Question 3 of 7: Three-Winding Transformer — Currents, Voltages, Powers and Efficiency
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2017 — 16-Elec-B7 Power Systems Engineering. Open-book, 3 hours, seven problems of equal value (25 points each); any five constitute a complete paper, and only the first five in the answer book are marked. All seven are solved here, because the set is a study resource rather than a sitting.
Reference texts. This exam code follows M. E. El-Hawary, Electrical Power Systems: Design and Analysis (IEEE Press) — the notation is his throughout: the bundle-conductor mean radii and long-line ABCD constants of Problem 1, the round-rotor power-angle relations of Problem 2, the three-winding equivalent circuit of Problem 3, the sequence-network fault reduction of Problem 6 and the equal-area criterion of Problem 7. Corroborating references: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, 6th ed. (ch. 4–5 transmission-line parameters, ch. 6 power flow, ch. 7–9 symmetrical components and faults, ch. 11 transient stability); J. J. Grainger and W. D. Stevenson, Power System Analysis; and S. J. Chapman, Electric Machinery Fundamentals, 5th ed., for the synchronous-machine and transformer chapters. Canadian practice references where protection and grounding are discussed: CSA C22.1 Canadian Electrical Code, Part I, CSA C22.3 for overhead systems, and IEEE C37 (protective relaying) and IEEE 142 Green Book (grounding).
Check: four judgement calls carried through the solutions, recorded here rather than hedged inside the answers.
(i) Problem 1 states neither a frequency nor a line length. Parts (a)–(c) need neither. Part (d) needs both, so 60 Hz is used (Canadian national exam) and the ABCD constants are given as closed-form functions of length, evaluated at a representative 400 km with a sensitivity table at 200 km and 600 km. The characteristic impedance and the surge-impedance loading do not depend on either quantity.
(ii) Problem 1 gives “bundle separation S” for a bundle whose figure shows eight sub-conductors on a circle. S is read as the distance between adjacent sub-conductors, so the bundle radius is $A = S/[2\sin(\pi/N)] = 0.6533\ \text{m}$ — the standard El-Hawary reading. If a marker instead reads S as the bundle diameter, every mean radius scales by the same factor and the method is unchanged.
(iii) Problem 5 says “the voltage at both sources is 1 p.u.” above a figure that draws three generators. All three are taken at 1.0 pu, which is the only reading that closes the network; the three internal e.m.f.s then merge into one node.
(iv) Problem 6: the zero-sequence source reactance is printed as $X_{0S}=j0.04$, possibly followed by a fifth digit. The value $j0.04$ is used. Reading it as $j0.045$ moves the zero-sequence Thévenin reactance from 0.19978 to 0.20051 pu and the fault current by 0.1 % — below the resolution of any answer here.
Problem 3: Three-Winding Transformer — Currents, Voltages, Powers and Efficiency (25 points)
Find. The primary current, the star-point (intermediate) voltage, the two load-side voltages referred to the primary, the three terminal apparent powers with their power factors, and the efficiency of the transformer under this loading.
Figure 3 — The star (T) equivalent circuit of the three-winding transformer, all quantities referred to the primary. The three branch impedances meet at the fictitious internal node V₀; the primary current is simply the sum of the two load currents because no magnetising branch is modelled.
Approach. The star equivalent has no shunt branch, so Kirchhoff’s current law at the internal node gives $I_1$ directly; one voltage drop then gives $V_0$, and one more drop along each load branch gives $V_2$ and $V_3$. Complex power at each terminal follows from $S=VI^{*}$, and the efficiency is the ratio of output to input real power. All phasors are r.m.s. and the calculation is per phase.
Part (a) — apply KCL at the star point. With no magnetising branch in the model, the primary current is the phasor sum of the two secondary-side currents:$$I_1=I_2+I_3=55\angle-30^\circ+45\angle-40^\circ.$$Resolving into rectangular form, $55\angle-30^\circ=47.6314-j27.5000$ and $45\angle-40^\circ=34.4723-j28.9254$, so $I_1=82.1037-j56.4254$ and$$\boxed{I_1=99.623\angle-34.499^\circ\ \text{A}.}$$The resultant angle lies between the two component angles and nearer the larger current, as it must.
Part (b) — drop the primary branch impedance. The internal node sits one impedance behind the terminal:$$V_0=V_1-I_1Z_p=410\angle0^\circ-(82.1037-j56.4254)(0.025+j0.08).$$The product is $(82.1037)(0.025)+(56.4254)(0.08)+j[(82.1037)(0.08)-(56.4254)(0.025)]=6.5666+j5.1590$, so $V_0=403.4334-j5.1590$ and$$\boxed{V_0=403.466\angle-0.732^\circ\ \text{V}.}$$Note the drop is dominated by its real part even though $Z_p$ is mainly reactive — the current lags, so the reactive drop projects onto the reference axis.
Part (c) — drop each load branch in turn. The two load-side voltages, both referred to the primary, are$$V_2=V_0-I_2Z_s,\qquad V_3=V_0-I_3Z_t.$$With $I_2Z_s=(47.6314-j27.5000)(0.025+j0.07)=3.1158+j2.6472$ and $I_3Z_t=(34.4723-j28.9254)(0.025+j0.08)=3.1758+j2.0347$,$$\boxed{V_2=400.394\angle-1.117^\circ\ \text{V},\qquad V_3=400.322\angle-1.029^\circ\ \text{V}.}$$
Check the regulation implied by those numbers. The secondary sees a voltage drop of $410-400.394=9.61\ \text{V}$, i.e. 2.34 per cent of the primary, and the tertiary 2.36 per cent. That the two land so close together is a coincidence of the data: the tertiary carries less current but through a larger reactance and at a worse power factor, and the two effects cancel almost exactly.
Part (d) — complex power at each terminal. Using $S=VI^{*}$ with the currents as defined (into the transformer at the primary, out of it at the two loads),$$\begin{aligned}S_1&=V_1I_1^{*}=33\,662+j23\,134\ \text{VA},\\ S_2&=V_2I_2^{*}=19\,282+j10\,637\ \text{VA},\\ S_3&=V_3I_3^{*}=14\,006+j11\,330\ \text{VA}.\end{aligned}$$The apparent powers are the magnitudes, and the power factor at each terminal is $\cos(\angle V-\angle I)$.
Evaluate the apparent powers and power factors. Taking magnitudes and angles of the three complex powers gives$$\boxed{\begin{aligned}|S_1|&=40.846\ \text{kVA},&\ \mathrm{pf}_1&=0.8241\ \text{lagging},\\|S_2|&=22.022\ \text{kVA},&\ \mathrm{pf}_2&=0.8756\ \text{lagging},\\|S_3|&=18.014\ \text{kVA},&\ \mathrm{pf}_3&=0.7775\ \text{lagging}.\end{aligned}}$$Every power factor is lagging because both load currents were specified with negative angles. The primary power factor is worse than the secondary and better than the tertiary, which is the expected blend — it is the weighted combination of the two loads plus the reactive absorption of the windings themselves.
Part (e) — efficiency from the real-power balance. Efficiency is output over input real power:$$\eta=\frac{P_2+P_3}{P_1}=\frac{19\,282.3+14\,005.7}{33\,662.4}=\frac{33\,288.0}{33\,662.4},$$so that$$\boxed{\eta=0.98888=98.89\ \text{per cent}.}$$
Confirm the loss by an independent route. The real loss must equal the $I^{2}R$ dissipation in the three branches:$$\begin{aligned}P_{\text{loss}}&=|I_1|^2R_p+|I_2|^2R_s+|I_3|^2R_t\\&=(99.623)^2(0.025)+(55)^2(0.025)+(45)^2(0.025)\\&=248.12+75.63+50.63=374.37\ \text{W},\end{aligned}$$which matches $P_1-P_2-P_3=33\,662.4-33\,288.0=374.4\ \text{W}$ to the last figure. The reactive balance closes in the same way: $23\,134-10\,637-11\,330=1167\ \text{var}$ absorbed by the three leakage reactances.
