Question

A balanced Y-connected generator with terminal voltage${\mathit{V}}_{\mathbf{b}\mathbf{c}}\mathbf{=}\mathbf{200}\mathbf{\angle }\mathbf{0}\mathbf{°}\mathit{V}$volts is connected to a balanced $\mathit{\Delta }$load whose impedance is $\mathbf{10}\mathbf{\angle }\mathbf{40}\mathbf{°}$ohms per phase. The line impedance between the source and load is $\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{\angle }\mathbf{80}\mathbf{°}$ohm for each phase. The generator neutral is grounded through an impedance of ohms. The generator sequence impedances are given by${\mathit{Z}}_{\mathbf{g}\mathbf{0}}\mathbf{=}\mathit{j}\mathbf{7}$, ${\mathit{Z}}_{\mathbf{g}\mathbf{1}}\mathbf{=}\mathit{j}\mathbf{15}$, and${\mathit{Z}}_{\mathbf{g}\mathbf{2}}\mathbf{=}\mathit{j}\mathbf{10}$ ohms. Draw the sequence networks for this system and determine the sequence components of the line currents.

Short answer

The zero-sequence impedance is shown below.

The negative sequence impedance is shown below.

The positive sequence impedance is shown below.

The positive sequence component of the line current is.$30.99\angle 74.94°\text{\hspace{0.33em}A}$

Step-by-step solution

Write the given data from the question:

The terminal voltage of $\text{Y}$-connected generator.$V_{bc}=200\angle 0°\text{\hspace{0.33em}V}$

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

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

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

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

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

The $\Delta$-connected load impedance$Z_{\Delta }=10\angle 40°\frac{\Omega }{\text{phase}}$.

 Step 2: Determine the formulas to calculate the sequence component of the line currents.

The equation to calculate the impedance from$\Delta$ -connected to -$\text{Y}$connected load is given as follows.

role="math" localid="1656330820137" $Z_Y=\frac{Z_{\Delta }}{3}$ …… (1)

The equation to calculate the per phase terminal voltage is given as follows.

$V_b=\frac{V_{bc}}{3}\angle -30°V$ …… (2)

The equation to calculate the positive sequence current is given as follows.

$I_1=\frac{V_1}{Z_{L1}+Z_Y}$ …… (3)

Here, ${\mathit{V}}_{\mathbf{1}}$is the per phase voltage,${\mathit{Z}}_{\mathbf{L}\mathbf{1}}$ is the impedance of line and is the positive sequence current.

Calculate the line current and draw the sequence network for the system.

Calculate the impedance of $\text{Y}$-connected load from -connected load.

Substitute$10\angle 40°\frac{\Omega }{\text{phase}}$for $Z_{\Delta }$into equation (1).

$\begin{array}{l}{Z}_Y=\frac{10\angle 40°}{3}\\ Z_Y=3.33\angle 40°\Omega \end{array}$

Calculate the per phase voltage.

Substitute$200\angle 0°\text{\hspace{0.33em}V}$for$V_{bc}$into equation (2).

$\begin{array}{l}{V}_b=\frac{200\angle -30}{\sqrt{3}}\\ V_b=115.47\angle -30\text{\hspace{0.33em}A}\end{array}$

Since the supply and the load are balanced, the zero and negative sequence network will not have the source voltage, and only the positive sequence network will have the source voltage.

The load is connected into delta and the neutral will be open. Therefore, open circuit will be there in the load side zero sequence network. The generator is grounded with the impedance of .$j5\Omega$There will be impedance of$3Z_n$the source side zero sequence network.

Therefore, the zero-sequence impedance is shown below.

Therefore, the negative sequence impedance is shown below.\

Calculate the positive sequence terminal voltage.

$\begin{array}{l}{V}_1=V_b\\ V_1=115.47\angle -30°\text{\hspace{0.33em}V}\end{array}$

Therefore, the positive sequence impedance is shown below.

Apply the Kirchhoff’s voltage law in positive sequence network.

$V_1-I_1{Z}_{L1}-I_1\left(\frac{Z_{\Delta }}{3}\right)=0$

Substitute $Z_Y$for$\frac{Z_{\Delta }}{\text{3}}$into above equation.

$\begin{array}{c}{V}_1-I_1{Z}_{L1}-I_1{Z}_Y=0\\ I_1{Z}_{L1}+I_1{Z}_Y=V_1\\ I_1\left(Z_{L1}+Z_Y\right)=V_1\\ I_1=\frac{V_1}{Z_{L1}+Z_Y}\end{array}$

Calculate the positive sequence component of the line current

Substitute$115.47\angle -30°\text{\hspace{0.33em}V}$for$V_1$,$0.5\angle 80°\Omega$for$Z_{L1}$and$3.33\angle 40°\Omega$into equation(3).

$\begin{array}{l}{I}_1=\frac{115.47\angle -30°}{0.5\angle 80°+3.33\angle 40°}\\ I_1=\frac{115.47\angle -30°}{3.726\angle 44.94°}\\ I_1=30.99\angle 74.94°\text{\hspace{0.33em}A}\end{array}$

Hence the positive sequence component of the line current is$30.99\angle 74.94°\text{\hspace{0.33em}A}$.