Question 3 of 7: Single-phase transformer: parameters from a load test, and input power from a known impedance

Problem 3 — Single-phase transformer: parameters from a load test, and input power from a known impedance (20 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. Two independent single-phase transformer problems on the same 2300/230 V frame (turns ratio $a = 10$), the no-load branch neglected throughout so that the equivalent circuit is the series arm $R_{eq} + jX_{eq}$ alone.

Find. (a) $R_{eq}$ and $X_{eq}$ referred to the high-voltage side; (b) the active power drawn at the primary terminals.

Part (a) — parameters from the loading test

Approach. The efficiency fixes the total copper loss and hence $R_{eq}$ directly; the primary-voltage magnitude is then supposed to fix $X_{eq}$ through the phasor equation $V_1 = V_2^{\prime} + I(R_{eq}+jX_{eq})$. That second step is where the printed data fail, and the answer says so explicitly rather than forcing a number.

1. Part (a) — convert the rating into an output power and a current. At full rated kVA and 0.85 power factor the delivered real power is $P_{out}=500\,\text{kVA}\times 0.85=425.0\ \text{kW}$, and the current, referred to the high-voltage side, is $$|I|=\frac{S}{V_1^{rated}}=\frac{500\,000}{2300}=217.391\ \text{A}$$ (the same current referred to the low-voltage side is 2173.9 A, ten times larger, and either frame gives the same per-unit answer).
2. Turn the efficiency into a copper loss. With the no-load branch neglected there is no core loss, so every watt of loss is $I^{2}R_{eq}$: $$\begin{aligned} P_{in} &= \frac{P_{out}}{\eta}=\frac{425\,000}{0.93}=456989.2\ \text{W} \\ P_{loss} &= P_{in}-P_{out}=31989.2\ \text{W} \end{aligned}$$
3. Recover the equivalent resistance. Dividing the loss by the square of the referred current, $$R_{eq}=\frac{P_{loss}}{|I|^{2}}=\frac{31989.2}{(217.391)^{2}}=\boxed{0.6769\ \Omega\ \text{(HV side)}}$$ On the transformer base $Z_{base}=2300^{2}/500\,000=10.580\ \Omega$ this is $0.0640$ pu, i.e. 6.4 per cent — an implausibly resistive transformer, which is the first hint that the stated efficiency is not that of a real 500 kVA unit. Referred to the low-voltage side the same resistance is $R_{eq}/a^{2}=0.006769\ \Omega$.
4. Set up the phasor equation for the reactance. Taking the referred secondary voltage as reference, $V_2^{\prime}=2300\angle 0^{\circ}$ V and $I=217.391\angle -31.788^{\circ}=184.783-j114.518$ A, the sending-end magnitude must satisfy $$\left|2300+(184.783-j114.518)(R_{eq}+jX_{eq})\right|=2400\ \text{V}$$ which, with $R_{eq}$ now known, is one quadratic in $X_{eq}$.
5. Solve it — and find no admissible root. Expanding gives $47259.0X^{2}+526782.7X+127011.9=0$, whose roots are $X_{eq}=-0.2466\ \Omega$ and $X_{eq}=-10.9002\ \Omega$. Both are negative, i.e. capacitive, and a transformer leakage reactance cannot be. The reason is visible without the algebra: with $X_{eq}=0$ the resistive drop alone already gives $$|V_1|=\left|2300+(184.783-j114.518)(0.6769)\right|=2426.3\ \text{V}$$ so no non-negative reactance can bring the primary down to the stated 2400 V.
6. Diagnose the inconsistency and state the defensible answer. The paper supplies four data — rated kVA, power factor, efficiency and primary voltage — for two unknowns, and at a lagging power factor they cannot all hold. Two of the three usable pairs give a clean answer, and both are reported: (i) trusting the efficiency, $R_{eq}=0.6769\ \Omega$ and the smallest admissible reactance is $X_{eq}\to 0$, for which the primary would have to be $2426.3$ V rather than 2400 V; (ii) keeping every printed number but reading the power factor as 0.85 leading, the same quadratic has the positive root $X_{eq}=0.2466\ \Omega$ and both the 0.93 efficiency and the 2400 V primary are satisfied exactly. Reading (i) is the one to submit, because it changes none of the printed words; reading (ii) is quoted so that a marker working from the setter’s own arithmetic is served.

Part (b) — input power from a known series impedance

Approach. Refer the load to the high-voltage side, find the current, add the copper loss to the delivered real power. No iteration and no assumption about the primary voltage is needed, because the receiving-end voltage and the load are both fixed.

1. Part (b) — refer the receiving end to the high-voltage side. With $a = 2300/230 = 10$, the 225 V secondary appears on the primary side as $$V_2^{\prime}=aV_2=10\times 225=2250\ \text{V}$$ The apparent power is unchanged by referral, so $S=400$ kVA still.
2. Find the load current in the referred frame. $$|I|=\frac{S}{V_2^{\prime}}=\frac{400\,000}{2250}=177.7778\ \text{A}$$ at an angle of $-31.788^{\circ}$ to $V_2^{\prime}$, the 0.85 lagging load angle.
3. Compute the copper loss in the series arm. Only the resistive part of $Z$ dissipates: $$P_{loss}=|I|^{2}R=(177.7778)^{2}(0.25)=7901.2\ \text{W}$$
4. Add it to the delivered power. The load takes $P_{out}=400\,000\times 0.85=340000$ W, so $$\boxed{P_{in}=P_{out}+P_{loss}=340.0+7.901=347.90\ \text{kW}}$$
5. Confirm the answer the long way. The sending-end phasor is $V_1=V_2^{\prime}+IZ=2325.2+j37.0$ V, that is $2325.5\angle 0.9124^{\circ}$ V, and $\operatorname{Re}\{V_1I^{*}\}=347901.2$ W — the same figure by an independent route. The implied efficiency of this loading is $97.73$ per cent, and the reactive input is $223.36$ kvar.