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Pancyprian TS Elec. Eng. III ΘΚ study materials
Pancyprian TS Elec. Eng. III ΘΚ study resources

Pancyprian TS Elec. Eng. III ΘΚ: Electrical Machines and Transformers

Pancyprian TS Elec. Eng. III ΘΚ · Work through your exam, one topic at a time.

All guide sections

Electrical Machines and Transformers

This domain tests whether you can predict machine behaviour from a small number of equations and from standard test data: transformer losses and regulation, induction-motor slip and torque, synchronous-machine excitation, and DC-machine speed control. Cyprus questions typically combine a nameplate (for example 50 kVA, 11000/400 V, Dyn11) with an efficiency or starting-current calculation.

Core rules

  • Transformer EMF: E = 4.44 f N Φm. Turns ratio a = N1/N2 = V1/V2; current ratio reverses it.
  • The open-circuit test (on the LV side, rated voltage) gives iron loss and the magnetising branch; the short-circuit test (reduced voltage, rated current) gives full-load copper loss and the series impedance, from which the percentage impedance and regulation follow.
  • Efficiency η = output/(output + iron loss + copper loss). Copper loss varies with load squared, so maximum efficiency occurs when copper loss equals iron loss, at load fraction x = √(Piron/Pcu,FL).
  • Induction motor: Ns = 120f/p, slip s = (Ns − N)/Ns. Torque is proportional to sV²/R2 near standstill; direct-on-line starting current is typically 5–8 times full load, and star-delta reduces starting torque and current to one third.
  • Synchronous machine: excitation controls reactive power, the governor controls real power; parallel connection needs matching voltage, frequency, phase sequence and phase angle.
  • DC machine: E = kΦω (back EMF), T = kΦIa; speed control by armature voltage or field flux.

Worked example

A 50 kVA, 11000/400 V transformer has iron loss 400 W and full-load copper loss 1200 W. Full-load current on the LV side is 50000/(√3 × 400) = 72.2 A. At full load and 0.8 power factor the output is 40 kW, total loss is 1.6 kW, so η = 40/41.6 = 96.2%. Maximum efficiency is at x = √(400/1200) = 0.577, i.e. 57.7% load, where output is 23.1 kW and loss 0.8 kW, giving η = 23.1/23.9 = 96.7%.

Traps that cost marks

  • Assuming maximum efficiency occurs at full load.
  • Using kVA instead of kW as the output in an efficiency calculation.
  • Quoting slip as a percentage of rotor speed instead of of synchronous speed.
  • Forgetting that star-delta starting is only possible on a motor with six terminals rated for delta operation.
  • Ignoring the 30° phase shift implied by a Dyn11 vector group when paralleling transformers.