Question

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

Problem 3.13

A single-phase $\mathbf{50}\mathbf{kVA},\mathbf{2400}/\mathbf{240}-\mathbf{volt}$, $\mathbf{60}-\mathbf{Hz}$distribution transformer has a I-ohm equivalent leakage reactance and a$\mathbf{5000}\mathbf{ohm}$ 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 $239.76\text{\hspace{0.17em}V}$.

Step-by-step solution

determine the formulas to calculate the open circuit secondary voltage.

The base value of the primary winding is given by,

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

Leakage impedance in per unit is given by,

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

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

Calculate the value of the open circuit secondary voltage.

Let assume the base value of voltage is $V_{pu1}=1\angle 0°$

Calculate base value of primary winding by using equation 1,

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

$\begin{array}{c}{Z}_{eq.p.u}=\frac{1}{115.2}\\ Z_{eq.p.u}=8.86\times {10}^{-3}\text{\hspace{0.17em}p.u}\end{array}$

Calculate the magnetizing reactance by using equation (3)

$\begin{array}{l}{Y}_{eq.p.u}=\frac{5000}{115.2}\\ Y_{eq.p.u}=43.40\text{\hspace{0.17em}p.u}\end{array}$

The voltage drop across the primary winding is given by

$\begin{array}{l}{E}_{p.u1}=(1\angle 0°)\left(\frac{43.40}{8.86\times {10}^{-3}+43.40}\right)\\ E_{p.u1}=0.99\angle 0°\end{array}$

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

The open circuit voltage on primary side,

$\begin{array}{l}{V}_{2.p.u}=E_{1.p.u}\\ V_{2.p.u}=0.99\angle 0°\end{array}$

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

$\begin{array}{l}{V}_2=0.999\angle 0°(240\angle 0°)\\ V_2=239.76\text{\hspace{0.17em}V}\end{array}$

Hence the open circuit secondary voltage is $239.76\text{\hspace{0.17em}V}$.