Question 3 of 7: Frequency Change and the Cantilever Transformer Model

Question 3: Frequency Change and the Cantilever Transformer Model (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) — a 50 Hz transformer run at 60 Hz. The governing statement is the transformer EMF equation, $V \approx 4.44\,f\,N\,\Phi_{max}$. At a fixed applied voltage, peak flux is inversely proportional to frequency, so raising the frequency from 50 to 60 Hz drops the flux to $50/60 = 83.3\ \%$ of its design value. Everything else follows from that one line, and the practical shorthand is the volts-per-hertz ratio: the machine was designed for $V/f$ at 50 Hz and is now being run at five sixths of it.

Flux and magnetising requirement. With the core working 17 % further down its B–H curve, the core is de-saturated and the magnetising current falls more than proportionately — the relation is strongly non-linear near the knee. In circuit terms the magnetising reactance $X_m = 2\pi f L_m$ rises both because $f$ has risen and because the reduced flux places the core on a higher-permeability part of the curve, so $I_\phi = V/X_m$ falls on both counts. The magnetising reactive demand therefore drops.

Reactive power overall. Pulling the other way, every leakage reactance also scales with frequency, $X_{eq} \propto f$, rising by 20 %. At unchanged load current the leakage reactive demand $I^2X_{eq}$ therefore rises by 20 %, and the voltage regulation worsens. In a distribution transformer the magnetising branch dominates at light load and the leakage branch at full load, so the net reactive requirement falls at light load and rises at heavy load.

Losses. Copper loss $I^2R$ is essentially unchanged, since the load and hence the current are unchanged (skin and proximity effects add a percent or so at 60 Hz). Core loss splits: hysteresis loss follows $k_h f B_{max}^{n}$ with $n \approx 1.6$–2.0, and since $B_{max} \propto 1/f$ the product falls roughly as $f^{1-n}$, i.e. by about 15 % here; eddy-current loss follows $k_e f^2 B_{max}^2$, which is independent of frequency at constant applied voltage because the two effects cancel exactly. Total core loss therefore falls modestly.

Efficiency. With copper loss constant and core loss slightly reduced, the efficiency at any given loading improves by a few tenths of a percent. The transformer is also thermally and magnetically safe in this direction. The reverse operation is the dangerous one: a 60 Hz transformer on a 50 Hz supply is asked to carry 20 % more flux, drives the core into saturation, and draws a grossly distorted magnetising current with an enormous peak — which is why nameplate ratings are always quoted with their design frequency and why derating by the $V/f$ ratio is mandatory when a machine is moved to a lower-frequency system.

Given. A single-phase distribution transformer, all parameters referred to the 2400 V side.

Find. Primary voltage, primary current, input real and reactive power, primary power factor and efficiency at this loading.

Approach. In the cantilever model the excitation branch hangs across the primary terminals, so the series arm carries the load current alone. Work outward from the known secondary: find $I_2$, add the series drop to get $V_1$, put $V_1$ across the excitation branch to get $I_0$, and sum the currents. Nothing iterates.

1. Part (b)(i) — the referred load current. Referring the 240 V secondary to the 2400 V side gives $V_2' = 2400\angle 0^\circ$ V. The load draws $$I_2 = \frac{S_2}{V_2'}\angle -\arccos(0.8) = \frac{40\,000}{2400}\angle -36.87^\circ = 16.667\angle -36.87^\circ\ \text{A} = 13.333 - j10.000\ \text{A}$$ Referring rather than working on the low-voltage side is what keeps a single set of impedances valid throughout.
2. Add the series drop to reach the primary terminals. The whole of $I_2$ flows in $R_{eq} + jX_{eq}$, so $$V_1 = V_2' + I_2(R_{eq} + jX_{eq}) = 2400 + (13.333 - j10.000)(0.5 + j3.0)$$ $$= 2400 + (36.667 + j35.000) = 2436.67 + j35.00\ \text{V}$$ $$\boxed{V_1 = 2436.92\angle 0.823^\circ\ \text{V}}$$ That is a voltage regulation of $(2436.92 - 2400)/2400 = 1.538\ \%$ — unremarkable for a distribution unit at 80 % of rating.
3. Part (b)(ii) — the excitation current, then the primary current. This is the step the cantilever topology makes trivial: $V_1$ sits directly across $R_c \parallel jX_m$, so $$I_0 = V_1\!\left(\frac{1}{R_c} + \frac{1}{jX_m}\right) = (2436.67 + j35.00)\left(3.3333\times10^{-5} - j2.2222\times10^{-4}\right)$$ $$I_0 = 0.0890 - j0.5403 = 0.5476\angle -80.65^\circ\ \text{A}$$ The excitation current is almost purely magnetising, as the ratio $R_c/X_m = 6.7$ implies. Then $$I_1 = I_2 + I_0 = 13.422 - j10.540\ \text{A}$$ $$\boxed{I_1 = 17.066\angle -38.14^\circ\ \text{A}}$$ Note the excitation branch adds only 0.55 A to a 16.7 A load current but shifts the phase angle by 1.3° further into lag, because what it adds is almost entirely reactive.
4. Parts (b)(iii) and (iv) — input complex power. $$S_1 = V_1 I_1^{*} = (2436.67 + j35.00)(13.422 + j10.540)$$ $$\boxed{P_1 = 32.337\ \text{kW}}\qquad\boxed{Q_1 = 26.153\ \text{kvar}}$$
5. Part (b)(v) — primary power factor. The apparent power is $|S_1| = 41.589$ kVA, so $$\mathrm{pf}_1 = \frac{P_1}{|S_1|} = \frac{32.337}{41.589}$$ $$\boxed{\mathrm{pf}_1 = 0.7775\ \text{lagging}}$$ It is worse than the 0.80 of the load, and it must be: the transformer adds its own magnetising and leakage reactive demand on top of the load's, while adding almost nothing to the real power. Lagging is confirmed by $\angle V_1 = +0.82^\circ$ sitting ahead of $\angle I_1 = -38.14^\circ$.
6. Part (b)(vi) — efficiency. The output is the real power delivered to the load, $P_2 = 40\,000 \times 0.8 = 32.000$ kW, so $$\eta = \frac{P_2}{P_1} = \frac{32.000}{32.337}$$ $$\boxed{\eta = 98.96\ \%}$$
7. Close the loop on both balances. The losses can be built independently from the two dissipative elements: $$P_{core} = \frac{|V_1|^2}{R_c} = \frac{2436.92^2}{30\,000} = 197.95\ \text{W}, \qquad P_{cu} = |I_2|^2R_{eq} = 16.667^2(0.5) = 138.89\ \text{W}$$ Their sum, 336.84 W, reproduces $P_1 - P_2 = 32\,336.84 - 32\,000$ exactly. The reactive side closes the same way: the load takes $40\,000\sin 36.87^\circ = 24\,000$ var, the magnetising branch $|V_1|^2/X_m = 1319.68$ var and the leakage arm $|I_2|^2X_{eq} = 833.33$ var, totalling 26 153.0 var, which is $Q_1$. Two independent closures of this kind are worth more than re-checking the arithmetic.