Question

Two three-phase generators supply a three-phase load through separate three-phase lines The load absorbs$\mathbf{30}\mathbf{KW}$at $\mathbf{0}\mathbf{.}\mathbf{8}$power factor lagging. The line impedance is $\left(1.4+j1.6\right)\mathbf{\Omega }$per phase between generator ${\mathbf{G}}_{\mathbf{1}}$and the load, and $\left(0.8+j1\right)$n per phase between generator and the load. If generator ${\mathbf{G}}_{\mathbf{2}}$suppliesat$\mathbf{15}\mathbf{KW}\mathbf{}\mathbf{at}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{8}$power factor lagging, With a terminal voltage of $\mathbf{460}\mathbf{}\mathbf{V}$line-to-line, determine (a) the voltage at the load terminals, (b) the voltage at the terminals of generator and (c) the real and reactive power supplied by generator ${\mathbf{G}}_{\mathbf{2}}$. Assume balanced

Short answer

(a) The line-to-line voltage at the load terminal is$375.7\angle -2.{73}^{\circ }$.

(b) The line-to-line voltage at the terminal of generator is ${\mathrm{G}}_2\mathrm{is}449.65\angle 0.{63}^{\circ }\mathrm{V}$.

(c) The real and reactive power supplied by the generator are and${\mathrm{S}}_{\mathrm{G}2}\mathrm{are}20.12\mathrm{kW}$.

Step-by-step solution

Calculate the voltage at the load terminal.

(a)

The line-to-line voltage of generator 1,

${\mathrm{v}}_{\mathrm{LL}}=460\mathrm{V}$

The phase voltage of generator 1,

$V_{G1}=\frac{460}{\sqrt{3}}\angle 0^{\circ }$

The single line diagram can be drawn as${\mathrm{v}}_{\mathrm{LL}}=460\mathrm{V}$

The expression for the real power

${\mathbf{P}}_{\mathbf{G}\mathbf{1}}\mathbf{=}\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{LL}}{\mathbf{I}}_{\mathbf{L}}\left(p.f\right)$

$I_L=\frac{P_{G1}}{\sqrt{3}{V}_{LL}\left(p.f\right)}\angle -{\cos }^{-1}\left(p.f\right)$

Substitute the values and solve as

$$\begin{gathered} I_L=\frac{15\times {10}^3}{\sqrt{3}\times 460\times 0.8}\angle -{\cos }^{-1}\left(0.8\right) \\ {I}_L=23.53\angle -36.{87}^{\circ } \end{gathered}$$

Apply the Kirchhoff’s voltage law in the left side of the loop,

$V_L=V_{G1}-\left(1.4+j1.6\right)I_{G1}$

Substitute the values and solve as

$$\begin{gathered} V_L=\frac{460}{\sqrt{3}}\angle 0^{\circ }\left(1.4+j1.6\right)\left(23.53\angle -36.{87}^{\circ }\right) \\ {V}_L=265.58\angle 0-(2.216\angle 48.{81}^{\circ })\left(23.53\angle -36.{87}^{\circ }\right) \\ {V}_L=265.58\angle 0-50.02\angle 11.94 \\ {V}_L=216.9\angle -2.73 \end{gathered}$$

The line-to-line voltage at the load terminal

$$\begin{gathered} {\left(V_L\right)}_{LL}=\sqrt{3}{V}_L \\ {\left(V_L\right)}_{LL}=\sqrt{3}\times 216.9\angle -2.{73}^{\circ } \\ {\left(V_L\right)}_{LL}=375.7\angle 2.{73}^{\circ } \end{gathered}$$

Hence, the line-to-line voltage at the load terminal is $375.7\angle 2.{73}^{\circ }$.

Calculate the voltage at the terminal of the generator.

(b)

The apparent power can be calculated as

$$\begin{gathered} \mathrm{S}=\frac{{\mathrm{P}}_{\mathrm{L}}}{\mathrm{p}.\mathrm{f}} \\ \mathrm{S}=\frac{30\times {10}^3}{0.8} \\ \mathrm{S}=37.5\times {10}^3\mathrm{VA} \end{gathered}$$

The complex power can be calculated as

$$\begin{gathered} {\mathrm{S}}_{\mathrm{L}}=37.5\times {10}^3\angle {\cos }^{-1}\left(0.8\right) \\ {\mathrm{S}}_{\mathrm{L}}=37.5\angle 36.{87}^{\circ }\mathrm{kVA} \end{gathered}$$

The complex power can be calculated as

$$\begin{gathered} S_L={\sqrt{3\left(V_L\right)}}_{LL}I{\text{'}}_L \\ I{\text{'}}_L=\frac{S_L}{\sqrt{3}{\left(V_L\right)}_{LL}} \end{gathered}$$

Substitute the values and solve as

$$\begin{gathered} {I^{*}}_L=\frac{37.5\times {10}^3\angle 36.{87}^{\circ }}{\sqrt{3}\times 375.7\angle -2.{73}^{\circ }} \\ {I^{*}}_L=57.62\angle 39.6^{\circ } \\ {I^{*}}_L=57.62\angle -39.6^{\circ }A \end{gathered}$$

The current through the generator can be calculated as

$$\begin{gathered} I_{G2}=I_L-I_{G1} \\ {I}_{G2}=57.63\angle -39.6^{\circ }-23.53\angle 36.{87}^{\circ } \\ {I}_{G2}=34.12\angle -41.{49}^{\circ }A \end{gathered}$$

Apply the Kirchhoff’s voltage law in right hand side loop,

$$\begin{gathered} V_{G2}=V_L+\left(0.8+j1\right)V_{G2} \\ {V}_{G2}=216.9\angle -2.{73}^{\circ }-\left(0.8+j1\right)\left(34.12\angle -41.{49}^{\circ }\right) \\ {V}_{G2}=216.9\angle -2.{73}^{\circ }-1.28\angle 51.{34}^{\circ }\left(34.12\angle -41.{49}^{\circ }\right) \\ {V}_{G2}=216.9\angle -2.{73}^{\circ }-43.67\angle 9.{45}^{\circ } \end{gathered}$$

Solve further as

$V_{G2}=259.61\angle -0.{63}^{\circ }V$

The line-to-line voltage at the terminal of generator

$$\begin{gathered} {\left(V_{G2}\right)}_{LL}=\sqrt{3}{V}_{G2} \\ {\left(V_{G2}\right)}_{LL}=\sqrt{3\times 2}59.61\angle 0.{63}^{\circ }V \\ {\left(V_{G2}\right)}_{LL}=449.65\angle -0.{63}^{\circ }V \end{gathered}$$

Hence the line-to-line voltage at the terminal of generator $G_2$is $449.65\angle -0.{63}^{\circ }V$.

Calculate the real and reactive power supplied by generatorG2 .

(c)

The complex power supplied by the generator $G_2$can be calculated as

$$\begin{gathered} S_{G2}=\sqrt{3}{\left(V_{G2}\right)}_{LL}{I}_{G2}^{*} \\ {S}_{G2}=\sqrt{3}\times \left(449.65\angle -0.{63}^{\circ }\right)\times \left(34.14\angle 41.{48}^{\circ }\right) \\ {S}_{G2}=26.6\angle 40.{86}^{\circ } \end{gathered}$$

Convert the power into cartesian form

$S_{G2}=\left(20.12+j17.\right)kVA$

The real power supplied by the generator$G_2$

$$\begin{gathered} P=Re\left(S_{G2}\right) \\ P=20.12KW \end{gathered}$$

The reactive power supplied by the generator$G_2$

$$\begin{gathered} \mathrm{Q}=\mathrm{Im}\left({\mathrm{S}}_{\mathrm{G}2}\right) \\ \mathrm{Q}=17.4\mathrm{kvar} \end{gathered}$$

Hence, the real and reactive power supplied by the generator $S_{G2}$are $20.12kW$and $17.4\mathrm{kvar}$.