Question

For a three-bus power system, assume is the slack with a per unit voltage of $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\angle }\mathbf{0}\mathbf{°}$, bus 2 is a PQ bus with a per unit load of $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{+}\mathit{j}\mathbf{0}\mathbf{.}\mathbf{5}$, and bus 3 is a PV bus with 1.0 per unit generation and a 1.0 voltage set point. The per unit line impedances are $\mathit{j}\mathbf{0}\mathbf{.}\mathbf{1}$between buses 1 and 2, j0.4 between buses 1 and 3, and j0.2 between buses 2 and 3. Using a flat start, use the Newton-Raphson approach to determine the first iteration Phasor volt-ages at buses 2 and 3.

Short answer

The first iteration of Newton-Raphson method, the value of voltages at buses 2 and 3

$x\left(1\right)=\left[\begin{array}{c}{\delta }_2\left(1\right)\\ {\delta }_3\left(1\right)\\ V_2\left(1\right)\end{array}\right]=\left[\begin{array}{c}-0.1142857\\ 0.05714285\\ 1.0333333\end{array}\right]\begin{array}{c}\mathrm{radians}\\ \mathrm{radians}\\ \mathrm{per}\mathrm{unit}\end{array}$

Step-by-step solution

Write the given data from the question.

A bus 1 is the slack with a per unit voltage of $1.0\angle 0°$.

A bus 2 is a PQ bus with a per unit load of$2.0+\mathrm{j}0.5$.

A bus 3 is a PV bus with 1.0 per unit generation and a 1.0 voltage set point.

The per unit line impedances are j0.1 between buses 1 and 2.

The per unit line impedances are j0.4 between buses 1 and 3 , and j0.2 between buses 2 and 3 .

Determine the formula of mismatch real and reactive power.

Write the formula of reactive voltage controlled bus.

${\mathbf{Q}}_{\mathbf{3}}\mathbf{=}{\mathbf{V}}_{\mathbf{3}}\left[V_1{Y}_{31}\sin \left({\delta }_3‐{\delta }_1‐{\theta }_{31}\right)+Y_{32}{V}_2\sin \left({\delta }_3‐{\delta }_2‐{\theta }_{32}\right)+Y_{33}{V}_3\sin \left(‐{\theta }_{33}\right)\right]$……. (1)

Here, $V_1,V_2\mathrm{and}{V}_3$are the voltage of bus 1,2 and 3, $y_{32},y_{32}\mathrm{and}{y}_{33}$are the bus admittance,${\theta }_{31},{\theta }_{32}\mathrm{and}{\theta }_{33}$ are angle between bus voltage.

Determine the the first iteration Phasor volt-ages at buses 2 and 3 .

As reference of section 6.4 in the textbook.

Calculate the elements of the matrix, ${\overline{Y}}_{Bus}$.

The diagonal elements, $Y_{kk}$, are the sum of admittance connected to bus $k$.

The off-diagonal elements are,

$Y_{kn}=-\left(sumofadmit\tan cesconnectedbetweenkandn\right)$

The bus admittance matrix is,

$$\begin{gathered} {\overline{Y}}_{Bus}=\left[\begin{array}{ccc}-j12.5& +j10.0& +j2.5\\ +j10.0& -j15.0& +j5.0\\ +j2.5& +j5.0& -j7.5\end{array}\right] \\ {\delta }_2\left(0\right)={\delta }_2\left(0\right)=0° \\ {V}_2\left(0\right)=1.0 \end{gathered}$$

Determine $∆y\left(0\right)$by using following formulae:

localid="1655961746854" $P_k=V_k\sum _{n-1}^N{Y}_{kn}{V}_n\cos \left({\delta }_k-{\delta }_n-{\theta }_{kn}\right)$ …… (2)

Here, $P_k$is net real power, $V_k$is voltage magnitude,$Y_{kn}$is an off-diagonal element, $V_n$isthe$N$ vector of bus voltages, ${\delta }_k$is phase angle, ${\delta }_n$is swing bus.

$Q_k=V_k\sum _{n-1}^N{Y}_{kn}{V}_n\sin \left({\delta }_k-{\delta }_n-{\theta }_{kn}\right)$ …… (3)

Here,$P_k$is net real power, $V_k$is voltage magnitude,$Y_{kn}$is an off-diagonal element, $V_n$isthe$N$ vector of bus voltages,${\delta }_k$is phase angle, ${\delta }_n$is swing bus.

$P_2\left(X\right)=V_2\left[\begin{array}{c}{V}_1{Y}_{21}\cos \left({\delta }_2-{\delta }_1-{\theta }_{21}\right)+V_2{Y}_{22}\cos \left({\delta }_2-{\delta }_2-{\theta }_{22}\right)\\ +V_3{Y}_{23}\cos \left({\delta }_2-{\delta }_3-{\theta }_{23}\right)\end{array}\right]$ …… (4)

Substitute 1.0 for $V_2$, 1.0 for $V_1$, 10 for $Y_{21}$, 1.0 for, $-15$for $Y_{22}$, 1.0 for $V_3$, and 5 for$Y_{23}$into equation (4).

$$\begin{gathered} P_2\left(X\right)=1.0\left[\cos \left(-90°\right)-15\cos \left(-90°\right)+5\cos \left(-90°\right)\right] \\ =0 \end{gathered}$$

Similarly,

localid="1655962750553" $$\begin{gathered} P_3\left(X\right)=2.5\cos \left(-90°\right)-15\cos \left(-90°\right)+5\cos \left(-90°\right) \\ =0 \end{gathered}$$

Similarly,

$$\begin{gathered} P_3\left(X\right)=2.5\cos \left(-90°\right)+5\cos \left(-90°\right) \\ =0 \end{gathered}$$

$Q_2\left(X\right)=V_2\left[\begin{array}{c}{V}_1{Y}_{21}\sin \left({\delta }_2-{\delta }_1-{\theta }_{21}\right)+V_2{Y}_{22}\sin \left({\delta }_2-{\delta }_2-{\theta }_{22}\right)\\ +V_3{Y}_{23}\sin \left({\delta }_2-{\delta }_3-{\theta }_{23}\right)\end{array}\right]$ …… (5)

Substitute 1.0 for $V_2$, 1.0 for $V_1$, 10 for $Y_{21}$, 1.0 for, -15 for $\hspace{0.17em}{Y}_{22}$, 1.0 for $V_3$, and 5 for$Y_{23}$into equation (5).

$$\begin{gathered} Q_2\left(X\right)=1.0\left[10\sin \left(-90°\right)-15\sin \left(-90°\right)+\sin \left(-90°\right)\right] \\ =0 \\ ∆y\left(0\right)=\left[\begin{array}{c}{P}_2-P_2\left(X\right)\\ P_3-P_3\left(X\right)\\ Q_2-Q_2\left(X\right)\end{array}\right]=\left[\begin{array}{c}-2.0-0\\ 1.0-0\\ -0.5-0\end{array}\right]=\left[\begin{array}{c}-2.0\\ 1.0\\ -0.5\end{array}\right] \end{gathered}$$

Determine$J\left(0\right)$.

localid="1655963678610" $$\begin{gathered} J{1}_{22}=\frac{\partial P_2}{\partial {\delta }_2} \\ =-V_2{Y}_{23}{V}_3\sin \left({\delta }_2-{\delta }_3-{\theta }_{23}\right) \\ =\left(1.0\right)\left(5\right)\sin \left(-90°\right) \\ =-5 \end{gathered}$$

Here,$J{1}_{22}$is Jacobian matrix of partial derivatives first form.

localid="1655963861831" $$\begin{gathered} J{1}_{32}=\frac{\partial P_3}{\partial {\delta }_3} \\ =-V_2{Y}_{32}{V}_2\sin \left({\delta }_3-{\delta }_2-{\theta }_{32}\right) \\ =\left(1\right)\left(5\right)\left(1\right)\sin \left(-90°\right) \\ =-5 \end{gathered}$$

Here, $J{1}_{32}$is Jacobian matrix of partial derivatives first form.

$$\begin{gathered} J{1}_{33}=\frac{\partial P_3}{\partial {\delta }_3} \\ =-V_3{Y}_{31}{V}_1\sin \left({\delta }_3-{\delta }_1-{\theta }_{31}\right)+Y_{32}{V}_2\sin \left({\delta }_3-{\delta }_2-{\theta }_{32}\right) \\ =-1.0\left[\left(2.5\right)\left(1\right)\sin \left(-90°\right)+\left(5\right)\left(1\right)\sin \left(-90°\right)\right] \\ =7.5 \end{gathered}$$

Here,$J{1}_{33}$is Jacobian matrix of partial derivatives first form.

$$\begin{gathered} J{2}_{22}=\frac{\partial P_2}{\partial V_2} \\ =V_2{Y}_{22}\cos \left({\theta }_{22}\right)+Y_{21}{V}_1\cos \left({\delta }_2-{\delta }_1-{\theta }_{21}\right)+Y_{23}{V}_2\cos \left(-{\theta }_{22}\right)+Y_{23}{V}_3\cos \left({\delta }_2-{\delta }_3-{\theta }_{23}\right) \\ =0 \end{gathered}$$

Here, localid="1655964818739" $J{2}_{22}$is Jacobian matrix of partial derivatives second form.

localid="1655964576802" $$\begin{gathered} J{2}_{22}=\frac{\partial P_2}{\partial V_2} \\ =V_3{Y}_{32}\cos \left({\delta }_2-{\delta }_1-{\theta }_{21}\right) \\ =0 \end{gathered}$$

Here, $J{2}_{32}$is Jacobian matrix of partial derivatives second form.

localid="1655965276120" $$\begin{gathered} J{3}_{22}=\frac{\partial Q_3}{\partial {\delta }_2} \\ =V_2\left[Y_{21}{V}_1\cos \left({\delta }_2-{\delta }_1-{\theta }_{23}\right)+Y_{23}{V}_3\cos \left({\delta }_2-{\delta }_3-{\theta }_{23}\right)\right] \\ =0 \end{gathered}$$

Here, $J{3}_{22}$is Jacobian matrix of partial derivatives third form.

localid="1655965236686" $$\begin{gathered} J{3}_{23}=\frac{\partial Q_2}{\partial {\delta }_3} \\ =-V_2{Y}_{23}{V}_3\cos \left({\delta }_2-{\delta }_1-{\theta }_{23}\right) \\ =0 \end{gathered}$$

Here, $J{3}_{23}$is Jacobian matrix of partial derivatives third form.

$$\begin{gathered} J{4}_{22}=\frac{\partial Q_2}{\partial V_2} \\ =-V_2{Y}_{22}\sin \left({\theta }_{22}\right)+Y_{22}{V}_2\sin \left({\delta }_2-{\delta }_1-{\theta }_{21}\right)+Y_{22}{V}_2\sin \left({\delta }_2-{\delta }_3-{\theta }_{23}\right) \\ =\left(-1\right)\left(15\right)\sin \left(-90°\right)+\left[\begin{array}{c}\left(10\right)\left(1\right)\sin \left(-90°\right)+\left(15\right)\left(1\right)\sin \left(-90°\right)\\ +\left(5\right)\left(1\right)\sin \left(-90°\right)\end{array}\right] \\ =-15 \end{gathered}$$

Here, $J{4}_{22}$is Jacobian matrix of partial derivatives fourth form.

$J\left(0\right)=\left[\begin{array}{cc}J1& J2\\ J3& J4\end{array}\right]=\left[\begin{array}{ccc}15& -5& 0\\ -5& 7.5& 0\\ 0& 0& -15\end{array}\right]\mathrm{per}\mathrm{unit}$

Solve the equation, $J∆\chi =∆y$with gauss elimination method.

localid="1655965984611" $\left[\begin{array}{ccc}15& -5& 0\\ -5& 7.5& 0\\ 0& 0& -15\end{array}\right]\left[\begin{array}{c}∆{\delta }_2\\ ∆{\delta }_3\\ ∆V_2\end{array}\right]=\left[\begin{array}{c}-2.0\\ 0.3333\\ -0.5\end{array}\right]$

Multiply row one with $\frac{-5}{15}$and subtract the first row from second row to make the coefficient matrix as upper triangular.

$\left[\begin{array}{ccc}15& -5& 0\\ 0& 5.8333& 0\\ 0& 0& -15\end{array}\right]\left[\begin{array}{c}∆{\delta }_2\\ ∆{\delta }_3\\ ∆V_2\end{array}\right]=\left[\begin{array}{c}-2.0\\ 0.3333\\ -0.5\end{array}\right]$

Use back substitution.

$$\begin{gathered} ∆V_2=\frac{-0.5}{-15} \\ =0.033333 \\ ∆{\delta }_3=\frac{0.333333}{5.833333} \\ =0.0571428 \\ ∆{\delta }_2=\frac{-2.0+5\times 0.0571428}{15} \\ =-0.1142857 \end{gathered}$$

Determine $\chi \left(1\right)$.

localid="1655966475857" $$\begin{gathered} \chi \left(1\right)=\left[\begin{array}{c}{\delta }_2\left(1\right)\\ {\delta }_3\left(1\right)\\ V_2\left(1\right)\end{array}\right] \\ =\chi \left(0\right)+∆\chi \\ =\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]+\left[\begin{array}{c}-0.1142857\\ 0.05714285\\ 0.0333333\end{array}\right] \\ =\left[\begin{array}{c}-0.1142857\\ 0.05714285\\ 1.0333333\end{array}\right]\begin{array}{c}\mathrm{radians}\\ \mathrm{radians}\\ \mathrm{per}\mathrm{unit}\end{array} \end{gathered}$$

Check the value of $Q_{G3}$.

Substitute 1 for $V_3$, 2.5 for $V_1$, 1 for $Y_{31}$, 5 for $Y_{32}$, 1.0333333 for $V_2$, 7.5 for$Y_{33}$ and 1 for$V_3$ into equation (1).

$$\begin{gathered} Q_3=1\left[\begin{array}{c}\left(2.5\right)\left(1\right)\sin \left(0.05714-\frac{\mathrm{\pi }}{2}\right)+5\times 1.0333333\sin \left(0.05714+0.11429-\frac{\mathrm{\pi }}{2}\right)\\ +7.5\left(1\right)\sin \frac{\mathrm{\pi }}{2}\end{array}\right] \\ =1\left[-2.4959-5.0909+7.5\right] \\ =-0.0868\mathrm{per}\mathrm{unit} \end{gathered}$$

Thus, $Q_{G3}$is $-0.0868$per unit which is within the limit $\left[-5.0+5.0\right]$.

Therefore, the bus 3 remains a is the voltage-controlled bus. After the first iteration of and, the value of$\chi$ is,

$x\left(1\right)=\left[\begin{array}{c}{\delta }_2\left(1\right)\\ {\delta }_3\left(1\right)\\ V_2\left(1\right)\end{array}\right]=\left[\begin{array}{c}-0.1142857\\ 0.05714285\\ 1.0333333\end{array}\right]\begin{array}{c}\mathrm{radians}\\ \mathrm{radians}\\ \mathrm{per}\mathrm{unit}\end{array}$