Question

Consider the simplified electric power system shown in Figure 6.22 for which the power flow solution can be obtained without resorting to iterative techniques. (a) Compute the elements of the bus admittance matrix ${\mathbf{Y}}_{\mathbf{bus}}$. (b) Calculate the phase angle localid="1655380756919" ${\mathbf{\delta }}_{\mathbf{2}}$by using the real power equation at bus 2 (voltage-controlled bus). (c) Determine localid="1655380759322" $\left|\begin{array}{c}{V}_3\end{array}\right|$and localid="1655380762395" ${\mathbf{\delta }}_{\mathbf{3}}$by using both the real and reactive power equations at bus 3 (load bus). (d) Find the real power generated at bus 1 (swing bus). (e) Evaluate the total real power losses in the system.

Short answer

(a) The $Y_{bus}$matrix of the system is $\left[\begin{array}{ccc}5-j10& -2+\mathrm{j}4& -3+j6\\ -2+j4& 2-\mathrm{j}4& 0\\ -3+\mathrm{j}6& 0& 3-\mathrm{j}6\end{array}\right]\mathrm{pu}.$

(b) The phase angle at bus 2 is $15.{18}^{\circ }$.

(c) The value of the voltage magnitude is $0.9686\mathrm{pu}$and phase angle is

$-15.{03}^{\circ }\mathrm{or}248.{17}^{\circ }$at bus 3.

(d) The real power at bus 1 is $0.386\mathrm{pu}$.

(e) The total power loss in the system is $0.386\mathrm{pu}$.

Step-by-step solution

Write the given data from the question.

The admittance between the bus 1 and 2, $Y_{12}=2-j4$

The admittance between the bus 1 and 3m, $Y_{13}=3-j6$

Voltage on bus 1, $\left|V_1\right|=1\angle 0$

The real power on bus 2,$P_2=1.5\mathrm{pu}$

Voltage on bus 2, $\left|V_2\right|=1\mathrm{pu}$

The real power on bus 3, $P_3=-1.5\mathrm{pu}$

The reactive power on bus 3, $Q_3=0.8\mathrm{pu}$

Number of the buses,$N=3$

Determine the formulas to calculate the admittance matrix, phase angleδ2, voltage and phase angle on bus 3, real and reactive power on bus 1 and total real power losses in the system.

The equation to calculate the diagonal element of the $Y_{\mathrm{bus}}$matrix of the system is

given as follows.

${\mathbf{Y}}_{\mathbf{kk}}$= sum of the admittance connected to bus $\mathit{k}$. ......(1)

The equation to calculate the non-diagonal element of the $Y_{\mathrm{bus}}$matrix of the system

is given as follows.

${\mathbf{Y}}_{\mathbf{kn}}$= (-sum of admittance connected between bus $\mathit{k}\mathbf{}\mathbf{and}\mathbf{}\mathit{n}$) ......(2)

The equation to calculate the real power at any bus$k$is given as follows.

localid="1655380801411" ${\mathbf{P}}_{\mathbf{k}}\mathbf{=}{\mathbf{V}}_{\mathbf{k}}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{N}}{\mathbf{Y}}_{\mathbf{kn}}{\mathbf{V}}_{\mathbf{n}}\mathbf{cos}\mathbf{(}{\mathbf{\delta }}_{\mathbf{k}}\mathbf{-}{\mathbf{\delta }}_{\mathbf{n}}\mathbf{-}{\mathbf{\theta }}_{\mathbf{kn}}\mathbf{)}$ ......(3)

Here, localid="1655380810565" ${\mathbf{P}}_{\mathbf{k}}$is the real power supplied to the bus localid="1655380805473" $\mathit{k}\mathbf{,}{\mathbf{V}}_{\mathbf{k}}$, is the voltage magnitude at

bus localid="1655380817007" $\mathit{k}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{V}}_{\mathbf{n}}$,voltage magnitude at bus localid="1655380813737" $\mathit{n}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{\delta }}_{\mathbf{k}}$is the phase angle at bus localid="1655380821222" $\mathit{k}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{\delta }}_{\mathbf{k}}$is

the phase angle at bus localid="1655380828432" $\mathit{n}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{\theta }}_{\mathbf{kn}}$is the angle of admittance localid="1655380824861" ${\mathbf{Y}}_{\mathbf{k}\mathbf{n}}$and is the

number of buses.

The equation to calculate the real power at any bus localid="1655380832166" $k$is given as follows.

localid="1655380836735" ${\mathbf{Q}}_{\mathbf{k}}\mathbf{=}{\mathbf{V}}_{\mathbf{k}}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{N}}{\mathbf{Y}}_{\mathbf{kn}}{\mathbf{V}}_{\mathbf{n}}\mathbf{cos}\mathbf{(}{\mathbf{\delta }}_{\mathbf{k}}\mathbf{-}{\mathbf{\delta }}_{\mathbf{n}}\mathbf{-}{\mathbf{\theta }}_{\mathbf{kn}}\mathbf{)}$ ......(4)

The equation to calculate the total power losses in the system is given as follows.

localid="1655380840659" $$\begin{gathered} {\mathbf{P}}_{\mathbf{loss}}\mathbf{=}{\mathbf{P}}_{\mathbf{gen}}\mathbf{-}{\mathbf{P}}_{\mathbf{load}} \\ {\mathbf{P}}_{\mathbf{loss}}\mathbf{=}{\mathbf{P}}_{\mathbf{1}}\mathbf{+}{\mathbf{P}}_{\mathbf{2}}\mathbf{+}{\mathbf{P}}_{\mathbf{3}} \end{gathered}$$ ......(5)

Here,localid="1655380845375" ${\mathbf{P}}_{\mathbf{gen}}$is the total generated real power in the system and localid="1655380850348" ${\mathbf{P}}_{\mathbf{load}}$is the real

power of the load.

Calculate the admittance matrix of the system.

(a)

Calculate the admittance of matrix $Y_{11}$.

By using the equation (1).

$$\begin{gathered} Y_{11}=j4+3-j6 \\ {Y}_{11}=5-j10 \\ {Y}_{11}=11.18\angle -63.{43}^{\circ }\mathrm{pu} \end{gathered}$$

Calculate the admittance of matrix $Y_{22}$.

By using the equation (1).

$$\begin{gathered} Y_{22}=2-j4\mathrm{pu} \\ {Y}_{22}=4.472\angle -63.{43}^{\circ }\mathrm{pu} \end{gathered}$$

Calculate the admittance of matrix $Y_{33}$.

By using the equation (1).

$$\begin{gathered} Y_{33}=3-j6\mathrm{pu} \\ {Y}_{33}=6.71\angle -63.{43}^{\circ }\mathrm{pu} \end{gathered}$$

Calculate the admittance of matrix $Y_{12}$and $Y_{21}$.

By using the equation (2).

$$\begin{gathered} Y_{12}=Y_{21} \\ {Y}_{12}=-(2-j4) \\ {Y}_{12}=-2+j4 \\ {Y}_{12}=4.47\angle -116.{56}^{\circ }\mathrm{pu} \end{gathered}$$

Calculate the admittance of matrix $Y_{23}$and $Y_{32}$.

By using the equation (2).

$$\begin{gathered} Y_{23}=Y_{32} \\ {Y}_{23}=0\mathrm{pu} \end{gathered}$$

Calculate the admittance of matrix $Y_{13}$and $Y_{31}$.

By using the equation (2).

$$\begin{gathered} Y_{13}=Y_{31} \\ {Y}_{13}=-(3-j6) \\ {Y}_{13}=-3+j6 \\ {Y}_{13}=6.71\angle -116.{56}^{\circ }\mathrm{pu} \end{gathered}$$

Hence the $Y_{\mathrm{bus}}$matrix of the system is $\left[\begin{array}{ccc}5-j10& -2+\mathrm{j}4& -3+j6\\ -2+j4& 2-\mathrm{j}4& 0\\ -3+\mathrm{j}6& 0& 3-\mathrm{j}6\end{array}\right]\mathrm{pu}.$

Calculate the phase angle δ2at bus 2.

(b)

Calculate the real phase angle ${\delta }_2$at bus by using the real power equation.

Substitute 2 for $k$and 3 for $N$into equation (3).

$$\begin{gathered} {\mathbf{P}}_{\mathbf{2}}\mathbf{=}{\mathbf{V}}_{\mathbf{2}}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{3}}{\mathbf{Y}}_{\mathbf{2}\mathbf{n}}{\mathbf{V}}_{\mathbf{n}}\mathbf{cos}\mathbf{(}{\mathbf{\delta }}_{\mathbf{2}}\mathbf{-}{\mathbf{\delta }}_{\mathbf{n}}\mathbf{-}{\mathbf{\theta }}_{\mathbf{2}\mathbf{n}}\mathbf{)} \\ {\mathbf{P}}_{\mathbf{2}}\mathbf{=}{\mathbf{V}}_{\mathbf{2}}\mathbf{\left[}{\mathit{Y}}_{\mathbf{21}}{\mathit{V}}_{\mathit{1}}\mathbf{cos}\mathit{\right(}{\mathbf{\delta }}_{\mathit{2}}\mathit{-}{\mathbf{\delta }}_{\mathit{1}}\mathit{-}{\mathbf{\theta }}_{\mathit{2}\mathit{1}}\mathit{)}\mathit{+}{\mathit{Y}}_{\mathit{22}}{\mathit{V}}_{\mathit{2}}\mathbf{cos}\mathit{(}{\mathbf{\delta }}_{\mathit{2}}\mathit{-}{\mathbf{\delta }}_{\mathit{2}}\mathit{-}{\mathbf{\theta }}_{\mathit{22}}\mathit{)}\mathit{+}{\mathit{Y}}_{\mathit{23}}{\mathit{V}}_{\mathit{1}}\mathbf{cos}\mathit{(}{\mathbf{\delta }}_{\mathit{2}}\mathit{-}{\mathbf{\delta }}_{\mathit{3}}\mathit{-}{\mathbf{\theta }}_{\mathit{23}}\mathit{)} \\ {\mathbf{P}}_{\mathit{2}}\mathit{=}{\mathbf{V}}_{\mathit{2}}\mathit{[}{\mathit{Y}}_{\mathit{21}}{\mathit{V}}_{\mathit{1}}\mathbf{cos}\mathit{(}{\mathbf{\delta }}_{\mathit{2}}\mathit{-}{\mathbf{\delta }}_{\mathit{1}}\mathit{-}{\mathbf{\theta }}_{\mathit{21}}\mathit{)}\mathit{+}{\mathit{Y}}_{\mathit{22}}{\mathit{V}}_{\mathit{2}}\mathbf{cos}\mathbf{(}\mathbf{-}{\mathit{\theta }}_{\mathit{22}}\mathbf{)}\mathit{+}{\mathbf{Y}}_{\mathbf{23}}{\mathbf{V}}_{\mathbf{1}}\mathbf{cos}\mathit{(}{\mathbf{\delta }}_{\mathbf{2}}\mathit{-}{\mathbf{\delta }}_{\mathbf{3}}\mathit{-}{\mathbf{\theta }}_{\mathbf{23}}\mathit{)} \end{gathered}$$

Substitute 1.5pu for $P$,1.1pu for $V_2$,$116.{56}^{\circ }$for ${\theta }_{21}$,$-63.{43}^{\circ }$for${\theta }_{22}$ and $0$for ${\delta }_1$.

$$\begin{gathered} 1.5=\left\{\begin{array}{c}1.1[4.47(1)\cos \left({\delta }_2-0-116.{56}^{\circ }\right)+4.47(1.1)\cos \left(-\left(-63.{43}^{\circ }\right)\right)\\ +\left(0\right)\left(1\right)\cos \left({\delta }_2-{\delta }_3-{\delta }_{23}\right)\end{array}\right\} \\ 1.5=1.1\times 4.47\left[\cos \left({\delta }_2-0-116.{56}^{\circ }\right)+(1.1)\cos \left(63.{43}^{\circ }\right)\right] \\ 1.5=4.917\left[\cos \left({\delta }_2-116.{56}^{\circ }\right)+0.4920\right] \\ 0.3050=\cos \left({\delta }_2-116.{56}^{\circ }\right)+0.4920\le \end{gathered}$$

Solve further as,

$0.3050=\cos \left({\mathrm{\delta }}_2-116.{56}^{\circ }\right)+0.4920$

$$\begin{gathered} \cos \left({\mathrm{\delta }}_2-116.{56}^{\circ }\right)=0.3050-0.4920 \\ {\delta }_2-116.{56}^{\circ }={\cos }^{-1}\left(-0.187\right) \\ {\delta }_2-116.{56}^{\circ }=\pm 100.{77}^{\circ } \end{gathered}$$

Solve further as,

${\delta }_2=15.{18}^{\circ },217.{34}^{\circ }$

Hence, the phase angle ${\delta }_2$at bus 2 is $15.{18}^{\circ }$

Calculate the voltage and phase angle at bus 3.

(c)

Calculate the real power at bus 3,

Substitute 3 for $k$into equation (3),

$$\begin{gathered} {\mathrm{P}}_3={\mathrm{V}}_3\sum _{\mathrm{n}=1}^3{\mathrm{Y}}_{3\mathrm{n}}{\mathrm{V}}_{\mathrm{n}}\cos ({\mathrm{\delta }}_2-{\mathrm{\delta }}_{\mathrm{n}}-{\mathrm{\theta }}_{3\mathrm{n}}) \\ {\mathrm{P}}_3={\mathrm{V}}_3\left[Y_{31}{V}_{\mathit{1}}\cos \mathit{\right(}{\mathrm{\delta }}_3\mathit{-}{\mathrm{\delta }}_{\mathit{1}}\mathit{-}{\mathrm{\theta }}_{\mathit{31}}\mathit{)}\mathit{+}{Y}_{\mathit{32}}{V}_{\mathit{2}}\cos \mathit{(}{\mathrm{\delta }}_3\mathit{-}{\mathrm{\delta }}_{\mathit{2}}\mathit{-}{\mathrm{\theta }}_{\mathit{32}}\mathit{)}\mathit{+}{Y}_{\mathit{33}}{V}_3\cos \mathit{(}{\mathrm{\delta }}_3\mathit{-}{\mathrm{\delta }}_{\mathit{3}}\mathit{-}{\mathrm{\theta }}_{\mathit{33}}\mathit{)} \\ {\mathrm{P}}_3\mathit{=}{\mathrm{V}}_3\mathit{[}{Y}_{\mathit{31}}{V}_{\mathit{1}}\cos \mathit{(}{\mathrm{\delta }}_3\mathit{-}{\mathrm{\delta }}_{\mathit{1}}\mathit{-}{\mathrm{\theta }}_{\mathit{31}}\mathit{)}\mathit{+}{Y}_{\mathit{32}}{V}_{\mathit{2}}\cos \mathit{(}{\mathrm{\delta }}_3\mathit{-}{\mathrm{\delta }}_3\mathit{-}{\mathrm{\theta }}_{32})\mathit{+}{\mathrm{Y}}_{33}{\mathrm{V}}_3(-{\mathrm{\theta }}_{\mathit{22}}) \end{gathered}$$

Substitute role="math" localid="1655381731955" $-1.5\mathrm{pu}$for role="math" localid="1655381741105" $P_3$,$1.1\mathrm{pu}$for $V_2$, $1\mathrm{pu}$for $V_1$, $6.71$for $Y_{31}$, 0 for $Y_{32}$, for $6.71$, for $Y_{33}$, $0$for ${\delta }_1$,$116.{56}^{\circ }$for ${\theta }_{31}$,$-63.{43}^{\circ }$for ${\theta }_{33}$and $0$for ${\theta }_{32}$into above equation.

$$\begin{gathered} -1.5=V_3\left\{\begin{array}{c}6.71\left(1\right)\cos \left({\delta }_3-0-116.{56}^{\circ }\right)+\left(0\right)\times 1.1\times \cos \left({\delta }_3-{\delta }_2-0\right)\\ +6.17V_3\cos \left(-\left(-63.{43}^{\circ }\right)\right)\end{array}\right\} \\ -1.5=V_{3}1.1\times 4.47\left[\cos \left({\delta }_2-116.{56}^{\circ }\right)+6.17V_3\cos \left(63.{43}^{\circ }\right)\right] \\ -1.5=6.17V_3\left[\cos \left({\delta }_3-116.{56}^{\circ }\right)+0.447V_3\right] \\ \frac{-1.5}{6.17V_3}=\cos \left({\delta }_3-116.{56}^{\circ }\right)+0.447V_3 \end{gathered}$$

Solve further as,

$\cos \left({\delta }_3-116.{56}^{\circ }\right)=\frac{-1.5}{6.71V_3}-0.447V_3$ ......(6)

Calculate the reactive power at bus 3,

Substitute 3 for $k$into equation (4),