Question 3 of 7: Transformer Behaviour on a Changed Supply Frequency; Performance from the Cantilever Equivalent Circuit

Question 3: Transformer Behaviour on a Changed Supply Frequency; Performance from the Cantilever Equivalent Circuit (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.

Part (a) — a 50 Hz transformer operated at 60 Hz, load unchanged.

Everything follows from the EMF equation $V\approx4.44\,N\,f\,\phi_{max}$. If the same rated voltage is applied at a frequency twenty per cent higher, the peak core flux must fall in inverse proportion, to $50/60=0.833$ of its design value. That is the safe direction: the core moves further away from saturation, and none of the thermal or mechanical ratings of the winding are threatened. (The dangerous case is the reverse one, a 60 Hz transformer energised at 50 Hz, where the flux would rise by twenty per cent and drive the core hard into saturation.)

Reactive-power requirement. The magnetising reactance is proportional to frequency, $X_m=2\pi fL_m$, so at constant applied voltage the magnetising current falls as $V/f$, again to about eighty-three per cent, and the magnetising reactive power $V^{2}/X_m$ falls in the same proportion. Because the core is less saturated the incremental permeability is also a little higher, so the real reduction is slightly greater than the linear estimate. Working against this, the leakage reactance $X_{eq}$ is likewise proportional to frequency and therefore rises by twenty per cent, so the reactive power absorbed in the series path, $I^{2}X_{eq}$, increases by twenty per cent at unchanged load current. In a distribution transformer the magnetising branch dominates at light load and the leakage term dominates near full load, so the net change in reactive demand is a clear reduction at light load and roughly a wash at full load.

Losses and efficiency. Copper loss is unchanged, because the load and hence the current are unchanged. Core loss splits into two parts that behave differently. Hysteresis loss follows $P_h\propto fB_{max}^{n}$ with $n$ near 1.6 to 2.0; substituting $B_{max}\propto V/f$ gives $P_h\propto V^{n}f^{1-n}$, so at constant voltage it falls, by about seventeen per cent for $n=2$. Eddy-current loss follows $P_e\propto f^{2}B_{max}^{2}\propto V^{2}$ and is therefore essentially unchanged. Total core loss thus decreases and total loss decreases with it, so efficiency improves slightly — on a unit of the size treated in parts (b) to (d), by roughly a tenth of a percentage point.

Primary power factor and regulation. Two opposing effects again. The smaller exciting current removes lagging reactive current at the primary terminals, which improves the power factor; the larger leakage reactance adds lagging reactive absorption in the series arm, which worsens it, and also worsens voltage regulation on a lagging load because the $I X_{eq}\sin\phi$ term in the approximate regulation expression grows by twenty per cent. At light and medium load the exciting current dominates and the primary power factor improves; at heavy lagging load the leakage term wins and both the power factor and the regulation deteriorate marginally. A separate practical caution is that the transformer, if it carries cooling fans or an oil pump driven from an auxiliary supply, will see those motors run twenty per cent faster, which is usually beneficial for cooling.

Given. A 50 kVA, 2400/240 V, 60 Hz single-phase distribution transformer with all parameters referred to the high-voltage side, feeding a stated secondary load.

Find. The primary voltage and current, the active and reactive power drawn at the primary, and the efficiency and primary power factor of the transformer at this load.

[Figure not reproduced: Figure 3.1 — the cantilever equivalent circuit of Figure (1) of the examination paper, all parameters referred to the high-voltage side. See the official exam paper.]

Approach. In the cantilever model the entire excitation branch hangs across the primary terminals, so the load current passes through $R_{eq}+jX_{eq}$ alone. Working from the known secondary voltage outward, add the series drop to find $V_p$, then compute the excitation current from $V_p$ and add it to the referred load current. No iteration is required.

1. Part (b) — refer the load to the high-voltage side. The turns ratio is $a=2400/240=10$, so the referred secondary voltage is $aV_r=2400\,/\,0^{\circ}$ V, taken as the reference phasor. The secondary current is $I_s=40\,000/240=166.667$ A, and referring it to the primary divides it by $a$: $$\frac{I_s}{a}=16.6667\,/\,{-36.8699^{\circ}}=13.3333-j10.0000\ \text{A}$$ The angle is $\arccos(0.8)$ and it is negative because the load is lagging.
2. Add the series-arm drop to reach the primary terminals. In the cantilever circuit the whole of $I_s/a$ flows through the series arm, so $$V_p=aV_r+(R_{eq}+jX_{eq})\frac{I_s}{a}=2400+(1.5+j2.0)(16.6667\,/\,{-36.8699^{\circ}})$$ The series impedance is $2.5\,/\,53.1301^{\circ}\ \Omega$, so the drop is $41.6667\,/\,16.2602^{\circ}=40.000+j11.667$ V. Hence $$\boxed{V_p=2440.00+j11.667=2440.03\,/\,0.2740^{\circ}\ \text{V}}$$ The regulation implied is $(2440.03-2400)/2400=1.67$ per cent, which is a believable figure for a distribution transformer at eighty per cent of rating.
3. Compute the two excitation currents from that primary voltage. Because the excitation branch is connected across $V_p$ itself, both of its currents follow in one line each: $$I_c=\frac{V_p}{R_c}=\frac{2440.00+j11.667}{30\,000}=0.081333+j0.000389\ \text{A}$$ $$I_m=\frac{V_p}{jX_m}=\frac{2440.00+j11.667}{j4500}=0.002593-j0.542222\ \text{A}$$ Their sum is $I_{h+e}=0.083926-j0.541833=0.548295\,/\,{-81.195^{\circ}}$ A, a current of about 0.55 A against a rated primary current of 20.8 A — some 2.6 per cent, which is typical.
4. Primary current. Kirchhoff's current law at the primary node gives $$I_p=\frac{I_s}{a}+I_{h+e}=(13.3333+0.083926)+j(-10.0000-0.541833)$$ $$\boxed{I_p=13.4173-j10.5418=17.0632\,/\,{-38.1564^{\circ}}\ \text{A}}$$ The exciting current has pushed the primary current angle about 1.3 degrees further into lag than the referred load current, which is the whole of its effect here.
5. Part (c) — complex power at the primary terminals. $$S_p=V_pI_p^{*}=(2440.00+j11.667)(13.4173+j10.5418)$$ Multiplying out, the real part is $2440.00(13.4173)-11.667(10.5418)$ and the imaginary part is $2440.00(10.5418)+11.667(13.4173)$, giving $$\boxed{P_p=32.6151\ \text{kW}\qquad Q_p=25.8786\ \text{kVAr}}$$ with $|S_p|=41.6347$ kVA.
6. Confirm the input power by an independent loss audit. The output is $P_{out}=40\,000\times0.8=32\,000$ W. The two losses are the core loss $|V_p|^{2}/R_c=(2440.03)^{2}/30\,000=198.46$ W and the copper loss $|I_s/a|^{2}R_{eq}=(16.6667)^{2}(1.5)=416.67$ W, a total of 615.12 W. Adding this to the output reproduces $32\,615.12$ W exactly, which validates the whole phasor calculation. The same audit works on the reactive side: $|V_p|^{2}/X_m+|I_s/a|^{2}X_{eq}+Q_{load}=1323.05+555.56+24\,000=25\,878.6$ VAr.
7. Part (d) — efficiency. $$\eta=\frac{P_{out}}{P_{in}}=\frac{32\,000}{32\,615.12}=\boxed{\eta=0.98114\ \ (98.11\ \text{per cent})}$$ The loss split is instructive: copper loss is more than twice the core loss at this load, so the unit is being worked past its maximum-efficiency point, which occurs where the two are equal — here at about $\sqrt{198.46/416.67}=0.69$ of the present load current, that is around 27.6 kVA.
8. Primary power factor. The angle between $V_p$ and $I_p$ is $0.2740^{\circ}-(-38.1564^{\circ})=38.4304^{\circ}$, so $$\boxed{\text{p.f.}_{p}=\cos(38.4304^{\circ})=0.7834\ \text{lagging}}$$ which agrees with $P_p/|S_p|=32\,615.12/41\,634.71=0.78336$. It is noticeably poorer than the load's own 0.800 because the exciting current and the leakage reactance both add lagging reactive power that the source must supply on top of the load's 24 kVAr.