By Chief Engineer level the thermodynamics question has stopped being about a cycle on a p-V diagram — it is about defending a number to a superintendent, in tonnes of fuel and kilowatts that actually reach the shaft.
Every Chief-level thermodynamics question starts from the same place: fuel goes in as chemical energy, and only part of it leaves the flywheel as shaft work. The rest has to go somewhere, and where it goes is the substance of the answer — not the cycle diagram, the balance sheet.
For a modern slow-speed diesel plant, typical splits run something like 45–50 % to shaft power, 25–30 % to exhaust gas, 5–8 % to jacket cooling water, 5–8 % to charge air cooling, 2–3 % to lubricating oil cooling, and the remainder to radiation and unaccounted losses. None of these figures is fixed by any rule — they shift with load, ambient conditions and engine design — but the structure of the balance is what a Chief-level answer is built on.
The reason the balance matters commercially is that every one of those loss streams is a candidate for recovery, or at least for a decision about whether recovery is worth the capital. Jacket water at 80–90 °C is too cool to do much useful work but can preheat fuel or run a fresh-water evaporator; charge air cooling water sits at a similar temperature. Exhaust gas, by contrast, arrives hot enough — 250–400 °C at the turbocharger outlet — to raise steam, and that is where the serious recoverable energy sits.
The heat balance is not academic bookkeeping — it is the working tool for sizing a waste heat recovery plant and for putting a number, in kilowatts or tonnes of fuel a day, on any proposed efficiency measure.
The exhaust stream carries roughly a quarter to a third of the fuel energy, but it is low-grade heat compared with the combustion temperature it came from, and only part of it can be pulled out usefully. The limit is not the boiler design — it is chemistry at the cold end of the gas path.
Marine fuel contains sulphur, and when exhaust gas is cooled enough, sulphur trioxide in the gas combines with water vapour to form sulphuric acid, which condenses onto the coolest metal it can find — the last row of economiser tubes, or the boiler's cold end. Once that acid condenses it attacks the tube surface directly, and the corrosion this causes is far more expensive than the fuel saved by chasing the last few degrees of heat recovery. The temperature at which this starts, the acid dew point, sits roughly in the 130–160 °C band and rises as the fuel's sulphur content rises — a heavier, higher-sulphur fuel pushes the floor up, and a switch to a low-sulphur fuel for emission control areas lowers it again.
Where a power turbine is fitted, it taps the exhaust stream ahead of the economiser and extracts shaft or electrical power directly from the gas expansion, rather than raising steam from it. That means the power turbine and the economiser are competing for the same finite gas energy: more extraction upstream leaves a smaller temperature drop available downstream, and sizing one without reference to the other is a common design and exam error.
An engine's rated output on the builder's data sheet is not a fact about the engine alone — it is a fact about the engine at a stated set of reference conditions, typically 25 °C combustion air, 25 °C low-temperature cooling water and a specified fuel calorific value. Change any of those and the same engine, at the same fuel rack setting, delivers a different number.
Air temperature acts through density. Hotter air is less dense, so for a turbocharger delivering roughly the same volumetric flow at a given speed, the mass of air — and therefore the mass of oxygen available to burn fuel — falls as inlet temperature rises. Because combustion is smoke-limited, fuelling is trimmed to match the air actually available, so less air means less fuel burned means less power, even though nothing mechanical has changed.
Sea water temperature acts on a different path: a warmer sea raises the temperature of the low-temperature cooling water, which raises the charge air temperature leaving the charge air cooler and, on some plants, the scavenge air temperature. The effect compounds with a hot ambient air condition rather than replacing it — a ship trading in the tropics faces both a hotter compressor inlet and a warmer charge air cooler at the same time.
The commonest error in a real performance dispute is comparing a tropical performance report against the builder's trials figure without correcting for both air temperature and sea temperature — the ship can look like it is underperforming when it is simply running hotter than the reference condition.
However cleverly a heat exchanger network is arranged, there is a ceiling on how much work can be extracted from a heat source at temperature T_h rejecting to a sink at T_c, and no amount of engineering ingenuity moves that ceiling — only changing T_h or T_c does.
This is why, once the diesel engine itself is running close to its own practical efficiency limit, the next gain does not come from trying to squeeze more work out of the engine cycle directly. It comes from adding a second, lower-temperature cycle underneath it — a steam turbine or similar unit running on the engine's waste exhaust heat, rejecting to sea water. The two cycles work between different temperature bands, so their ceilings stack: the diesel cycle claims the high-temperature ceiling, and the bottoming cycle claims whatever is left between the exhaust temperature and the sea.
In practice a real bottoming plant achieves only a fraction of its Carnot ceiling — turbine stage losses, pressure drops and the temperature differences needed to actually drive heat transfer all eat into it — but the ceiling still sets the outer limit on what is worth investing in. A wider gap between boiler and condenser temperature, from higher steam pressure or a better condenser vacuum, raises the ceiling; nothing else does.
Heat exchanger duty in a boiler, economiser or cooler is governed by the same relationship regardless of which fluid is hot and which is cold: the rate of heat transfer depends on the surface area available, how well that surface conducts heat from one side to the other, and how large a temperature difference is driving it.
The log-mean temperature difference is used rather than a simple average because the temperature gap between the two fluids usually changes shape along the exchanger — for a counter-flow economiser the gas cools steadily while the water side may be heating, or evaporating at a near-constant temperature, so the driving temperature difference at the gas inlet end is quite different from that at the gas outlet end.
U, the overall heat transfer coefficient, is the term that erodes in service. Soot and unburnt carbon build up on the gas side, scale forms on the water side, and both act as insulating layers that reduce U even though A has not changed. This is why an economiser's performance is tracked over time against its gas-side and water-side temperatures rather than assumed constant from commissioning — a fall in recovered duty at the same load and gas flow is usually a fouling problem, not a design problem, and the fix is cleaning, not a bigger unit.
None of the preceding analysis earns its keep until it is converted into the form the question — and the superintendent — actually wants: money, or its shipping equivalent, tonnes of fuel per day.
A kilowatt saved anywhere in the plant, whether from waste heat recovery displacing an auxiliary boiler, a cleaned economiser recovering more duty at the same gas flow, or a corrected understanding that a "low" trials figure was actually just a tropical ambient effect, converts to tonnes of fuel through specific fuel oil consumption and the hours the saving applies for. Getting the units right — kW, not MW; g/kWh, not kg/kWh — is where marks are lost as often as in the physics.
Before quoting any efficiency figure in that answer, state its boundary. Engine efficiency, plant efficiency and ship (propulsive) efficiency are three different numbers built from three different boundaries — engine alone, engine plus waste heat recovery, or the whole ship including hull and propeller condition — and quoting one where the question asked for another is a common way to lose marks despite doing the arithmetic correctly.
The calculation is not finished when the kilowatts are found — it is finished when they are converted to tonnes per day, with the efficiency boundary stated, because that is the form the decision actually gets made in.
The three examples below are deliberately layered — each needs two or more reasoning steps, not a single formula substitution, because that is the level real Chief-level questions are set at.
A Chief Engineer is assessing the exhaust gas economiser on a slow-speed main engine running 3.5 % sulphur fuel. Exhaust gas leaves the turbocharger and enters the economiser at 340 °C, with a mass flow of 20 kg/s and a specific heat of 1.05 kJ/kg·K. The engineer sets the economiser gas outlet at 190 °C, clear of the acid dew point for this fuel grade. The recovered heat displaces an oil-fired auxiliary boiler of 90 % efficiency burning fuel of 42,000 kJ/kg calorific value. Find the exhaust heat recovered and the fuel saved per day.
Exhaust mass flow ṁ_g = 20 kg/s Specific heat c_p = 1.05 kJ/kg·K Economiser gas inlet T_in = 340 °C Economiser gas outlet T_out = 190 °C Fuel sulphur content = 3.5 % (acid dew point ≈155 °C) Auxiliary boiler efficiency η_b = 90 % Auxiliary boiler fuel calorific value = 42,000 kJ/kg
Find the exhaust heat recovered and the fuel saved per day
Check the outlet temperature against the acid dew point first.
At 3.5 % sulphur the acid dew point for this fuel sits around 155 °C; taking the economiser gas outlet down to 190 °C leaves a 35 °C margin, enough to keep the cold-end tube metal above the temperature where sulphuric acid starts to condense out of the gas.
Recovered exhaust heat is the sensible heat the gas stream gives up between the economiser inlet and outlet.
This recovered duty means the auxiliary boiler no longer has to fire to supply the same heat.
Express that avoided firing as an equivalent fuel input, allowing for boiler efficiency.
Convert the avoided firing rate to a mass of fuel per day using the boiler fuel's calorific value.
AnswerExhaust heat recovered ≈3150 kW, saving about 7.2 tonnes of auxiliary boiler fuel per day.
The trap: chasing extra recovery by dropping the economiser outlet closer to the dew point — a few more percent of Q_exh bought that way is repaid many times over in cold-end corrosion and tube replacement.
A main engine is trials-rated at 7200 kW with the turbocharger compressor drawing air at a reference condition of 25 °C and 1000 mbar. The same engine is now running in the tropics with compressor inlet air at 45 °C and the same barometric pressure. The turbocharger delivers an essentially constant volumetric flow of 5.0 m³/s at this speed, and fuel injection is trimmed to match the air actually available (smoke-limited operation). Estimate the charge-air mass flow at each condition and the power the engine can be expected to develop in the tropics.
Trials rated power P_rated = 7200 kW at T_1 = 25 °C (298 K) Tropical inlet temperature T_2 = 45 °C (318 K) Barometric pressure (both cases) P = 100 kPa Gas constant for air R = 0.287 kJ/kg·K Compressor volumetric flow V̇ ≈ 5.0 m³/s (assumed constant at this speed)
Estimate the charge-air mass flow at each condition and the power the engine can be expected to develop in the tropics
Find the air density at each condition from the ideal gas law.
Multiply each density by the (assumed constant) volumetric flow to get the mass flow the compressor can actually deliver at each condition.
Express the shortfall as a percentage, then apply it to the rated power.
The smoke limit means fuelling, and so output, follows the air actually available.
AnswerCharge-air mass flow falls by about 6.3 %; expect roughly 6750 kW available in the tropics, a derate of about 450 kW against the trials figure.
The trap: correcting for air temperature alone and forgetting that a higher sea-water temperature also raises the charge-air temperature downstream of the cooler — the derate under full tropical conditions (hot air and hot sea together) is worse than this air-only calculation suggests.
An exhaust gas boiler on a large containership raises steam using 4000 kW of recovered exhaust heat, feeding a turbo-generator. Boiler (evaporation) temperature is 127 °C and the condenser, cooled by sea water, runs at 47 °C. The manufacturer states the real cycle achieves about half of the ideal Carnot ceiling once turbine and cycle losses are allowed for, and the generator is 95 % efficient. The set is intended to displace a diesel generator running at an SFOC of 210 g/kWh for 24 hours a day. Find the electrical output and the fuel saved per day.
Recovered thermal duty to steam generator Q_in = 4000 kW Boiler (evaporation) temperature T_h = 127 °C (400 K) Condenser temperature T_c = 47 °C (320 K) Real cycle achieves ≈50 % of the Carnot ceiling Generator efficiency = 95 % Displaced diesel generator SFOC = 210 g/kWh, running 24 h/day
Find the electrical output and the fuel saved per day
Find the absolute ceiling first, using absolute temperature.
No arrangement of heat exchangers or turbine stages can beat this, whatever the hardware.
Apply the manufacturer's stated utilisation of that ceiling to get the real thermal efficiency, then convert to mechanical shaft power.
Take out the generator loss to reach electrical output at the switchboard.
Convert the avoided diesel generator running to a fuel saving using the standard kW-to-tonnes conversion.
AnswerAbout 380 kW electrical from the turbo-generator, saving roughly 1.9 tonnes of diesel generator fuel per day.
The trap: using Celsius directly in the Carnot formula — 1 − 47/127 gives a meaningless 63 %, more than three times the real 20 % ceiling; the formula only works with absolute temperature.
η_th = W/Q_inState the boundary: engine, plant or shipQ_exh = ṁ_g·c_p·(T_in − T_out)c_p ≈ 1.05 kJ/kg·K for exhaust gasAcid dew point ≈ 130–160 °CRises with fuel sulphur; sets the economiser outlet floorη_Carnot = 1 − T_c/T_hAbsolute temperatures only; the ceiling for any bottoming cycleη_Rankine = (h₁ − h₂)/(h₁ − h_f)Turbo-generator working on recovered steamQ = U·A·ΔT_lmEconomiser/cooler duty; U falls as tubes foulPV = mRTCharge air mass flow from receiver or compressor conditionsFuel/day = P × SFOC × 24/10⁶P in kW, SFOC in g/kWh, answer in tonnesSFOC correction to ISO25 °C air, 25 °C LT water, 42 700 kJ/kg LCVṁ_air ≈ ρ·V̇ = (P/RT)·V̇Compressor delivers ~constant volumetric flow; density does the rest