Question 15.13: Regulation and Efficiency Calculations Find the percentage r...

First, we draw the circuit as shown in Figure 15.30. Notice that we have placed the magnetizing reactance X_m and core loss resistance R_c on the left-hand side of R₁ and X₁, because this makes the calculations a bit simpler and is sufficiently accurate. We assume a zero phase reference for the load voltage. It is customary in power-system engineering to take the values of phasors as the rms values (rather than the peak values) of the currents and voltages. Thus, as a phasor, we have

\mathrm{V_{load}}=240\angle 0^\circ \mathrm{~V~rms}  

For rated load (20 kVA), the load current is

I_2=\frac{20\mathrm{~kVA}}{240\mathrm{~V}}=83.33\mathrm{~A~rms}

The load power factor is

\mathrm{power~factor}=\cos{(\theta)}=0.8

Solving, we find that

\theta =36.87^\circ

Thus, the phasor load current is

\mathrm{I}_2 =83.33 \angle -36.87^\circ\mathrm{~A~rms}

where the phase angle is negative because the load was stated to have a lagging power factor.
The primary current is related to the secondary current by the turns ratio:

\mathrm{I}_1=\frac{N_2}{N_1}\mathrm{I}_2=\frac{1}{10}\times 83.33 \angle -36.87^\circ =8.333\angle -36.87^\circ \mathrm{~A~rms}

Next, we can compute the voltages:

\mathrm{V}_2=\mathrm{V_{load}}+(R_2+jX_2) \mathrm{I}_2

\\=240+(0.03+j0.07)83.33\angle -36.87^\circ \\=240+6.345\angle 29.93^\circ=245.50+j3.166\mathrm{~V~rms}

The primary voltage is related to the secondary voltage by the turns ratio:

\mathrm{V}_1=\frac{N_1}{N_2}\mathrm{V}_2=10\times (245.50+j3.166)\\=2455.0+j31.66\mathrm{~V~rms}

Now, we can compute the source voltage:

\mathrm{V}_s=\mathrm{V}_1+(R_1+jX_1) \mathrm{I}_1

\\=2455.0+j31.66+(3+j6.5)\times (8.333\angle -36.87^\circ)\\=2508.2\angle 1.37^\circ \mathrm{~V~rms}

Next, we compute the power loss in the transformer:

P_{\mathrm{loss}}=\frac{V^2_s}{R_c}+I^2_1R_1+I_2^2R_2\\=62.91+208.3+208.3\\=479.5\mathrm{~W}

The power delivered to the load is given by

P_{\mathrm{load}}=V_{\mathrm{load}}I_2\times \mathrm{power~factor}\\=20\mathrm{~kVA}\times 0.8=16,000\mathrm{~W}

The input power is given by

P_{\mathrm{in}}=P_{\mathrm{load}}+P_{\mathrm{loss}}\\=16,000+479.5=16,479.5\mathrm{~W}

At this point, we can compute the power efficiency:

\mathrm{efficiency}=\left( 1-\frac{P_{\mathrm{loss}}}{P_{\mathrm{in}}} \right)\times 100\% \\= \left( 1-\frac{ 479.5}{16,479.5 } \right) \times 100\% =97.09\%

Next, we can determine the no-load voltages. Under no-load conditions, we have

I_1=I_2=0

\\V_1=V_s=2508.2

\\V_{\mathrm{no-load}}=V_2=V_1\frac{N_2}{N_1}=250.82\mathrm{~V~rms}

Finally, the percentage regulation is

percent regulation =\frac{ V_{\mathrm{no-load}}- V_{\mathrm{load}}}{ V_{\mathrm{load}}}\times 100\%\\=\frac{250.82-240}{240}\times 100\%\\=4.51\%