Question

Three single-phase two-winding transformers, each rated $\mathbf{3}\mathbf{kVA}\mathbf{,}\mathbf{}\mathbf{220}\mathbf{/}\mathbf{110}\mathbf{}\mathbf{volt}\mathbf{,}\mathbf{}\mathbf{60}\mathbf{Hz}$, with

a 0.10 per unit leakage reactance, are connected as a three-phase extended$\mathbf{∆}$autotransformer bank, as shown in Figure 3.36 The low-voltage A winding has a 110 volt rating. (a) Draw the positive- sequence phasor diagram and show that the high-voltage winding has a 479.5 volt rating. (b) A three-phase load connected to the low-voltage terminals absorbs 6kW at 110 volts

and at 0.8 power factor lagging. Draw the per-unit impedance diagram and calculate the voltage and current at the high-voltage terminals. Assume positive-sequence operation.

Short answer

(a) Therefore, the positive diagram referred to high voltage side is

(b) Therefore, the per unit diagram

The current and voltage at high voltage side are$13.1\angle -36.{86}^{\circ }\mathrm{pu}\mathrm{and}1.039\angle 2.{825}^{\circ }\mathrm{pu}$.

Step-by-step solution

Write the given data by the question.

The power rating of three phase transformer, S = 3kVA

Voltage rating of three phase transformer, V = 220 / 110 V

The frequency of three phase transformer, f = 60 Hz

Leakage reactance,${\mathrm{X}}_{\mathrm{pu}}=0.10\mathrm{pu}$

Power absorbed by the load,${\mathrm{S}}_{\mathrm{L}}=6\mathrm{kW}$

Power factor of load, pf = 0.8 lagging

Load connected to low voltage winding.

Determine the formula to calculate the diagram value and voltage and current to hight voltage terminals.

The expression for current flowing through high voltage is given by,

$\mathit{I}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}}{{\mathbf{V}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}}$ ….. (1)

The expression to calculate the base value of impedance is given by,

${\mathit{Z}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}^{\mathbf{2}}}{{\mathbf{S}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}}$ ….. (2)

The expression to calculate the per unit value of the quantity is given by,

$\mathbf{X}\mathbf{=}{\mathbf{Z}}_{\mathbf{base}}\mathbf{\times }{\mathbf{X}}_{\mathbf{pu}}$ ….. (3)

The expression for kVA rating of autotransformer is given by,

${\mathbf{S}}_{\mathbf{Auto}}\mathbf{=}{\mathbf{E}}_{\mathbf{H}}{\mathbf{I}}_{\mathbf{H}}$ ….. (4)

Here,${\mathbf{S}}_{\mathbf{Auto}}$is the kVA rating of auto transformer,${\mathbf{E}}_{\mathbf{H}}$high voltage of autotransformer

and${\mathbf{I}}_{\mathbf{H}}$is the current on high voltage winding.

The expression for base current for high voltage winding is given by,

${\mathbf{I}}_{\mathbf{baseH}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{base}}}{{\left(V_{\mathrm{baseH}}\right)}_{\mathbf{LL}}}$ ….. (5)

Here,${\mathbf{I}}_{\mathbf{baseH}}$is the high voltage winding current, and${\left(V_{\mathrm{baseH}}\right)}_{\mathbf{LL}}$is the base line-to-line voltage for high voltage winding.

The expression for base current for low voltage winding is given by,

${\mathbf{I}}_{\mathbf{baseX}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{auto}}}{{\left(V_{\mathrm{baseX}}\right)}_{\mathbf{LL}}}$ ….. (6)

Here,${\mathbf{I}}_{\mathbf{baseX}}$is the low voltage winding current, and${\left(V_{\mathrm{baseX}}\right)}_{\mathbf{LL}}$is the base line-to-line voltage for low voltage winding.

The expression for actual value of the low voltage winding current is given by,

${\mathbf{I}}_{\mathbf{x}}\mathbf{=}\frac{{{\mathbf{S}}_{\mathbf{L}\mathbf{\angle }\mathbf{COS}}}^{\mathbf{-}\mathbf{1}}\left(\mathrm{pf}\right)}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{baseX}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{pf}}$ ….. (7)

The expression to calculate voltage at high voltage side is given by,

${\mathbf{V}}_{\mathbf{H}}\mathbf{=}{\mathbf{V}}_{\mathbf{X}}\mathbf{+}{\mathbf{jX}}_{\mathbf{eq}}{\mathbf{I}}_{\mathbf{x}}$ ….. (8)

Draw the positive sequence phasor diagram.

The positive sequence diagram referred to high voltage side is shown below

Draw the impedance diagram and calculate the voltage and current at high voltage terminal.   

Calculate the current flowing through high voltage.

Substitute$3kVA$forlocalid="1655900316292" ${\mathrm{S}}_{\mathrm{base},}220\mathrm{V}$,for$V_{base}$into equation (1).

$$\begin{gathered} I_H=\frac{3200}{220} \\ {I}_H=13.6\mathrm{A} \end{gathered}$$

Calculate the base impedance.

Substitute$3\mathrm{kVA}$for$S_{base}$,$220\mathrm{V}$for${\mathrm{V}}_{\mathrm{base}}$into equation (2).

$$\begin{gathered} Z_{base}=\frac{{220}^2}{3000} \\ {Z}_{base}=\frac{48400}{3000} \\ {Z}_{base}=16.13\mathrm{\Omega } \end{gathered}$$

Calculate the leakage reactance

Substitute$16.13\mathrm{\Omega }$forlocalid="1655900369545" ${\mathrm{Z}}_{\mathrm{base}}$andlocalid="1655900377025" $0.1\mathrm{pu}$forlocalid="1655900384188" ${\mathrm{X}}_{\mathrm{pu}}$into equation (3).

localid="1655900394435" $$\begin{gathered} \mathrm{X}=16.13\times 0.10 \\ \mathrm{X}=1.613\mathrm{\Omega } \end{gathered}$$

While converting the two-winding transformer into autotransformer, the primary voltage will remain the same and secondary winding voltage will be the sum of the primary and secondary voltage of two-winding transformer.

localid="1655900425559" $$\begin{gathered} E_x=E_1 \\ {E}_x=110\mathrm{V} \end{gathered}$$

Then, the high voltage of autotransformer.

localid="1655900439412" $$\begin{gathered} E_H=110+220 \\ {E}_H=330\mathrm{V} \end{gathered}$$

Calculate the power rating of autotransformer.

Substitutelocalid="1655900446989" $330\mathrm{V}$forlocalid="1655900453849" ${\mathrm{E}}_{\mathrm{H}}$and 13.6 A forlocalid="1655900460317" $I_{\mathrm{H}}$into equation (4).

localid="1655900465499" $$\begin{gathered} S_{Auto}=330\times 13.6 \\ {S}_{Auto}=4.488\mathrm{kV} \end{gathered}$$

Calculate the base current for high voltage winding.

Substitutelocalid="1655900477999" $4.488\mathrm{kV}$forlocalid="1655900471493" $S_{auto}$andlocalid="1655900484004" $330\mathrm{V}$forlocalid="1655900488990" ${\left({\mathrm{V}}_{\mathrm{baseH}}\right)}_{\mathrm{LL}}$into equation (5).

localid="1655900495763" $$\begin{gathered} I_{baseH}=\frac{4.488}{330} \\ {I}_{baseH}=13.6\mathrm{A} \end{gathered}$$

Calculate the base current for low voltage winding.

Substitutelocalid="1655900500668" $4.488\mathrm{kVA}$forlocalid="1655900505918" $S_{auto}$and 110 V forlocalid="1655900525508" ${\left(V_{baseX}\right)}_{LL}$into equation (6).

localid="1655900532004" $$\begin{gathered} I_{baseH}=\frac{4.488}{110} \\ {I}_{baseH}=40.8\mathrm{A} \end{gathered}$$

Calculate the base impedance of autotransformer.

Substitutelocalid="1655900536969" $4.488\mathrm{kVA}$forlocalid="1655900541792" $S_{Auto}$,330 V for${\left(V_{baseH}\right)}_{LL}$into equation (2).

localid="1655900572705" $$\begin{gathered} {\left(Z_{base}\right)}_{auto}=\frac{{330}^2}{4488} \\ {\left(Z_{base}\right)}_{auto}=24.26\mathrm{\Omega } \end{gathered}$$

Calculate per unit leakage reactance.

Substitutelocalid="1655900593384" $24.26\mathrm{\Omega }$forlocalid="1655900586482" ${\left(Z_{base}\right)}_{auto}$andlocalid="1655900660551" $1.613\mathrm{\Omega }$forlocalid="1655900667867" $X$into equation (3).

localid="1655900602308" $$\begin{gathered} {\left(X_{pu}\right)}_{auto}=\frac{1.613}{24.26} \\ {\left(X_{pu}\right)}_{auto}=0.066\mathrm{pu} \end{gathered}$$

The per unit equivalent diagram is shown below,

Calculate actual value of the load current.

Substitute$6000\mathrm{W}$for$S_L$, 0.8 for pf and 110 V for${\mathrm{V}}_{\mathrm{baseX}}$into equation (7).$$\begin{gathered} I_x=\frac{6000}{\sqrt{3}\times 110\times 0.8}\angle {\cos }^{-1}\left(0.8\right) \\ {I}_x=39.36\angle -36.{38}^{\circ }\mathrm{A} \end{gathered}$$

Calculate per unit value of the current.

$$\begin{gathered} I_{xpu}=\frac{39.39\angle -36.{86}^{\circ }}{40.8}\angle {\cos }^{-1}\left(0.8\right) \\ {I}_{xpu}=0.966\angle -36.{38}^{\circ }\mathrm{pu} \end{gathered}$$

The current at voltage side can be calculated as

$$\begin{gathered} I_{Hpu}=0.966\angle -36.{86}^{\circ }\times 13.6 \\ {I}_{Hpu}=13.1\angle -36.{86}^{\circ }\mathrm{pu} \end{gathered}$$

Calculate the voltage at high voltage side.

Substitute 1 pu for ${\mathrm{V}}_{\mathrm{X}}$, 0.66$\Omega$for ${\left(X_{pu}\right)}_{auto}$and $0.966\angle -36.{86}^{\circ }$for $I_{xpu}$into

equation (8).

$$\begin{gathered} V_H=1.0+j\left(0.066\right)\left(0.966\angle -36.{86}^{\circ }\right) \\ {V}_H=1.039\angle 2.{825}^{\circ }\mathrm{pu} \end{gathered}$$

Hence, the current and voltage at high voltage side are$13.1\angle -36.{86}^{\circ }\mathrm{pu}$and $1.039\angle 2.{825}^{\circ }\mathrm{pu}$.