Chapter 13, Problem 51E

A transformer whose nameplate reads 2300/230 V, 25 kVA operates with primary and secondary voltages of 2300 V and 230 V rms, respectively, and can supply 25 kVA from its secondary winding. If this transformer is supplied with 2300 V rms and is connected to secondary loads requiring 8 kW at unity PF and 15 kVA at 0.8 PF lagging, (a) what is the primary current? (b) How many kilowatts can the transformer still supply to a load operating at 0.95 PF lagging?

To determine

Find the value of primary current for the given data.

Answer to Problem 51E

The value of primary current for the given data is $\underset{\_}{9.5\angle -24.23°\text{A}}$.

Explanation of Solution

Given data:

Primary voltage of the transformer $\left(V_1\right)$ is 2300 V rms.

Secondary voltage of the transformer $\left(V_2\right)$ is 230 V rms.

Apparent power can supplied by the transformer from its secondary winding $\left(S\right)$ is $25\text{kVA}$.

Required average power for load 1 $\left(P_1\right)$ is $8\text{kW}$.

Power factor of the load 1 $\left({\text{PF}}_1\right)$ is 1 (unity power factor).

Required apparent power for load 2 $\left(S_2\right)$ is $15\text{kVA}$.

Power factor of the load 2 $\left({\text{PF}}_2\right)$ is 0.8 lagging.

Formula used:

Write the expression transformer ratio as follows:

$a=\frac{V_2}{V_1}$        (1)

Here,

$V_1$ is the rms voltage across the primary winding of the transformer and

$V_2$ is the rms value of the voltage across the load impedance.

Write the expression transformer ratio as follows:

$a=\frac{I_1}{I_2}$        (2)

Here,

$I_1$ is the rms current through the primary winding of the transformer, and

$I_2$ is the rms value of the current through the secondary winding of the transformer.

Write the expression for current in terms of average power and voltage as follows:

$I=\frac{P}{V}\angle -{\cos }^{-1}\left(\text{PF}\right)$        (3)

Here,

$P$ is the average power,

$V$ is the rms value of the voltage,

PF is the power factor.

Write the expression for current in terms of apparent power and voltage as follows:

$I=\frac{S}{V}\angle -{\cos }^{-1}\left(\text{PF}\right)$        (4)

Here,

$S$ is the apparent power and.

Calculation:

Modify the expression in Equation (3) for the current drawn by the load 1 as follows:

${\left(I_2\right)}_{\text{load-1}}=\frac{P_1}{V_2}\angle -{\cos }^{-1}\left({\text{PF}}_1\right)$

Substitute $8\text{kW}$ for $P_1$, 230 V for $V_2$, and 1 for ${\text{PF}}_1$ to obtain the value of current drawn by the load 1.

$\begin{array}{c}{\left(I_2\right)}_{\text{load-1}}=\frac{8\text{kW}}{230\text{V}}\angle -{\cos }^{-1}\left(1\right)\\ =\frac{8\times {10}^3\text{W}}{230\text{V}}\angle 0°\\ =34.7826\angle 0°\text{A}\end{array}$

Modify the expression in Equation (4) for the current drawn by the load 2 as follows:

${\left(I_2\right)}_{\text{load-2}}=\frac{S_2}{V_2}\angle -{\cos }^{-1}\left({\text{PF}}_2\right)$

Substitute $15\text{kVA}$ for $S_2$, 230 V for $V_2$, and 0.8 for ${\text{PF}}_2$ to obtain the value of current drawn by the load 2.

$\begin{array}{c}{\left(I_2\right)}_{\text{load-2}}=\frac{15\text{kVA}}{230\text{V}}\angle -{\cos }^{-1}\left(0.8\right)\\ =\frac{15\times {10}^3\text{VA}}{230\text{V}}\angle -36.87°\\ =65.2173\angle -36.87°\text{A}\end{array}$

As the secondary current of the transformer is the sum of the currents drawn by the two loads, write the expression for secondary current of the transformer as follows:

$I_2={\left(I_2\right)}_{\text{load-1}}+{\left(I_2\right)}_{\text{load-2}}$

Substitute $34.7826\angle 0°\text{A}$ for ${\left(I_2\right)}_{\text{load-1}}$ and $65.2173\angle -36.87°\text{A}$ for ${\left(I_2\right)}_{\text{load-2}}$ to obtain the value of secondary current of the transformer.

$\begin{array}{c}{I}_2=34.7826\angle 0°\text{A}+65.2173\angle -36.87°\text{A}\\ =\left(34.7826+j0\right)\text{A}+\left(52.1737-j39.1304\right)\text{A}\\ =\left(86.9563-j39.1304\right)\text{A}\\ =95.3550\angle -24.23°\text{A}\end{array}$

Substitute 2300 V for $V_1$ and 230 V for $V_2$ in Equation (1) to obtain the value of transformer ratio.

$\begin{array}{c}a=\frac{230\text{V}}{2300\text{V}}\\ =\frac{1}{10}\end{array}$

Rearrange the expression in Equation (2) for the primary current of the transformer as follows:

$I_1=a{I}_2$

Substitute $\left(\frac{1}{10}\right)$ for $a$ and $95.3550\angle -24.23°\text{A}$ for $I_2$ to obtain the value of primary current of the transformer.

$\begin{array}{c}{I}_1=\left(\frac{1}{10}\right)\left(95.3550\angle -24.23°\text{A}\right)\\ =9.53550\angle -24.23°\text{A}\\ \cong 9.5\angle -24.23°\text{A}\end{array}$

Conclusion:

Thus, the value of primary current for the given data is $\underset{\_}{9.5\angle -24.23°\text{A}}$.

To determine

Calculate the average power can be supplied by the transformer to a load of 0.95 lagging along with the existing loads.

Answer to Problem 51E

The average power can be supplied by the transformer to a load of 0.95 lagging is $\underset{\_}{2.93\text{kW}}$.

Explanation of Solution

Given data:

Consider the load operating at 0.95 lagging as load 3.

Power factor of the load 3 $\left({\text{PF}}_3\right)$ is 0.95 lagging.

Formula used:

Write the expression for maximum rated secondary current of the transformer as follows:

${\left(I_2\right)}_{\text{max rated}}=\frac{S}{V_2}$        (5)

Write the expression for average power in terms of voltage, current, and power factor as follows:

$P=\left|V\right|\left|I\right|\left(\text{PF}\right)$        (6)

Calculation:

Substitute $25\text{kVA}$ for $S$ and 230 V for $V_2$ in Equation (5) to obtain the value of maximum rated secondary current of the transformer.

$\begin{array}{c}{\left(I_2\right)}_{\text{max rated}}=\frac{25\text{kVA}}{230\text{V}}\\ =\frac{25\times {10}^3\text{VA}}{230\text{V}}\\ =108.6956\text{A}\end{array}$

From the given data, write the expression for current drawn by the load 3 as follows:

${\left(I_2\right)}_{\text{load-3}}=\left|{\left(I_2\right)}_{\text{load-3}}\right|\angle -{\cos }^{-1}\left({\text{PF}}_3\right)\text{A}$

Substitute 0.95 for ${\text{PF}}_3$ as follows:

$\begin{array}{c}{\left(I_2\right)}_{\text{load-3}}=\left|{\left(I_2\right)}_{\text{load-3}}\right|\angle -{\cos }^{-1}\left(0.95\right)\text{A}\\ =\left|{\left(I_2\right)}_{\text{load-3}}\right|\angle -18.19°\text{A}\end{array}$

Write the expression for secondary current of the transformer by considering load 3 as follows:

$I_2={\left(I_2\right)}_{\text{load-1}}+{\left(I_2\right)}_{\text{load-2}}+{\left(I_2\right)}_{\text{load-3}}$

Substitute $34.7826\angle 0°\text{A}$ for ${\left(I_2\right)}_{\text{load-1}}$, $65.2173\angle -36.87°\text{A}$ for ${\left(I_2\right)}_{\text{load-2}}$, and $\left|{\left(I_2\right)}_{\text{load-3}}\right|\angle -18.19°\text{A}$ for ${\left(I_2\right)}_{\text{load-3}}$ to obtain the value of secondary current of the transformer.

$\begin{array}{c}{I}_2=34.7826\angle 0°\text{A}+65.2173\angle -36.87°\text{A}+\left|{\left(I_2\right)}_{\text{load-3}}\right|\angle -18.19°\text{A}\\ =\left\{\begin{array}{l}\left(34.7826+j0\right)\text{A}+\left(52.1737-j39.1304\right)\text{A}+\\ \left[\left|{\left(I_2\right)}_{\text{load-3}}\right|\cos \left(-18.19°\right)+j\left|{\left(I_2\right)}_{\text{load-3}}\right|\sin \left(-18.19°\right)\right]\text{A}\end{array}\right\}\\ =\left\{\begin{array}{l}\left(34.7826+j0\right)\text{A}+\left(52.1737-j39.1304\right)\text{A}+\\ \left[\left|{\left(I_2\right)}_{\text{load-3}}\right|\left(0.95\right)+j\left|{\left(I_2\right)}_{\text{load-3}}\right|\left(-0.3121\right)\right]\text{A}\end{array}\right\}\\ =\left\{\left[86.9563+0.95\left|{\left(I_2\right)}_{\text{load-3}}\right|\right]-j\left[39.1304+0.3121\left|{\left(I_2\right)}_{\text{load-3}}\right|\right]\right\}\text{A}\end{array}$

The absolute value of the new secondary current of the transformer is equal to the maximum rated secondary current of the transformer. Therefore, write the expression as follows:

$\left|I_2\right|={\left(I_2\right)}_{\text{max rated}}$

Substitute $\left\{\left[86.9563+0.95\left|{\left(I_2\right)}_{\text{load-3}}\right|\right]-j\left[39.1304+0.3121\left|{\left(I_2\right)}_{\text{load-3}}\right|\right]\right\}\text{A}$ for $I_2$ and 108.6956 A for ${\left(I_2\right)}_{\text{max rated}}$ as follows:

$\begin{array}{l}\left|\left\{\left[86.9563+0.95\left|{\left(I_2\right)}_{\text{load-3}}\right|\right]-j\left[39.1304+0.3121\left|{\left(I_2\right)}_{\text{load-3}}\right|\right]\right\}\text{A}\right|=108.6956\text{A}\\ \sqrt{{\left[86.9563+0.95\left|{\left(I_2\right)}_{\text{load-3}}\right|\right]}^2+{\left[39.1304+0.3121\left|{\left(I_2\right)}_{\text{load-3}}\right|\right]}^2}=\text{108}\text{.6956}\end{array}$

Simplify the expression as follows:

$\begin{array}{l}{\left[86.9563+0.95\left|{\left(I_2\right)}_{\text{load-3}}\right|\right]}^2+{\left[39.1304+0.3121\left|{\left(I_2\right)}_{\text{load-3}}\right|\right]}^2={\left(\text{108}\text{.6956}\right)}^2\\ \left(\begin{array}{l}7561.3981+0.9025{\left|{\left(I_2\right)}_{\text{load-3}}\right|}^2+165.2169\left|{\left(I_2\right)}_{\text{load-3}}\right|+\\ 1531.1882+0.0974{\left|{\left(I_2\right)}_{\text{load-3}}\right|}^2+24.4251\left|{\left(I_2\right)}_{\text{load-3}}\right|\end{array}\right)=11814.7334\\ {\left|{\left(I_2\right)}_{\text{load-3}}\right|}^2+189.642\left|{\left(I_2\right)}_{\text{load-3}}\right|-2722.1471=0\end{array}$

Simplify the quadratic expression and find the value of current drawn by the load 3 as follows:

$\begin{array}{c}\left|{\left(I_2\right)}_{\text{load-3}}\right|=\frac{-189.642\pm \sqrt{{\left(189.642\right)}^2-\left(4\right)\left(1\right)\left(-2722.1471\right)}}{\left(2\right)\left(1\right)}\text{A}\\ =13.4063\text{A or }-203.0483\text{A}\end{array}$

Ignore the negative value and consider the positive value for the current drawn by the load 3 as follows:

$\left|{\left(I_2\right)}_{\text{load-3}}\right|=13.4063\text{A}$

Modify the expression in Equation (6) for the average power that can be supplied for the load 3 along with the two loads as follows:

$P_3=\left|V_2\right|\left|{\left(I_2\right)}_{\text{load-3}}\right|\left({\text{PF}}_3\right)$

Substitute $13.4063\text{A}$ for $\left|{\left(I_2\right)}_{\text{load-3}}\right|$, 230 V for $V_2$, and 0.95 for ${\text{PF}}_3$ to obtain the average power that can be supplied for the load 3 along with the two loads.

$\begin{array}{c}{P}_3=\left(230\text{V}\right)\left(13.4063\text{A}\right)\left(0.95\right)\\ =2929.2765\text{W}\\ \cong 2930\text{W}\\ \cong 2.93\text{kW}\end{array}$

Conclusion:

Thus, the average power can be supplied by the transformer to a load of 0.95 lagging is $\underset{\_}{2.93\text{kW}}$.