Thermodynamics and Marine Engineering I: The Laws, Gas Power Cycles, IC Engines, Refrigeration and Air Conditioning
1. The first law, ideal-gas processes and the steady flow energy equation
For a closed system undergoing a cycle, the first law states that the net heat supplied equals the net work done, ∮δQ = ∮δW. For a closed system undergoing a change of state, the difference Q − W depends only on the end states, so it is the change of a property, the internal energy U: Q − W = ΔU (heat added to the system and work done by it positive). For an ideal gas U depends on temperature alone, ΔU = mc_vΔT. The work of a quasi-static expansion of a moving boundary is W = ∫p dV, the area under the process on a p–V diagram, so work, like heat, is a path function and not a property.
| Process | Law | Work w | Heat q |
|---|---|---|---|
| Constant volume (isochoric) | v = constant | 0 | c_v(T₂ − T₁) |
| Constant pressure (isobaric) | p = constant | p(v₂ − v₁) = R(T₂ − T₁) | c_p(T₂ − T₁) |
| Isothermal | pv = constant | RT ln(v₂/v₁) | equal to w |
| Reversible adiabatic (isentropic) | pv^γ = constant | R(T₁ − T₂)/(γ − 1) | 0 |
| Polytropic | pvⁿ = constant | R(T₁ − T₂)/(n − 1) | w(γ − n)/(γ − 1) |
An isothermal expansion of 1 kg of air at 300 K to five times its volume does w = RT ln 5 = 0.287 × 300 × 1.609 = 138.6 kJ/kg of work and takes in the same heat, because its internal energy does not change. A polytropic compression of 1 kg of air with n = 1.3 from 300 K to 400 K needs w = 0.287 × (300 − 400)/0.3 = −95.7 kJ, and the heat is q = w(γ − n)/(γ − 1) = −95.7 × 0.25 = −23.9 kJ: heat is rejected even though the gas gets hotter, because the work put in exceeds the rise in internal energy (c_v ΔT = 71.8 kJ).
In an open system, fluid crossing the boundary must be pushed in and out, and the work of doing so, the flow energy pv per unit mass, combines with u into the enthalpy h = u + pv. For one-dimensional steady flow the steady flow energy equation (SFEE) per unit mass is q − w_s = (h₂ − h₁) + (V₂² − V₁²)/2 + g(z₂ − z₁). Applied to a nozzle (no work, no heat) it gives V₂ = √[2(h₁ − h₂) + V₁²], so an enthalpy drop of 200 kJ/kg from negligible inlet speed gives √(2 × 200 000) = 632.5 m/s; to an adiabatic turbine or compressor, w_s = h₁ − h₂; to a boiler or condenser, q = h₂ − h₁; and to a throttle such as an expansion valve, h₂ = h₁.
2. The second law, Carnot, the Clausius inequality and entropy
The Kelvin–Planck statement: no device operating in a cycle can produce work while exchanging heat with a single reservoir — every heat engine must reject heat. The Clausius statement: no cyclic device can transfer heat from a colder to a hotter body without work input. The two are equivalent: a violation of either lets one build a violation of the other. A process is reversible if system and surroundings can both be restored; real processes are irreversible because of friction, heat transfer across a finite temperature difference, unrestrained expansion, mixing and inelastic deformation.
Corollaries of the second law (Carnot’s theorems): no engine working between two reservoirs can be more efficient than a reversible one between the same reservoirs, and all reversible engines between them have the same efficiency. That efficiency depends only on the reservoir temperatures, which defines the absolute (thermodynamic) temperature scale, Q_H/Q_L = T_H/T_L. The Carnot engine — two reversible isotherms and two reversible adiabats — has η = 1 − T_L/T_H, 62.5% between 800 K and 300 K. Run backwards it is a Carnot refrigerator, COP_R = T_L/(T_H − T_L), 260/40 = 6.5 between 260 K and 300 K, or a heat pump, COP_HP = T_H/(T_H − T_L) = COP_R + 1 = 7.5.
The Clausius inequality ∮δQ/T ≤ 0 holds for every cycle, with equality only for a reversible one; ∮δQ/T > 0 is impossible. Because ∮δQ_rev/T = 0, the integral δQ_rev/T between two states is path-independent, and it defines the property entropy, dS = δQ_rev/T. For any process dS ≥ δQ/T, and the entropy of an isolated system never decreases — the increase-of-entropy principle. For an ideal gas, Δs = c_p ln(T₂/T₁) − R ln(p₂/p₁) = c_v ln(T₂/T₁) + R ln(v₂/v₁); heating air at constant pressure until its absolute temperature doubles raises its entropy by 1.005 ln 2 = 0.697 kJ/(kg·K).
3. Gas power cycles: Carnot, Stirling, Ericsson, Brayton, Otto, Diesel and dual
The air-standard analysis replaces the working fluid by air as an ideal gas, combustion by heat addition and exhaust by heat rejection, so that each engine becomes a cycle of simple processes whose efficiency depends on a few ratios. The Stirling cycle (two isotherms joined by two constant-volume processes) and the Ericsson cycle (two isotherms joined by two constant-pressure processes) both reach the Carnot efficiency when a perfect regenerator passes the heat of one connecting process to the other. The Brayton (Joule) cycle — isentropic compression, constant-pressure heat addition, isentropic expansion, constant-pressure heat rejection — is the gas-turbine cycle, with η = 1 − 1/r_p(γ−1)/γ, 48.2% at a pressure ratio of 10.
The reciprocating-engine cycles differ in how heat is added. The Otto cycle adds it at constant volume: η = 1 − 1/rγ−1, r being the compression ratio, 56.5% at r = 8. The Diesel cycle adds it at constant pressure up to the cut-off ratio ρ = v₃/v₂: η = 1 − [1/rγ−1]·(ρ^γ − 1)/[γ(ρ − 1)]. With r = 16 and ρ = 2: 1/160.4 = 0.3299, (21.4 − 1)/(1.4 × 1) = 1.639/1.4 = 1.171, so η = 1 − 0.3299 × 1.171 = 61.4%. The dual (limited-pressure) cycle adds part of the heat at constant volume (pressure ratio α) and part at constant pressure: η = 1 − [1/rγ−1]·(αρ^γ − 1)/[(α − 1) + γα(ρ − 1)], which reduces to Otto at ρ = 1 and Diesel at α = 1. The mean effective pressure is the net work per cycle divided by the swept volume, p_m = W_net/V_s — 1.2 kJ from 0.002 m³ is 600 kPa — the constant pressure that would do the same work in one stroke, and the fair measure of an engine’s output for its size.
| Basis of comparison | Order of efficiency | Why |
|---|---|---|
| Same compression ratio and heat input | Otto > dual > Diesel | Otto adds all its heat at the highest temperature the compression allows, at top dead centre |
| Same maximum pressure and temperature | Diesel > dual > Otto | The Diesel then has the highest compression ratio and rejects the least heat |
| Practical engines | CI engines more efficient than SI | CI engines run at r = 14–22, SI engines are limited by knock to about 8–12 |
4. Internal-combustion engines: SI and CI, two and four stroke, combustion and knock
IC engines are classified by ignition (spark ignition, SI, or compression ignition, CI), by working cycle (four-stroke or two-stroke), by fuel, by cooling (water or air), by cylinder arrangement (in-line, V, opposed-piston) and by speed. The four-stroke engine takes one power stroke in two revolutions — suction, compression, power, exhaust — with valves timed by a camshaft at half crank speed. The two-stroke engine completes the cycle in one revolution: the exhaust and the admission of fresh charge, scavenging, happen together near bottom dead centre through ports or valves, so it gives nearly twice the power for its size but with some loss of charge and a poorer exchange of gases. In the SI engine a premixed fuel–air charge is compressed and ignited by a spark; power is controlled by throttling the charge. In the CI engine air alone is compressed to a temperature above the fuel’s self-ignition point and the fuel is injected near the end of compression; power is controlled by the quantity of fuel, with no throttle.
Stages of combustion. In the SI engine: an ignition lag (preparation phase) while the spark kernel grows; flame propagation across the chamber, the main pressure rise; and after-burning as the flame reaches the walls. In the CI engine: an ignition delay while the injected fuel atomises, vaporises and mixes; rapid (uncontrolled) combustion of all the fuel prepared during the delay, giving a steep pressure rise; controlled combustion of fuel burning as it is injected; and after-burning. Knock in the SI engine is detonation: the unburnt end gas ahead of the flame self-ignites, producing pressure waves, a metallic ping, overheating and damage. It is made worse by a high compression ratio, high inlet temperature and pressure, advanced spark timing, a long flame path and a low-octane fuel, and prevented by high-octane fuel, retarding the spark, a compact chamber with the plug central, turbulence and cooling the end gas. Diesel knock is the opposite problem: too long a delay lets too much fuel accumulate before it ignites, and it is prevented by a high-cetane fuel, a high compression ratio, a high inlet temperature and good atomisation — exactly the conditions that promote SI knock.
5. Refrigeration and air conditioning
The simple vapour-compression system has four components: the compressor raises the refrigerant vapour from evaporator to condenser pressure (1→2), the condenser rejects heat and condenses it (2→3), the expansion (throttle) valve drops its pressure at constant enthalpy (3→4, h₄ = h₃), and the evaporator takes in heat from the cold space as the liquid boils (4→1). Its COP = (h₁ − h₄)/(h₂ − h₁), refrigerating effect over compressor work; with h₁ = 400, h₂ = 440 and h₃ = h₄ = 250 kJ/kg the COP is 150/40 = 3.75. The vapour-absorption system replaces the compressor by an absorber, a pump and a generator: the vapour is absorbed into a solution (ammonia in water, or water in lithium bromide), pumped as a liquid, and driven off again by heat. It needs little work but much heat, so its COP is lower, but it can run on waste heat — steam or exhaust gas on a ship — and has almost no moving parts.
Air-conditioning principles. Moist air is described by its dry-bulb temperature, its specific humidity ω = 0.622 p_v/(p − p_v) kg of vapour per kg of dry air, its relative humidity φ = p_v/p_sat, its dew point (the temperature at which the vapour begins to condense on cooling) and its wet-bulb temperature, all read together on the psychrometric chart; at a vapour pressure of 2 kPa and 101.325 kPa total, ω = 0.622 × 2/99.325 = 0.0125 kg/kg. The processes are sensible heating and cooling (ω constant, a horizontal line on the chart); humidification and dehumidification (ω changes); cooling and dehumidification, by a coil below the dew point, where the vapour condenses — the summer process, characterised by the coil’s bypass factor; heating and humidification, the winter process; cooling and humidification, the evaporative cooler, along a line of nearly constant wet-bulb temperature; and heating and dehumidification, by a chemical (desiccant) dehumidifier. In adiabatic mixing of two streams, mass and energy balances put the mixture’s ω and enthalpy at the mass-weighted means, and its dry-bulb temperature very nearly so: 2 kg/s at 30 °C mixed with 1 kg/s at 15 °C gives about 25 °C.
Cooling and heating load calculation adds up the heat that the plant must remove (or supply) to hold the conditioned space at its design state: heat conducted through the walls, decks and glass; solar gain; the sensible and latent heat of occupants; lights and equipment; and the fresh-air load, the enthalpy difference of the outside air brought in for ventilation. The loads are split into sensible heat (changing temperature) and latent heat (changing moisture), and their ratio, the sensible heat factor SHF = sensible/(sensible + latent), fixes the slope of the line along which the supply air must be delivered. On a ship the machinery spaces, galleys and accommodation have very different loads, and the fresh-air load in tropical waters is often the largest single item.
Key takeaways
- Q − W = ΔU for a closed system; W = ∫p dV; isothermal w = RT ln(v₂/v₁); polytropic w = R(T₁ − T₂)/(n − 1) with q = w(γ − n)/(γ − 1).
- SFEE q − w = Δh + ΔV²/2 + gΔz: nozzle V = √(2Δh) with Δh in J/kg; throttle h₂ = h₁.
- Carnot η = 1 − T_L/T_H, COP_R = T_L/(T_H − T_L), COP_HP = COP_R + 1; ∮δQ/T ≤ 0; Δs = c_p ln(T₂/T₁) − R ln(p₂/p₁).
- Otto 1 − 1/rγ−1; Diesel 1 − (ρ^γ − 1)/[γ(ρ − 1)rγ−1]; Brayton 1 − 1/r_p(γ−1)/γ; p_m = W_net/V_s; same r: Otto > dual > Diesel.
- VCR COP = (h₁ − h₄)/(h₂ − h₁); ω = 0.622p_v/(p − p_v); cooling-and-dehumidifying needs a coil below the dew point; SI knock is cured by high octane, CI knock by high cetane.
Practice questions (24)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
In a constant-volume process of a closed system, the heat supplied is equal to
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Answer: A — the increase in internal energy
With dV = 0 no boundary work is done, so Q − 0 = ΔU. The heat equals the enthalpy rise in a constant-PRESSURE process; zero heat describes an adiabatic one.One kilogram of air (R = 0.287 kJ/(kg·K)) expands isothermally and reversibly at 300 K to five times its initial volume. The work done (in kJ), to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 138.6
W = mRT ln(V₂/V₁) = 1 × 0.287 × 300 × ln 5 = 86.1 × 1.6094 = 138.6 kJ, and the heat taken in is the same since ΔU = 0. Using log₁₀ 5 gives 60.2 kJ.One kilogram of air (R = 0.287 kJ/(kg·K), γ = 1.4) is compressed reversibly according to pv1.3 = constant from 300 K to 400 K. The magnitude of the heat rejected (in kJ), to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 23.9
W = R(T₁ − T₂)/(n − 1) = 0.287 × (−100)/0.3 = −95.67 kJ; Q = W(γ − n)/(γ − 1) = −95.67 × 0.1/0.4 = −23.9 kJ. Check: ΔU = c_vΔT = (0.287/0.4) × 100 = 71.75 kJ, and Q = ΔU + W = 71.75 − 95.67 = −23.9 kJ. Taking the process as adiabatic gives zero.Steam expands through an adiabatic nozzle with an enthalpy drop of 200 kJ/kg, entering with negligible velocity. The exit velocity (in m/s), to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 632.5
SFEE with q = w = 0 and V₁ ≈ 0: V₂ = √(2Δh) = √(2 × 200 000) = 632.5 m/s. Leaving Δh in kJ/kg gives 20 m/s.Which statements about the steady flow energy equation are correct?
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Answer: A — across a throttling valve the enthalpy is unchanged; B — for an adiabatic turbine with negligible kinetic and potential energy changes, the work is h₁ − h₂; C — flow energy pv is part of the enthalpy
A throttle does no work and exchanges no heat, so h₂ = h₁; a turbine’s work is the enthalpy drop; h = u + pv includes the flow work. A boiler does no shaft work — it is the HEAT supplied that equals the enthalpy rise.A Carnot engine works between reservoirs at 800 K and 300 K. Its thermal efficiency (in per cent), to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 62.5
η = 1 − T_L/T_H = 1 − 300/800 = 0.625 = 62.5%. Using Celsius temperatures (527 °C and 27 °C) gives 94.9%, which is wrong: the ratio must be of absolute temperatures.A Carnot refrigerator maintains a cold space at 260 K while rejecting heat to surroundings at 300 K. Its coefficient of performance, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 6.5
COP_R = T_L/(T_H − T_L) = 260/40 = 6.5. The heat-pump COP between the same temperatures is T_H/(T_H − T_L) = 7.5; the engine efficiency form 1 − T_L/T_H = 0.133 is a different quantity.Air (c_p = 1.005 kJ/(kg·K)) is heated at constant pressure until its absolute temperature doubles. The change in its specific entropy (in kJ/(kg·K)), to three decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.697
Δs = c_p ln(T₂/T₁) − R ln(p₂/p₁) = 1.005 ln 2 − 0 = 1.005 × 0.6931 = 0.697 kJ/(kg·K). Using c_v (0.718) gives 0.498, which is the constant-volume answer.For a certain cycle the cyclic integral ∮δQ/T is found to be negative. The cycle is
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Answer: A — possible and irreversible
The Clausius inequality ∮δQ/T ≤ 0: zero for a reversible cycle, negative for an irreversible one, and a positive value would make the cycle impossible. The first law says nothing about δQ/T.Which statements about the second law are correct?
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Answer: A — a violation of the Clausius statement implies a violation of the Kelvin–Planck statement; B — all reversible engines between the same two reservoirs have the same efficiency; C — heat transfer across a finite temperature difference is irreversible
The two statements are equivalent; Carnot’s corollary makes the reversible efficiency a function of the reservoir temperatures only; finite-ΔT heat transfer generates entropy. An isolated system’s entropy never decreases.An air-standard Otto cycle has a compression ratio of 8. Taking γ = 1.4, its thermal efficiency (in per cent), to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 56.5
η = 1 − 1/rγ−1 = 1 − 8−0.4 = 1 − 0.4353 = 0.5647 = 56.5%. Using γ instead of γ − 1 as the exponent gives 94.6%.An air-standard Diesel cycle has a compression ratio of 16 and a cut-off ratio of 2. Taking γ = 1.4, its thermal efficiency (in per cent), to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 61.4
η = 1 − [1/rγ−1](ρ^γ − 1)/[γ(ρ − 1)] = 1 − (1/3.031)(2.639 − 1)/(1.4 × 1) = 1 − 0.3299 × 1.1707 = 1 − 0.3862 = 0.6138 = 61.4%. The Otto cycle at the same r would give 67.0%; dropping the γ in the denominator gives 45.9%.An ideal Brayton (Joule) cycle has a pressure ratio of 10. Taking γ = 1.4, its thermal efficiency (in per cent), to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 48.2
η = 1 − 1/r_p(γ−1)/γ = 1 − 10−0.2857 = 1 − 0.5179 = 48.2%. Treating 10 as a volume ratio (as in Otto) gives 60.2%.An engine cycle produces a net work of 1.2 kJ per cycle with a swept volume of 0.002 m³. Its mean effective pressure (in kPa) is ____.
Numerical answer — type the value.
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Answer: 600
p_m = W_net/V_s = 1.2 kJ/0.002 m³ = 600 kPa. Dividing by the total cylinder volume instead of the swept volume understates it.For the same compression ratio and the same heat input, the air-standard cycles in decreasing order of thermal efficiency are
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Answer: A — Otto, dual, Diesel
At a fixed r the Otto cycle adds all its heat at the highest temperature and rejects the least, so it is the most efficient. The order reverses — Diesel first — when the cycles are compared at the same maximum pressure and temperature.The Stirling and Ericsson cycles, each with a perfect regenerator, have a thermal efficiency
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Answer: A — equal to that of a Carnot cycle between the same temperatures
With perfect regeneration all external heat is added and rejected isothermally at T_H and T_L, so η = 1 − T_L/T_H, the Carnot value. No cycle can exceed Carnot between the same temperatures.Which of the following increase the tendency of a spark-ignition engine to knock?
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Answer: A — raising the compression ratio; B — advancing the spark timing; C — raising the inlet air temperature
Higher compression ratio, earlier spark and hotter inlet air all raise the end-gas temperature and pressure, bringing its self-ignition before the flame arrives. A higher octane number means greater resistance to self-ignition, so it reduces knock.Knock in a compression-ignition engine is reduced by
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Answer: A — a fuel of high cetane number, which shortens the ignition delay
CI knock comes from a long ignition delay that lets fuel accumulate; high cetane shortens the delay. High octane, a lower compression ratio and colder inlet air all lengthen the delay and make diesel knock worse.A four-stroke engine running at 1200 rpm has, in each cylinder, a number of power strokes per minute equal to
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Answer: A — 600
A four-stroke cycle takes two revolutions, so there are N/2 = 600 power strokes per minute per cylinder. A two-stroke engine at the same speed would have 1200.In a simple vapour-compression refrigerator the enthalpies are 400 kJ/kg at the compressor inlet, 440 kJ/kg at the compressor outlet and 250 kJ/kg at the condenser outlet. Its coefficient of performance, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 3.75
Throttling keeps h₄ = h₃ = 250; refrigerating effect h₁ − h₄ = 150 kJ/kg; compressor work h₂ − h₁ = 40 kJ/kg; COP = 150/40 = 3.75. Using the condenser heat 190 in the numerator gives the heat-pump COP 4.75.Moist air at a total pressure of 101.325 kPa has a partial pressure of water vapour of 2 kPa. Its specific humidity (in kg per kg of dry air), to four decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.0125
ω = 0.622 p_v/(p − p_v) = 0.622 × 2/99.325 = 0.01252 ≈ 0.0125. Dividing by the total pressure instead of the dry-air pressure gives 0.0123.A stream of 2 kg/s of air at 30 °C is mixed adiabatically with 1 kg/s of air at 15 °C, both of similar humidity. The dry-bulb temperature of the mixture (in °C), taken as the mass-weighted mean, is ____.
Numerical answer — type the value.
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Answer: 25
t = (2 × 30 + 1 × 15)/3 = 75/3 = 25 °C. The simple average 22.5 °C ignores the flow rates.To cool air and remove moisture from it in a summer air-conditioning plant, the cooling coil must be at a temperature
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Answer: A — below the dew point of the entering air
Moisture condenses only on a surface below the dew point. A coil between the dew point and the dry-bulb temperature cools the air sensibly with no change in specific humidity.Which statements comparing vapour-absorption and vapour-compression refrigeration are correct?
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Answer: A — the absorption system replaces the compressor by an absorber, a pump and a generator; B — the absorption system can be driven by waste heat; C — the absorption system usually has the lower COP
Absorption pumps a liquid solution instead of compressing vapour, so its mechanical work is small; its energy input is heat, which may be waste steam or exhaust, and its COP (heat-based) is lower. It needs LESS mechanical work, not more.