Question

A single-phase $\mathbf{10}\mathit{k}\mathit{V}\mathit{A}\mathbf{,}\mathbf{2300}\mathbf{/}\mathbf{230}\mathit{v}\mathit{o}\mathit{l}\mathit{t}\mathbf{,}\mathbf{60}\mathbf{-}\mathit{H}\mathit{z}$two-winding distribution transformer is connected as an autotransformer to step up the voltage from 2300 to$\mathbf{2530}$volts. (a) Draw a schematic diagram of this arrangement, showing all voltages and currents when delivering full load at rated voltage (b) Find the 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 this kVA rating is transformed by magnetic induction? (c) The following data are obtained from tests carried out on the transformer When it is connected as a two-winding transformer:

Open-circuit test with the low-voltage terminals excited:
Applied voltagelocalid="1656747038872" $\mathbf{=}\mathbf{230}\mathit{V}$ , input current localid="1656747042832" $\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{5}\mathit{A}$, input power localid="1656747046762" $\mathbf{=}\mathbf{70}\mathit{W}$

Short-circuit test with the high-voltage terminals excited:
Applied voltage localid="1656747050183" $\mathbf{=}\mathbf{120}\mathit{V}$, input current localid="1656747053614" $\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{5}\mathit{A}$, input power localid="1656747057853" $\mathbf{=}\mathbf{240}\mathit{W}$

Based on the data, compute the efficiency of the autotransformer corresponding to full load, rated voltage, andlocalid="1656747062178" $\mathbf{0}\mathbf{.}\mathbf{8}$ power factor lagging. Comment on Why the efficiency is higher as an autotransformer than as a two-winding transformer.

Short answer

a) The arrangement of the autotransformer is shown below

(b) The kVA rating of the autotransformer is $110\text{\hspace{0.17em}kVA}$and power delivered by magnetic induction is$100\text{\hspace{0.17em}kVA}$.

(c) The efficiency of the autotransformer is $99.66\%$and more than the two winding of the transformer

Step-by-step solution

Write the given data by the question.

Two winding distribution transformers $\text{kVA}$rating $S=10\text{\hspace{0.17em}kVA}$.

Two winding distribution transformers voltage rating$=2300/230\text{\hspace{0.17em}volt}$

Frequency$f=60\text{\hspace{0.17em}Hz}$

Autotransformer voltage rating$=2300/2530\text{\hspace{0.17em}volt}$

Consider the open circuit test results are

Applied voltage$=230\text{\hspace{0.17em}V}$

The value of the input current$=4.5\text{\hspace{0.17em}A}$

The value of the input power$=70\text{\hspace{0.17em}W}$

Consider the short circuit test results are,

The appliedvoltage $=120\text{\hspace{0.17em}V}$.

The input current $=4.5\text{\hspace{0.17em}A}$.

The input power $=240\text{\hspace{0.17em}W}$.

Determine the formulas to draw the diagram, to calculate the permissible  kVA rating of autotransformer and  kVA rating which is transferred by induction

The expression for primary winding current is given by,

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

Here,${\mathit{I}}_{\mathbf{1}}$is the primary winding current, $\mathit{S}$ is the power rating of two winding transformer and ${\mathit{E}}_{\mathbf{1}}$is the primary voltage of two winding transformer.

The expression for secondary winding current is given by,

${\mathit{I}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{S}}{{\mathbf{E}}_{\mathbf{2}}}$…… (2)

Here, ${\mathit{I}}_{\mathbf{2}}$is the secondary winding current, $\mathit{S}$is the power rating of two winding transformer and${\mathit{E}}_{\mathbf{2}}$is the secondary voltage of two winding transformer.

The expression to calculate the KVA rating of the autotransformer is given by,

${\mathit{S}}_{\mathbf{H}}\mathbf{=}{\mathit{E}}_{\mathbf{H}}{\mathit{I}}_{\mathbf{H}}$…… (3)

Here, ${\mathit{E}}_{\mathbf{H}}$is the secondary voltage of autotransformer, ${\mathit{S}}_{\mathbf{H}}$is the power rating of autotransformer and ${\mathit{I}}_{\mathbf{H}}$is the secondary current of autotransformer.

The expression to calculate the electrical power delivered is given by,

${\mathit{S}}_{\mathbf{e}}\mathbf{=}{\mathit{S}}_{\mathbf{H}}\mathbf{-}\mathit{S}$…… (4)

Here, ${\mathit{S}}_{\mathbf{e}}$ is the electrical power and ${\mathit{S}}_{\mathbf{H}}$ is the autotransformer power.

The expression to calculate the cupper loss with high voltage terminal excited is given by,

${\mathit{P}}_{\mathbf{c}\mathbf{u}}\mathbf{=}{\left(\frac{I_1}{I_{sc}}\right)}^{\mathbf{2}}{\mathit{P}}_{\mathbf{s}\mathbf{c}}$…… (5)

Here, ${\mathit{I}}_{\mathbf{s}\mathbf{c}}$is the short circuit current and ${\mathit{P}}_{\mathbf{s}\mathbf{c}}$is the short circuit losses.

The expression to calculate the power factor for two winding transformer is given by

$\mathit{c}\mathit{o}\mathit{s}\mathit{\varphi }\mathbf{=}\frac{\mathbf{P}}{\mathbf{V}\mathbf{I}}$…… (6)

The expression to calculate the efficiency of two winding transformer is given by,

$\mathit{\eta }\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{o}\mathbf{u}\mathbf{t}}}{{\mathbf{P}}_{\mathbf{o}\mathbf{u}\mathbf{t}}\mathbf{+}{\mathbf{P}}_{\mathbf{l}\mathbf{o}\mathbf{s}\mathbf{s}}}$…… (7)

Here, $\mathit{\eta }$is the efficiency of two winding transformer, ${\mathit{P}}_{\mathbf{o}\mathbf{u}\mathbf{t}}$is the output power, ${\mathit{P}}_{\mathbf{l}\mathbf{o}\mathbf{s}\mathbf{s}}$is the losses of two winding transformer.

The expression to calculate the efficiency of autotransformer transformer is given by,

$\mathit{\eta }\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{P}{\mathbf{\text{'}}}_{\mathbf{o}\mathbf{u}\mathbf{t}}}{\mathbf{P}{\mathbf{\text{'}}}_{\mathbf{o}\mathbf{u}\mathbf{t}}\mathbf{+}\mathbf{P}{\mathbf{\text{'}}}_{\mathbf{l}\mathbf{o}\mathbf{s}\mathbf{s}}}$…… (8)

Here, $\mathit{\eta }\mathbf{\text{'}}$is the efficiency of autotransformer, $\mathit{P}{\mathbf{\text{'}}}_{\mathbf{o}\mathbf{u}\mathbf{t}}$is the output power, $\mathit{P}{\mathbf{\text{'}}}_{\mathbf{l}\mathbf{o}\mathbf{s}\mathbf{s}}$is the losses of autotransformer.

Step 3:Draw the schematic diagram of the given autotransformer arrangement.

(a)

Calculate the primary winding voltage of two winding transformer by using the equation (1),

$\begin{array}{l}{I}_1=\frac{10\times {10}^3}{2300}\\ I_1=4.348\text{\hspace{0.17em}A}\end{array}$

Calculate the secondar winding voltage of two winding transformer by using the equation (2),

$\begin{array}{l}{I}_2=\frac{10\times {10}^3}{230}\\ I_2=43.48\text{\hspace{0.17em}A}\end{array}$

The primary winding voltage of the autotransformer and two winding transformers will remain the same.

$E_x=E_1$

The secondary voltage of the autotransformer will be the sum of the primary and secondary voltage of the two-winding transformer.

$\begin{array}{l}{E}_H=E_1+E_2\\ E_H=2300+230\\ E_H=2530\text{\hspace{0.17em}V}\end{array}$

The primary current of autotransformer will the sum of the primary and secondary current of the two-winding transformer.

$\begin{array}{l}{I}_x=I_1+I_2\\ I_x=4.348+43.48\\ I_x=47.828\text{\hspace{0.17em}A}\end{array}$

Hence the arrangement of the autotransformer is shown below.

Calculate the kVA rating of the autotransformer and kVA rating delivered by magnetic induction

(b)

Calculate the KVA rating of the transformer by using the (3),

$\begin{array}{l}{S}_H=2530\times 43.48\\ S_H=110\text{\hspace{0.17em}kVA}\end{array}$

Calculate the electrical delivered power by using the equation (4),

$\begin{array}{l}{S}_H=110-10\\ S_H=100\text{\hspace{0.17em}kVA}\end{array}$

Hence the kVA rating of the autotransformer is $110\text{\hspace{0.17em}kVA}$ and power delivered by magnetic induction is $100\text{\hspace{0.17em}kVA}$

Calculate the efficiency of the autotransformer and two winding transformer and comment the reasonof autotransformer have higher efficiency than two winding transformers.

(c)

The core losses of the transformer are given by open circuit test that is ${\text{P}}_c=70\text{\hspace{0.17em}W}$,

The cupper losses of the transformer can be calculated by using the equation (5).

$\begin{array}{l}{P}_{cu}={\left(\frac{4.348}{4.5}\right)}^2\times 240\\ P_{cu}=0.9335\times 240\\ P_{cu}=224.06\text{\hspace{0.17em}W}\end{array}$

Calculate the power factor for two-winding transformer by using the equation (6).

$\begin{array}{l}\cos \varphi =\frac{70}{230\times 0.45}\\ \cos \varphi =0.676\end{array}$

Calculate the efficiency of the two-winding transformer by using the equation (7).

$\begin{array}{l}\eta =\left(\frac{10\times {10}^3\times 0.676}{10\times {10}^3\times 0.676+224.06+70}\right)\times 100\\ \eta =\left(\frac{6760}{7054.06}\right)\times 100\\ \eta =95.8\%\end{array}$

Calculate the efficiency of the autotransformer by using the equation (8).

$\begin{array}{l}\eta \text{'}=\left(\frac{110\times {10}^3\times 0.8}{10\times {10}^3\times 0.8+224.06+70}\right)\times 100\\ \eta \text{'}=\left(\frac{88000}{88294.06}\right)\times 100\\ \eta \text{'}=99.66\%\end{array}$

From the above the autotransformer has the higher efficiency because of the power factor