathrm{is}\mathrm{reactive}\mathrm{power}. \\ \mathrm{Substitute}-0.6\mathrm{for}{\mathrm{Q}}_2\mathrm{and}-0.703\mathrm{for}{\mathrm{Q}}_2\left(2\right)\mathrm{into}\mathrm{equation}\left(10\right) \\ ∆{\mathrm{Q}}_2\left(2\right)=-0.6-\left(-0.703\right) \\ =0.103\mathrm{p}.\mathrm{u}. \end{gathered}$$

localid="1655958152481" $$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{power}\mathrm{mismatch}\mathrm{is}\mathrm{within}\mathrm{the}\mathrm{convergence}\mathrm{limit}\mathrm{of}0.1\mathrm{MVA} \\ \mathrm{Therefore},\mathrm{the}\mathrm{final}\mathrm{values}\mathrm{are}, \\ {\mathrm{\delta }}_2=-11.4° \\ {\mathrm{V}}_2=0.902\mathrm{p}.\mathrm{u} \end{gathered}$$