Question 15.13: Find the percentage regulation and power efficiency for the ...
Find the percentage regulation and power efficiency for the transformer of Table 15.1 for a rated load having a lagging power factor of 0.8.
Table 15.1. Circuit Values of a 60-Hz 20-kVA 2400/240-V Transformer Compared with Those of an Ideal Transformer
| Element Name | Symbol | Ideal | Real |
| Primary resistance | R_1 | 0 | 3 Ω |
| Secondary resistance | R_2 | 0 | 0.03 Ω |
| Primary leakage reactance | X_1 = ωL_1 | 0 | 6.5 Ω |
| Secondary leakage reactance | X_2 = ωL_2 | 0 | 0.07 Ω |
| Magnetizing reactance | X_m = ωL_m | ∞ | 15 kΩ |
| Core-loss resistance | R_C | ∞ | 100 kΩ |
In this example, we are taking the rms values of currents and voltages (rather than peak values) as the phasor magnitudes. This is often done by power-distribution engineers. We will clearly indicate when phasors represent rms values rather than peak values.
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 R_1 and X_1, 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
\textbf{V}_{load} = 240∠0° V rms
For rated load (20 kVA), the load current is
I_2 = \frac{20 kVA}{ 240 V} = 83.33 A rms
The load power factor is
power factor = cos(θ) = 0.8
Solving, we find that
θ = 36.87°
Thus, the phasor load current is
\textbf{I}_2 = 83.33∠−36.87° 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:
\textbf{I}_{1} = \frac{N_2}{ N_1}\textbf{I}_2 =\frac{1}{10} × 83.33 ∠−36.87° ∠ = 8.333∠−36.87° A rms
Next, we can compute the voltages:
\textbf{V}_2 = \textbf{V}_{load} + (R_2 + jX_2)\textbf{I}_2
= 240 + (0.03 + j0.07)83.33∠−36.87°
= 240 + 6.346∠29.93°
= 245.50 + j3.166 V rms
The primary voltage is related to the secondary voltage by the turns ratio:
\text{V}_{1} = \frac{N_1}{ N_2}\textbf{V}_2 = 10 × (245.50 + j3.166)
= 2455.0 + j31.66 V rms
Now, we can compute the source voltage:
\textbf{V}_s = \textbf{V}_1 + (R_1 + jX_1)I_1
= 2455.0 + j31.66 + (3 + j6.5) × (8.333∠−36.87°)
= 2508.2∠1.37° V rms
Next, we compute the power loss in the transformer:
P_{loss} = \frac{V^2_s} {R_c} + I^2_1R_1 + I^2_2R_2
= 62.91 + 208.3 + 208.3
= 479.5 W
The power delivered to the load is given by
P_{load} = V_{load}I_2 × power factor
= 20 kVA × 0.8 = 16,000 W
The input power is given by
P_{in} = P_{load} + P_{loss}
= 16,000 + 479.5 = 16,479.5 W
At this point, we can compute the power efficiency:
efficiency = \left(1 − \frac{P_{loss}} {P_{in}}\right) × 100%
= \left(1 − \frac{479.5} {16,479.5}\right) × 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_{no-load} = V_2 = V_1 \frac{N_2}{ N_1} = 250.82 V rms
Finally, the percentage regulation is
percent regulation = \frac{V_{no-load} − V_{load}} {V_{load}} × 100%
= \frac{250.82 − 240}{ 240} × 100%
= 4.51%
