Question

Consider two interconnected voltage sources connected by a line of

impedance$\mathbf{Z}\mathbf{}\mathbf{=}\mathbf{}\mathbf{jX}\mathbf{}\mathbf{\Omega }$, as shown in Figure 2.27.

(a) Obtain expressions forlocalid="1655190041824" ${\mathbf{P}}_{\mathbf{12}}$andlocalid="1655190045880" ${\mathbf{Q}}_{\mathbf{12}}$.

(b) Determine the maximum power transfer and the condition for it to occur.

Short answer

(a) Therefore, the expression for real power $P_{12}$is $\frac{V_1{V}_2}{X}$$\sin \left({\delta }_1-{\delta }_2\right)$and the

expression for the reactive power $Q_{12}$is $\frac{V_1}{X}\left[\begin{array}{cc}{V}_1-& V_{2}c\cos ({\delta }_1-{\delta }_2)\end{array}\right]$.

(b) The condition for the maximum power transfer is ${\delta }_1-{\delta }_2={90}^{\circ }$and maximum power is $\frac{V_1{V}_2}{X}$.

Step-by-step solution

Calculate the expression for P12 & Q12.

(a)

Apply the Kirchhoff’s voltage law in the given circuit.

$$\begin{gathered} {\mathbf{V}}_{\mathbf{1}}\mathbf{=}{\mathbf{V}}_{\mathbf{2}}\mathbf{+}{\mathbf{I}}_{\mathbf{12}}\mathbf{Z} \\ {\mathbf{I}}_{\mathbf{12}}\mathbf{Z}\mathbf{=}{\mathbf{V}}_{\mathbf{1}}\mathbf{-}{\mathbf{V}}_{\mathbf{2}} \\ {\mathbf{I}}_{\mathbf{12}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{1}}\mathbf{-}{\mathbf{V}}_{\mathbf{2}}}{\mathbf{Z}} \end{gathered}$$

Substitute the values and solve as

$$\begin{gathered} I_{12}=\frac{V_1\angle {\delta }_1-V_2\angle {\delta }_2}{X\angle {90}^{\circ }} \\ {I}_{12}=\frac{V_1\angle {\delta }_1-{90}^{\circ }}{X}-\frac{V_1\angle {\delta }_1-{90}^{\circ }}{X} \end{gathered}$$

The complex power can be calculated as

$$\begin{gathered} S_{12}=V_1{I}_{12}^{*} \\ {S}_{\mathit{12}}=V_{\mathit{1}}\mathit{\angle }{\delta }_{\mathit{1}}{\mathit{(}\frac{V_{\mathit{1}}}{X}\mathit{\angle }{\delta }_{\mathit{1}}\mathit{-}{\mathit{90}}^{\mathit{\circ }}\mathit{-}\frac{V_{\mathit{2}}}{X}\mathit{\angle }{\delta }_{\mathit{2}}\mathit{-}{\mathit{90}}^{\mathit{\circ }}\mathit{)}}^{*} \\ {S}_{\mathit{12}}={\mathrm{V}}_{\mathit{1}}\mathit{\angle }{\mathrm{\delta }}_{\mathit{1}}\mathit{(}\frac{V_{\mathit{1}}}{X}\mathit{\angle }\mathit{-}{\delta }_{\mathit{1}}\mathit{+}{\mathit{90}}^{\mathit{\circ }}\mathit{-}\frac{V_{\mathit{2}}}{X}\mathit{\angle }\mathit{-}{\delta }_{\mathit{2}}\mathit{+}{\mathit{90}}^{\mathit{\circ }}\mathit{)} \\ {\mathrm{S}}_{12}=\frac{{\mathrm{V}}_1^2}{X}\angle {90}^{\circ }-\frac{V_{\mathit{1}}{V}_{\mathit{2}}}{X}\angle {\mathrm{\delta }}_1-{\mathrm{\delta }}_2+90(1) \end{gathered}$$

The complex power can be written in terms of real and reactive power as

$S_{\mathit{12}}=P_{\mathit{12}}+j{Q}_{\mathit{12}}\left(2\right)$

By equating the equation (1) and equation (2), the real power supplied from source 1 is $P_{\mathit{12}}$

$$\begin{gathered} P_{\mathit{12}}=\frac{V_{\mathit{1}}^{\mathit{2}}}{X}\cos {90}^{\circ }-\frac{V_{\mathit{1}}{V}_{\mathit{2}}}{\mathrm{X}}\cos \left(({\mathrm{\delta }}_1-{\mathrm{\delta }}_2+{90}^{\circ }\right) \\ {P}_{\mathit{12}}\mathit{=}0-\frac{V_{\mathit{1}}{V}_{\mathit{2}}}{X}\cos \left(({\mathrm{\delta }}_1-{\mathrm{\delta }}_2+{90}^{\circ }\right) \\ {P}_{\mathit{12}}=\frac{V_{\mathit{1}}{V}_{\mathit{2}}}{X}\sin \left({\mathrm{\delta }}_1-{\mathrm{\delta }}_2\right) \end{gathered}$$

Hence the expression for $P_{12}$is $\frac{V_1{V}_2}{X}\sin \left({\mathrm{\delta }}_1-{\mathrm{\delta }}_2\right)$
.

The reactive power supplied from source 1 is $Q_{\mathit{12}}$

$$\begin{gathered} Q_{\mathit{12}}=\frac{V_{\mathit{1}}^{\mathit{2}}}{X}\sin {90}^{\circ }-\frac{V_{\mathit{1}}{V}_{\mathit{2}}}{X}\sin \left({\mathrm{\delta }}_1-{\mathrm{\delta }}_2+90\right) \\ {\mathrm{Q}}_{\mathit{12}}=\frac{{\mathrm{V}}_{\mathit{1}}^{\mathit{2}}}{\mathrm{X}}-\frac{{\mathrm{V}}_{\mathit{1}}{\mathrm{V}}_{\mathit{2}}}{\mathrm{X}}\cos \left({\mathrm{\delta }}_1-{\mathrm{\delta }}_2\right) \\ {\mathrm{Q}}_{\mathit{12}}=\frac{V_{\mathit{1}}}{X}\left[V_{\mathit{1}}\mathit{-}{V}_{\mathit{2}}\mathit{cos}\begin{array}{cc}({\mathrm{\delta }}_1-& {\mathrm{\delta }}_2)\end{array}\right] \end{gathered}$$

Thus, the expression for the reactive power $Q_{\mathit{12}}$is $\frac{V_{\mathit{1}}}{X}\left[V_{\mathit{1}}\mathit{-}{V}_{\mathit{2}}cos\begin{array}{cc}({\mathrm{\delta }}_1-& {\mathrm{\delta }}_2)\end{array}\right]$.

Step 2: Calculate the maximum power transfer and the condition for it to occur.

(b)

For the maximum power transfer, the power should be maximum.

$P_{\mathit{12}}=\frac{V_{\mathit{1}}{V}_{\mathit{2}}}{X}\sin \left({\mathrm{\delta }}_1-{\mathrm{\delta }}_2\right)$

For power to be maximum $\sin \left({\delta }_1-{\delta }_2\right)$should ne maximum.

Therefore

${\delta }_1-{\delta }_2={90}^{\circ }$

So, for the maximum power

$$\begin{gathered} P_{\mathit{12}}=\frac{V_1{V}_2}{X}\sin \left(90\right) \\ {P}_{12}=\frac{V_1{V}_2}{X} \end{gathered}$$

Thus, the condition for the maximum power transfer is ${\delta }_1-{\delta }_2={90}^{\circ }$and maximum power is $\frac{V_1{V}_2}{X}$.