Question

In a three-phase system, a synchronous generator supplies power to a $200$-volt synchronous motor through a line having an impedance of$\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{\angle }\mathbf{80}\mathbf{°}$ohm per phase. The motor draws $\mathbf{5}\mathit{k}\mathit{W}$at $\mathbf{0}\mathbf{.}\mathbf{8}$pf leading and at rated voltage. The neutrals of both the generator and motor are grounded through impedances of ohms. The sequence impedances of both machines are$Z_{g0}=j7$,$Z_{g1}=j15$, and$Z_{g2}=j10$ohms. Draw the sequence networks for this system and find the line-to-line voltage at the generator terminals. Assume balanced three-phase operation.

Short answer

Therefore, zero, negative and positive sequence network are shown below.

The line to line voltage at generator terminalis$193.5\text{\hspace{0.33em}V}$ .

Step-by-step solution

Write the given data from the question:

The voltage,$V=200\text{\hspace{0.33em}volts}$

The generator neutral impedance.$Z_n=j5\Omega$

The zero-sequence impedance of generator.$Z_{g0}=j7\Omega$

The positive sequence impedance of generator.$Z_{g1}=j15\Omega$

The negative sequence impedance of generator.$Z_{g2}=j10\Omega$

The line impedance.$Z_L=0.5\angle 80°\Omega$

Motor power.$P=5\text{\hspace{0.33em}kW}$

Power factor$pf=0.8$leading.

Determine the formulas to calculate the line to line voltage at generator terminal.

The equation to calculate the current through motor is given as follows.

${\mathit{I}}_{\mathbf{m}\mathbf{1}}\mathbf{=}\frac{\mathbf{P}\mathbf{\angle }\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}\left(pf\right)}{\mathbf{V}\mathbf{\times }\mathbf{p}\mathbf{f}}$ …… (1)

The equation to calculate the line to line voltage at generator terminal is given as follows.

$\begin{array}{l}{\mathbf{V}}_{\mathbf{g}\mathbf{a}\mathbf{n}}\mathbf{=}{\mathbf{V}}_{\mathbf{g}\mathbf{1}}\\ {\mathbf{V}}_{\mathbf{g}\mathbf{a}\mathbf{n}}\mathbf{=}\frac{\mathbf{V}}{\sqrt{\mathbf{3}}}\mathbf{+}{\mathbf{I}}_{\mathbf{m}}{\mathbf{Z}}_{\mathbf{L}}\end{array}$ …… (2)

Calculate the line to line voltage at terminal of generator and draw the sequence network.

Since the neutral of generator and motor are grounded with the impedance of$j5\Omega$. Therefore the$3Z_n$will be there in the zero sequence of generator and motor network.

Therefore, zero, negative and positive sequence network are shown below.

Calculate the current through motor.

Substitute$5\text{\hspace{0.33em}kW}$for $P$, $0.8$for$pf$and $200\text{\hspace{0.33em}V}$for$V$into equation (1).

$\begin{array}{l}{I}_m=\frac{5000\times \angle {\cos }^{-1}\left(0.8\right)}{200\sqrt{3}\times 0.8}\\ I_m=18.04\angle 36.87°\text{\hspace{0.33em}A}\end{array}$

Calculate the voltage at generator terminal

Substitute$200\text{\hspace{0.33em}V}$for$V$, $18.04\angle 36.87°\text{\hspace{0.33em}A}$for $I_m$and$0.5\angle 80°\Omega$for $Z_L$into equation (2).

$\begin{array}{l}{V}_{gan}=\frac{200}{\sqrt{3}}\angle 0°+\left(18.04\angle 36.87\right)\times \left(0.5\angle 80°\right)\\ V_{gan}=115.47-4.077+j8.046\\ V_{gan}=111.39+j8.046\\ V_{gan}=111.7\angle 4.131°\text{\hspace{0.33em}V}\end{array}$

Calculate the line to line voltage at generator terminal.

$\begin{array}{l}{V}_{gan}=\sqrt{3}\times 111.7\\ V_{gan}=193.5\text{\hspace{0.33em}V}\end{array}$

Hence, the line to line voltage at generator terminalis$193.5\text{\hspace{0.33em}V}$.