Broad and shallow. The marks sit in pumps, pipe losses, heat exchangers and the standard cycles — not in machinery description, which is where most candidates spend their time.
A pump adds energy to a fluid, and that energy is bookkept as head — energy per unit weight, measured conveniently in metres.
Total head has to include everything the pump works against. Forgetting the suction-side static lift, or the friction in either line, is the usual source of an under-sized pump on paper.
The duty point is where the pump curve crosses the system curve. The pump curve falls as flow rises; the system curve rises, as static head plus friction that grows with the square of flow. They meet at exactly one point, and that is what the installation actually delivers — not what the pump is nameplated for.
This is why throttling a discharge valve reduces flow: it steepens the system curve, moving the intersection to the left. It is also why throttling wastes energy, and why a variable-speed drive is better — it moves the pump curve instead of adding an artificial resistance.
A pump does not have a flow rate. A pump and a system together have a flow rate, at the point where their curves cross.
Liquid boils when the local pressure falls to its vapour pressure. The lowest pressure in a pump is at the impeller eye, and if it drops that far the liquid flashes to vapour, the bubbles collapse violently further along the impeller, and the pump loses head, erodes and makes a noise like gravel.
Net positive suction head available is how much head the system offers at the eye above vapour pressure:
NPSH required is a property of the pump, published as a curve rising with flow. The rule is simply NPSHa > NPSHr, with a margin — usually at least 0.5 to 1.0 m.
Three situations make trouble, and all three are examinable:
Friction loss in a straight pipe comes from Darcy–Weisbach:
Note the V² term. Doubling the flow through a fixed pipe quadruples the friction — which is why velocity limits, not just diameters, appear in piping design rules, and why a system curve is a parabola.
Bends, valves and fittings are handled either as an equivalent length of straight pipe or through a loss coefficient:
On a real marine system the fittings often dominate. A sea water cooling line with a strainer, several bends, a butterfly valve and a non-return valve can easily lose more head in the fittings than in all the straight pipe.
The governing equation is short, and nearly every question is about getting the mean temperature difference right.
ΔT1 and ΔT2 are the temperature differences at the two ends of the exchanger — and which streams face each other depends entirely on the arrangement.
Where an outlet temperature is unknown, LMTD requires iteration and the ε–NTU method is faster: effectiveness is actual heat transfer over the maximum thermodynamically possible, and it is tabulated against NTU = UA/Cmin.
The overall coefficient U is a chain of resistances in series, and fouling adds one more:
In service, fouling is the term that grows. At constant duty it shows up as a rising terminal temperature difference — the practical early warning that a cooler needs cleaning.
Refrigeration. The vapour compression cycle is read off a pressure–enthalpy diagram: compression, condensation, expansion through the valve, evaporation. Coefficient of performance is useful effect over work in.
The heat pump result is not a trick: the condenser rejects both the heat absorbed and the work put in, so the useful output is larger by exactly the work.
Prime mover cycles. For GATE purposes, know the three air-standard efficiencies and how they move.
Two things follow. Efficiency always rises with compression ratio. And for the same compression ratio the Otto cycle is more efficient than the Diesel — the Diesel cycle wins in practice only because it can run at a much higher compression ratio without knocking.
A small but reliably examined corner of the syllabus.
Power factor is the ratio P/S. A poor power factor means more current for the same useful power — larger cables, higher losses, and generators limited by current before they are limited by kilowatts.
Two shipboard specifics worth knowing. First, load sharing between alternators is controlled by two independent devices: the governor sets real power in kW, the automatic voltage regulator sets reactive power in kVAr. Adjusting one to correct the other is the classic mistake.
Second, marine low-voltage distribution is normally insulated from the hull rather than earthed, so a single earth fault does not interrupt supply — it raises an alarm. The danger is the second fault on another phase, which becomes a short circuit through the hull. That is why earth fault indication is monitored and acted on rather than acknowledged.
Pumps, affinity laws and a heat exchanger — three questions that between them cover where most of the marks in this topic sit.
A centrifugal pump delivers 250 m³/h of sea water. It draws from a tank whose surface is 2.5 m below the pump centreline, through a suction line with 1.2 m of friction loss, and discharges 18 m above the pump against 4.0 m of discharge friction. Pump efficiency is 72 %. Find the shaft power, and check the NPSH available against a required NPSH of 3.0 m. Atmospheric pressure 101.3 kPa, vapour pressure 1.7 kPa, ρ = 1025 kg/m³.
Q = 250 m³/h, η = 0.72 Suction lift 2.5 m, suction friction 1.2 m Static discharge 18 m, discharge friction 4.0 m p_atm = 101.3 kPa, p_v = 1.7 kPa, ρ = 1025 kg/m³
Find the shaft power, and check the NPSH available against a required NPSH of 3.0 m
Convert the flow to SI before anything else.
Total head is everything the pump must overcome: the static lift on the suction side.
The static rise on the discharge side, and the friction in both lines.
Hydraulic power is the rate of doing work on the fluid.
Shaft power follows from the pump efficiency.
Now the NPSH check.
Available NPSH is the head at the impeller eye above the vapour pressure — atmospheric pressure on the tank surface, minus the lift because the pump is above the liquid, minus suction friction, minus vapour pressure.
Compare with what the pump needs, and note the margin.
A design point worth stating: if the same pump were handling condensate at 80 °C.
Vapour pressure would rise to about 47 kPa, costing roughly 4.7 m of head, and the margin would nearly vanish. Hot liquids are where NPSH problems live.
AnswerP_shaft = 24.9 kW; NPSH available = 6.20 m against 3.0 m required — acceptable
The trap: the sign of the static suction term. A pump below the liquid level has a flooded suction and the head is added; a pump above it is lifting, and the head is subtracted. Get that sign wrong and a cavitating installation looks perfectly healthy on paper.
A centrifugal pump running at 1750 rpm delivers 120 m³/h at 32 m head, absorbing 16 kW. It is to be run at 1500 rpm. Find the new duty and power. Then find what speed would be needed to deliver 140 m³/h, and comment.
N₁ = 1750 rpm: Q₁ = 120 m³/h, H₁ = 32 m, P₁ = 16 kW N₂ = 1500 rpm; same impeller
Find the new duty and power
The affinity laws relate the three quantities to speed.
Each with a different power. They follow from the velocity triangles: flow goes with tip speed, head with tip speed squared, and power is the product of the two.
Speed ratio first.
It is used three times, so compute it once.
Apply each law.
Note the leverage.
A 14 % speed reduction has cut the power by 37 %. This is the entire argument for variable-speed drives on pumps that spend most of their life throttled.
Now the speed needed for 140 m³/h.
Using the flow law in reverse.
And the comment, which is the examinable half.
The affinity laws describe the pump, not the system. The pump will only deliver 140 m³/h at 2042 rpm if the system curve happens to demand 43.6 m at that flow. If the system is mostly static head, its curve is nearly flat and the operating point will land somewhere else entirely.
AnswerAt 1500 rpm: 102.9 m³/h, 23.5 m, 10.1 kW. For 140 m³/h: 2042 rpm, 43.6 m, 25.4 kW
The trap: applying the same exponents to a change of impeller diameter. Trimming an impeller does not scale the same way as changing speed — the exit angle and passage width stay fixed — and using the speed laws for a diameter change overstates the effect.
A fresh water cooler must reject 1800 kW. Fresh water enters at 55 °C and leaves at 40 °C. Sea water enters at 28 °C and leaves at 38 °C. The overall heat transfer coefficient is 2200 W/m²K. Find the required area for counterflow, and compare with parallel flow.
Q = 1800 kW, U = 2200 W/m²K Hot: 55 °C → 40 °C Cold: 28 °C → 38 °C
Find the required area for counterflow, and compare with parallel flow
Counterflow first.
The streams run in opposite directions, so the hot inlet faces the cold outlet and the hot outlet faces the cold inlet.
Log mean temperature difference.
It is a logarithmic mean rather than an arithmetic one because the temperature difference decays exponentially along the exchanger.
Required area.
Now parallel flow, where both streams enter at the same end.
The hot inlet now faces the cold inlet, and the hot outlet faces the cold outlet.
Area needed in parallel flow.
| Counterflow | Parallel flow | |
|---|---|---|
| ΔT at end 1 | 17 K | 27 K |
| ΔT at end 2 | 12 K | 2 K |
| ΔT_lm | 14.36 K | 9.60 K |
| Area required | 57.0 m² | 85.2 m² |
There is a second, harder point.
In parallel flow the cold outlet can never exceed the hot outlet — here they are only 2 K apart and the exchanger is running out of driving force. Counterflow has no such limit: the cold stream can leave hotter than the hot stream leaves, which parallel flow can never achieve at any area.
Finally, a service observation.
Fouling shows up as a falling U, which appears as a rising terminal temperature difference at fixed duty — the first evidence of a dirty cooler, well before any alarm.
AnswerCounterflow: ΔT_lm = 14.36 K, A = 57.0 m². Parallel flow needs 85.2 m² — 50 % more
The trap: pairing the wrong ends. Writing 55 − 28 and 40 − 38 for a counterflow exchanger gives 9.6 K instead of 14.4 K and oversizes the cooler by half. Sketch the two streams with arrows before writing any temperatures down.
H = static + friction + velocity headInclude both suction and discharge sidesP_hyd = ρgQH, P_shaft = P_hyd/ηQ in m³/sNPSH_a = p_a/ρg ± h_s − h_f − p_v/ρgPlus for flooded suction, minus for a liftQ∝N, H∝N², P∝N³Affinity laws — speed only, not impeller trimN_s = N√Q/H^(3/4)Specific speed selects the impeller typeh_f = f·L·V²/(2gD)f = 64/R_e when laminar; Moody chart otherwiseQ = U·A·ΔT_lmΔT_lm = (ΔT₁−ΔT₂)/ln(ΔT₁/ΔT₂)1/U = 1/h_i + x/k + 1/h_o + R_fFouling is the term that grows in serviceCOP_ref = Q_e/W, COP_hp = COP_ref + 1Carnot ceiling = T_c/(T_h − T_c)P = √3·V_L·I_L·cos φGovernor sets kW, AVR sets kVAr