Question

The transformer of Problem 3.16 is supplying a rated load of 50kVAat a rated secondary voltage of 240Vand at 0.8 power factor lagging. Neglect the transformer exciting current. (a) Determine the input terminal voltage of the transformer on the high-voltage side (b) Sketch the corresponding phasor diagram. (c) If the transformer is used as a step-down transformer at the load end of a feeder whose impedance is $\left(0.5+j2.0\right)\mathbf{}\mathbf{\Omega }$, find the voltage${\mathbf{V}}_{\mathbf{s}}$ and the power factor at the sending end of the feeder.

Short answer

(a) Therefore, the input terminal voltage of the transformer is $2449.93\angle 0.34°\mathrm{V}$.

(b) The phasor diagram of the transformer is given below:

(c) The sending end voltage of feeder is$2483.52\angle 0.95°\mathrm{V}$ and power factor is 0.79 .

Step-by-step solution

Write the given data from the question.

The power rating ofthetransformer is 50kVA .

The high voltage winding, $V_1=2400\mathrm{V}$.

The low voltage winding, $V_2=240\mathrm{V}$.

The resistance of high voltage winding, $R_1=0.75\mathrm{\Omega }$.

The resistance of high low winding, $R_2=0.0075\mathrm{\Omega }$.

The leakage reactance of high voltage winding, $X_1=1\mathrm{\Omega }$.

The leakage reactance of low voltage winding, $X_2=0.01\mathrm{\Omega }$.

The exciting admittance on the low voltage side$=0.003-j0.02S$

The secondary power factor is 0.8 lagging.

Feeder impedance,$Z=\left(0.5+j2\right)\mathrm{\Omega }$

Determine the formulas to calculate the equivalent impedance to draw the equivalent circuit and calculate the input terminal voltage, and phasor of the transformer.

The equation for the impedance of low voltage winding is as follows:

${\mathbf{Z}}_{\mathbf{2}}\mathbf{=}{\mathbf{R}}_{\mathbf{2}}\mathbf{+}{\mathbf{jX}}_{\mathbf{2}}$ ….. (1)

Theequation forthe impedance of low voltage winding to high voltage winding isas follows:

${\mathbf{Z}}_{\mathbf{2}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{Z}}_{\mathbf{2}}{\left(\frac{N_1}{N_2}\right)}^{\mathbf{2}}$ ….. (2)

Here,${\mathbf{Z}}_{\mathbf{2}}^{\mathbf{\text{'}}}$is the low voltage winding impedance,referred toas thehigh voltage side,${\mathbf{N}}_{\mathbf{1}}$ is the number of turnsontheprimary side,and ${\mathbf{N}}_{\mathbf{2}}$is the number of turns in the secondary winding.

Theequationof the impedance of high voltage winding isas follows:

${\mathbf{Z}}_{\mathbf{1}}\mathbf{=}{\mathbf{R}}_{\mathbf{1}}\mathbf{+}{\mathbf{jX}}_{\mathbf{1}}$ ….. (3)

Theequationfor exciting admittance isas follows:

${\mathbf{Y}}_{\mathbf{o}}\mathbf{=}{\mathbf{G}}_{\mathbf{c}}\mathbf{+}{\mathbf{jB}}_{\mathbf{m}}$ ….. (4)

Here, ${\mathbf{Y}}_{\mathbf{o}}$is the low voltage side admittance,${\mathbf{G}}_{\mathbf{c}}$ is the exciting conductance,${\mathbf{B}}_{\mathbf{m}}$and is the exciting susceptance.

Theequationfor exciting susceptance to refer high voltage side isas follows:,

${\mathbf{Y}}_{\mathbf{o}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{Y}}_{\mathbf{o}}{\left(\frac{N_2}{N_1}\right)}^{\mathbf{2}}$ ….. (5)

Here,${\mathbf{Y}}_{\mathbf{o}}^{\mathbf{\text{'}}}$ is the low voltage winding impedance referredtoas thehigh voltage side.

Theequationfor feeder sending end voltage isas follows:

${\mathbf{V}}_{\mathbf{s}}\mathbf{=}{\mathbf{V}}_{\mathbf{1}}\mathbf{+}{\mathbf{I}}_{\mathbf{1}}\mathbf{Z}$ ….. (6)

Here,${\mathbf{V}}_{\mathbf{s}}$is the feeder sending voltage.

Determine the equivalent circuit of the transformer.

(a)

Calculate the low voltage side impedance using equation (1),

$Z_2=\left(0.0075+j0.01\right)\mathrm{\Omega }$

Calculate the low voltage impedance referred toas thehigh voltage side using equation (2).

$$\begin{gathered} Z_2=\left(0.0075+j0.01\right){\left(\frac{2400}{240}\right)}^2 \\ {Z}_2=\left(0.0075+j0.01\right)\times 100 \\ {Z}_2=\left(0.75+j1\right)\mathrm{\Omega } \end{gathered}$$

Calculate the high voltage side impedance using equation (1).

$Z_1=\left(0.75+j1\right)\mathrm{\Omega }$

Calculate the exciting impedance usingequation (4).

$Y_o=\left(0.003+j0.02\right)\mathrm{S}$

Calculate the exciting impedance referred toas thehigh voltage side using equation (5).

$$\begin{gathered} Y_o=\left(0.003+j0.02\right){\left(\frac{240}{2400}\right)}^2 \\ {Y}_o=\left(0.003+j0.02\right)\times \frac{1}{100} \\ {Y}_o=\left(0.03+j0.2\right)\mathrm{mS} \end{gathered}$$

The equivalent circuit of the transformer, referred to as the high voltage side, is shown below.

Neglect the exciting current and redraw the equivalent circuit of the transformer.

The current$I_2^{\text{'}}$can be calculated from the above circuit as follows:

$$\begin{gathered} I_2^{\text{'}}=\frac{50\times {10}^3}{2400}\angle -{\cos }^{-1}\left(0.8\right) \\ {I}_2^{\text{'}}=20.83\angle -36.86°\mathrm{A} \end{gathered}$$

Apply the Kirchhoff’s in the above circuit,and calculate the voltage $V_1$asfollows:

$$\begin{gathered} V_1=2400\angle 0°+I_1\left(R_{eq}+j{X}_{eq}\right) \\ {V}_1=2400\angle 0°+20.83\angle -36.86°\left(1.5+j2\right) \\ {V}_1=2400\angle 0°+\left(20.83\angle -36.86\right)\left(2.5\angle 53.13°\right) \\ {V}_1=2449.93\angle 0.34°\mathrm{V} \end{gathered}$$

Hence, the input terminal voltage of the transformer is $2449.93\angle 0.34°\mathrm{V}$.

Determine the phasor diagram of the equivalent circuit of the transformer.

(b)

The phasor diagram is shown below.

Determine the voltage and power factor at the sending end of the feeder.

(c)

Calculate the feeder sending end voltage using equation (6),

$$\begin{gathered} V_s=2449.93\angle 0.34°+20.83\angle -36.86°\left(0.5+j2\right) \\ {V}_s=2449.93\angle 0.34°\left(+20.83\angle -36.86\right)\left(2.06\angle 75.96\right) \\ {V}_s=2449.93\angle 0.34°+42.90\angle 39.1° \\ {V}_s=2483.52\angle 0.95°\mathrm{V} \end{gathered}$$

The power factor isthecosine of the angle between the voltage and current.

Calculate the power factor asfollows:

$$\begin{gathered} PF=\cos \left(0.95+36.86\right) \\ PF=\cos \left(37.81\right) \\ PF=0.79 \end{gathered}$$

Hence, the sending end voltage of the feeder is$2483.52\angle 0.95°\mathrm{V}$ , and the power factor is 0.79 .