Question

A $23/230kV$step-up transformer feeds a three-phase transmission line, which in turn supplies a$150MVA,0.8$ lagging power factor load through a step-down$230/23kV$ transformer. The impedance of the line and transformers at $230kV$is $18+j60\Omega$. Determine the tap setting for each transformer to maintain the voltage at the load at .

Short answer

The change in the percentage of the primary voltage of the load side transformer is$10.25\%$ .

Step-by-step solution

Step 1:  Write the given values by the questions.

The load MVA rating.$S_L=150\text{\hspace{0.17em}MVA}$

The load voltage$V_L=23\text{\hspace{0.17em}kV}$.

The power factor of the load$\cos \varphi =0.8\text{\hspace{0.17em}lagging}$.

The impedance$Z=18+j60\Omega$.

Step 2: Determine the formulas to calculate tap setting to maintain the voltage at the load.

The expression to calculate the load current is given by,

$I_L=\frac{S_L}{\sqrt{3}{V}_L\angle co{s}^{-1}\varphi }$ …… (1)

The expression to calculate the line current is given by,

$I_{line}=\frac{V_L}{V_H}\times I_L$ …… (2)

Here,$V_H$is the high voltage.

The expression for the voltage drops due to line and transformer impedance is given by,

$\Delta V=I_{line}\left(18+j60\right)$ …… (3)

The expression to calculate the primary voltage of load side transformer is given by,

$V_{p2}=V_{s1}-\Delta V$ …… (4)

Here, localid="1655288477762" $V_{s1}$is secondary side voltage of the supply side transformer.

The expression to calculate the change in percentage of primary voltage or the tap of the tap setting of the load side transformer is given by,

$\Delta =\frac{V_{s1}-V_{p2}}{V_{s1}}\times 100$ …… (5)

Here, $V_{p2}$is the primary voltage of load side transformer.

Step 3:  Determine the tap setting for each transformer to maintain the voltage at the load at23 kV .

$23\text{\hspace{0.17em}kV}$The schematic of the circuit is shown below.

Calculate the value of load current.

Substitute $150\text{\hspace{0.17em}MVA}$for$150\text{\hspace{0.17em}MVA}$and$23\text{\hspace{0.17em}kV}$for$V_L$into equation (1),

$\begin{array}{l}{I}_L=\frac{150\times {10}^6}{\sqrt{3}\times 23\times {10}^3\angle {\cos }^{-1}0.8}\\ I_L=3.76533\times {10}^3\angle -36.87\\ I_L=3765.33\times {10}^3\angle -36.87\text{\hspace{0.17em}A}\end{array}$

Calculate the line current.

Substitute$23\text{\hspace{0.17em}kV}$ for$V_L$ ,$230\text{\hspace{0.17em}kV}$for$V_H$ and$3765.33\times {10}^3\angle -36.87\text{\hspace{0.17em}A.}$ for $I_L$into equation (2).

$\begin{array}{l}{I}_{line}=\frac{23}{230}\times 3765.33\times {10}^3\angle -36.87\text{\hspace{0.17em}A}\\ I_{line}=376.533\times {10}^3\angle -36.87\text{\hspace{0.17em}A}\end{array}$

Calculate the voltage drop.

Substitute$376.533\times {10}^3\angle -36.87\text{\hspace{0.17em}A}$for$I_{line}$and$18+j60\Omega$ for $Z$into equation (3).

$\begin{array}{l}\Delta V=376.533\times {10}^3\angle -36.87\left(18+j60\right)\\ \Delta V=\left(376.533\times {10}^3\angle -36.87\right)\left(62.64\angle 73.3°\right)\\ \Delta V=23586.02\angle 36.43°\text{\hspace{0.17em}V}\end{array}$

Calculate the primary voltage of load side transformer.

Substitute the $230\text{\hspace{0.17em}kV}$for $V_{s1}$and$23.586\text{\hspace{0.17em}kV}$for$\Delta V$into equation (4) and

$\begin{array}{l}{V}_{p2}=230-23.585\\ V_{p2}=206.413\text{\hspace{0.17em}kV}\end{array}$

Calculate the change in the percentage of the primary voltage of the load side transformer.

Substitute$206.413\text{\hspace{0.17em}kV}$for $V_{p2}$&$230\text{\hspace{0.17em}kV}$for$V_{s1}$into equation (5).

$\begin{array}{l}\Delta =\frac{230-206.413}{230}\times 100\\ \Delta =10.25\%\end{array}$

Hence the change in the percentage of the primary voltage of the load side transformer is$10.25\%$.