Every item in this chapter is selected against a duty point with a defined margin, not a rule of thumb, so the questions below work the sizing reasoning through from first principles — pumps, heat exchangers, HVAC, steering gear and compressed air alike.
Every rotodynamic pump on board is chosen against a duty point — a flow rate and a head — read off the intersection of the pump's characteristic curve and the system's resistance curve. Centrifugal pumps dominate seawater cooling, ballast, bilge and general-service duties because their smooth, continuous delivery suits a resistance curve that rises with the square of flow; positive-displacement pumps (screw, gear, reciprocating) take over wherever the fluid is viscous, the delivery must stay near-constant regardless of back-pressure, or the duty is metering — fuel injection, lube-oil transfer, hydraulic power packs.
Matching flow and head at the design point is only half the selection. The other half is checking that the pump will not cavitate there, which means comparing the net positive suction head available (NPSHa) — what the installation actually offers at the impeller eye — against the net positive suction head required (NPSHr), a property of the pump itself read off the maker's curve at the duty flow, not at shut-off.
The mechanism matters as much as the arithmetic. As liquid accelerates into the impeller eye, the local static pressure falls below the pressure at the suction flange. If that local pressure reaches the liquid's vapour pressure at its running temperature, vapour bubbles form; carried into the higher-pressure region further round the impeller, they collapse violently against the vanes, and the pitting that follows is cavitation damage. NPSHa has to clear NPSHr by a margin — commonly a metre or so, set by the pump maker — precisely because that collapse starts before the flow visibly stalls or the head curve visibly droops.
NPSHa is a property of the installation — elevation, friction, vapour pressure — while NPSHr is a property of the pump. Selection means checking the two against each other at the actual duty point, not assuming a flooded suction is automatically safe.
A duty point also has to be checked across the pump's expected operating range, not only at the design flow. A ballast pump stripping the last of a tank runs close to shut-off, where head rises and NPSHr can rise with it on some curves; a fire pump running two hydrants instead of one runs at a different flow and head altogether. Selecting for one point and ignoring the rest of the curve is how pumps end up cavitating in service conditions nobody actually checked on paper.
Heat exchanger sizing rests on one relationship: the duty a unit can transfer is the product of an overall heat-transfer coefficient, the surface area available, and a mean temperature difference driving the heat across that surface.
The log-mean, not the arithmetic mean, is correct because the temperature difference between the two streams changes continuously along the exchanger — decaying roughly exponentially with surface area — and the log-mean is exactly the average that a plain arithmetic mean of the two ends would overstate. Getting ΔT_lm right depends first on pairing the terminal temperatures to the actual flow arrangement: in a counterflow exchanger the hot inlet sits opposite the cold outlet, and the hot outlet sits opposite the cold inlet, because the streams pass each other travelling in opposite directions. Pair them the parallel-flow way by mistake and every individual temperature is still correct, but the ΔT_lm — and therefore the area — comes out wrong.
Counterflow is worth understanding, not just naming, because it is the only arrangement in which the cold stream can leave hotter than the hot stream leaves — impossible in parallel flow, where both streams converge toward the same temperature from opposite starting points. For the same duty and the same terminal temperatures, counterflow always gives the larger ΔT_lm, and a larger ΔT_lm means less area is needed for the same Q — which is why counterflow, or near-counterflow, arrangements dominate wherever surface area is expensive or space is tight, as it usually is on board.
The overall coefficient U is not a fixed property of the exchanger; it is built up from the film coefficients on each side plus the wall and any fouling layer, and it is the weakest of these resistances that dominates.
Where an outlet temperature is not fixed by the process — a plate cooler whose water outlet depends on how the whole system settles, for instance — the LMTD method needs an assumed outlet temperature to start from, and the ε–NTU method is the more direct route: it works from the exchanger's actual size and the two fluids' capacity rates to predict both outlet temperatures directly, rather than needing one as an input.
Accommodation and machinery-space air conditioning is a psychrometric problem before it is a refrigeration problem: comfort and condensation both depend on dry bulb temperature and moisture content together, not on temperature alone. The same dry bulb temperature can feel comfortable or oppressive depending on how much water vapour the air is carrying, which is why a design condition for accommodation spaces is always stated as a dry bulb temperature and a relative humidity together, not a temperature on its own.
A psychrometric chart plots dry bulb temperature against moisture content, with lines of constant relative humidity, wet bulb temperature and dew point overlaid. The dew point is the temperature at which air, cooled at constant moisture content, would just start to condense — and it is the reason cooling coils dehumidify as well as cool. Any coil surface running below the air's dew point strips moisture out of the airstream as condensate, whether or not dehumidification was the intention.
Cooling below the dew point is how dehumidification actually happens — a side effect of the coil temperature, not a separate process — which is why an over-cooled, over-dried airstream sometimes has to be reheated before it reaches the space.
That reheat requirement is a consequence of having only one coil temperature to work with: to pull enough moisture out of humid air, the coil often has to run colder than the temperature actually wanted in the space, so the air is reheated — by a trim heater, or by mixing with warmer recirculated air — back up to the design dry bulb after the moisture has already been removed. Sizing the cooling coil for the latent load (moisture removal) and the sensible load (temperature drop) separately, then checking whether reheat is needed to hit both targets together, is the core of an HVAC calculation — a coil sized on sensible load alone can dehumidify far more, or far less, than the space actually needs.
A refrigeration plant and a heat pump are the same machine described from opposite ends: both move heat from a cold reservoir to a warm one by doing work on a refrigerant that evaporates at low pressure and condenses at high pressure, and both are judged by how much heat is moved per unit of work put in.
Because energy is conserved, the heat rejected at the condenser equals the heat picked up at the evaporator plus the work put in, and that simple bookkeeping links the two coefficients of performance together: a heat pump's COP is always exactly one more than the refrigeration COP of the same cycle running between the same two temperatures, because the numerator for heating (heat rejected) exceeds the numerator for cooling (heat absorbed) by precisely the work input.
The Carnot cycle sets the ceiling neither real machine can reach: for a reversible cycle working between an evaporating temperature T_c and a condensing temperature T_h (both in kelvin), COP_ref cannot exceed T_c/(T_h − T_c). The practical consequence is that COP falls as the temperature lift — the gap the compressor works across — widens, which is why a provisions plant holding a deep-freeze space at a low temperature, with a high condensing temperature to reject to in a hot engine room, will always run at a markedly worse COP than an air-conditioning plant lifting heat across a much smaller gap. Selecting compressor size and heat-exchanger area against the worst-case lift, not the average one, is what keeps the plant capable on the hottest day in the warmest port, not merely on the test bed.
Steering gear is one of the few areas of the syllabus where the requirement is genuinely prescriptive rather than a matter of engineering judgement, and the conditions attached to the figures are as much a part of the requirement as the figures themselves. The main steering gear must be able to put the rudder over from 35° on one side to 30° on the other in 28 seconds, with the ship at its maximum ahead service speed and at its deepest seagoing draught — both conditions matter, because a gear that meets the timing at half speed or at a lighter draught, where the hydrodynamic torque on the rudder is smaller, has not actually demonstrated compliance.
Auxiliary steering gear exists to cover the case where part of the main gear is unavailable, and its requirement is correspondingly less demanding: rudder movement from 15° on one side to 15° on the other in 60 seconds, at half the ship's maximum ahead service speed or 7 knots, whichever is greater. The lower speed condition reflects that auxiliary steering is a degraded-mode capability, not a full-power one — the ship is expected to be making way cautiously, not at full service speed, while relying on it.
A steering gear timing figure quoted without its speed and draught conditions is an incomplete answer — the torque the gear must overcome depends on both, and the requirement is the figure and the conditions together.
Sizing the gear itself works back from the maximum rudder torque expected at the design speed and angle, through the hydraulic ram or rotary-vane geometry, to the pump capacity and relief-valve setting needed to move that torque within the time limit — with margin held in the pump capacity, not in the timing, since the timing is fixed by the requirement rather than open to trade-off.
Deck machinery — windlass, mooring winches, cargo winches and cranes — is sized against a duty condition rather than a single peak load: a windlass has to develop its rated pull continuously at a specified chain speed while heaving the anchor and cable off the seabed, and momentarily exceed that to break the anchor out of the ground, and both figures come from the anchor and cable weight the ship is required to carry, not from an arbitrary factor applied to whatever motor happens to be on the shelf. Mooring winches are sized similarly against the line pull and speed needed to bring the vessel alongside against wind and current, with a slipping clutch or relief valve protecting the line, the winch and the deck fittings if the load stalls.
Compressed air for starting the main engine works on the same size-against-the-duty logic as the deck machinery, but the duty here is a fixed number of consecutive starts rather than a heave rate. The air receiver has to hold enough free air, at the pressure it is charged to, to deliver a defined number of starts without the compressor topping it up in between — twelve consecutive starts for a reversible, direct-drive engine, because reversing under air alone can take several attempts, and six for a non-reversible engine, whose air demand is only for getting the engine turning in one direction.
A receiver's volume is a means to an end — it exists to deliver a required number of starts across a working pressure range — not a number chosen for its own sake, which is why sizing has to start from the starts requirement and work backward to a volume, not the other way round.
Only the air between the fully-charged pressure and the lowest pressure that still gives a reliable start counts toward that requirement; air below the minimum starting pressure is still sitting in the receiver but is no use to the engine, which is the detail most often skipped when a receiver is sized on volume alone.
The three problems below carry the reasoning from the sections above into numbers — a cooler sized from first principles, a starting-air receiver sized from the starts requirement rather than a guessed volume, and a feed pump checked for NPSH at its actual duty point.
A shell-and-tube lube-oil cooler is being sized for a diesel generator. Lube oil enters the cooler at 60 °C and must leave at 45 °C, at a flow rate of 4 kg/s with cp = 2.1 kJ/kg·K. Cooling water is available at 20 °C, flowing counterflow to the oil at 3 kg/s with cp = 4.2 kJ/kg·K. The overall heat-transfer coefficient for this duty is taken as 850 W/m²K. Find the water outlet temperature, the log-mean temperature difference, and the heat-transfer area required, including a 20% fouling margin.
Lube oil: flow 4 kg/s, cp = 2.1 kJ/kg·K, enters at 60°C, leaves at 45°C Cooling water (counterflow): flow 3 kg/s, cp = 4.2 kJ/kg·K, enters at 20°C Overall coefficient assumed for this duty: U = 850 W/m²K Fouling/design margin to be applied to the clean area: 20%
Find the water outlet temperature, the log-mean temperature difference, and the heat-transfer area required, including a 20% fouling margin
First find the duty from the oil side.
Since both its flow and its temperature drop are known.
The cooling water carries the same duty.
So its temperature rise follows from an energy balance on the water side.
Pair the terminal temperatures to the counterflow arrangement.
The hot inlet sits opposite the cold outlet, and the hot outlet sits opposite the cold inlet, because the two streams travel in opposite directions along the exchanger.
With both terminal differences in hand.
The log-mean temperature difference follows directly.
The clean-surface area comes straight from Q = U·A·ΔT_lm.
A fouling margin is then added because 1/U only grows once the cooler is in service.
AnswerWater leaves at 30°C, ΔT_lm ≈ 27.4°C, required area ≈ 5.4 m² clean, ≈ 6.5 m² with a 20% fouling margin — select the next standard cooler size above this.
The trap: pairing the terminals as if the exchanger were parallel-flow (hot inlet against cold inlet, hot outlet against cold outlet) gives the wrong ΔT₁ and ΔT₂ for a unit that is actually counterflow — every individual temperature used is still correct, but the resulting LMTD, and therefore the area, is not.
A vessel's main engine is a reversible, direct-drive diesel, requiring 12 consecutive starts without recharging the starting-air receivers. The engine builder states that each start consumes 6 m³ of free air (referred to 1.0 bar). The receivers are charged to 30 bar, and the lowest pressure at which a reliable start can still be made is 12 bar. Two receivers of approximately equal capacity are to be fitted. Determine the total receiver volume required and the size of each receiver.
Main engine: reversible, direct-drive — 12 consecutive starts required Air consumption per start (builder's data): 6 m³ of free air, referred to 1.0 bar Receiver charged pressure: 30 bar Minimum pressure for a reliable start: 12 bar Capacity to be split across two receivers of approximately equal size
Determine the total receiver volume required and the size of each receiver
Work from the number of starts actually required, not from a volume chosen by eye.
The free-air quantity needed sets everything else that follows.
Only the air between the charged pressure.
The lowest pressure that still gives a reliable start is usable — treating the blow-down as isothermal, each cubic metre of receiver volume yields (P₁ − P₂)/P₀ cubic metres of free air.
Divide the total free-air requirement by that yield to get the receiver volume.
Then split it across the two receivers to be fitted.
AnswerFit two air receivers of 2.0 m³ each (4.0 m³ total), charged to 30 bar.
The trap: sizing a receiver on a round volume figure that 'looks about right' rather than back-calculating from the required number of starts and the usable pressure range — a receiver sized that way can pass a first start and still fail the consecutive-starts count the requirement actually sets.
A boiler feed pump draws suction from a deaerator that operates at saturation conditions, so the water's vapour pressure at the pump suction equals the deaerator's operating pressure. The deaerator water level sits 5 m above the pump centreline, and the suction pipe friction loss at the duty flow is 0.7 m. The pump's NPSH required at this flow, from the maker's curve, is 3.6 m, and the maker specifies a minimum margin of 0.5 m. Determine the NPSH available and whether the installation is acceptable.
Deaerator operates at saturation: tank pressure = water vapour pressure Deaerator water level above pump centreline: h_s = 5 m Suction pipe friction loss at duty flow: h_f = 0.7 m NPSH required at duty flow (maker's curve): 3.6 m Minimum NPSH margin required by the maker: 0.5 m
Determine the NPSH available and whether the installation is acceptable
Start from the general expression for NPSH available.
Which compares the pressure at the pump suction with the liquid's vapour pressure.
Because the deaerator is at saturation.
Its operating pressure and the water's vapour pressure are the same number — that term cancels, and only elevation and friction are left protecting the pump.
Substitute the given elevation and friction figures.
Compare against what the pump needs.
Including the maker's required margin, before accepting the installation.
AnswerNPSHa (4.3 m) clears NPSHr plus margin (4.1 m) by only 0.2 m — the installation is acceptable, but the margin is thin.
The trap: assuming a gravity-fed, above-pump suction is automatically safe from cavitation — in a saturated system the operating pressure buys nothing, since it cancels against the vapour pressure, so a fouled strainer, a part-shut valve, or a slightly hotter deaerator can erase the whole margin.
NPSHa = (P_atm−P_v)/(ρg) + h_s − h_fCompare to NPSHr at duty flow, with the maker's marginQ = U·A·ΔT_lmΔT_lm=(ΔT₁−ΔT₂)/ln(ΔT₁/ΔT₂); pair terminals to the actual flow arrangement1/U = 1/h_i + x/k + 1/h_o + R_fFouling is the term that grows in serviceε–NTUUse when an outlet temperature is not fixed by the processSteering: 35° to 30° in 28 sMain gear, at max ahead service speed and deepest draughtAuxiliary steering: 15° to 15° in 60 sAt half service speed or 7 kn, whichever is greaterStarting air: 12 / 6 startsReversible direct-drive / non-reversible engineUsable free air = V_R×(P₁−P₂)/P₀Only pressure between charge and minimum-start pressure countsCOP_ref = Q_e/W; COP_hp = COP_ref+1Carnot ceiling T_c/(T_h−T_c)Dew pointCoil surface below dew point ⇒ dehumidification as a side effect