Question

Using the transformer ratings as base quantities, work Problem 3.13 in per unit.

Problem 3.13

A single-phase$\mathbf{50}\mathbf{}\mathbf{kVA}$,$\mathbf{2400}\mathbf{/}\mathbf{240}\mathbf{-}\mathbf{volt}$,$\mathbf{60}\mathbf{-}\mathbf{Hz}$distribution transformer has a I-ohm equivalent leakage reactance and a$\mathbf{5000}\mathbf{}\mathbf{ohm}$ magnetizing reactance referred to the high-voltage side If rated voltage is applied to the high-voltage winding, calculate the open-circuit secondary voltage Neglect${\mathbf{I}}^{\mathbf{2}}\mathbf{R}$ and${{\mathbf{G}}_{\mathbf{c}}}^{\mathbf{2}}\mathbf{V}$ losses. Assume equal series leakage reactance’s for the primary and the referred secondary.

Short answer

Therefore, the open circuit secondary voltage is $239.76\mathrm{V}$.

Step-by-step solution

determine the formulas to calculate the open circuit secondary voltage.

The base value of the primary winding is given by,

${\mathrm{Z}}_{\mathrm{base}1}=\frac{{{\mathrm{V}}^2}_{\mathrm{base}1}}{{\mathrm{S}}_{\mathrm{base}}}$ ……(1)

Leakage impedance in per unit is given by,

${\mathrm{Z}}_{\mathrm{eq}.\mathrm{p}.\mathrm{u}}=\frac{{\mathrm{Z}}_{\mathrm{eq}1}}{{\mathrm{Z}}_{\mathrm{base}1}}$ ……(2)

The magnetizing admittance in per unit is given by,

${\mathrm{Y}}_{\mathrm{eq}.\mathrm{p}.\mathrm{u}}=\frac{{\mathrm{Y}}_{\mathrm{O}1}}{{\mathrm{Z}}_{\mathrm{base}1}}$ ……(3)

The open circuit secondary voltage is given by

${\mathrm{V}}_2={\mathrm{V}}_{2.\mathrm{p}.\mathrm{u}}{\mathrm{V}}_{\mathrm{base}}$ ……(4)

Calculate the value of the open circuit secondary voltage.

$239.76\mathrm{V}$Let assume the base value of voltage is ${\mathrm{V}}_{\mathrm{pu}1}=1\angle 0^0$

Calculate base value of primary winding by using equation 1,

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{base}1}=\frac{{2400}^2}{50\times {10}^3} \\ {\mathrm{Z}}_{\mathrm{base}1}=\frac{5760000}{50000} \\ {\mathrm{Z}}_{\mathrm{base}1}=115.2\mathrm{\Omega } \end{gathered}$$

Calculate the leakage impedance by using the equation (2),

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{eq}.\mathrm{p}.\mathrm{u}}=\frac{1}{115.2} \\ {\mathrm{Z}}_{\mathrm{eq}.\mathrm{p}.\mathrm{u}}=8.86\times {10}^{-3}\mathrm{p}.\mathrm{u}. \end{gathered}$$

Calculate the magnetizing reactance by using equation (3)

$$\begin{gathered} {\mathrm{Y}}_{\mathrm{eq}.\mathrm{p}.\mathrm{u}}=\frac{5000}{115.2} \\ {\mathrm{Y}}_{\mathrm{eq}.\mathrm{p}.\mathrm{u}}=43.40\mathrm{p}.\mathrm{u}. \end{gathered}$$

The voltage drop across the primary winding is given by

$$\begin{gathered} {\mathrm{E}}_{\mathrm{p}.\mathrm{u}1}=\left(1\angle 0^0\right)\left(\frac{43.40}{8.86\times {10}^{-3}+43.40}\right) \\ {\mathrm{E}}_{\mathrm{p}.\mathrm{u}1}=0.99\angle 0^0 \end{gathered}$$

In per unit system quantities remains the same both the sides of the transformer.

The open circuit voltage on primary side,

localid="1655275647495" $$\begin{gathered} {\mathrm{V}}_{2.\mathrm{p}.\mathrm{u}}={\mathrm{E}}_{1.\mathrm{p}.\mathrm{u}} \\ {\mathrm{V}}_{2.\mathrm{p}.\mathrm{u}}=0.99\angle 0^0 \end{gathered}$$

Calculate open circuit secondary voltage by using the equation (4),

$$\begin{gathered} {\mathrm{V}}_2=0.999\angle 0^0\left(240\angle 0^0\right) \\ {\mathrm{V}}_2=239.76\mathrm{V} \end{gathered}$$

Hence the open circuit secondary voltage is localid="1655276200268" $239.76\mathrm{V}$.