Question 3 of 7: Single-phase transformer — primary quantities, efficiency and regulation

Question 3: Single-phase transformer — primary quantities, efficiency and regulation (25 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A single-phase step-up transformer with all parameters referred to the primary, and a specified secondary operating point.

Find. The primary voltage and current, the active and reactive power drawn at the primary terminals, the primary power factor, the efficiency, and the voltage regulation.

[Figure not reproduced: Figure 3.1 — The equivalent circuit of Figure (1) of the paper, redrawn with the excitation branch across the primary terminals (El-Hawary's cantilever form). Because the shunt branch sees V̂ 1 and the series arm carries only Î 2 ′, the whole problem is closed-form: no it. See the official exam paper.]

Approach. Refer the secondary operating point to the primary, add the series drop to reach the primary terminal voltage, take the excitation current at that voltage, sum the two currents to obtain the primary current, and read every remaining quantity off the resulting complex power.

1. Refer the secondary quantities to the primary. For a $1:4$ step-up unit the referred voltage divides by four and the referred current multiplies by four: $$V_2'=\frac{V_2}{4}=\frac{200}{4}=50.0\ \text{V},\qquad I_2'=4I_2=4(5)=20.0\ \text{A}$$ Taking $V_2'$ as the phase reference and keeping the 0.8 lagging power factor, $$\hat V_2'=50.0\,\angle\,0^\circ\ \text{V},\qquad \hat I_2'=20.0\,\angle\,-36.870^\circ=16.00-j12.00\ \text{A}$$ Referring rather than working on the secondary side is what keeps every impedance in the problem at its stated value.
2. Add the series drop to reach the primary terminal voltage. Only the load current flows in the series arm, so $$\hat V_1=\hat V_2'+\hat I_2'\,(R_{eq}+jX_{eq}) =50.0+(16.00-j12.00)(0.05+j0.225)$$ The product evaluates to $3.500+j3.000$ V, giving $$\boxed{\hat V_1=53.500+j3.000=53.584\,\angle\,3.209^\circ\ \text{V}}$$ Note how a lagging load current turns a small series impedance into a mostly real voltage rise: the resistive drop adds in phase with $V_2'$, and so does the part of the reactive drop that is in quadrature with the current.
3. Find the excitation current at that voltage. The magnetising branch sits across the primary terminals in this cantilever form, so it sees the full $\hat V_1$: $$\hat I_\phi=\hat V_1\left(\frac{1}{R_c}-\frac{j}{X_M}\right) =(53.500+j3.000)\left(\frac{1}{75}-\frac{j}{20}\right) =0.8633-j2.6350\ \text{A}$$ whose magnitude is 2.7728 A. The in-phase part supplies the core loss and the quadrature part supplies the mutual flux; the latter dominates, as it always does in a transformer.
4. Sum the currents for the primary current. Kirchhoff at the primary node gives $$\hat I_1=\hat I_2'+\hat I_\phi=16.8633-j14.6350$$ so that $$\boxed{\hat I_1=22.328\,\angle\,-40.953^\circ\ \text{A}}$$ The excitation current has added 2.7728 A of mostly lagging current, which is why $|I_1|=22.328$ A exceeds the referred load current of 20.0 A and why the primary power factor will be poorer than the load's.
5. Take the primary complex power and the power factor. With $S_1=\hat V_1\hat I_1^{*}$, $$\boxed{P_1=858.28\ \text{W},\qquad Q_1=833.56\ \text{var}}$$ and $|S_1|=1196.44$ VA, so the primary power factor is $$\cos\phi_1=\frac{P_1}{|S_1|}=\frac{858.28}{1196.44} =\boxed{\cos\phi_1=0.7174\ \text{lagging}}\qquad(\phi_1=44.163^\circ)$$ The load is at 0.8 lagging but the transformer presents 0.7174 lagging to the supply — the magnetising current has degraded it by more than eight points, which on a lightly loaded unit is entirely normal.
6. Compute the efficiency. The output is the true secondary power, $$P_{\text{out}}=V_2I_2\cos\phi_2=(200)(5)(0.8)=800\ \text{W}$$ so $$\eta=\frac{P_{\text{out}}}{P_1}=\frac{800}{858.28}=\boxed{\eta=93.209\ \%}$$ The loss balance closes exactly, which is the check worth doing: copper loss $|I_2'|^{2}R_{eq}=20.00$ W plus core loss $|V_1|^{2}/R_c=38.283$ W equals 58.283 W, and $858.28-800=58.283$ W. That the core loss is nearly twice the copper loss says the unit is running well below the load at which its efficiency would peak.
7. Compute the voltage regulation. At no load the series arm carries no current, so the referred secondary voltage rises to the primary terminal voltage itself. Regulation is therefore $$\mathrm{VR}=\frac{|V_1|-|V_2'|}{|V_2'|}\times100\ \% =\frac{53.584-50.0}{50.0}\times100 =\boxed{\mathrm{VR}=7.168\ \%}$$ Expressed on the actual secondary side the same figure applies, since both voltages are scaled by the same turns ratio: the 200 V would rise to about 53.584 × 4 = 214.3 V if the load were removed with the supply held constant.