Question

A single-phase step-down transformer is rated $\mathbf{13}\mathbf{}\mathit{M}\mathit{V}\mathit{A}$, $\mathbf{6}\mathbf{}\mathit{k}\mathit{V}\mathbf{/}\mathbf{11}\mathbf{.}\mathbf{5}\mathbf{}\mathit{k}\mathit{V}$. With the $\mathbf{11}\mathbf{.}\mathbf{5}\mathbf{}\mathit{k}\mathit{V}$winding short-circuited, rated current flows when the voltage applied to the primary is $\mathbf{5}\mathbf{.}\mathbf{5}\mathbf{}\mathit{k}\mathit{V}$. The power input is read as $\mathbf{100}\mathbf{}\mathit{k}\mathit{W}$. 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

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

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

Write the formula of power absorbed by real power

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

Write the formula of equivalent impedance

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

Write the formula of equivalent leakage reactance

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

Determine the equivalent winding resistance and equivalent leakage reactance.

Determine rated current at $66kV$.

Substitute $13MVA$for$S_1$and $66kV$for $E_1$in equation (1).

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

Determine the equivalent resistance.

Substitute role="math" localid="1655817798997" $100kW$for P and $196.96A$for $I_1$in equation (2)

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

Determine the equivalent impedance

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

$$\begin{gathered} Z_{eq}=\frac{E_1^{\text{'}}}{I_1} \\ =\frac{5.5\times {10}^3}{196.96} \\ =27.92\Omega \end{gathered}$$

Determine the equivalent reactance

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

$$\begin{gathered} X_{eq}=\sqrt{{Z_{eq}}^2-R_{e{q}^2}} \\ =\sqrt{27.{92}^2-2.{57}^2} \\ =27.8\Omega \end{gathered}$$

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