Question

A single-phase $\mathbf{50}\mathbf{}\mathbf{kVA}$, $2400/240\mathrm{V}$, $\mathbf{60}\mathbf{}\mathbf{Hz}$ distribution transformer has a $\mathbf{1}\mathbf{}\mathbf{\Omega }$ equivalent leakage reactance and a $\mathbf{500}\mathbf{}\mathbf{\Omega }$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 reactances for the primary and the referred secondary.

Short answer

Therefore, the open-circuit secondary voltage is ${\mathrm{E}}_2=239.98\mathrm{V}$.

Step-by-step solution

Determine the formulas of, turns ratio primary and secondary winding induced voltages.

Write the formula of turns ratio.

$\mathrm{a}=\frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}$ ……. (1)

Write the formula of induced voltage at primary side.

${\mathrm{E}}_1={\mathrm{V}}_1\left(\frac{{\mathrm{jX}}_{\mathrm{m}}}{{\mathrm{jX}}_{\mathrm{m}}+{\mathrm{jX}}_{\mathrm{s}}}\right)$ ……. (2)

Write the formula of induced voltage at secondary side.

${\mathrm{E}}_2=\frac{{\mathrm{E}}_1}{\mathrm{a}}$ ……. (3)

Determine the open-circuit secondary voltage.

Determine turns ratio.

Substitute $2400$ for ${\mathrm{N}}_1$ and $240$ for ${\mathrm{N}}_2$in equation (1).

$\begin{array}{rcl}\mathrm{a}& =& \frac{2400}{240}\\ & =& 10\end{array}$

Draw an equivalent circuit as per the given data.

The shunt reactance is $\mathrm{j}5000\mathrm{\Omega }$and series leakage reactance is $\mathrm{j}1\mathrm{\Omega }$.

The leakage reactance at primary and secondary is same, therefore, the primary and secondary leakage reactance is,

$\begin{array}{rcl}{\mathrm{X}}_{\mathrm{eq}1}& =& {\mathrm{X}}_{\mathrm{eq}2}\\ & =& \frac{{\mathrm{X}}_{\mathrm{eq}}}{2}\\ & =& \frac{1}{2}\\ & =& 0.5\mathrm{\Omega }\end{array}$

Determine the primary winding induced voltage.

Substitute $\mathrm{j}5000\mathrm{\Omega }$for ${\mathrm{X}}_{\mathrm{m}}$, $2400\mathrm{V}$for ${\mathrm{V}}_1$and $\mathrm{j}0.5\mathrm{\Omega }$for ${\mathrm{X}}_{\mathrm{s}}$ in equation (2)

$\begin{array}{rcl}{\mathrm{E}}_1& =& 2400\times \frac{\mathrm{j}5000}{\mathrm{j}5000+\mathrm{j}0.5}\\ & =& 2399.8\mathrm{V}\end{array}$

Determinethe secondary winding induced voltage.

Substitute $2399.8\mathrm{V}$for ${\mathrm{E}}_1$and $10$for $\mathrm{a}$in equation (3)

$\begin{array}{rcl}{\mathrm{E}}_2& =& \frac{2399.8}{10}\\ & =& 239.98\end{array}$