Question

A single-phase, $\mathbf{50}\mathbf{kVA}\mathbf{,}\mathbf{2400}\mathbf{/}\mathbf{240}\mathbf{}\mathbf{V}\mathbf{,}\mathbf{}\mathbf{60}\mathbf{‐}\mathbf{Hz}$ distribution transformer has the following parameters:

Resistance of the 2400 - V winding:${\mathbf{R}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{75}\mathbf{}\mathbf{\Omega }$ ,

Resistance of the 240 - V winding:${\mathbf{R}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0075}\mathbf{}\mathbf{\Omega }$

Leakage reactance of the 2400 - V winding:${\mathbf{X}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\Omega }$ ,

Leakage reactance of the 240 - V winding:${\mathbf{X}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{}\mathbf{\Omega }$

Exciting admittance on the 240 - V side$=\mathbf{0}\mathbf{.}\mathbf{003}\mathbf{}\mathbf{‐}\mathbf{}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{}\mathbf{S}$

(a) Draw the equivalent circuit referred to the high-voltage side of the transformer.

(b) Draw the equivalent circuit referred to the low-voltage side of the transformer. Show the numerical values of impedances on the equivalent circuits.

Short answer

(a) Therefore, the equivalent circuit of the transformer is referred to asthe high voltage side.

(b) Therefore, the equivalent circuit is referred to as the low voltage side of the transformer.

Step-by-step solution

Write the data given in the question:

The power rating of the transformer is 50kVA .

The high voltage winding,$V_1=2400\mathrm{V}$.

The low voltage winding,$V_2=240\mathrm{V}$.

The resistance ofthehigh voltage winding,$R_1=0.75\mathrm{\Omega }$ .

The resistance ofthehigh low winding,$R_2=0.0075\mathrm{\Omega }$.

The leakage reactance ofthehigh voltage winding,$X_1=1\mathrm{\Omega }$ .

The leakage reactance ofthelow voltage winding,$X_2=0.01\mathrm{\Omega }$.

The exciting admittance on the low voltage side$=0.003-j0.02\mathrm{S}$.

Determine the formulas to calculate the equivalent impedance of the circuit to respective sides.

The expression for impedance of low voltage winding isas follows:

${\mathit{Z}}_{\mathbf{2}}\mathbf{=}{\mathbf{R}}_{\mathbf{2}}\mathbf{+}{\mathbf{jX}}_{\mathbf{2}}$ ….. (1)

The expressionofthe impedance of low voltage winding to high voltage winding isas follows:

${\mathit{Z}}_{\mathbf{2}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{Z}}_{\mathbf{2}}{\mathbf{\left(}\frac{{\mathbf{N}}_{\mathbf{1}}}{{\mathbf{N}}_{\mathbf{2}}}\mathbf{\right)}}^{\mathbf{2}}$ ….. (2)

Here,${\mathit{Z}}_{\mathbf{2}}^{\mathbf{\text{'}}}$is the low voltage winding impedance referred to high voltage side,${\mathbf{N}}_{\mathbf{1}}$is the number of turnsintheprimary side,and${\mathbf{N}}_{\mathbf{2}}$is the number of turns in the secondary winding.

The expression of the impedance of high voltage winding is givenbelow:

localid="1655272381641" ${\mathit{Z}}_{\mathbf{1}}\mathbf{=}{\mathbf{R}}_{\mathbf{1}}\mathbf{+}{\mathbf{jX}}_{\mathbf{1}}$ ….. (3)

The expression for exciting admittance is givenbelow:

${\mathbf{Y}}_{\mathbf{o}}\mathbf{=}{\mathbf{G}}_{\mathbf{c}}\mathbf{+}{\mathbf{jB}}_{\mathbf{m}}$ ….. (4)

Here, ${\mathbf{Y}}_{\mathbf{o}}$is the low voltage side admittance,${\mathbf{G}}_{\mathbf{c}}$is the exciting conductance,and ${\mathbf{B}}_{\mathbf{m}}$is the exciting susceptance.

The expression for exciting susceptance to refer high voltage side is givenbelow.

${\mathbf{Y}}_{\mathbf{o}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{Y}}_{\mathbf{o}}{\left(\frac{N_2}{N_1}\right)}^{\mathbf{2}}$ ….. (5)

Here,${\mathbf{Y}}_{\mathbf{o}}^{\mathbf{\text{'}}}$is the low voltage winding impedance referredtoas thehigh voltage side.

The expression for impedance of high voltage winding is givenbelow:

localid="1655272397796" ${\mathit{Z}}_{\mathbf{1}}\mathbf{=}{\mathbf{R}}_{\mathbf{1}}\mathbf{+}{\mathbf{jX}}_{\mathbf{1}}$ ….. (6)

The expression for referringtothe impedance of high voltage winding to low voltage windingisgivenbelow:

${\mathbf{Z}}_{\mathbf{1}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{Z}}_{\mathbf{1}}{\left(\frac{N_2}{N_1}\right)}^{\mathbf{2}}$ ….. (7)

Here, ${\mathbf{Z}}_{\mathbf{1}}^{\mathbf{\text{'}}}$is the high voltage impedance refer to low voltage side.

Determine the value of the equivalent impedance referred to asthe high voltage side and draw the equivalent diagram of the transformer.

(a)

Calculate the low voltage side impedance using equation (1).

$Z_2=\left(0.0075+j0.01\right)\mathrm{\Omega }$

Calculate the low voltage impedanceasthehigh voltage side using equation (2).

$$\begin{gathered} Z_2=\left(0.0075+\mathrm{j}0.01\right){\left(\frac{2400}{240}\right)}^2 \\ {Z}_2=\left(0.0075+j0.01\right)\times 100 \\ {Z}_2=\left(0.75+j1\right)\mathrm{\Omega } \end{gathered}$$

Calculate the high voltage side impedance using equation (1).

$Z_1=\left(0.75+j1\right)\mathrm{\Omega }$

Calculate the exciting impedance using equation (4).

$Y_o=\left(0.003+j0.02\right)\mathrm{S}$

Calculate the exciting impedanceashigh voltage side using equation (5).

$$\begin{gathered} Y_o^{\text{'}}=\left(0.003+j0.02\right){\left(\frac{240}{2400}\right)}^2 \\ {Y}_o^{\text{'}}=\left(0.003+j0.02\right)\times \frac{1}{100} \\ {Y}_o^{\text{'}}=\left(0.003+j0.02\right)\mathrm{mS} \end{gathered}$$

Therefore, the equivalent circuit of the transformer, referred to high voltage side, is shown below.

Determine the equivalent impedance referred to the low voltage side and draw the equivalent circuit of the transformer.

(b)

Calculate the high voltage side impedance using equation (6).

$Z_1=\left(0.75+j1\right)\mathrm{\Omega }$

Calculate the impedance ofthehigh voltage side,referred toas thelow voltage side,using equation (7).

$$\begin{gathered} Z_1^{\text{'}}=\left(0.75+j1\right)\times {\left(\frac{240}{2400}\right)}^2 \\ {Z}_1^{\text{'}}=\left(0.0075+j0.01\right)\mathrm{\Omega } \end{gathered}$$

Write the expression for the exciting admittance on the low voltage side asfollows:

$Y_o=G_c+j{B}_m$

Therefore, the equivalent circuit,referred to as the low voltage side of the transformer, is shown below.