Question 1

How do I calculate three-phase power consumption?

Accepted Answer

Three-phase active power: P (kW) = √3 × VL × IL × cos(φ) / 1000, where VL is line voltage (V), IL is line current (A), and cos(φ) is power factor. For a 400V, 100A, 0.85 PF load: P = 1.732 × 400 × 100 × 0.85 / 1000 = 58.8 kW. Three-phase is standard for industrial motors in South Africa, operating at 400V (line-to-line) or 230V (phase-to-neutral).

Question 2

What is power factor and why does it matter?

Accepted Answer

Power factor (PF) = Active Power (kW) / Apparent Power (kVA). It measures how efficiently electrical power is used. PF = 1.0 is ideal (purely resistive). Motors, transformers, and inductive loads typically have PF = 0.7–0.9. Low PF means higher current for the same useful power, causing: larger cables, transformers, and switchgear; higher electricity bills (some utilities charge kVA, not just kWh); voltage drops in distribution systems. Eskom and municipalities often penalise PF below 0.95.

Question 3

How do I use Ohm's Law for power calculations?

Accepted Answer

Ohm's Law: V = I × R. Power formulas derived from this: P = V × I = I² × R = V² / R. Example: A 230V heater with 10Ω resistance: I = V/R = 23A, P = V × I = 230 × 23 = 5,290W = 5.29 kW. For DC circuits and purely resistive AC loads (heaters, incandescent lights), PF = 1.0 so apparent power equals active power.

Question 4

How do I calculate electricity cost for industrial equipment?

Accepted Answer

Monthly cost (ZAR) = Power (kW) × Hours per day × Working days × Tariff (c/kWh) / 100. For a 37 kW compressor running 8h/day, 22 days/month at R2.50/kWh: Cost = 37 × 8 × 22 × 2.50 = R16,280/month. Eskom Megaflex and Miniflex tariffs vary by time-of-use. Peak-period rates are significantly higher (R4–R6/kWh) vs off-peak (R0.90–R1.50/kWh). VFDs can reduce motor power by 30–50% through speed control.

Question 5

How much capacitance is needed for power factor correction?

Accepted Answer

Required kVAR = P × (tan φ₁ - tan φ₂), where P is active power (kW), φ₁ is initial angle (arccos of current PF), and φ₂ is target angle (arccos of target PF). Example: 100 kW load at PF = 0.75, correct to 0.95: kVAR = 100 × (tan(arccos 0.75) - tan(arccos 0.95)) = 100 × (0.882 - 0.329) = 55.3 kVAR. Install capacitor banks to supply this reactive power locally, reducing kVA demand.

Question 6

What is the difference between kW, kVA, and kVAR?

Accepted Answer

kW (kilowatt) = Active power — does actual work (heat, motion). kVAR (kilovolt-ampere reactive) = Reactive power — stored/released by inductors and capacitors, does no useful work. kVA (kilovolt-ampere) = Apparent power = √(kW² + kVAR²). Power factor = kW/kVA. A 100 kVA transformer at 0.8 PF delivers only 80 kW of useful power. Equipment is rated in kVA because it must handle the full apparent power current.

Question 7

How do I calculate motor full load current?

Accepted Answer

Three-phase motor FLA = P / (√3 × V × PF × η), where P is rated power (W), V is rated voltage (V), PF is power factor, η is efficiency. Example: 37 kW, 400V motor, PF = 0.85, η = 0.92: FLA = 37,000 / (1.732 × 400 × 0.85 × 0.92) = 68.4A. Add 25% for starting current inrush (DOL starting). Use soft starters or VFDs to reduce starting current to 150–200% of FLA.

Question 8

What voltage levels are standard in South African industry?

Accepted Answer

South Africa (SANS/IEC): LV distribution: 400V three-phase (230V phase-to-neutral), 50 Hz. MV distribution: 6.6 kV, 11 kV, 22 kV, 33 kV. HV transmission: 66 kV, 132 kV, 220 kV, 275 kV, 400 kV, 765 kV. Industrial motors: Typically 400V (up to ~160 kW), 3.3 kV or 6.6 kV for larger motors. NRS 097 governs LV distribution standards. Always verify local utility requirements.

Motor Power	Voltage	Full Load Amps (A)	DOL Starting Current (A)	IEC Frame
1.5 kW	400V	3.5	21	90L
5.5 kW	400V	11.5	69	112M
11 kW	400V	21.5	129	132M
22 kW	400V	42	252	160M
37 kW	400V	68	408	200L
55 kW	400V	100	600	225M
75 kW	400V	137	822	250M
110 kW	400V	200	1200	280M

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