Question

Repeat Problem 6.43 except with the bus $\mathbf{2}$ real power load changed to $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{perunit}$.

Short answer

The first iteration of Newton-Raphson method, the value of voltages at bus $3$except with the bus $2$.

$\mathrm{x}\left(1\right)=\left[\begin{array}{c}{\mathrm{\delta }}_2\left(1\right)\\ {\mathrm{\delta }}_3\left(1\right)\\ {\mathrm{V}}_2\left(1\right)\end{array}\right]=\left[\begin{array}{c}-0.028571\\ 0.11429\\ 0.96667\end{array}\right]\begin{array}{c}\text{radians}\\ \text{radians}\\ \text{per\hspace{0.17em}unit}\end{array}$

Step-by-step solution

Write the given data from the question.

The is the slack with a per unit voltage of $\text{1.0}\angle \text{0°}$.

The is a PQ bus with a per unit load of $\text{2.0+j0.5}$.

The bus 3 is a PV bus with $\text{1.0perunit}$generation and a $\text{1.0}$voltage set point.

The per unit line impedances are $\text{j0.1}$between buses $1$ and $2$.

The per unit line impedances are $\text{j0.4}$between buses $1$and $3$, and $\text{j0.2}$between buses $2$and $3$.

Determine the formula of mismatch real and reactive power.

Write the formula of reactive voltage controlled bus.

${\mathbf{Q}}_3\mathbf{=}{\mathbf{V}}_3[{\mathbf{V}}_1{\mathbf{Y}}_{31}\mathbf{sin}({\mathbf{\delta }}_3\mathbf{-}{\mathbf{\delta }}_1\mathbf{-}{\mathbf{\theta }}_{31})\mathbf{+}{\mathbf{Y}}_{32}{\mathbf{V}}_2\mathbf{sin}({\mathbf{\delta }}_3\mathbf{-}{\mathbf{\delta }}_2\mathbf{-}{\mathbf{\theta }}_{32})\mathbf{+}{\mathbf{Y}}_{33}{\mathbf{V}}_3\mathbf{sin}(\mathbf{-}{\mathbf{\theta }}_{33})]$ ……. (1)

Here,${\mathrm{V}}_1,{\mathrm{V}}_2$ and$V_3$ are the voltage of bus$1,2$ and $3$,${\mathrm{y}}_{31},{\mathrm{y}}_{32}$ and${\mathrm{y}}_{33}$ are the bus admittance,${\mathrm{\theta }}_{31},{\mathrm{\theta }}_{32}$ and${\mathrm{\theta }}_{33}$ are angle between bus voltage.

Determine the the first iteration Phasor volt-ages at bus 3 except bus 2.

From reference of section 6.4 in the textbook.

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

The diagonal elements, ${\mathrm{Y}}_{\mathrm{kk}}$, are the sum of admittance connected to bus .

The off-diagonal elements are,

${\mathrm{Y}}_{\mathrm{kn}}=-\left(\mathrm{sum}\mathrm{of}\mathrm{admittances}\mathrm{connected}\mathrm{between}\mathrm{buses}\mathrm{k}\mathrm{and}\mathrm{n}\right)$

The bus admittance matrix is,

${\overline{\mathrm{Y}}}_{\mathrm{Bus}}=\left[\begin{array}{ccc}-\mathrm{j}12.5& +\mathrm{j}10.0& +\mathrm{j}2.5\\ +\mathrm{j}10.0& -\mathrm{j}15.0& +\mathrm{j}5.0\\ +\mathrm{j}2.5& +\mathrm{j}5.0& -\mathrm{j}7.5\end{array}\right]$

Write the expression for the load angle.

$\begin{array}{c}{\mathrm{\delta }}_2\left(0\right)={\mathrm{\delta }}_2\left(0\right)\\ =0°\end{array}$

Write the expression for the voltage.

$V_2\left(0\right)=1.0$

Determine$\mathrm{\Delta y}\left(0\right)$by using following formulae:

Write the expression for power.

${\mathrm{P}}_{\mathrm{k}}={\mathrm{V}}_{\mathrm{k}}\sum _{\mathrm{n}=1}^{\mathrm{N}}{\mathrm{Y}}_{\mathrm{kn}}{\mathrm{V}}_{\mathrm{n}}\cos ({\mathrm{\delta }}_{\mathrm{k}}-{\mathrm{\delta }}_{\mathrm{n}}-{\mathrm{\theta }}_{\mathrm{kn}})$ …… (2)

Here, ${\mathrm{P}}_{\mathrm{k}}$is net real power, ${\mathrm{V}}_{\mathrm{k}}$is voltage magnitude, ${\mathrm{Y}}_{\mathrm{kn}}$is an off-diagonal element, ${\mathrm{V}}_{\mathrm{n}}$is the $\mathrm{N}$vector of bus voltages, ${\mathrm{\delta }}_{\mathrm{k}}$is phase angle, ${\mathrm{\delta }}_{\mathrm{n}}$is swing bus.

Write the expression for reactive power.

${\mathrm{Q}}_{\mathrm{k}}={\mathrm{V}}_{\mathrm{k}}\sum _{\mathrm{n}=1}^{\mathrm{N}}{\mathrm{Y}}_{\mathrm{kn}}{\mathrm{V}}_{\mathrm{n}}\sin ({\mathrm{\delta }}_{\mathrm{k}}-{\mathrm{\delta }}_{\mathrm{n}}-{\mathrm{\theta }}_{\mathrm{kn}})$ …… (3)

Here, ${\mathrm{P}}_{\mathrm{k}}$is net real power, ${\mathrm{V}}_{\mathrm{k}}$is voltage magnitude, ${\mathrm{Y}}_{\mathrm{kn}}$is an off-diagonal element, $V_n$is the $N$vector of bus voltages, ${\mathrm{\delta }}_{\mathrm{k}}$is phase angle, ${\mathrm{\delta }}_{\mathrm{n}}$is swing bus.

Write the expression for power.

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

Substitute $1.0$for $V_2$, $1.0$for ${\mathrm{V}}_1$, $10$for ${\mathrm{Y}}_{21}$, $-15$for ${\mathrm{Y}}_{23}$,$1.0$for ${\mathrm{V}}_3$, and$5$for${\mathrm{Y}}_{23}$into equation (4).

$\begin{array}{c}{\mathrm{P}}_2\left(\mathrm{X}\right)=1.0[10\cos (-90°)-15\cos (-90°)+5\cos (-90°)]\\ =0\end{array}$

Similarly,

$\begin{array}{c}{\mathrm{P}}_3\left(\mathrm{X}\right)=2.5\cos (-90°)+5\cos (-90°)\\ =0\end{array}$

Write reactive power as,

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

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

$\begin{array}{c}{\mathrm{Q}}_2\left(\mathrm{X}\right)=1.0[10\sin (-90°)-15\sin (-90°)+5\sin (-90°)]\\ =0\end{array}$

The matrix is obtained as,

$\begin{array}{c}\mathrm{\Delta y}\left(0\right)=\left[\begin{array}{l}{\mathrm{P}}_2-{\mathrm{P}}_2\left(\mathrm{X}\right)\\ {\mathrm{P}}_3-{\mathrm{P}}_3\left(\mathrm{X}\right)\\ {\mathrm{Q}}_2-{\mathrm{Q}}_2\left(\mathrm{X}\right)\end{array}\right]\\ =\left[\begin{array}{l}-2.0+1.0-0\\ 1.0-0\\ -0.5-0\end{array}\right]\\ =\left[\begin{array}{l}-1.0\\ 1.0\\ -0.5\end{array}\right]\end{array}$

Determine $J{1}_{22}$.

$\begin{array}{c}J{1}_{22}=\frac{\partial P_2}{\partial {\delta }_2}\\ =-V_2[Y_{21}{V}_1\sin ({\delta }_2-{\delta }_1-{\theta }_{21})+Y_{23}{V}_3\sin ({\delta }_2-{\delta }_3-{\theta }_{23})]\\ =-1.0[10\left(1\right)\sin (-90°)+5\left(1\right)\sin (-90°)]\\ =15\end{array}$

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

Determine$J{1}_{23}$.

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

Determine $J{1}_{32}$.

$\begin{array}{c}J{1}_{32}=\frac{\partial P_3}{\partial {\delta }_2}\\ =V_3{Y}_{32}{V}_2\sin ({\delta }_3-{\delta }_2-{\theta }_{32})\\ =\left(1\right)\left(5\right)\left(1\right)\sin (-90°)\\ =-5\end{array}$

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

Determine $J{1}_{33}$.

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

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

Determine $J{2}_{22}$.

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

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

Determine $J{2}_{32}$.

$\begin{array}{c}J{2}_{32}=\frac{\partial P_3}{\partial V_2}\\ =V_3{Y}_{32}\cos ({\delta }_3-{\delta }_2-{\theta }_{32})\\ =0\end{array}$

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

Determine $J{3}_{22}$.

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

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

Determine $J{3}_{23}$.

$\begin{array}{c}J{3}_{23}=\frac{\partial Q_2}{\partial {\delta }_3}\\ =-V_2{Y}_{23}{V}_3\cos ({\delta }_2-{\delta }_1-{\theta }_{23})\\ =0\end{array}$

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

Determine $J{4}_{22}$.

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

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

Determine $J{4}_{22}$.

$\mathrm{J}\left(0\right)=\left[\begin{array}{cc}\mathrm{J}1& \mathrm{J}2\\ \mathrm{J}3& \mathrm{J}4\end{array}\right]=\left[\begin{array}{ccc}15& -5& 0\\ -5& 7.5& 0\\ 0& 0& 15\end{array}\right]\text{\hspace{0.17em}per\hspace{0.17em}unit}$

Consider Gauss elimination method to solve the equation,$\mathrm{J\Delta x}=\mathrm{\Delta y}$

$\left[\begin{array}{ccc}15& -5& 0\\ -5& 7.5& 0\\ 0& 0& 15\end{array}\right]\left[\begin{array}{c}\Delta {\delta }_2\\ \Delta {\delta }_3\\ \Delta V_2\end{array}\right]=\left[\begin{array}{c}-1.0\\ 1.0\\ -0.5\end{array}\right]$

Multiply row one with $\frac{-5}{15}$. To make coefficient matrix upper triangular subtract first row with the second. The equation becomes as follows.

$\left[\begin{array}{ccc}15& -5& 0\\ 0& \frac{35}{6}& 0\\ 0& 0& 15\end{array}\right]\left[\begin{array}{c}\mathrm{\Delta \delta }\left(2\right)\\ \mathrm{\Delta \delta }\left(3\right)\\ {\mathrm{\Delta V}}_2\end{array}\right]=\left[\begin{array}{c}-1.0\\ \frac{2}{3}\\ -0.5\end{array}\right]$

Use back substitution.

$\begin{array}{c}\Delta V_2=\frac{-0.5}{15}\\ =-0.033333\end{array}$

$\begin{array}{c}\mathrm{\Delta \delta }\left(3\right)=\frac{0.66666}{5.833333}\\ =0.11429\end{array}$

$\begin{array}{c}\Delta \delta \left(2\right)=\frac{-1.0+5\times 0.11429}{15}\\ =-0.02857\end{array}$

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

$\begin{array}{c}\mathrm{\chi }\left(1\right)=\left[\begin{array}{c}{\mathrm{\delta }}_2\left(1\right)\\ {\mathrm{\delta }}_3\left(1\right)\\ {\mathrm{\delta }}_2\left(1\right)\end{array}\right]\\ =\mathrm{\chi }\left(0\right)+\mathrm{\Delta \chi }\\ =\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]+\left[\begin{array}{c}-0.02857\\ 0.11429\\ -0.0333333\end{array}\right]\\ =\left[\begin{array}{c}-0.02857\\ 0.11429\\ 0.96667\end{array}\right]\begin{array}{c}\text{radians}\\ \text{radians}\\ \text{Per\hspace{0.17em}unit}\end{array}\end{array}$

Check the value of ${\mathrm{Q}}_{\mathrm{G}3}$.

Substitute$1$for $V_3$, $2.5$for $V_1$, $1$for $Y_{31}$,$5$for $Y_{32}$,$0.96667$for $V_2$,$7.5$for$Y_{33}$and$1$for$V_3$into equation (1).

$\begin{array}{c}{\mathrm{Q}}_3=1\left[\begin{array}{l}(2.5)\left(1\right)\sin \left(0.11429-\frac{\mathrm{\pi }}{2}\right)+5\times 0.96667\sin \left(0.11429+0.028571-\frac{\mathrm{\pi }}{2}\right)\\ +7.5\left(1\right)\sin \frac{\mathrm{\pi }}{2}\end{array}\right]\\ =0.2322\text{\hspace{0.17em}per\hspace{0.17em}unit}\end{array}$

$\begin{array}{c}{Q}_{G3}=Q_3+Q_{13}\\ =0.2322+0\\ =0.2322\text{\hspace{0.17em}per\hspace{0.17em}unit}\end{array}$

Thus,$Q_{G3}$ is$0.2322\text{\hspace{0.17em}per\hspace{0.17em}unit}$ which is within the limit $[-5.0+5.0]$.

Therefore, The bus$3$ is the voltage-controlled bus. The value of$\chi$ obtained after first iteration of Newton-Raphson method is,

$\mathrm{x}\left(1\right)=\left[\begin{array}{c}{\mathrm{\delta }}_2\left(1\right)\\ {\mathrm{\delta }}_3\left(1\right)\\ {\mathrm{V}}_2\left(1\right)\end{array}\right]=\left[\begin{array}{c}-0.028571\\ 0.11429\\ 0.96667\end{array}\right]\begin{array}{c}\text{radians}\\ \text{radians}\\ \text{per\hspace{0.17em}unit}\end{array}$