Question 2 of 5: Transformers

Question 2: Transformers (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.

Parts (a)–(c) — Concept questions

(a) A transformer cannot operate on DC. Transformer action depends on Faraday's law: the secondary emf is \(e_2=-N_2\,d\Phi/dt\), which requires a time-varying flux. A steady DC excitation produces a constant flux, so after the initial switch-on transient there is no changing flux and therefore no induced secondary voltage. Worse, with DC the primary is limited only by its own small winding resistance (there is no back-emf and no inductive reactance to the steady component), so it draws a very large current and overheats. A transformer is inherently an AC device.

(b) A secondary short-circuit reflects a very large current into the primary. A short on the secondary removes the load impedance, so the only thing limiting current is the transformer's small equivalent (leakage) impedance. The secondary current rises to a high fault value, and because the ampere-turns must balance (\(N_1I_1\approx N_2I_2\)) the primary current rises in proportion, \(I_1\approx I_2/a\) plus the small magnetizing current. Unless the primary is protected by a fuse or breaker, this fault current will overheat and damage the windings; this is exactly the condition deliberately (and briefly) created in the short-circuit test, where a much-reduced applied voltage is used to hold the current to rated value.

(c) Heating is essentially the same for resistive and inductive loads of equal VA. A transformer is rated in volt-amperes, not watts, precisely because its losses depend on voltage and current magnitude rather than on the load power factor. The core (iron) loss is fixed by the applied voltage and frequency, which are the same in both cases; the copper loss is \(I^2R_{eq}\), and for the same VA at the same voltage the current magnitude is identical whether the load is resistive or inductive. Since both loss components are unchanged, the total heating is approximately the same — a purely inductive load at the rated VA heats the transformer as much as a resistive load at the same VA, even though it delivers no real power to the load.

Part (d) — Equivalent-circuit calculations

Given. Single-phase transformer, 300 kVA, 11 kV/2.2 kV, 60 Hz; parameters referred to the HV (11 kV) side:

Find. (i) no-load current as % of full-load; (ii) core loss; (iii) no-load power factor; (iv) full-load copper loss; (v) voltage regulation with \(Z_{load}=16\angle60^\circ\ \Omega\) on the LV side.

Approach. The full-load HV current is \(S/V_{HV}\). The no-load current is the magnetizing-branch current at rated voltage; core loss is \(V^2/R_c\). Full-load copper loss is \(I_{FL}^2R_{eq}\). Voltage regulation is obtained from the approximate circuit referred to the LV side.

1. Full-load and no-load currents (HV side). $$I_{FL}=\frac{S}{V_{HV}}=\frac{300\,000}{11\,000}=27.27\text{ A}.$$ The shunt-branch components at rated voltage are \(I_c=V/R_c=11\,000/57\,600=0.191\text{ A}\) and \(I_m=V/X_m=11\,000/16\,340=0.673\text{ A}\), so $$I_0=\sqrt{I_c^2+I_m^2}=\sqrt{0.191^2+0.673^2}=0.700\text{ A}.$$ As a percentage of full load, \(\dfrac{I_0}{I_{FL}}=\dfrac{0.700}{27.27}=\boxed{2.57\%}.\)
2. No-load power loss (core loss). Only the core-loss resistance dissipates power at no load: $$P_{core}=\frac{V^{2}}{R_c}=\frac{(11\,000)^2}{57\,600}=2.10\text{ kW}.$$
3. No-load power factor. The no-load real power over the no-load apparent power (equivalently \(I_c/I_0\)): $$\text{pf}_0=\frac{P_{core}}{V\,I_0}=\frac{2101}{11\,000(0.700)}=0.273\text{ lagging.}$$
4. Full-load copper loss. $$P_{cu}=I_{FL}^{2}\,R_{eq}=(27.27)^2(2.784)=2.07\text{ kW}.$$
5. Voltage regulation (approximate circuit, referred to LV). Refer the series impedance to the LV side with \(a=5\): \(R_{eq}'=2.784/25=0.1114\ \Omega\), \(X_{eq}'=8.45/25=0.338\ \Omega\). The LV load current is $$I_2=\frac{V_2}{Z_{load}}=\frac{2200}{16\angle60^\circ}=137.5\angle{-60^\circ}\text{ A} \quad(\text{pf }0.5\text{ lagging}).$$ With the rated secondary voltage \(V_2=2200\angle0^\circ\) as reference, the primary voltage referred to the LV side is $$E_2=V_2+I_2\,(R_{eq}'+jX_{eq}') =2200+(137.5\angle{-60^\circ})(0.1114+j0.338)=2248\angle0.25^\circ\text{ V},$$ so $$\text{VR}=\frac{|E_2|-V_2}{V_2}=\frac{2248-2200}{2200}=\boxed{2.18\%}.$$