Question

Rework Problem 3.14 if the transformer is delivering rated load at rated secondary voltage and at (a) unity power factor, (b) 0.8power factor leading. Compare the results with those Of Problem 3.14.

Short answer

(a) The primary voltage of the transformer is$2421.32\angle 1.23°\mathrm{V}$.

The voltage at the sending end of the feeder is$2443.44\angle 2.20°\mathrm{V}$.

The real and reactive powers delivered to the sending end of the feeder are

$50.85\mathrm{kW}\mathrm{and}1.95\mathrm{kVA}$_,respectively.

(b) The primary voltage of the transformer is$2386.04\angle 1.30°\mathrm{V}$.

The voltage at the sending end of the feeder is $2379.21\angle 2.41°\mathrm{V}$.

The real and reactive powers delivered to the sending end of the feeder are$40.85\mathrm{kW}\mathrm{and}28.04\mathrm{kVA}$_,respectively.

Step-by-step solution

Write the given data in the question.

The rating of the transformer is $50\mathrm{kVA},2400/240‐\mathrm{volt},60‐\mathrm{Hz}$.

Feeder impedance,$Z_2=\left(1+j2\right)\mathrm{ohm}$_.

The transformer impedance,$Z_T=\left(1+j2.5\right)\mathrm{ohm}$_.

The power factor of the transformer,$PF=0.8$lagging.

The transformer circuit with the feeder is shown below.

Determine the formulas to calculate the voltage at the primary transformer terminal, the voltage at the sending of the feeder and real and reactive power delivered to the sending end of the feeder.

The expression for the current in the secondary winding is as follows:

${\mathbf{I}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{r}}}{{\mathbf{V}}_{\mathbf{2}}}\mathbf{\angle }\mathbf{-}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\left(P.F\right)$ …… (1)

Here,${\mathbf{I}}_{\mathbf{2}}$is the secondary winding current,${\mathbf{P}}_{\mathbf{r}}$is the power rating of the transformer, and${\mathbf{V}}_{\mathbf{2}}$is the secondary terminal voltage.

The expression for primary winding current is as follows:,

${\mathbf{I}}_{\mathbf{1}}\mathbf{=}\left(\frac{N_2}{N_1}\right){\mathbf{I}}_{\mathbf{2}}$ …… (2)

Here,${\mathbf{I}}_{\mathbf{1}}$is the primary winding current,${\mathbf{I}}_{\mathbf{2}}$is the secondary winding current,${\mathbf{N}}_{\mathbf{1}}$is the number of turns in the primary winding, and${\mathbf{N}}_{\mathbf{2}}$is the number of turns in the secondary winding.

The expression for the primary winding voltage is as follows:

${\mathbf{E}}_{\mathbf{1}}\mathbf{=}\left(\frac{N_1}{N_2}\right){\mathbf{E}}_{\mathbf{2}}$ …… (3)

Here,${\mathbf{E}}_{\mathbf{1}}$is the primary winding voltage, and${\mathbf{E}}_{\mathbf{2}}$is the secondary winding voltage.

By applying the Kirchhoff’s law, the expression for the primary voltage across the transformer is as follows:

${\mathbf{V}}_{\mathbf{1}}\mathbf{=}{\mathbf{E}}_{\mathbf{1}}\mathbf{+}{\mathbf{I}}_{\mathbf{1}}{\mathbf{Z}}_{\mathbf{T}}$ …… (4)

The expression for feeder terminal voltage is calculated as follows:

${\mathbf{V}}_{\mathbf{s}}\mathbf{=}{\mathbf{E}}_{\mathbf{1}}\mathbf{+}{\mathbf{I}}_{\mathbf{1}}{\mathbf{Z}}_{\mathbf{2}}$ …… (5)

The expression for power delivered to the feeder is calculated as follows:

$\mathbf{S}\mathbf{=}{\mathbf{V}}_{\mathbf{s}}{\mathbf{I}}_{\mathbf{1}}^{\mathbf{*}}$ …… (6)

Calculate the primary voltage of the transformer, voltage at sending end of the feeder, and real and reactive powers delivered to sending end for the unit power factor.

(a)

Calculate the secondary wing current using equation (1).

$$\begin{gathered} I_2=\frac{50\times {10}^3}{240}\angle {\cos }^{-1}\left(0\right) \\ {I}_2=208.33\angle 0°\mathrm{A} \end{gathered}$$

Calculate the current in the primary winding using equation (2).

$$\begin{gathered} I_1=\left(\frac{240}{2400}\right)\times 208.33\angle 0° \\ {I}_1=0.1\times 208.33\angle 0 \\ {I}_1=20.83\angle -0°\mathrm{A} \end{gathered}$$

Calculate the voltage across the secondary winding using equation (3).

$$\begin{gathered} E_1=\left(\frac{2400}{240}\right)\times 240\angle 0° \\ {E}_1=2400\angle 0°\mathrm{V} \end{gathered}$$

Calculate the voltage across the primary winding of the transformer using equation (4).

localid="1655264106979" $$\begin{gathered} V_1=\left(20.83\angle 0°\right)\left(1+\mathrm{j}2.5\right)+2400\angle 0° \\ {V}_1=\left(20.83\angle 0°\right)\left(2.69\angle 68.19°\right)+2400\angle 0° \\ {V}_1=56.03\angle 68.19+2400\angle 0° \\ {V}_1=2421.32\angle 1.23°\mathrm{V} \end{gathered}$$

Hence, the primary voltage of the transformer is$2421.32\angle 1.23°\mathrm{V}$.

Calculate the sending end voltage of the feeder using equation (5),

localid="1655264402514" $$\begin{gathered} V_s=2421.32\angle 1.23°+20.83\angle 0°\left(1+j2\right) \\ {V}_s=2421.32\angle 1.23°+\left(46.58\angle 63.43°\right) \\ {V}_s=2443.44\angle 2.20°\mathrm{V} \end{gathered}$$

Hence, the voltage at the sending end of the feeder is .

Calculate the power delivered to the feeder using equation (6).

$$\begin{gathered} S=\left(2443.44\angle 2.20°\right)+\left(20.83\angle 0°\right) \\ S=2490\angle 2.20°+\left(20.83\angle 0°\right) \\ S=50896.86\angle 2.20\mathrm{VA}° \end{gathered}$$

The active power can be calculated as follows:

$$\begin{gathered} P=50896.86\cos \left(2.20\right) \\ P=50.85\mathrm{kW} \end{gathered}$$

The reactive power can be calculated as follows:

$$\begin{gathered} Q=50896.86\sin \left(38.02\right) \\ Q=1.95\mathrm{kVA} \end{gathered}$$

Hence, the real and reactive power delivered to the sending end of the feeder is $50.85\mathrm{kW}\mathrm{and}1.95\mathrm{kVA}$_,respectively.

Calculate the primary voltage of the transformer, voltage at sending end of the feeder, and real and reactive power delivered to sending end for  leading power factor.

(b)

Calculate the secondary wing current using equation (1).

$$\begin{gathered} I_2=\frac{50\times {10}^3}{240}\angle {\cos }^{-1}\left(0.8\right) \\ {I}_2=208.33\angle 36.86°\mathrm{A} \end{gathered}$$

Calculate the current in the primary winding using equation (2).

$$\begin{gathered} I_1=\left(\frac{240}{2400}\right)\times 208.33\angle 36.86° \\ {I}_1=0.1\times 208.33\angle 36.86 \\ {I}_1=20.83\angle 36.86°\mathrm{A} \end{gathered}$$

Calculate the voltage across the secondary winding using equation (3).

$$\begin{gathered} E_1=\left(\frac{2400}{240}\right)\times 240\angle 0° \\ {E}_1=2400\angle 0°\mathrm{V} \end{gathered}$$

Calculate the voltage across the primary winding of the transformer using equation (4).

$$\begin{gathered} V_1=\left(20.83\angle 36.86°\right)\left(1+j2.5\right)+2400\angle 0° \\ {V}_1=\left(20.83\angle 36.86°\right)\left(2.69\angle 68.19°\right)+2400\angle 0° \\ {V}_1=56.03\angle 105.05+2400\angle 0° \\ {V}_1=2386.04\angle 1.30°\mathrm{V} \end{gathered}$$

Hence, the primary voltage of the transformer is$2386.04\angle 1.30°\mathrm{V}$.

Calculate the sending end voltage of the feeder using equation (5).

$$\begin{gathered} V_s=2386.04\angle 1.30°+20.83\angle 36.86°\left(1+j2\right) \\ {V}_s=2386.04\angle 1.30°+\left(20.83\angle 36.86\right)\left(2.23\angle 63.43°\right) \\ {V}_s=22386.04\angle 1.30°+46.57\angle 100.30° \\ {V}_s=2379.21\angle 2.41°\mathrm{V} \end{gathered}$$

Hence, the voltage at the sending end of the feeder is$2379.21\angle 2.41°\mathrm{V}$ .

Calculate the power delivered to the feeder using equation (6),

$$\begin{gathered} S=\left(2379.21\angle 2.41°\right){\left(20.83\angle 36.86°\right)}^{*} \\ S=\left(2379.21\angle 2.41°\right)\left(20.83\angle 36.86°\right) \\ S=49558.93\angle -34.47°\mathrm{VA} \end{gathered}$$

The active power can be calculated as follows:

$$\begin{gathered} P=49558.93\cos \left(-34.47\right) \\ P=40.85\mathrm{kW} \end{gathered}$$

The reactive power can be calculated as follows:

$$\begin{gathered} Q=49558.93\sin \left(38.02\right) \\ Q=28.04\mathrm{kVA} \end{gathered}$$

Hence, the real and reactive power delivered to the sending end of the feeder is $40.85\mathrm{kW}\mathrm{and}28.04\mathrm{kVA}$ _,respectively.