Mechanical Equipment and Components (Pumps, Compressors, Turbines, Engines)
This domain tests whether you can predict how rotating equipment behaves away from its design point: where a pump will actually operate, why a compressor surges, what limits a turbine, and how engines lose efficiency at part load. Expect affinity-law problems, NPSH calculations and fault-to-cause matching.
Pumps
- Affinity laws: Q proportional to N, H proportional to N^2, P proportional to N^3. Trimming an impeller follows the same laws using diameter.
- The operating point is where the pump curve crosses the system curve. Throttling moves the system curve; a VFD moves the pump curve.
- NPSHa = (ps - pv)/(rho x g) + z - hf. It must exceed the pump's NPSHr with margin, or the pump cavitates: noise, falling head, vibration, impeller erosion.
- Run near best efficiency point. Far left causes recirculation and radial loading; far right causes cavitation and motor overload.
Compressors and turbines
- Positive displacement (reciprocating, screw, vane) suits high pressure and low flow; dynamic (centrifugal, axial) suits large flows.
- Reciprocating machines need clearance-volume control; multi-stage intercooling pulls the path toward isothermal and cuts power.
- Surge is a dynamic-machine instability where flow reverses. Anti-surge recycle opens as flow approaches the surge line.
- Steam turbines: a condenser lowers exhaust pressure and raises output; gland seals keep air out, because air ingress destroys vacuum.
Engines
- Four-stroke diesel: intake, compression, power, exhaust, with ignition by compression heat. Efficiency falls at light load; jacket-water and exhaust heat recovery matter.
- Gas turbines: compressor, combustor, power turbine. Output drops as ambient temperature rises because mass flow falls.
Worked example
A pump delivers 60 L/s at 35 m head. Hydraulic power = rho x g x Q x H = 1000 x 9.81 x 0.060 x 35 = 20.6 kW. At 72% pump efficiency, shaft power = 20.6/0.72 = 28.6 kW. Doubling speed gives 120 L/s and 140 m head, and shaft power rises eightfold to about 229 kW. That third-power relationship is why overspeeding a motor or fitting a larger impeller can overload a drive that looked adequately sized.
Traps
- Using discharge pressure instead of head, then mixing pressure units between pump curve and system curve.
- Forgetting vapour pressure when the liquid is hot condensate - the classic deaerator extraction pump cavitation.
- Confusing NPSHr (a pump property from the curve) with NPSHa (a system property you calculate).
- Ignoring how steeply centrifugal compressor discharge temperature rises with pressure ratio.