Transformers

A transformer is a static machine that transfers electrical energy between circuits by mutual induction, changing voltage and current levels at the same frequency. It is the most common machine in any power system, from grid substation transformers to 11 kV/400 V distribution transformers. This topic covers construction and accessories, no-load and loaded operation, the equivalent circuit, open- and short-circuit tests, losses and efficiency, three-phase connections, parallel operation, rating and temperature rise.

Key concepts

• Principle: an alternating flux in the core links both windings; $E = 4.44fN\Phi_m$. Works only on AC (on DC the primary would draw a huge current and burn).
• Core type vs shell type:

Core type	Shell type
Windings surround the limbs	Core surrounds the windings
Single magnetic path (two limbs)	Two parallel flux paths (central limb carries full flux)
Easier to repair/insulate; preferred for HV	Better mechanical strength, less leakage; used for LV/high current

• Core: CRGO silicon steel laminations (0.25–0.35 mm) insulated by varnish/oxide to limit eddy currents; stepped cross-section for circular coils.
• Accessories:
  • Tank: holds core-winding and oil; cooling tubes/radiators.
  • Transformer oil: mineral oil for insulation and cooling; good oil has high breakdown voltage (BDV, tested across a 2.5 mm gap), low moisture and low acidity.
  • Conservator: small drum above the tank that lets oil expand/contract, keeping the main tank full and limiting oil–air contact.
  • Breather: silica gel (blue when dry, pink when moist) dries the air entering the conservator.
  • Buchholz relay: gas-actuated relay between tank and conservator; alarm for slow gas (incipient faults), trip for oil surge. Fitted on larger conservator-type transformers.
  • Explosion vent / pressure relief device: diaphragm that bursts to release pressure during a severe internal fault.
  • Bushings: porcelain/condenser-type insulators bringing leads out of the tank.
  • Others: tap changer, temperature indicators (oil and winding), oil level gauge, cooling fans/pumps.
• No-load operation: primary draws a small $I_0$ (2–5% of rated), lagging V by nearly 90°. Components: magnetising $I_\mu = I_0\sin\phi_0$ and core-loss $I_w = I_0\cos\phi_0$. No-load pf is low (≈0.1–0.3).
• On load: secondary current produces opposing MMF; primary draws extra current $I_2' = I_2 N_2/N_1$ so core flux stays nearly constant from no load to full load.
• Equivalent circuit: series $R_1, X_1$ and $R_2', X_2'$ (leakage), shunt $R_0 \parallel X_0$ (core loss and magnetising). Referred quantities: $R_2' = R_2/K^2$ when referred to primary, where $K = N_2/N_1$.
• Tests:
  • Open-circuit test (on LV side, HV open, rated voltage): gives core (iron) loss and $R_0, X_0$.
  • Short-circuit test (on HV side, LV shorted, reduced voltage ≈ 5–10% to circulate rated current): gives full-load copper loss and $R_{eq}, X_{eq}, Z_{eq}$ (% impedance).
  • Others: Sumpner's (back-to-back) heat-run test, polarity, ratio, insulation resistance, oil BDV.
• Losses: iron loss — hysteresis $P_h \propto B_m^{1.6}f$ (reduced by silicon steel) and eddy current $P_e \propto B_m^2f^2t^2$ (reduced by thin laminations); constant at constant V, f — and copper loss ($I^2R$, varies with load²). No mechanical losses, so efficiency is very high (95–99.5%).
• Maximum efficiency occurs when copper loss = iron loss, i.e. at load fraction $x = \sqrt{P_i/P_{cu,FL}}$.
• All-day efficiency = output kWh / input kWh over 24 h; important for distribution transformers which are energised all day but lightly loaded, so they are designed with low iron loss.
• Voltage regulation: $\%R \approx \dfrac{I(R\cos\phi \pm X\sin\phi)}{V}\times100$ (+ lagging, − leading). It can be zero or negative at leading pf.
• Three-phase connections: Y–Y, Δ–Δ, Y–Δ (step-down at substations), Δ–Y (step-up at generating stations, and distribution 11 kV Δ / 400 V Y with earthed neutral). Δ–Y and Y–Δ give 30° phase shift. Open-delta (V–V) gives 57.7% of Δ–Δ capacity.
• Parallel operation conditions: same voltage ratio, same polarity, same phase sequence and phase displacement (vector group) — essential; equal % impedance and similar X/R ratio — for proper load sharing.
• Rating is in kVA (not kW) because losses depend on V and I, not on load pf.
• Temperature rise: limited by insulation class; heat must be removed by cooling — ONAN, ONAF, OFAF, OFWF etc. (O = oil, A = air, N = natural, F = forced, W = water).

High-yield points

• $E = 4.44fN\Phi_m$; core flux depends on V/f.
• OC test (LV side) gives iron loss; SC test (HV side, reduced voltage) gives full-load copper loss and $Z_{eq}$.
• Maximum efficiency when copper loss = iron loss, at $x = \sqrt{P_i/P_{cu,FL}}$.
• Open-delta (V–V) capacity = 57.7% of Δ–Δ; Δ–Y and Y–Δ give a 30° shift.
• Paralleling essentials: same ratio, polarity, phase sequence and vector group.

Formulas

$E_1 = 4.44 f N_1 \Phi_m,\qquad K = \frac{N_2}{N_1} = \frac{E_2}{E_1} = \frac{I_1}{I_2}$

$\eta = \frac{xS\cos\phi}{xS\cos\phi + P_i + x^2P_{cu}}$ x = fraction of full load, S = rated VA, $P_i$ = iron loss, $P_{cu}$ = full-load copper loss.

Worked example

A 100 kVA transformer has iron loss 1 kW and full-load copper loss 1.44 kW. At what load is efficiency maximum, and what is it at unity pf?

1. $x = \sqrt{1/1.44} = 0.833$ → 83.3 kVA.
2. Output = 83.3 kW; losses = 1 + 1 = 2 kW.
3. $\eta_{max} = 83.3/85.3 = 97.66\%$.

Exam tips

• OC test → iron loss (LV side); SC test → copper loss (HV side).
• Max efficiency when Cu loss = iron loss — independent of pf, but its value is highest at unity pf.
• Buchholz relay works only for internal faults in oil-filled transformers with conservators.
• Polarity and vector group must match for paralleling; unequal % impedance only causes unequal sharing.
• kVA rating; $E = 4.44fN\Phi_m$; flux depends on V/f.

Exam-style questions

From Calmon's question bank, which has 25 questions on this topic.

1. Transformer oil in a power transformer mainly serves to provide:

• A. Insulation and cooling
• B. Lubrication of the core
• C. Arc quenching in bushings only
• D. Magnetic shielding

Answer: A. Mineral transformer oil insulates the windings and carries heat away from the core and windings to the tank and radiators.

2. If a 230 V, 50 Hz transformer primary is connected to a 230 V DC supply, the most likely result is:

• A. The core loss becomes very large but current stays normal
• B. The secondary gives 230 V DC
• C. The winding draws a very large current and may burn out, as there is no back emf
• D. The transformer works normally with lower losses

Answer: C. With DC there is no changing flux, so no back emf; the current is limited only by the small winding resistance and becomes dangerously high.

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