Question

Rework Example 3 .12 for a $\mathbf{+}\mathbf{10}\mathit{\%}\mathbf{}$tap, providing an increase for the high-voltage winding.

Short answer

(a) Therefore, the per-unit equivalent circuit is shown below.

(b) Therefore, the per-unit equivalent circuit is shown below.

Step-by-step solution

Write the given data from the question

The power rating of the transformer,$S=1000\text{\hspace{0.17em}MVA}$.

The voltage rating of the transformer,$V=13.8\text{\hspace{0.17em}kV\hspace{0.17em}}\Delta \text{/345\hspace{0.17em}kV\hspace{0.17em}Y}$.

The equivalent impedance of the transformer,$Z_{eq}=j0.10\text{\hspace{0.17em}pu}$.

The base power rating of the system,$S_{base}=500\text{\hspace{0.17em}MVA}$.

The base voltage of low voltage winding,$V_{baseXLL}=13.8\text{\hspace{0.17em}kV}$.

The base voltage of high voltage winding, $V_{baseHLL}=345\text{\hspace{0.17em}kV}$.

Determine the formulas to calculate the parameter of the equivalent circuit per unit.

The relationship between the rated $V_{1rated}\&V_{2rated}$from the 3.8.1 is given as follows:

$V_{1rated}=a_t{V}_{2rated}$ …… (1)

Here,$a_t$is assumed to be either real or complex.

The relationship between the rated $V_{base1}\&V_{base2}$from the 3.8.2 is given as follows

$V_{base1}=b{V}_{base2}$ …… (2)

Here, $b$is assumed to be either real or complex.

The turn ratio of the transformer is givenas follows:

$c=\frac{a_t}{b}$ …… (3)

Theequationto calculate the new per unit impedance isas follows:

$Z_{punew}=Z_{puold}\left(\frac{S_{base}}{S}\right)$ …… (4)

The relation between the equivalent admittance and impedance is givenas follows:

$Y_{eq}=\frac{1}{Z_{punew}}$ …… (5)

Theequationto calculate the per unit value isas follows:

$perunit=\frac{Actualvalue}{Basevalue}$ …… (6)

Theequationfor the series inductance isas follows:

$L_s=c{Y}_{eq}$ …… (7)

Theequationfor the shunt capacitance isas follows:

$C_{sh}=\left(1-c\right)Y_{eq}$ …… (8)

Theequationfor the shunt inductance isas follows:

$L_{sh}=\left({\left|c\right|}^2-c\right)Y_{eq}$ …… (9)

Draw the equivalent circuit of the system for the rated tap setting

(a)

Calculate the value of $a_t$.

Substitute $13.8\text{\hspace{0.17em}kV}$for $V_{1\text{\hspace{0.17em}rated}}$, and$345\text{\hspace{0.17em}kV}$for$V_{2\text{\hspace{0.17em}rated}}$into equation (1).

$\begin{array}{l}{a}_t=\frac{13.8}{345}\\ a_t=0.04\end{array}$

Calculate the value of$b$.

Substitute$13.8\text{\hspace{0.17em}kV}$for $V_{base1}$, and $345\text{\hspace{0.17em}kV}$for$V_{base2}$into equation (2).

$\begin{array}{l}b=\frac{13.8}{345}\\ b=0.04\end{array}$

Calculate the turns ratio of the transformer.

Substitute $0.04$for $a_t$and $0.04$for $b$into equation (3).

$\begin{array}{l}c=\frac{0.04}{0.04}\\ c=1\end{array}$

Calculate the value of new per unit impedance.

Substitute$j0.10\text{\hspace{0.17em}pu}$for $Z_{puold}$, $500\text{\hspace{0.17em}MVA}$for $S_{base}$,and $1000\text{\hspace{0.17em}MVA}$for$S$into equation (4).

$\begin{array}{l}{Z}_{punew}=j0.10\left(\frac{500}{1000}\right)\\ Z_{punew}=j0.10\times 0.5\\ Z_{punew}=j0.05\text{\hspace{0.17em}pu}\end{array}$

The per unit equivalent circuit is shown below:

Determine the equivalent circuit for tap setting.

(b)

The tapping is provided on the high voltage side. So, under this condition,the high voltage side voltage is givenas follows:

$\begin{array}{l}{\left(V_{2rated}\right)}_{+10\%}=345\left(\frac{10}{100}\right)+345\\ {\left(V_{2rated}\right)}_{+10\%}=379.5\text{\hspace{0.17em}kV}\end{array}$

The new rating of the transformer after applying the tap on the high voltage side is .

$13.8\text{\hspace{0.17em}kV/379.5\hspace{0.17em}kV}$

Calculate the value of $a_t$.

Substitute $13.8\text{\hspace{0.17em}kV}$for $V_{1\text{\hspace{0.17em}rated}}$and $379.5\text{\hspace{0.17em}kV}$for $V_{2\text{\hspace{0.17em}rated}}$into equation (1).

$\begin{array}{l}{a}_t=\frac{13.8}{379.5}\\ a_t=0.036\end{array}$

Calculate the value of $b$.

Substitute $13.8\text{\hspace{0.17em}kV}$for $V_{base1}$and $345\text{\hspace{0.17em}kV}$for $V_{base2}$into equation (2).

$\begin{array}{l}b=\frac{13.8}{345}\\ b=0.04\end{array}$

Calculate the turns ratio of the transformer.

Substitute$0.036$for$a_t$and $0.04$for $b$into equation (3).

$\begin{array}{l}c=\frac{0.036}{0.04}\\ c=0.9\end{array}$

Calculate the equivalent admittance.

Substitute$j0.05\text{\hspace{0.17em}pu}$for $Z_{punew}$into equation (5).

$Y_{eq}=\frac{1}{j0.05}$

Calculate the series inductance.

Substitute$0.9$for$c$and $\frac{1}{j0.05}$for$Y_{eq}$into equation (7).

$\begin{array}{c}{L}_s=0.9\times \frac{1}{j0.05}\\ L_s=-j18\text{\hspace{0.17em}pu}\end{array}$

Calculate the shunt capacitance.

Substitute $0.9$for $c$and $\frac{1}{j0.05}$for into equation (8).

$\begin{array}{c}{C}_{sh}=\left(1-0.9\right)\times \frac{1}{j0.05}\\ C_{sh}=-j2\text{\hspace{0.17em}pu}\end{array}$

Calculate the shunt inductance.

Substitute$0.9$for $c$and $\frac{1}{j0.05}$for $Y_{eq}$into equation (8).

$\begin{array}{c}{L}_{sh}=\left({\left|0.9\right|}^2-0.9\right)\times \frac{1}{j0.05}\\ L_{sh}=j1.8\text{\hspace{0.17em}pu}\end{array}$

Calculate the per unit voltage ofthehigh voltage sideusing equation (5).

$\begin{array}{l}{V}_{H1}=\frac{379.5}{345}\\ V_{H1}=1.1\text{\hspace{0.17em}pu}\end{array}$

Calculate the per unit voltage ofthelow voltage sideusing equation (5).

$\begin{array}{l}{V}_{X1}=\frac{13.8}{13.8}\\ V_{X1}=1\text{\hspace{0.17em}pu}\end{array}$

The per unit equivalent circuit is shown below