Question

A single-phase two-winding transformer rated $\mathbf{90}\mathit{M}\mathit{V}\mathit{A}\mathbf{,}\mathbf{80}\mathbf{/}\mathbf{120}\mathbf{kV}\mathbf{,}$is to be connected as an autotransformer rated$\mathbf{80}\mathbf{/}\mathbf{200}\mathbf{kV}$ . Assume that the transformer is ideal. (a) Draw a schematic diagram of the ideal transformer connected as an autotransformer, showing the voltages, currents,and dot notation for polarity. (b) Determine the permissible kVA rating of the autotransformer if the winding currents and voltages are not to exceed the rated values as a two-winding transformer. How much of the kVA rating is transferred by magnetic induction?

Answer

Short answer

(a) The schematic of the autotransformer.

(b) The kVA rating of the autotransformer is$150\text{\hspace{0.17em}MVA}$ and power delivered by magnetic induction is$90\text{\hspace{0.17em}MVA}$ .

Step-by-step solution

Write the given data by the question

The power rating of two winding transformer,$S=90\text{\hspace{0.17em}MVA}$

Voltage rating of two winding transformer,$V=80/120\text{\hspace{0.17em}kV}$

Determine the formulas to calculate the value to draw the schematic of the autotransformer and kVA rating of the autotransformer.

The expression to determine the current in the winding is given by,

$I=\frac{S}{V}$…… (1)

The primary and secondary winding of the two-winding transformer will be equal to the input current of autotransformer.

$I_x=I_1+I_2$…… (2)

Here,$I_x$is the secondary current of the autotransformer, $I_1$is the primary current of two-winding transformer and is the secondary current of two-winding transformer.

The expression for input power of autotransformer is given by,

$S_{in}=V_1{I}_H$…… (3)

Here$S_{in}$,is the input kVA and $V_1$is the input voltage and $I_H$input current of the autotransformer.

The expression for output power of autotransformer is given by,

$S_{out}=V_x{I}_x$…… (4)

Here,$S_{out}$is the output kVA and$V_x$ is the output voltage and$I_H$ is output current of the autotransformer.

Draw the schematic of the auto transformer. 

(a)

Calculate the current is low voltage winding by using the equation (1).

$\begin{array}{l}{I}_1=\frac{90\times {10}^6}{80\times {10}^3}\\ I_1=1125\text{\hspace{0.17em}A}\end{array}$

Calculate the current is high voltage winding by using the equation (1).

$\begin{array}{l}{I}_2=\frac{90\times {10}^6}{120\times {10}^3}\\ I_2=750\text{\hspace{0.17em}A}\end{array}$

Calculate the input current of the autotransformer by using the equation (2).

$\begin{array}{l}{I}_x=1125+750\\ I_x=1875\text{\hspace{0.17em}A}\end{array}$

The output current will be the same as the primary current of the two-winding of the transformer.

$\begin{array}{l}{I}_H=I_1\\ I_H=1125\text{\hspace{0.17em}A}\end{array}$

Hence the schematic of the autotransformer is shown below.

 Step 4: Calculate the kVA rating of the transformer and power delivered by magnetic induction.

(b)

Calculate the input kVA of the autotransformer by using the equation (3).

$\begin{array}{l}{S}_{in}=80\times {10}^3\times 1875\\ S_{in}=150\text{\hspace{0.17em}MVA}\end{array}$

Calculate the output kVA of the autotransformer by using the equation (3).

$\begin{array}{l}{S}_{in}=750\times {10}^3\times 200\\ S_{in}=150\text{\hspace{0.17em}MVA}\end{array}$

The kVA transferred by the magnetic induction would the same as the kVA rating of the two-winding transformer.

Therefore, the kVA rating of the autotransformer is $150\text{\hspace{0.17em}MVA}$and power delivered by magnetic induction is $90\text{\hspace{0.17em}MVA}$.