Power Eng 2nd Class study materials
Power Eng 2nd Class study resources

Power Eng 2nd Class: Electrical Systems and Motors

Power Eng 2nd Class · Work through your exam, one topic at a time.

All guide sections

Electrical Systems and Motors

The electrical paper is aimed at the operator and supervisor, not the electrician: you must compute motor load, recognise protection and starting arrangements, understand variable-frequency drives, and know what makes an arc flash hazardous. Expect one three-phase power calculation and several fault-recognition items.

Fundamentals

  • Three-phase power: kW = 1.732 x V(line) x I(line) x power factor / 1000.
  • Synchronous speed = 120 x frequency / poles. Slip = (ns - nr)/ns, typically 1-5% at full load.
  • Power factor correction capacitors reduce line current and kVA but never reduce kW. Over-correcting a motor that can be driven by its load risks over-voltage.

Motors and starting

  • Direct-on-line starting draws 5-8 times full-load current. Star-delta reduces starting current to one third and torque to one third. Soft starters and VFDs ramp voltage and frequency.
  • VFDs bring cable-length limits, bearing currents, harmonics and reduced cooling at low speed; motors on VFD service often need separately driven fans and inverter-duty insulation.
  • Protection: overload (thermal or electronic), short circuit (fuse or MCCB), earth fault, phase loss, under/over-voltage, winding temperature.
  • Insulation resistance testing: compare with previous readings and temperature-corrected limits. A falling trend predicts failure better than any single value.

Distribution and safety

  • Transformers, switchgear, motor control centres, earthing/grounding. Arc-flash incident energy depends on available fault current and clearing time; the label defines the PPE category. Work de-energised.
  • Lockout: isolate, lock, tag, prove dead, ground where required.
  • Standby generators: match voltage, frequency and phase angle before paralleling, and verify reverse-power protection.

Worked example

A 480 V three-phase motor draws 60 A at 0.85 power factor. Input kW = 1.732 x 480 x 60 x 0.85/1000 = 42.4 kW. Against a 45 kW nameplate, that is about 94% load - acceptable. Correcting power factor to 0.95 drops line current to 42,400/(1.732 x 480 x 0.95) = 53.7 A, which relieves cables and the starter, but the motor's kW, losses and heating are unchanged.

Traps

  • Substituting phase voltage for line voltage in the root-three formula.
  • Meggering a motor without isolating the VFD, which damages the drive.
  • Believing capacitors cut energy consumption rather than kVA demand and line losses.
  • Assuming a motor at nameplate kW is at nameplate current; power factor and efficiency decide that.