Energy Efficiency in Electrical Utilities
Energy Efficiency in Electrical Utilities
This domain examines motors, transformers, distribution systems, power factor, harmonics, pumps, fans, compressed air and lighting. Expect calculations on load factor, demand, losses and the affinity laws.
Core rules
- Motor input power = output kW ÷ efficiency. Loading below about 50% usually worsens efficiency and power factor.
- For fans and pumps, flow varies with speed, head with speed squared, and power with speed cubed, so modest speed reduction yields large savings.
- Power-factor correction reduces kVA demand and I²R losses but does not reduce the real kWh delivered to the load.
- Transformer loss = no-load loss + load² × load loss. Harmonics increase neutral current, capacitor stress and motor heating.
Worked example
A plant draws 500 kW at 0.80 PF. To raise PF to 0.95, capacitor kVAR = P × (tan φ1 − tan φ2). φ1 = cos⁻¹0.80 = 36.87°, tan = 0.750; φ2 = cos⁻¹0.95 = 18.19°, tan = 0.329. Required kVAR = 500 × (0.750 − 0.329) = 210.5 kVAR. Select a stepped 250 kVAR bank rather than one fixed unit, and check harmonics before installation. The reasoning shows that kW stays at 500; kVA falls from 625 to 526, cutting demand charges and upstream losses.
Traps
Using kW instead of kVA in PF formulas; oversizing capacitors and causing leading PF; assuming PF correction cuts kWh; ignoring variable-speed drive harmonics; and applying affinity laws to systems with high static head.