Question

For the transformer in Problem 3.10, the open-circuit test with $\mathbf{11}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{kV}$applied results in a power input of $\mathbf{65}\mathbf{}\mathbf{kW}$ and a current of $\mathbf{30}\mathbf{}\mathbf{A}$ . Compute the values for${\mathbf{G}}_{\mathbf{c}}$and ${\mathbf{B}}_{\mathbf{m}}$in siemens referred to the high-voltage winding. Compute the efficiency of the transformer for a load of $\mathbf{10}\mathbf{}\mathbf{MW}$ at 0.8 p.f. lagging at rated voltage.

Short answer

Therefore, the conductance${\mathrm{G}}_{\mathrm{c}}$ is $14.9\times {10}^{-6}\mathrm{S}$ and the susceptance ${\mathrm{B}}_{\mathrm{m}}$ is $77.79\times {10}^{-6}\mathrm{S}$.

The efficiency of the transformer for a load of $10\mathrm{MW}$ and 0.8 power factor is 98.38 %.

Step-by-step solution

Determine the formulas of turns ratio, admittance, conductance, susceptance and efficiency.

Write the formula of turns ratio.

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

Write the formula of admittance reffered to primary

$\mathrm{Y}=\frac{{\mathrm{I}}_2}{{\mathrm{V}}_2}\frac{1}{{\mathrm{a}}^2}$ ……. (2)

Write the formula of conductance.

${\mathrm{G}}_{\mathrm{c}}=\frac{{\mathrm{P}}_{\mathrm{c}}}{{\left({\mathrm{V}}_2\mathrm{a}\right)}^2}$ ……. (3)

Write the formula of susceptance.

${\mathrm{B}}_{\mathrm{m}}=\sqrt{{\mathrm{Y}}^2-{{\mathrm{G}}_{\mathrm{C}}}^2}$ ……. (4)

Write the formula of efficiency in percentage

$\mathrm{\eta }=\frac{{\mathrm{P}}_{\mathrm{out}}}{{\mathrm{P}}_{\mathrm{out}}+{\mathrm{P}}_{\mathrm{losses}}}$ ……. (5)

Determine the conductance and susceptance of middle branch in a transformer.

Determine turns ratio.

Substitute $66$for ${\mathrm{N}}_1$and $11.5$for ${\mathrm{N}}_2$in equation (1).

$\begin{array}{rcl}\mathrm{a}& =& \frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}\\ & =& \frac{66}{11.5}\\ & =& 5.74\end{array}$

Determine the admittance reffered to primary winding.

Substitute $30\mathrm{A}$for ${\mathrm{I}}_2$, $11.5\times {10}^3\mathrm{V}$for ${\mathrm{V}}_2$ and $5.74$for $\mathrm{a}$in equation (2)

role="math" localid="1655205614655" $\begin{array}{rcl}\mathrm{Y}& =& \frac{30}{11.5\times {10}^3}\times \frac{1}{{\left(5.74\right)}^2}\\ & =& 79.2\times {10}^{-6}\mathrm{S}\end{array}$

Determine the conductance.

Substitute$65\mathrm{kW}$for ${\mathrm{P}}_{\mathrm{c}}$, $11.5\times {10}^3\mathrm{V}$ for ${\mathrm{V}}_2$ and $5.74$ for $\mathrm{a}$ in equation (3)

$\begin{array}{rcl}{G}_c& =& \frac{65\times {10}^3}{{\left(11.5\times {10}^3\right)}^2}\times \frac{1}{5.{74}^2}\\ & =& 14.9\times {10}^{-6}S\end{array}$

Determinethe susceptance.

Substitute $14.9\times {10}^{-6}\mathrm{S}$ for ${\mathrm{G}}_{\mathrm{c}}$and $79.2\times {10}^{-6}\mathrm{S}$for $\mathrm{Y}$in equation (4).

$\begin{array}{rcl}{\mathrm{B}}_{\mathrm{m}}& =& \sqrt{{\left(79.2\times {10}^6\right)}^2-{\left(14.9\times {10}^{-6}\right)}^2}\\ & =& 77.79\times {10}^{-6}\mathrm{S}\end{array}$

Determine the efficiency of transformer.

Deterine the efficiency of transformer.

Substitute $10\mathrm{MW}$for ${\mathrm{P}}_{\mathrm{out}}$and $\left(100+65\right)\mathrm{kW}$for ${\mathrm{P}}_{\mathrm{loss}}$in equation (5)

$\begin{array}{rcl}\eta & =& \frac{10\times {10}^6}{10\times {10}^6+\left(100+65\right)\times {10}^3}\times 100\\ & =& 98.38\%\end{array}$