Question

Three identical Y-connected resistors of$\mathbf{1}\mathbf{\angle }\mathbf{0}\mathbf{°}$per unit form a load bank that is supplied from the low-voltage localid="1656322566335" $\mathbf{\text{Y}}$-side of a $\mathit{Y}\mathbf{-}\mathit{\Delta }$transformer. The neutral of the load is not connected to the neutral of the system. The positive- and negative-sequence currents flowing toward the resistive load are given by

localid="1656754777422" ${\mathit{I}}_{\mathbf{a}\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{\angle }\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{°}\mathit{p}\mathit{e}\mathit{r}\mathit{u}\mathit{n}\mathit{i}\mathit{t}{\mathit{I}}_{\mathbf{a}\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{\angle }\mathbf{250}\mathbf{°}\mathit{p}\mathit{e}\mathit{r}\mathit{u}\mathit{n}\mathit{i}\mathit{t}$

and the corresponding voltages on the low-voltage Y-side of the transformer are

localid="1656754827543" ${\mathit{V}}_{\mathbf{a}\mathbf{n}\mathbf{,}\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{\angle }\mathbf{45}\mathbf{°}\mathit{p}\mathit{e}\mathit{r}\mathit{u}\mathit{n}\mathit{i}\mathit{t}$(Line-to-neutral voltage base)

localid="1656754914888" ${\mathit{V}}_{\mathbf{an}\mathbf{,}\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{\angle }\mathbf{250}\mathbf{°}\mathit{p}\mathit{e}\mathit{r}\mathit{u}\mathit{n}\mathit{i}\mathit{t}$(Line-to-neutral voltage base)

Determine the line-to-line voltages and the line currents in per unit on the high-voltage side of the transformer. Account for the phase shift.

Short answer

The line to line voltages$V_{ab},V_{bc}$ and$V_{ca}$ are $0.3787\angle 100.9°$, $0.8476\angle -36.72°$and$0.6224\angle 167.48°$ respectively. The line currents are$I_{ab},I_{bc}$ and$I_{ca}$ are$0.3145\angle 29.05°$ , $0.7447\angle -64.85°$and $0.7811\angle 136.99°$respectively

Step-by-step solution

Write the given data from the question.

Write the positive and negative sequence current in per flowing the resistive load,

$\begin{array}{l}{I}_{a1}=1\angle 4.5°\text{\hspace{0.33em}pu}\\ I_{a2}=0.5\angle 250°\text{\hspace{0.33em}pu}\end{array}$

Write the voltage across the resistive load.

$\begin{array}{l}{V}_{a1}=1\angle 45°\text{\hspace{0.17em}pu}\\ {\text{V}}_{a2}=0.5\angle 250°\text{\hspace{0.33em}pu}\end{array}$

The star connected resistive load =$1.0\angle 0°\text{\hspace{0.33em}pu}$

Determine the formulas to calculate the line currents and line to line voltages.

The equation to calculate the phase currents at high voltage side is given as follows.

$\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{I}_{0H}\\ I_{1H}\\ I_{2H}\end{array}\right]$ …… (1)

The equation to calculate the line currents at high voltage side is given as follows.

$\left[\begin{array}{c}{I}_{ab}\\ I_{bc}\\ I_{ca}\end{array}\right]=\sqrt{3}\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]$ …… (2)

The equation to calculate the phase voltage at high voltage side is given as follows.

$\left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{V}_{0H}\\ V_{1H}\\ V_{2H}\end{array}\right]$ …… (3)

The equation to calculate the line voltages at high voltage side is given as follows.

$\left[\begin{array}{c}{V}_{ab}\\ V_{bc}\\ V_{ca}\end{array}\right]=\sqrt{3}\left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]$ …… (4)

calculate the line to line voltages and line currents at hight voltage side.

For the positive sequence, high voltage side current lead to low voltage side current by $30°$.

The current at high voltage side.

$\begin{array}{l}{I}_{1H}=\left(1\angle 4.5°\right)\left(1\angle 30°\right)\\ I_{1H}=1\angle 34.5°\text{\hspace{0.33em}pu}\end{array}$

For the positive sequence, high voltage side current lag to low voltage side current by$30°$.

The current at high voltage side.

$\begin{array}{l}{I}_{2H}=\left(0.5\angle 250°\right)\left(1\angle -30°\right)\\ I_{2H}=0.5\angle 220°\text{\hspace{0.33em}pu}\\ I_{2H}=0.5\angle -140°\text{\hspace{0.33em}pu}\end{array}$

Since the neutral is not ground on high voltage side the zero-sequence current is zero.

Calculate the phase current on high voltage side.

Substitute$0$for$I_0$,$1\angle 34.5°\text{\hspace{0.33em}pu}$for$I_{1H}$and $0.5\angle -140°\text{\hspace{0.33em}pu}$for$I_{2H}$into equation (1).

$\begin{array}{l}\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle -120°& 1\angle 120°\\ 1& 1\angle 120°& 1\angle -120°\end{array}\right]\left[\begin{array}{c}0\\ 1\angle 34.5°\\ 0.5\angle -140°\end{array}\right]\\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}0+1\angle 34.5°+0.5\angle -140°\\ 0+\left(1\angle 34.5°\right)\left(1\angle -120°\right)+\left(0.5\angle -140°\right)\left(1\angle 120°\right)\\ 0+\left(1\angle 34.5°\right)\left(1\angle 120°\right)+\left(0.5\angle -140°\right)\left(1\angle -120°\right)\end{array}\right]\\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}1\angle 34.5°+0.5\angle -140°\\ \left(1\angle -85.5°\right)+\left(0.5\angle -20°\right)\\ \left(1\angle 154.5°\right)+\left(0.5\angle -260°\right)\end{array}\right]\\ \left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{c}0.1681\angle 29.05°\\ 0.4300\angle -64.85°\\ 0.451\angle 136.99°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the line current.

Substitute $0.1681\angle 29.05°\text{\hspace{0.33em}pu}$for $I_a$, $0.4300\angle -64.85°\text{\hspace{0.33em}pu}$for $I_b$and$0.451\angle 136.99°\text{\hspace{0.33em}pu}$for $I_c$into equation (2).

$\begin{array}{l}\left[\begin{array}{c}{I}_{ab}\\ I_{bc}\\ I_{ca}\end{array}\right]=\sqrt{3}\left[\begin{array}{c}0.1681\angle 29.05°\\ 0.4300\angle -64.85°\\ 0.451\angle 136.99°\end{array}\right]\\ \left[\begin{array}{c}{I}_{ab}\\ I_{bc}\\ I_{ca}\end{array}\right]=\left[\begin{array}{c}0.2913\angle 29.05°\\ 0.7447\angle -64.85°\\ 0.7811\angle 136.99°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

The positive sequence voltage at high voltage side.

$\begin{array}{l}{V}_{1H}=\left(1\angle 45°\right)\left(1\angle 30°\right)\\ V_{1H}=1\angle 75°\text{\hspace{0.33em}pu}\end{array}$

The negative sequence voltage at high voltage side.

$\begin{array}{l}{V}_{2H}=\left(0.5\angle 250°\right)\left(1\angle -30°\right)\\ V_{2H}=0.5\angle 220°\text{\hspace{0.33em}pu}\\ V_{2H}=0.5\angle -140°\text{\hspace{0.33em}pu}\end{array}$

The zero-sequence voltage at high voltage side.

$V_{0H}=0\text{\hspace{0.33em}pu}$

Calculate the phase voltage on high voltage side.

Substitute$0$for$V_0$, $1\angle 75°\text{\hspace{0.33em}pu}$for$V_{1H}$and$0.5\angle -140°\text{\hspace{0.33em}pu}$for$V_{2H}$into equation (3).

$\begin{array}{l}\left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& 1\angle -120°& 1\angle 120°\\ 1& 1\angle 120°& 1\angle -120°\end{array}\right]\left[\begin{array}{c}0\\ 1\angle 75°\\ 0.5\angle -140°\end{array}\right]\\ \left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}0+1\angle 75°+0.5\angle -140°\\ 0+\left(1\angle 75°\right)\left(1\angle -120°\right)+\left(0.5\angle -140°\right)\left(1\angle 120°\right)\\ 0+\left(1\angle 75°\right)\left(1\angle 120°\right)+\left(0.5\angle -140°\right)\left(1\angle -120°\right)\end{array}\right]\\ \left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}1\angle 75°+0.5\angle -140°\\ \left(1\angle -45°\right)+\left(0.5\angle -20°\right)\\ \left(1\angle 195°\right)+\left(0.5\angle -260°\right)\end{array}\right]\\ \left[\begin{array}{c}{V}_a\\ V_b\\ V_c\end{array}\right]=\left[\begin{array}{c}0.2187\angle 100.9°\\ 0.4894\angle -36.72°\\ 0.3594\angle 167.48°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Calculate the line to line voltages.

Substitute$0.2187\angle 100.9°\text{\hspace{0.33em}pu}$for $V_a$, $0.4894\angle -36.72°\text{\hspace{0.33em}pu}$for $V_b$and$0.3594\angle 167.48°\text{\hspace{0.33em}pu}$for $V_c$into equation (2).

$\begin{array}{l}\left[\begin{array}{c}{V}_{ab}\\ V_{bc}\\ V_{ca}\end{array}\right]=\sqrt{3}\left[\begin{array}{c}0.2187\angle 100.9°\\ 0.4894\angle -36.72°\\ 0.3594\angle 167.48°\end{array}\right]\\ \left[\begin{array}{c}{V}_{ab}\\ V_{bc}\\ V_{ca}\end{array}\right]=\left[\begin{array}{c}0.3787\angle 100.9°\\ 0.8476\angle -36.72°\\ 0.6224\angle 167.48°\end{array}\right]\text{\hspace{0.33em}pu}\end{array}$

Hence, the line to line voltages$V_{ab},V_{bc}$ and$V_{ca}$ are$0.3787\angle 100.9°$ , $0.8476\angle -36.72°$and$0.6224\angle 167.48°$ respectively. The line currents are$I_{ab},I_{bc}$ and $I_{ca}$are ,$0.3145\angle 29.05°$ $0.7447\angle -64.85°$and $0.7811\angle 136.99°$respectively.