Question

A single-phase step-down transformer is rated $\mathbf{13}\mathbf{MVA}\mathbf{,}\mathbf{}\raisebox{1ex}{$\mathbf{66}\mathbf{kV}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{11}\mathbf{.}\mathbf{5}\mathbf{kV}$}\right.$. With the $\mathbf{11}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{kV}$winding short-circuited, rated current flows when the voltage applied to the primary is $\mathbf{5}\mathbf{.}\mathbf{5}\mathbf{kV}$. The power input is read as 100 kW. Determine${\mathit{R}}_{\mathbf{e}\mathbf{q}\mathbf{1}}$and${\mathit{X}}_{\mathbf{e}\mathbf{q}\mathbf{1}}$ in ohms referred to the high-voltage winding.

Short answer

The equivalent winding resistance is $2.57\Omega$ and equivalent leakage reactance is$27.8\Omega$ .

Step-by-step solution

Determine the formulas of apparent power, real power and equivalent impedance

Write the formula of apparent power

${\mathbf{S}}_{\mathbf{1}}\mathbf{=}{\mathbf{E}}_{\mathbf{1}}{\mathbf{I}}_{\mathbf{1}}$ ……. (1)

Write the formula of power absorbed by real power

$\mathbf{P}\mathbf{=}{\mathbf{I}}_{\mathbf{1}}^{\mathbf{2}}{\mathbf{R}}_{\mathbf{eq}}$ ……. (2)

Write the formula of equivalent impedance

${\mathbf{Z}}_{\mathbf{\infty }}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{1}}^{\mathbf{\text{'}}}}{{\mathbf{I}}_{\mathbf{1}}}$ ……. (3)

Write the formula of equivalent leakage reactance

${\mathbf{X}}_{\mathbf{eq}}\mathbf{=}\sqrt{{\left(Z_{\mathrm{eq}}\right)}^{\mathbf{2}}\mathbf{-}{\left(R_{\mathrm{eq}}\right)}^{\mathbf{2}}}$ ……. (4)

Determine the equivalent winding resistance and equivalent leakage reactance.

Determine rated current at 66 kV.

Substitute 13MVA for ${\mathrm{S}}_1$and 66 kV for ${\mathrm{E}}_1$ in equation (1).

$$\begin{gathered} S_1=E_1{I}_1 \\ 13\times {10}^6=\left(66\times {10}^3\right)I_1 \\ {I}_1=196.96\mathrm{A} \end{gathered}$$

Determine the equivalent resistance.

Substitute 100 kW for Pand 196.96A for $I_1$ in equation (2)

$$\begin{gathered} P={\left(I_1\right)}^2{R}_{eq} \\ 100\times {10}^3={\left(196.96\right)}^2{R}_{eq} \\ {R}_{eq}=\frac{100\times {10}^3}{{\left(196.96\right)}^2} \\ {R}_{eq}=2.57\Omega \end{gathered}$$

Determine the equivalent impedance

Substitute 5.5 kV for $E_1^{\text{'}}$ and 196.96A for $I_1$ in equation (3)

$\begin{array}{rcl}{\text{Z}}_{eq}& =& \frac{E_1^{\text{'}}}{I_1}\\ & =& \frac{5.5\times {10}^3}{196.96}\\ & =& 27.92\Omega \end{array}$

Determine the equivalent reactance

Substitute $27.92\Omega$for $Z_{eq}$and $2.57\Omega$for $R_{eq}$in equation (4)

$\begin{array}{rcl}{X}_{eq}& =& \sqrt{Z_{eq}^2-R_{eq}^2}\\ & =& \sqrt{27.{92}^2-2.{57}^2}\\ & =& 27.8\Omega \end{array}$

Therefore, the equivalent winding resistance is $2.57\Omega$ and equivalent leakage reactance is$27.8\Omega$ .