Electrical Fundamentals and Plant Services
What this domain tests
This domain covers the electrical principles and plant services that a power engineer must understand: motors, generators, transformers, protection, and the auxiliary services that support the plant.
Core rules to have at your fingertips
- Ohm's law and power: V = IR; P = VI = I²R = V²/R. Three-phase power P = √3·V_L·I_L·cos φ.
- AC motors: induction motors are the workhorses; slip = (N_s − N)/N_s, where N_s = 120f/p. Synchronous motors run at synchronous speed and can correct power factor.
- Generators: synchronous generators; excitation controls voltage; governor controls frequency.
- Transformers: V1/V2 = N1/N2 = I2/I1; losses are copper (I²R) and iron (hysteresis and eddy current).
- Protection: overcurrent, earth fault, differential, and distance protection. Circuit breakers interrupt fault current; fuses are one-shot.
- Plant services: compressed air, cooling water, HVAC, and emergency power (diesel generator, UPS).
Worked example
A 400 V, three-phase induction motor draws 50 A at a power factor of 0.85. Efficiency is 92 %.
Input power = √3 × 400 × 50 × 0.85 = 29 445 W ≈ 29.4 kW. Output power = 29.4 × 0.92 = 27.1 kW.
If the motor has 4 poles and is supplied at 50 Hz, synchronous speed N_s = 120 × 50 / 4 = 1 500 rpm. If it runs at 1 470 rpm, slip = (1 500 − 1 470)/1 500 = 0.02, i.e. 2 %.
Traps that cost marks
- Forgetting the √3 factor in three-phase power.
- Using line voltage where phase voltage is required (star vs delta).
- Confusing slip with efficiency.
- Assuming a fuse can be reset - it cannot.
- Ignoring power factor when calculating current.
Exam strategy
For motor questions, show input power, output power, and speed/slip separately. For protection questions, state the fault type and the device that clears it.