Question

Three single-phase, two-winding transformers, each rated $\mathbf{450}\mathit{M}\mathit{V}\mathit{A}$, $\mathbf{20}\mathbf{/}\mathbf{288}\mathbf{.}\mathbf{7}\mathit{k}\mathit{V}$, with leakage reactance,${\mathit{X}}_{\mathbf{eq}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathit{p}\mathit{u}$are connected to form a three-phase bank. The high-voltage windings are connected in Y with a solidly grounded neutral. Draw the per-unit equivalent circuit if the low-voltage windings are connected (a) in delta with American standard phase shift or (b) in Y with an open neutral. Use the transformer ratings as base quantities. Winding resistances and exciting current are neglected.

Short answer

(a) The per-unit diagram for delta connection is drawn below:

(b) The per-unit diagram for star connection is drawn below:

Step-by-step solution

Determine phase shift between two winding transformer.

In the first condition he secondary winding has delta connection, in the second conditionthe secondary winding has star connection.

For the $\mathit{Y}\mathbf{-}\mathit{\Delta }$two winding transformer, the phase shift is given by ${\mathit{e}}^{j30}$ .

For the $\mathit{Y}\mathbf{-}\mathit{Y}$ two winding transformer, there is no phase shift.

Write the formula of three-phase power.

${\mathit{S}}_{\mathbf{3}\mathbf{-}\mathbf{ph}}\mathbf{=}\mathbf{3}{\mathit{S}}_{\mathbf{1}\mathbf{-}\mathbf{ph}}$ ……. (1)

Here,$S_{1-ph}$is the single-phase power.

Write the formula of line-to-line voltage.

localid="1655725477287" ${\mathit{V}}_{\mathbf{L}}\mathbf{=}\sqrt{\mathbf{3}}{\mathit{V}}_{\mathbf{ph}}$ ……. (2)

Here,$V_{ph}$is the phase voltage.

Write the formula of base current.

localid="1655725489731" ${\mathit{I}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{base}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{base}}}$ ……. (3)

Here,$S_{base}$is the base power and $V_{base}$is the base voltage.

Write the formula of actual current.

localid="1655725504332" $I_{\mathbf{ph}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{3}\mathbf{-}\mathbf{ph}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{L}}}$ ……. (4)

Here,$S_{3-ph}$is the three-phase power and $V_L$ is the line voltage.

Write the formula of per-unit current at primary.

localid="1655725518823" ${\mathit{I}}_{\mathbf{p}\mathbf{u}}\mathbf{=}\frac{{\mathbf{I}}_{\mathbf{p}\mathbf{h}}}{{\mathbf{I}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}}$ ……. (5)

Here,$I_{pu}$is per-unit current.

Write the formula of per-unit current at secondary.

localid="1655725532963" ${\mathit{V}}_{\mathbf{X}\mathbf{1}}\mathbf{=}{\mathit{V}}_{\mathbf{H}\mathbf{1}}\mathbf{+}{\mathit{I}}_{\mathbf{X}\mathbf{1}}{\mathit{X}}_{\mathbf{p}\mathbf{u}}$ ……. (6)

Here, $I_{X1}$is per-unit current at secondary, $X_{pu}$ is leakage reactance and $V_{H1}$is voltage at high-voltage side.

Draw the single line diagram for star-delta two winding transformer.

(a)

Determine the three-phase power.

Substitute $450MVA$for $S_{1-ph}$in equation (1).

$\begin{array}{c}{S}_{3-ph}=3\times 450MVA\\ =1350MVA\end{array}$

Determine the line voltage.

Substitute $288.7kV$for $V_{ph}$ in equation (2).

$\begin{array}{c}{V}_L=\sqrt{3}\times 288.7kV\\ =500kV\end{array}$

Determine the line voltage.

Substitute $288.7kV$for $V_{ph}$ in equation (2).

$\begin{array}{c}{V}_L=\sqrt{3}\times 288.7kV\\ =500kV\end{array}$

Determine the base current

Substitute $500kV$for$V_{base}$ and $1350kV$for$S_{base}$ in equation (3).

$\begin{array}{c}{I}_{base}=\frac{1350kV}{\sqrt{3}\times 500kV}\\ =1.55kA\end{array}$

Determine the actual current.

Substitute $500kV$for $V_L$ and $1350kV$for $S_{3-ph}$ in equation (4).

$\begin{array}{c}{I}_{ph}=\frac{1350kV}{\sqrt{3}\times 500kV}\\ =1.55kA\end{array}$

Determine the per-unit current.

Substitute $1.55kA$ for $I_{ph}$and $1.55kA$for $I_{base}$in equation (5).

$\begin{array}{c}{I}_{pu}=\frac{1.55kA}{1.55kA}\\ =1\angle 0°pu\end{array}$

Therefore, $I_{H1}=1\angle 0°pu$.

For the $Y-\Delta$ two winding transformer, the phase shift is given by$e^{j30}$.

Therefore, current at secondary is $I_{X1}=1\angle -30°pu$

Consider high-voltage in per-unit as $V_H=1\angle 0°pu$.

Determine the per-unit voltage at secondary.

Substitute $1\angle 0°pu$for $V_H$, $1\angle -30°pu$ for $I_{X1}$ and $0.1pu$for $X_{eq}$in equation (6).

$\begin{array}{c}{V}_{X1}=1\angle 0°pu+\left(1\angle -30°pu\right)\left(0.1pu\right)\\ =1.05\angle 4.7°pu\end{array}$

Draw the single line diagram for $Y-\Delta$ transformer, with leakage reactance $X_{eq}=0.1pu$and phase shift $e^{j30}$

The transformer turn ratio will be $e^{j30}:1$.

Therefore, diagram is drawn for $Y-\Delta$connection.

Draw the single line diagram for star-star two winding transformer.

(b)

For the $Y-Y$ two winding transformer, there is no phase shift.

Therefore, current at secondary is $I_{X1}=1\angle 0°pu$.

Determine the per-unit voltage at secondary.

Substitute $1\angle 0°pu$for $V_H$, $1\angle 0°pu$ for $I_{X1}$and$0.1pu$for $X_{eq}$in equation (6).

$\begin{array}{c}{V}_{X1}=1\angle 0°pu+\left(1\angle 0°pu\right)\left(0.1pu\right)\\ =1.00\angle 5.71°pu\end{array}$

Draw the single line diagram for $Y-Y$transformer, with leakage reactance $X_{eq}=0.1pu$and phase shift $0$.

Therefore, diagram is drawn for $Y-Y$connection.