22-Elec-B7 Power Systems Engineering · December 2018

Question 3 of 7: Transformer Losses and the Cantilever Equivalent Circuit

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2018 — 16-Elec-B7 Power Systems Engineering, 3 hours, open book, any non-communicating calculator permitted. The paper prints seven questions, headed “Problem 1” to “Problem 7”; the rubric states that any five questions constitute a complete paper and that all questions are of equal value. Every one of the seven is solved in full here, because the set is a study resource rather than a three-hour sitting. Sub-part point values are quoted as printed; note that the printed splits total 25 points on Problems 1, 2, 4, 5 and 6, 20 points on Problem 3, and Problem 7 carries no printed split at all — under the “equal value” rubric each question is worth one fifth of the paper regardless.

Reference texts. M. E. El-Hawary, Electrical Power Systems: Design and Analysis, rev. ed., IEEE Press / Wiley — the reference text named in the Engineers Canada syllabus for this examination code, and the source of the “cantilever” transformer model used in Problem 3. Supporting texts: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, 6th ed.; J. J. Grainger and W. D. Stevenson, Power System Analysis; S. J. Chapman, Electric Machinery Fundamentals, 5th ed. Canadian practice context (protection philosophy, grounding) follows CSA and IEEE C37 relaying practice as applied by Canadian transmission utilities.

Check — two places where the printed data are not internally consistent, and what was done about each.

(1) Problem 1. The pair A = 0.95∠2°, B = 195∠87.5° forces a shunt parameter with a large real part (Re C = 3.00 × 10⁻⁴ S). No real transmission line has a shunt conductance of that size — on a physical line C lands within about a degree of +90°. The consequence is a transmission efficiency of only 80.9 %, of which about 156 MW of the 236 MW loss is charged to that fictitious conductance. The arithmetic below is exactly what the printed constants give and is what the marker will be looking for; the physical caveat is stated so the low efficiency is not read as a slip.

(2) Problem 6. Table (2) lists the line negative-sequence reactance as 0.320 pu against a positive-sequence 0.30 pu. A transmission line is a static, symmetric element, so physically X₋ must equal X₊. The printed 0.320 is used throughout (the paper governs). Taking 0.30 instead moves the answer to Problem 6(b) from 1.4169∠145.58° pu to 1.4212∠146.04° pu — a change of 0.3 %, so neither reading changes any conclusion.

Question 3: Transformer Losses and the Cantilever Equivalent Circuit (20 marks)

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.

Part (a) — how frequency acts on each loss mechanism. A transformer's losses divide into those that live in the core and those that live in the windings, and frequency touches them by entirely different routes.

Hysteresis loss follows the Steinmetz relation $P_h = k_h f B_{max}^{n}$ with the exponent $n$ between about 1.6 and 2.0. Taken alone that expression says the loss rises in direct proportion to frequency, but the qualifier matters: at a fixed applied voltage the peak flux density is set by Faraday's law, $B_{max} \propto V/(f N A)$, so raising the frequency at constant voltage lowers the flux density in the same proportion. The two effects fight, and the hysteresis loss actually falls roughly as $f^{-(n-1)}$, i.e. as $1/f^{0.6}$ to $1/f$. Raising frequency at constant volts per hertz instead holds $B_{max}$ fixed, and then the loss really does rise linearly with $f$.

Eddy-current loss in the core obeys $P_e = k_e f^{2} t^{2} B_{max}^{2}$, where $t$ is the lamination thickness. At constant applied voltage the $f^{2}$ and the $B_{max}^{2} \propto 1/f^{2}$ cancel exactly, so the eddy loss is essentially independent of frequency and depends only on the applied voltage and the lamination design. This is the reason a 50 Hz transformer operated at 60 Hz on the same voltage does not overheat its core — the hysteresis component actually drops and the eddy component is unchanged.

Winding (copper) loss is $I^{2}R$ and at first sight has no frequency term at all. It acquires one through skin effect and proximity effect: the a.c. resistance rises with the square root of frequency once the conductor dimension becomes comparable with the skin depth $\delta = \sqrt{2\rho/(\omega\mu)}$. At 50–60 Hz the skin depth in copper is about 9 mm, so the effect is negligible in small units but is real in large transformers with heavy conductors, which is why those windings are subdivided into transposed strands. Stray loss in the tank, clamps and core bolts, driven by leakage flux, also scales with roughly $f^{2}$ at fixed current.

The practical upshot for design is the reason aircraft and switch-mode power supplies use 400 Hz and higher: at fixed volts per hertz a higher frequency permits a proportionally smaller core, so the transformer shrinks dramatically — provided the lamination thickness (or a ferrite core) is reduced to keep the eddy loss in check.

Given.

Quantity	Value
Rating, voltages, frequency	25 kVA, 2200/220 V, 60 Hz, single-phase
Turns ratio $a = 2200/220$	10
Series arm (HV referred) $R_{eq}$, $X_{eq}$	$5.4\ \Omega$, $21\ \Omega$
Magnetising reactance $X_m$	$20\,000\ \Omega$
Core-loss resistance $R_c$	$37\,500\ \Omega$

Find. The secondary ammeter and wattmeter readings in a short-circuit test with 20 V on the secondary and the primary shorted; the primary ammeter and wattmeter readings in an open-circuit test with 2250 V on the primary; and the primary voltage when the transformer supplies 15 kVA at 220 V and 0.8 lagging.

El-Hawary's cantilever equivalent circuit. The whole excitation branch sits across the PRIMARY terminals and the series arm lies between that node and the referred secondary — which is what makes both tests closed-form rather than iterative.

Approach. The cantilever arrangement is the key to all three parts. Because the excitation branch is connected directly across the primary terminals rather than at an interior node, shorting the primary removes the excitation branch entirely from the circuit, and opening the secondary removes the series arm entirely. Each test therefore isolates one branch exactly, with no approximation.

Part (b) — refer the applied test voltage to the high-voltage side. The 20 V is applied at the 220 V terminals, so at the referred secondary node of Figure (1) it appears as $$a V_s = 10 \times 20 = 200\ \text{V}$$ With the primary short-circuited, $V_p = 0$; the excitation branch is connected across exactly that pair of terminals, so it is shorted out and carries no current whatever. Only the series arm is left.
Compute the current in the series arm. The series impedance and its magnitude are $$\begin{aligned}Z_{eq} &=5.4+j21\ \Omega \\ |Z_{eq}| &=\sqrt{5.4^{2}+21^{2}}=21.683\ \Omega\end{aligned}$$ so the referred current is $$\left|\frac{I_s}{a}\right|=\frac{200}{21.683}=9.2237\ \text{A}$$ The ammeter, however, sits on the secondary side and reads the actual secondary current, which is $a$ times larger: $$\boxed{|I_s| = 10 \times 9.2237 = 92.24\ \text{A}}$$ That is 0.81 of the rated secondary current of $25\,000/220 = 113.6$ A, a normal level for a short-circuit test.
Compute the wattmeter reading. Real power is invariant under referral, so it can be evaluated on either side. Using referred quantities, $$P_{sc}=\left|\frac{I_s}{a}\right|^{2}R_{eq}=(9.2237)^{2}(5.4)$$ $$\boxed{P_{sc}=459.4\ \text{W}}$$ The identical answer comes from the secondary side directly, since $R_{eq}$ referred to the low-voltage side is $5.4/a^{2} = 0.054\ \Omega$ and $(92.24)^{2}(0.054) = 459.4$ W. Because the excitation branch is out of circuit, this reading is pure copper loss — which is exactly the point of the test.
Part (c) — the open-circuit test isolates the excitation branch. With the secondary open the series arm carries no current at all, so the whole 2250 V appears across $R_c$ in parallel with $jX_m$. The two component currents are $$\begin{aligned}I_{R_c} &=\frac{2250}{37\,500}=0.06000\ \text{A} \\ I_m &=\frac{2250}{20\,000}=0.11250\ \text{A}\ \text{(lagging)}\end{aligned}$$ so the primary ammeter reads their phasor sum $$\begin{aligned}I_p &= 0.06000 - j0.11250\ \text{A} \\ \boxed{|I_p| = 0.1275\ \text{A}}\end{aligned}$$ at an angle of $-61.93^\circ$, i.e. a no-load power factor of 0.4706 lagging — characteristically poor, as it always is at no load.
Open-circuit wattmeter reading. The magnetising branch is purely reactive and consumes no real power, so the whole reading is core loss: $$\boxed{P_{oc}=\frac{V_p^{2}}{R_c}=\frac{2250^{2}}{37\,500}=135.0\ \text{W}}$$
Part (d) — convert the load to a referred current. At 15 kVA and 220 V the secondary current magnitude is $$|I_s|=\frac{15\,000}{220}=68.182\ \text{A}$$ and at 0.8 lagging its angle relative to the secondary voltage is $-36.87^\circ$. Referred to the high-voltage side, $$\frac{I_s}{a}=6.8182\angle -36.87^\circ = 5.4545 - j4.0909\ \text{A}$$ Taking the referred secondary voltage as reference, $a V_s = 10 \times 220 = 2200\angle 0^\circ$ V.
Walk the drop across the series arm to the primary terminals. The cantilever model gives $V_p$ in one step, with no contribution from the excitation branch because that branch hangs off the primary node itself: $$V_p=(R_{eq}+jX_{eq})\frac{I_s}{a}+aV_s$$ The drop term evaluates to $$\begin{aligned}(5.4+j21)(5.4545-j4.0909) &= 115.364+j92.455 \\ &= 147.84\angle 38.71^\circ\ \text{V}\end{aligned}$$ so that $$\boxed{V_p = 2315.36 + j92.45 = 2317.21\angle 2.287^\circ\ \text{V}}$$ The primary must therefore be held about 117 V, or 5.33 %, above nominal to hold rated voltage at this load — which is the transformer's voltage regulation at 0.8 lagging.

Quantity	Result
(b) Secondary ammeter, short-circuit test	92.24 A
(b) Secondary wattmeter, short-circuit test	459.4 W
(c) Primary ammeter, open-circuit test	0.1275 A at $-61.93^\circ$ (pf 0.4706 lagging)
(c) Primary wattmeter, open-circuit test	135.0 W
(d) Primary voltage at 15 kVA, 0.8 lagging	$2317.21\angle 2.287^\circ$ V (5.33 % regulation)

Topic. Transformer loss mechanisms and their frequency dependence; the cantilever (L-section) equivalent circuit and the two classical tests.

Key relations. $P_h = k_h f B_{max}^{n}$ and $P_e = k_e f^{2}t^{2}B_{max}^{2}$ with $B_{max} \propto V/f$ at constant applied voltage; $a = N_1/N_2$; short-circuit test with the primary shorted gives $|I| = aV_s/|Z_{eq}|$ and $P = |I/a|^{2}R_{eq}$; open-circuit test gives $P = V_p^{2}/R_c$ and $|I_p| = V_p\sqrt{1/R_c^{2}+1/X_m^{2}}$; loaded operation $V_p = Z_{eq}(I_s/a) + aV_s$.

Why this works. The exact equivalent circuit places the excitation branch between the primary leakage impedance and the referred secondary leakage impedance, which makes every calculation iterative because the current through the primary leakage arm is not the load current. The cantilever model moves the whole excitation branch out to the primary terminals and lumps both leakage impedances into a single series arm. The price is a small error in the no-load voltage drop; the reward is that a primary short kills the excitation branch and an open secondary kills the series arm, so each test isolates one branch exactly and every part of this question closes in one line. That is why El-Hawary adopts it, and why the question directs the candidate to Figure (1) rather than leaving the model open.

Common pitfall. The commonest error in part (b) is reporting the referred current, 9.22 A, as the ammeter reading. The instrument is physically on the 220 V side and reads 92.24 A; the factor of $a$ must be put back. A second trap is assuming the excitation branch still draws current during the short-circuit test — true for the exact circuit, false for the cantilever model the question specifies. In part (a), the trap is quoting “hysteresis loss is proportional to frequency” without stating whether the applied voltage or the volts-per-hertz ratio is being held constant; the two assumptions give opposite trends.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, ch. 4 (the power transformer, cantilever equivalent circuit, open- and short-circuit tests); Chapman, Electric Machinery Fundamentals, 5th ed., ch. 2 §2.5–2.7.

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