Agricultural Engineering · Engineering Sciences · also Thermodynamics and Heat Transfer (6th Deans' Committee), Thermal Engineering
Energy, work, heat and the cycles behind engines, refrigerators and dryers.
In the exams: GATE AG Section 3 (thermodynamic principles of IC engines); ICAR AIEEA-PG Unit III; NCEES FE thermodynamics and heat transfer 9–14 questions
Engineering thermodynamics is the science of energy and its conversions: how heat becomes work in an engine, how work moves heat uphill in a refrigerator, and how the properties of air, water and refrigerants change while they do it. It rests on a small number of laws that have never been observed to fail — the zeroth law (temperature), the first law (energy is conserved), and the second law (processes have a direction, and some energy is always degraded) — together with property data for real substances, gathered in steam tables, refrigerant charts and the psychrometric chart.
For an agricultural engineer thermodynamics is not abstract. The diesel engine in a tractor or pump set is a heat engine whose best possible efficiency is set by an air-standard cycle. A grain or spice dryer is a psychrometric process: air is heated, picks up moisture and leaves cooler and wetter. A cold store for potatoes, fruit or milk runs on the vapour-compression cycle, and its running cost is decided by the coefficient of performance. Rice parboiling, milk pasteurisation, retort sterilisation and many food plants run on steam, so steam tables and the steady-flow energy equation are daily tools.
The subject is examined in almost the same form everywhere. ICAR AIEEA-PG, GATE AG, the NCEES FE exam, the Philippine ABE licensure and engineering-council exams elsewhere all test the same core: property relations, first- and second-law balances, cycles and moist-air calculations. Standard SI units are used throughout; air is treated as an ideal gas with R = 0.287 kJ/kg·K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K and γ = 1.4 unless stated, which is the convention of most textbooks. Some textbooks and exam boards use slightly different values, so always use the values given in the question.
Thermodynamics begins by drawing a line around what is being studied. The quantity of matter or the region of space chosen for study is the system; everything outside it is the surroundings; the real or imaginary surface that separates them is the boundary. The boundary can be fixed, like the walls of a steel tank, or movable, like the face of a piston.
A closed system (also called a control mass) contains a fixed quantity of matter: mass does not cross the boundary, but energy can, as heat or work. The air trapped in a cylinder with its valves shut is a closed system. An open system (a control volume) is a region in space through which mass flows: a blower pushing air into a grain dryer, a pump, a nozzle, a heat exchanger or a compressor is analysed as a control volume, and its boundary is called the control surface. An isolated system exchanges neither mass nor energy with its surroundings; the universe, taken as system plus surroundings, is the standard example.
The choice is a matter of convenience, not of physics. The same diesel engine can be analysed as a closed system during the compression stroke (valves shut) and as a control volume over many cycles (air and fuel in, exhaust out).
A property is any measurable characteristic of a system — pressure, temperature, volume, mass, internal energy, enthalpy, entropy. Properties are either intensive, meaning independent of the size of the system (pressure, temperature, density), or extensive, meaning proportional to the mass (volume, total internal energy, total enthalpy). An extensive property divided by mass gives a specific property, such as specific volume v = V/m in m³/kg, which is intensive.
The state of a system is its condition as described by its properties. The state postulate says that the state of a simple compressible substance is fixed by two independent intensive properties. For air, pressure and temperature are independent; for water that is boiling, pressure and temperature are not independent (one fixes the other), so a second property such as quality or specific volume is needed.
A system is in thermodynamic equilibrium when it is simultaneously in thermal, mechanical, phase and chemical equilibrium: no unbalanced potentials exist inside it. A process is any change from one equilibrium state to another, and the series of states passed through is the path. A quasi-static (quasi-equilibrium) process proceeds so slowly that the system is always infinitesimally close to equilibrium; only such a process can be drawn as a continuous line on a property diagram. A cycle is a series of processes that returns the system to its initial state, so every property returns to its starting value and the net change of any property over a cycle is zero. Heat and work fail that test; they are path functions, not properties.
Pressure is the normal force exerted by a fluid per unit area, measured in pascals (1 Pa = 1 N/m²); engineering work uses kPa, bar (1 bar = 100 kPa) and MPa. The standard atmosphere is 101.325 kPa. Absolute pressure is measured from a perfect vacuum. Instruments usually read gauge pressure: P_abs = P_gauge + P_atm. Below atmospheric, instruments read vacuum pressure: P_abs = P_atm − P_vac.
Example: a pressure gauge on a steam line in a parboiling unit reads 250 kPa where the barometer reads 98 kPa. The absolute pressure is 250 + 98 = 348 kPa, and it is this value that is used to enter the steam tables.
Density ρ is mass per unit volume (kg/m³) and specific volume v is its reciprocal (m³/kg). Air at ordinary room conditions has a density of about 1.2 kg/m³, and liquid water about 1000 kg/m³; these round values are useful checks on any answer.
The zeroth law of thermodynamics states that if two bodies are each in thermal equilibrium with a third body, they are in thermal equilibrium with each other. It is what makes a thermometer meaningful: the thermometer is the third body. Temperature is the property that is equal for systems in thermal equilibrium.
A practical temperature scale needs a thermometric property that changes with temperature (length of a mercury column, resistance of platinum, voltage of a thermocouple, pressure of a constant-volume gas) and fixed reference points. Historically the Celsius scale used the ice point (0 °C) and steam point (100 °C) at one standard atmosphere. For a linear scale that reads a at the ice point and b at the steam point, a reading X corresponds to t = 100(X − a)/(b − a) °C.
The thermodynamic (absolute) temperature scale does not depend on any substance; it follows from the second law (Chapter 5). Its SI unit is the kelvin: T(K) = t(°C) + 273.15. The triple point of water, where ice, liquid and vapour coexist, is at 273.16 K (0.01 °C); since the 2019 revision of the SI the kelvin is defined by fixing the Boltzmann constant, and the triple point remains 273.16 K within measurement uncertainty. The Fahrenheit scale is related by t(°F) = 1.8 t(°C) + 32, and the absolute Rankine scale by T(R) = 1.8 T(K). A temperature difference of 1 K equals 1 °C, so specific heats in kJ/kg·K and kJ/kg·°C are numerically identical — but ratios of temperatures (efficiencies, ideal-gas laws) must always use kelvin.
The total energy E of a system is the sum of its kinetic energy (mV²/2), potential energy (mgz) and internal energy U. Internal energy is the energy stored at molecular level: the kinetic energy of molecules moving, rotating and vibrating (sensible energy), the energy tied up in phase (latent energy), and chemical and nuclear energy. Internal energy is a property; for an ideal gas it depends on temperature alone, a result demonstrated by Joule's free-expansion experiment. Its SI unit is the joule, and specific internal energy u is in kJ/kg.
A convenient combination property is enthalpy, H = U + PV, or per unit mass h = u + Pv. Enthalpy has no deeper physical meaning of its own for a closed system, but it appears naturally whenever a process occurs at constant pressure or a fluid flows across a boundary, which is why steam and refrigerant tables list h rather than u as their main energy property.
Work and heat are the two ways energy crosses the boundary of a closed system. Both are energy in transit; neither is stored in the system. Heat is energy transferred because of a temperature difference. Work is energy transferred by any other means, the classic test being that its sole effect on the surroundings could have been the raising of a weight.
The most important form in engines is boundary (displacement) work, done when a boundary moves against a pressure: W = ∫P dV for a quasi-static process. On a P–V diagram this integral is the area under the process curve, and because different paths between the same two states enclose different areas, work is a path function; the differential is written δW, not dW. At constant pressure W = P(V2 − V1): a gas at 200 kPa expanding from 0.1 m³ to 0.4 m³ does 200 × 0.3 = 60 kJ of work. At constant volume boundary work is zero. Other forms of work include shaft work (torque × angle turned), electrical work, paddle-wheel (stirring) work, spring work and surface-tension work.
This manual uses the classical engineering sign convention followed by most textbooks and exam papers: heat added to the system is positive and work done by the system is positive. Some modern texts count all energy entering as positive; the physics is identical but the signs in the first law change, so check the convention in each source. An adiabatic process is one with no heat transfer, either because the boundary is insulated or because the process is too fast for heat to flow; it is not the same as isothermal.
The first law of thermodynamics is the principle of conservation of energy. For a closed system executing a cycle, the net heat transfer equals the net work transfer: ∮δQ = ∮δW. For a closed system undergoing a process, Q − W = ΔE = ΔU + ΔKE + ΔPE, and when kinetic and potential energy changes are negligible, Q − W = ΔU. In words: heat added minus work done by the system equals the increase in stored energy.
A consequence is that a machine which produces work continuously without an energy input — a perpetual-motion machine of the first kind (PMM1) — is impossible. Any claim of an engine or pump whose output exceeds all its inputs violates the first law.
Sign discipline matters more than algebra. If 150 kJ of heat is added to a gas and it does 60 kJ of work, ΔU = 150 − 60 = +90 kJ. If a paddle wheel does 40 kJ of work on the gas in a rigid tank while 15 kJ of heat leaks out, W = −40 kJ and Q = −15 kJ, so ΔU = −15 − (−40) = +25 kJ.
For a cycle made of several processes, the first law is applied twice: the sum of the Q values equals the sum of the W values over the cycle, and within any one process Q − W = ΔU, with the ΔU values summing to zero around the cycle.
The specific heat at constant volume, cv = (∂u/∂T)v, is the energy needed to raise a unit mass by one degree at constant volume; the specific heat at constant pressure, cp = (∂h/∂T)p, is the same at constant pressure. For ideal gases du = cv dT and dh = cp dT for any process, not only constant-volume or constant-pressure ones, because u and h depend on temperature alone. cp is greater than cv because at constant pressure part of the heat supplied goes out as boundary work.
Example: heating 2 kg of air at constant pressure from 20 °C to 80 °C takes Q = m cp ΔT = 2 × 1.005 × 60 = 120.6 kJ. The internal energy rises by m cv ΔT = 2 × 0.718 × 60 = 86.2 kJ, and the difference, 34.4 kJ, leaves as boundary work, equal to m R ΔT.
For liquids and solids, which are nearly incompressible, cp and cv are practically equal and a single specific heat c is used: about 4.18 kJ/kg·K for liquid water near room temperature.
At low pressure and at temperatures well above the critical temperature, gases obey the ideal-gas equation of state: Pv = RT, PV = mRT, or on a molar basis PV = nR̄T, where R̄ = 8.314 kJ/kmol·K is the universal gas constant and R = R̄/M is the characteristic gas constant of a gas of molar mass M. For air (M ≈ 28.97 kg/kmol) R = 0.287 kJ/kg·K; for carbon dioxide (M = 44) R = 0.189 kJ/kg·K. P must be absolute and T in kelvin.
Example: the air in a 50 m³ headspace above grain in a sealed bin at 101.325 kPa and 30 °C has mass m = PV/RT = 101.325 × 50 / (0.287 × 303.15) = 58.23 kg. Using 30 instead of 303.15 K in the denominator gives a result ten times too large, a common slip.
Real gases deviate from ideal behaviour at high pressure and near saturation. The deviation is measured by the compressibility factor Z = Pv/RT; Z = 1 for an ideal gas, and generalised compressibility charts give Z as a function of reduced pressure and reduced temperature (actual values divided by critical values). Air in engines and dryers, and water vapour in atmospheric air, are treated as ideal. Steam in a boiler and refrigerant vapour near saturation are not ideal and must be taken from tables.
For an ideal gas the specific heats are linked by Mayer's relation, cp − cv = R, and their ratio is γ = cp/cv (written k in some texts). Combining the two gives cv = R/(γ − 1) and cp = γR/(γ − 1). For air, cp = 1.005 and cv = 0.718 kJ/kg·K give R = 0.287 kJ/kg·K and γ = 1.4.
Specific heats of real gases increase with temperature, so at combustion temperatures γ falls well below 1.4. Textbooks distinguish the 'cold-air-standard' assumption (constant specific heats evaluated at room temperature, the one used in most examination problems) from variable-specific-heat analysis using gas tables.
Constant volume (isochoric): P/T is constant; boundary work is zero; all heat changes internal energy, Q = m cv (T2 − T1). Heating a rigid cylinder of gas is the example.
Constant pressure (isobaric): V/T is constant; W = P(V2 − V1) = mR(T2 − T1); Q = m cp (T2 − T1) = ΔH. Heating air in a dryer's air heater at essentially atmospheric pressure is treated this way.
Isothermal: PV is constant; since u depends on T only, ΔU = 0 and Q = W. The work is W = mRT ln(V2/V1) = mRT ln(P1/P2). Example: compressing 0.5 kg of air isothermally at 300 K from 100 kPa to 500 kPa needs W = 0.5 × 0.287 × 300 × ln(100/500) = -69.3 kJ, that is 69.3 kJ of work input, and the same amount of heat must be rejected to keep the temperature constant. Using log base 10 instead of natural log is the usual error. Isothermal compression needs the least work of any compression between two pressures, which is why multistage compressors are intercooled toward it.
A reversible adiabatic (isentropic) process of an ideal gas with constant specific heats follows PV^γ = constant, which gives T2/T1 = (V1/V2)^(γ−1) = (P2/P1)^((γ−1)/γ). The work done is W = (P1V1 − P2V2)/(γ − 1) = mcv(T1 − T2), with Q = 0. On a P–V diagram an adiabatic line is steeper than an isothermal line through the same point, by the factor γ.
Example: air at 300 K and 100 kPa compressed isentropically in a diesel engine with compression ratio 16 reaches T2 = 300 × 16^0.4 = 909 K and P2 = 100 × 16^1.4 = 4,850 kPa. That temperature, above the self-ignition temperature of diesel fuel, is why a compression-ignition engine needs no spark plug.
Real compressions and expansions lie between isothermal and adiabatic and are modelled as polytropic, PV^n = constant. Special values of n recover the other processes: n = 0 constant pressure, n = 1 isothermal, n = γ reversible adiabatic, n → ∞ constant volume. The polytropic work is W = (P1V1 − P2V2)/(n − 1) = mR(T1 − T2)/(n − 1), and the heat transfer is Q = W (γ − n)/(γ − 1). For 1 kg of air compressed from 100 kPa and 300 K to 800 kPa with n = 1.3: T2 = 300 × 8^(0.3/1.3) = 484.8 K, W = 0.287 × (300 − 484.8)/0.3 = -176.8 kJ, and Q = -176.8 × (1.4 − 1.3)/0.4 = -44.2 kJ; the negative signs show work input and heat rejection to the cylinder walls.
Most equipment on a farm or in a food plant has fluid flowing through it: blowers, fans, pumps, compressors, heat exchangers, nozzles and valves. These are analysed as control volumes. Conservation of mass for a control volume says the rate of change of mass inside equals the mass flow in minus the mass flow out. In steady flow nothing inside changes with time, so mass flow in equals mass flow out; for a single stream, ṁ = ρAV is the same at inlet and outlet, where A is flow area and V mean velocity.
Pushing a fluid into a control volume takes work, because the incoming fluid must displace the fluid ahead of it against the pressure at the boundary. This flow work (or flow energy) is Pv per unit mass. A flowing fluid therefore carries u + Pv + V²/2 + gz per unit mass, and u + Pv is exactly the enthalpy h. This is why enthalpy, not internal energy, appears in every flow-process energy balance.
For a single-stream steady-flow device, the first law becomes the steady-flow energy equation (SFEE):
q − w = (h2 − h1) + (V2² − V1²)/2 + g(z2 − z1),
per unit mass, where q is heat added per kg and w is shaft work done by the fluid per kg. Multiplying by the mass flow rate ṁ gives the rate form, Q̇ − Ẇ = ṁ[Δh + ΔV²/2 + gΔz], in kW when ṁ is in kg/s and h in kJ/kg. The kinetic- and potential-energy terms come out in J/kg and must be divided by 1000 to be added to enthalpy in kJ/kg; forgetting this is the most common error. Kinetic energy matters only at high velocities (45 m/s from rest is about 1 kJ/kg).
Nozzles and diffusers have no shaft work and usually negligible heat transfer. A nozzle accelerates a fluid by dropping its enthalpy: V2 = √(2(h1 − h2) + V1²). Steam entering at negligible velocity with an enthalpy drop of 200 kJ/kg leaves at √(2 × 200 000) = 632 m/s. A diffuser does the reverse, slowing a fluid to raise its pressure.
Turbines and compressors: w = h1 − h2 (per kg) when heat loss and velocity changes are small; a turbine produces positive w, a compressor or fan absorbs it (negative w). Pumps handling liquids: with an incompressible liquid and no temperature change, the ideal work input is v(P2 − P1).
Throttling devices (partly open valves, capillary tubes, porous plugs) have no work, negligible heat transfer and small velocity change, so h1 = h2: throttling is an isenthalpic process. For an ideal gas this means no temperature change; for a real fluid the temperature usually drops, which is used in refrigeration.
Heat exchangers: no work; if the outer shell is insulated, the heat lost by the hot stream equals the heat gained by the cold stream, ṁh cp,h ΔTh = ṁc cp,c ΔTc. Mixing chambers: for adiabatic mixing of streams, Σṁh in = Σṁh out, so for one liquid with constant c the mixed temperature is the mass-weighted mean.
Example (dairy): a milk pasteuriser heats 1000 kg/h of milk from 4 °C to 72 °C. Taking c = 3.9 kJ/kg·K (milk values vary with fat and solids content), Q̇ = (1000/3600) × 3.9 × 68 = 73.7 kW. Leaving out the division by 3600 gives a number in kJ/h, not kW.
The first law says energy is conserved but not which way a process goes. A hot cup cools in a cold room; it never warms by drawing heat from the cooler room, though that would not violate the first law. Work can be turned wholly into heat by friction, but heat cannot be turned wholly into work in a cycle. The second law captures this direction and sets the upper limit on the performance of engines and refrigerators.
A heat engine is a device that operates in a cycle, receives heat QH from a high-temperature reservoir, converts part into net work W, and rejects the rest QL to a low-temperature reservoir. A thermal reservoir is a body so large that heat can be added or removed without changing its temperature — the atmosphere, a river, a furnace. The thermal efficiency is η = W/QH = 1 − QL/QH. A refrigerator or heat pump is the reverse: it takes work W to move heat QL from a cold space and reject QH to a warm one. Its performance is measured by the coefficient of performance, COP_R = QL/W for a refrigerator and COP_HP = QH/W for a heat pump; since QH = QL + W, COP_HP = COP_R + 1 for the same machine.
The Kelvin–Planck statement: it is impossible for any device operating in a cycle to receive heat from a single reservoir and produce a net amount of work. Every heat engine must reject some heat to a lower-temperature sink, so no engine can have a thermal efficiency of 100 percent. A machine that violated this — for example a ship driven by extracting heat from the ocean alone — would be a perpetual-motion machine of the second kind (PMM2).
The Clausius statement: it is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body. A refrigerator must be driven by work (or by heat from a higher-temperature source, as in absorption systems).
The two statements are equivalent: a device that violated one could be combined with an ordinary engine or refrigerator to violate the other.
A reversible process is one that can be reversed leaving no trace on either system or surroundings. It is an idealisation; all real processes are irreversible. The factors that make a process irreversible are called irreversibilities: friction, heat transfer across a finite temperature difference, unrestrained (free) expansion, mixing of different fluids, electrical resistance, inelastic deformation and chemical reactions such as combustion. An internally reversible process has no irreversibilities inside the system boundary; an externally reversible process has none in the surroundings.
Reversible processes matter because they deliver the maximum work (in work-producing devices) and require the minimum work (in work-consuming devices).
The Carnot cycle is a fully reversible cycle of four processes: reversible isothermal heat addition at TH, reversible adiabatic expansion from TH to TL, reversible isothermal heat rejection at TL, and reversible adiabatic compression back to TH. The Carnot principles state that (1) no engine working between two reservoirs can be more efficient than a reversible engine working between the same reservoirs, and (2) all reversible engines working between the same two reservoirs have the same efficiency, whatever the working fluid.
Because the reversible efficiency depends only on the reservoir temperatures, it defines a temperature scale independent of any substance, the Kelvin (thermodynamic) scale: QH/QL = TH/TL for a reversible cycle. Hence η_Carnot = 1 − TL/TH, COP_R,Carnot = TL/(TH − TL) and COP_HP,Carnot = TH/(TH − TL), all with temperatures in kelvin.
Examples: an engine between 527 °C (800 K) and 27 °C (300 K) can at best reach 1 − 300/800 = 62.5 percent. A reversible refrigerator keeping a store at −5 °C and rejecting heat at 35 °C has COP = 268.15/40 = 6.70, and the same machine used as a heat pump has COP = 308.15/40 = 7.70.
Any claim can be tested this way. An inventor who says an engine receives 100 kW from a source at 1000 K, rejects heat at 300 K and delivers 75 kW is claiming 75 percent against a Carnot limit of 70 percent: the claim is impossible.
For any cycle, ∮δQ/T ≤ 0, with equality only for a reversible cycle; this is the Clausius inequality. Because ∮δQ/T = 0 around every reversible cycle, the quantity δQ/T integrated along a reversible path depends only on the end states and so defines a property, entropy S: dS = (δQ/T)rev, in kJ/K, with specific entropy s in kJ/kg·K. Entropy changes are calculated along any convenient reversible path between the two states, even if the actual process was irreversible, because S is a property.
For a reversible process the heat transfer is Q = ∫T dS, the area under the process line on a T–s diagram, just as work is the area under a P–v line. A reversible adiabatic process therefore has constant entropy and is called isentropic. On a T–s diagram the Carnot cycle is a rectangle, and the enclosed area is the net work.
Combining the first law with the definition of entropy gives the Gibbs (T ds) relations: T ds = du + P dv and T ds = dh − v dP. They hold for any process between equilibrium states because they relate properties only.
For an ideal gas with constant specific heats: s2 − s1 = cv ln(T2/T1) + R ln(v2/v1) = cp ln(T2/T1) − R ln(P2/P1). For an isothermal expansion that doubles the volume of air, Δs = 0.287 ln 2 = 0.199 kJ/kg·K, positive because heat was added.
For incompressible liquids and solids, s2 − s1 = c ln(T2/T1). Heating 2 kg of water from 20 °C to 80 °C (c = 4.18 kJ/kg·K) raises its entropy by 2 × 4.18 × ln(353.15/293.15) = 1.557 kJ/K. Temperatures in the logarithm must be absolute; ln(80/20) gives a nonsense answer.
For any process, ΔS_system ≥ ∫δQ/T. The difference is the entropy generated by irreversibilities, S_gen ≥ 0. For an isolated system (or system plus surroundings) ΔS_isolated = S_gen ≥ 0: the entropy of the universe can only increase or, in the reversible limit, stay constant.
Example: 500 kJ of heat passes from a furnace gas at 800 K to drying air at 400 K. The hot body loses 500/800 = 0.625 kJ/K, the cold body gains 500/400 = 1.25 kJ/K, and the entropy generated is 0.625 kJ/K. The larger the temperature difference across which heat flows, the more entropy is generated and the more work potential is destroyed — the thermodynamic reason for designing heat exchangers and dryers with modest temperature differences where economics allows.
Not all heat can become work. The part of a heat quantity Q at temperature T that could be converted to work by a reversible engine rejecting heat to the surroundings at T0 is the available energy (exergy of heat), Q(1 − T0/T); the rest, Q T0/T, is unavailable energy. For 1000 kJ at 500 K with surroundings at 300 K, the available energy is 1000 × (1 − 300/500) = 400 kJ. The Gouy–Stodola theorem states that the work potential destroyed by irreversibility (the irreversibility I) equals T0 × S_gen. Second-law (exergetic) efficiency compares actual performance with the reversible limit, for example COP_actual/COP_Carnot for a refrigerator.
Real turbines, compressors, pumps and nozzles are nearly adiabatic but not reversible, so their exit entropy is higher than the inlet. Their performance is compared with an isentropic process between the same inlet state and exit pressure. Turbine isentropic efficiency ηT = actual work / isentropic work = (h1 − h2)/(h1 − h2s). Compressor and pump isentropic efficiency ηC = isentropic work / actual work = (h2s − h1)/(h2 − h1). Note the inversion: the ideal is on top for compressors and on the bottom for turbines.
Example: air at 300 K is compressed through a pressure ratio of 6 with ηC = 0.85. The isentropic exit temperature is 300 × 6^(0.4/1.4) = 500.6 K; the actual exit temperature is 300 + (500.6 − 300)/0.85 = 535.9 K. Multiplying the ideal rise by 0.85 instead of dividing is the classic mistake, and gives a temperature below the isentropic one, which is impossible for an adiabatic compressor.
A pure substance has a fixed chemical composition throughout; water, whether ice, liquid, vapour or a mixture of them, is a pure substance. Heating water at constant pressure passes through distinct states. Compressed (subcooled) liquid is liquid below its boiling temperature at that pressure. Saturated liquid is liquid about to boil. A saturated liquid–vapour (wet) mixture contains both phases at the saturation temperature. Saturated (dry saturated) vapour is vapour about to condense. Superheated vapour is vapour above the saturation temperature at that pressure. The temperature at which boiling occurs at a given pressure is the saturation temperature; the pressure at which a liquid boils at a given temperature is the saturation pressure. At one standard atmosphere, water boils at 100 °C; at 200 kPa it boils at about 120.2 °C, which is why pressure-cooking and steam sterilisation work.
The energy absorbed during boiling at constant pressure is the latent heat of vaporisation, hfg = hg − hf. It falls as pressure rises and becomes zero at the critical point, which for water is at about 22.06 MPa and 373.95 °C. Above the critical pressure there is no distinct boiling; liquid changes into vapour continuously. At the triple point (0.01 °C, 0.6117 kPa for water) solid, liquid and vapour coexist in equilibrium.
On T–v and P–v diagrams the saturated-liquid and saturated-vapour lines meet at the critical point and enclose a dome; states under the dome are wet mixtures, to its left compressed liquid, and to its right superheated vapour. Constant-pressure lines are horizontal inside the dome on a T–v diagram, because temperature stays constant during boiling. On the T–s diagram, used for cycles, the same dome appears, and heat added in a reversible process is the area under the path. The P–T diagram (phase diagram) shows the sublimation, fusion and vaporisation curves meeting at the triple point; water is unusual in that its fusion line slopes to the left, because ice is less dense than liquid water, so raising pressure lowers its melting point.
Steam tables, based today on the IAPWS-IF97 formulation, list properties in three groups. Saturation tables, indexed either by temperature or by pressure, give saturation pressure or temperature and the properties of saturated liquid (subscript f) and saturated vapour (subscript g), together with the differences (fg). Superheated tables give v, u, h and s at combinations of pressure and temperature. Compressed-liquid tables exist for high pressures; at moderate pressures a compressed liquid is approximated by saturated liquid at the same temperature (not the same pressure).
In the wet region the quality or dryness fraction x = mass of vapour / total mass fixes the state. Any specific property y (v, u, h or s) of the mixture is y = yf + x yfg. For wet steam at 200 kPa with x = 0.9 the tables give hf = 504.7 and hfg = 2201.6 kJ/kg, so h = 504.7 + 0.9 × 2201.6 = 2486.1 kJ/kg. At 100 kPa and x = 0.95, v = vf + x(vg − vf) = 1.609 m³/kg, close to x vg because vf is tiny.
To find which region a state is in, compare the given data with saturation values: at a given pressure, if T is below Tsat the water is compressed liquid, if above it is superheated; if h (or v or s) lies between the f and g values it is wet and x = (h − hf)/hfg. Water at 1 MPa and 150 °C is compressed liquid, because Tsat at 1 MPa is about 179.9 °C. Use the table supplied in the examination.
The Mollier chart plots enthalpy against entropy. It is the natural chart for flow devices because isentropic processes are vertical lines and the enthalpy change, which the SFEE needs, is read directly as a vertical distance. Lines of constant pressure, constant temperature (in the superheat region) and constant quality are drawn on it. In the wet region constant-pressure lines are straight, because their slope (∂h/∂s)p equals the saturation temperature, which is constant there. Throttling appears as a horizontal line (constant h).
The dryness of steam in a line is measured with a throttling calorimeter: a sample is throttled to a low pressure where it becomes superheated, its temperature and pressure are read, and because throttling is isenthalpic the line enthalpy equals the calorimeter enthalpy. Example: steam from a 1 MPa line throttled to 100 kPa and 120 °C has h = 2716.6 kJ/kg, so the line quality is x = (2716.6 − 762.7)/2014.4 = 0.970. Very wet steam cannot be superheated by throttling, so a separating calorimeter is placed first to remove most of the water (the combined separating-and-throttling calorimeter).
Rice parboiling, cooking and sterilising retorts, milk plants and small boilers all use saturated steam because it gives up its large latent heat at a constant, controllable temperature set by its pressure. In a jacketed kettle or a parboiling tank, steam condenses and each kilogram delivers approximately hfg of energy if the condensate leaves as saturated liquid. Wet steam delivers only x hfg.
Boiler duty is calculated with the SFEE: heat per kg = h_steam − h_feedwater. A small boiler producing 500 kg/h of dry saturated steam at 200 kPa from feedwater at 30 °C (h ≈ 125.7 kJ/kg) needs 500 × (2706.2 − 125.7)/3600 = 358 kW of heat into the water; dividing by a boiler efficiency and a fuel calorific value then gives fuel consumption. Returning hot condensate as feedwater lowers this duty.
Real engines burn fuel in a charge that changes composition and is exhausted each cycle. To make analysis possible, the air-standard assumptions replace this with a closed cycle of air as an ideal gas, all processes internally reversible, combustion replaced by heat addition from an external source, and exhaust replaced by heat rejection that returns the air to its initial state. With constant room-temperature specific heats these become cold-air-standard assumptions. The results overstate real efficiency but show the trends correctly.
Key terms: bore, stroke, top dead centre (TDC) and bottom dead centre (BDC), clearance volume Vc and swept (displacement) volume Vs. Compression ratio r = (Vc + Vs)/Vc = V_max/V_min. Mean effective pressure (MEP) is the constant pressure which, acting over the full stroke, would produce the same net work: MEP = W_net/(V_max − V_min). A cycle delivering 1.2 kJ per cycle from a 1.5-litre displacement has MEP = 1.2/0.0015 = 800 kPa.
The Otto cycle is the ideal model of the spark-ignition engine: isentropic compression 1–2, constant-volume heat addition 2–3, isentropic expansion 3–4, constant-volume heat rejection 4–1. Its thermal efficiency is η_Otto = 1 − 1/r^(γ−1), which depends only on the compression ratio and γ. For r = 8 and γ = 1.4, η = 1 − 8^(−0.4) = 56.5 percent.
Worked cycle: air at 100 kPa and 300 K, r = 8, heat added 1000 kJ/kg. T2 = 300 × 8^0.4 = 689.2 K; T3 = T2 + q/cv = 689.2 + 1000/0.718 = 2,082 K; net work = η q = 564.7 kJ/kg. Raising the compression ratio raises efficiency, but in a spark-ignition engine it is limited by knock (detonation — autoignition of the end gas ahead of the flame), which depends on fuel octane rating and chamber design.
The Diesel cycle models the compression-ignition engine used in tractors, pump sets and power tillers: isentropic compression, heat addition at constant pressure 2–3, isentropic expansion, and constant-volume rejection. Air alone is compressed to a temperature above the fuel's self-ignition point, fuel is injected near TDC, and combustion continues as the piston moves. The cut-off ratio rc = V3/V2 is the ratio of volumes at the end and start of heat addition.
η_Diesel = 1 − (1/r^(γ−1)) × (rc^γ − 1)/(γ(rc − 1)). The bracketed factor is always greater than 1, so at the same compression ratio a Diesel cycle is less efficient than an Otto cycle, and its efficiency falls as rc (the load) rises. For r = 16 and rc = 2: η = 61.4 percent, against 67.0 percent for an Otto cycle at r = 16. Diesel engines are nonetheless more efficient in practice because, free of knock, they run at much higher compression ratios (typically in the range of about 14–22, depending on design, against about 8–12 for spark-ignition engines).
Real compression-ignition engines add heat partly at nearly constant volume (the rapid burning of fuel injected during the ignition delay) and partly at nearly constant pressure. The dual cycle models this with a pressure ratio rp = P3/P2 for the constant-volume part and a cut-off ratio rc for the constant-pressure part: η_dual = 1 − (1/r^(γ−1)) × (rp rc^γ − 1)/[(rp − 1) + γ rp (rc − 1)]. It reduces to the Otto cycle when rc = 1 and to the Diesel cycle when rp = 1. For r = 14, rp = 1.5 and rc = 1.6, η = 62.5 percent.
Comparisons are a staple of exam papers. For the same compression ratio and the same heat input: η_Otto > η_dual > η_Diesel. For the same maximum pressure and maximum temperature (a comparison closer to the mechanical limits of a real engine): η_Diesel > η_dual > η_Otto. On a T–s diagram, the cycle with the higher mean temperature of heat addition is the more efficient.
The Brayton cycle models the simple gas-turbine plant: isentropic compression in a compressor, constant-pressure heat addition in a combustor, isentropic expansion in a turbine, and constant-pressure heat rejection. With pressure ratio rp = P2/P1, η_Brayton = 1 − 1/rp^((γ−1)/γ). For rp = 8, η = 44.8 percent. A large share of the turbine work is consumed by the compressor (the back-work ratio), so component efficiencies matter greatly. The same cycle run in reverse is the basis of air-cycle refrigeration.
The brake thermal efficiency of a real tractor diesel engine is well below its air-standard value. The reasons are: specific heats of the gases rise with temperature and the gases dissociate, lowering the effective γ; combustion takes finite time and is not exactly at constant volume or pressure; heat is lost to the cylinder walls and coolant; exhaust valves open before BDC (blowdown loss); pumping work is spent on intake and exhaust strokes; and mechanical friction and accessories absorb part of the indicated power.
A Carnot cycle using steam is impractical: its turbine would handle very wet steam that erodes blades, and its compressor would have to compress a two-phase mixture. The Rankine cycle, the ideal cycle of steam power plants, avoids both: water is pumped as liquid, heated to saturated or superheated vapour in the boiler at constant pressure, expanded isentropically in the turbine, and fully condensed at constant pressure in the condenser. Each component is analysed with the SFEE: pump work wp = v(P_boiler − P_condenser), boiler heat qin = h_turbine inlet − h_pump exit, turbine work wt = h_in − h_out, and efficiency η = (wt − wp)/qin.
Worked example: steam at 3 MPa and 400 °C (h = 3231.6 kJ/kg, s = 6.9233 kJ/kg·K) expands isentropically to a condenser at 10 kPa. At 10 kPa, sf = 0.6492, sfg = 7.4997, hf = 191.81 and hfg = 2392.1. Exit quality x = (6.9233 − 0.6492)/7.4997 = 0.8366; exit enthalpy = 2193.0 kJ/kg; turbine work = 1038.6 kJ/kg. Pump work = 0.00101 × (3000 − 10) = 3.02 kJ/kg, so heat added = 3036.7 kJ/kg and η = 34.1 percent. Pump work is under 1 percent of turbine work: the advantage of pumping a liquid.
The efficiency of the Rankine cycle rises when the average temperature of heat addition rises or that of heat rejection falls. Raising the boiler pressure and superheating the steam both raise the mean temperature of heat addition; superheating also reduces moisture at the turbine exit, whereas raising pressure alone increases it. Lowering the condenser pressure lowers the rejection temperature but is limited by the cooling-water temperature. Reheating (expanding in two stages with reheating between) allows high boiler pressure without excessive exit moisture, and regeneration (heating feedwater with steam bled from the turbine) raises the mean temperature of heat addition.
Most cold stores, milk chillers and refrigerated vehicles run on the vapour-compression refrigeration (VCR) cycle, a reversed Rankine cycle with a throttle in place of a turbine. The ideal cycle has four processes. (1) Evaporator: low-pressure liquid–vapour mixture absorbs heat from the stored produce at constant pressure and leaves as saturated (or slightly superheated) vapour. (2) Compressor: the vapour is compressed isentropically to the condenser pressure and becomes superheated. (3) Condenser: the vapour rejects heat to air or water at constant pressure and leaves as saturated (or subcooled) liquid. (4) Expansion device (capillary tube, thermostatic or electronic expansion valve): the liquid is throttled to evaporator pressure, h4 = h3, and part of it flashes to vapour.
The cycle is drawn on the pressure–enthalpy (p–h) chart, on which the evaporator and condenser are horizontal lines, throttling is a vertical line, and the compression follows a line of constant entropy. Reading enthalpies from the chart and applying the SFEE to each component gives the whole performance.
Refrigerating effect per kg = h1 − h4 (evaporator exit minus evaporator inlet). Compressor work per kg = h2 − h1. Condenser heat rejection per kg = h2 − h3 = refrigerating effect + compressor work. COP = (h1 − h4)/(h2 − h1). Refrigeration capacity is often stated in tonnes of refrigeration (TR): 1 TR is 211 kJ/min, about 3.517 kW, the rate of heat removal that freezes one short ton of water at 0 °C in 24 hours.
Example: for a cold store the chart gives h1 = 395, h2 = 425 and h3 = h4 = 250 kJ/kg. Refrigerating effect = 145 kJ/kg, compressor work = 30 kJ/kg, COP = 4.83. For a load of 10 TR (35.17 kW) the refrigerant flow is 35.17/145 = 0.2425 kg/s, the ideal compressor power 7.28 kW and the condenser rejects 42.4 kW.
Effects to remember: lowering the evaporator temperature or raising the condensing temperature lowers COP and capacity; subcooling the liquid leaving the condenser increases the refrigerating effect; superheating in the evaporator increases it slightly but also raises the compressor work and discharge temperature, and in practice a few degrees of superheat protect the compressor from liquid. Real cycles also have pressure drops in lines and non-isentropic compression.
Refrigerants are chosen for suitable pressures at working temperatures, high latent heat, chemical stability, safety and environmental impact. The chlorofluorocarbons (CFCs, such as R-12) and hydrochlorofluorocarbons (HCFCs, such as R-22) deplete stratospheric ozone and are phased out or being phased out under the Montreal Protocol, on schedules that differ between developed and developing countries. Hydrofluorocarbons (HFCs, such as R-134a and R-404A) have zero ozone-depletion potential but many have high global-warming potential and are being phased down under the Kigali Amendment of 2016. Natural refrigerants — ammonia (R-717), carbon dioxide (R-744) and hydrocarbons such as propane (R-290) and isobutane (R-600a) — are increasingly used. Ammonia, the traditional choice for large cold stores and dairies, is efficient but toxic, flammable within a range of concentrations, and attacks copper and its alloys. Refrigerant rules differ by country.
Psychrometry is the study of mixtures of dry air and water vapour. It is the thermodynamics of drying, cooling, ventilation, greenhouses, livestock housing and cold storage. Both components are treated as ideal gases, and by Dalton's law the total (barometric) pressure is the sum of the partial pressures: P = Pa + pv.
Dry-bulb temperature (DBT) is the air temperature read by an ordinary thermometer. Humidity ratio (specific humidity, moisture content) W is the mass of water vapour per kg of dry air: W = 0.622 pv/(P − pv), where 0.622 is the ratio of molar masses of water and air. Relative humidity φ is the ratio of the actual vapour pressure to the saturation pressure of water at the same dry-bulb temperature, φ = pv/psat(DBT). Degree of saturation μ is W/Ws, the ratio of the actual humidity ratio to that of saturated air at the same temperature; it is close to, but not equal to, φ. Dew-point temperature is the temperature at which the air becomes saturated when cooled at constant pressure and constant W, that is, the saturation temperature at pv. Quantities are expressed per kg of dry air, which stays constant through a process.
Example: air at 30 °C and 101.325 kPa with pv = 2.0 kPa has W = 0.622 × 2.0/(101.325 − 2.0) = 0.0125 kg/kg dry air. Since psat at 30 °C is about 4.247 kPa, φ = 47.1 percent, and the dew point, the temperature at which psat = 2.0 kPa, is about 17.5 °C.
The enthalpy of moist air per kg of dry air, with 0 °C as the reference for both dry air and liquid water, is h = 1.005 t + W(2501 + 1.88 t) kJ/kg dry air, where 2501 kJ/kg is approximately the latent heat of water at 0 °C and 1.88 kJ/kg·K the specific heat of water vapour. For the air above, h = 1.005 × 30 + 0.0125 × (2501 + 1.88 × 30) = 62.2 kJ/kg dry air; most of it is in the vapour term. The specific volume per kg dry air is v = Ra T/(P − pv) = 0.287 × 303.15/(101.325 − 2.0) = 0.876 m³/kg dry air, which converts a fan's volume flow into dry-air mass flow.
Wet-bulb temperature (WBT) is read by a thermometer whose bulb is covered by a wet wick in moving air. Evaporation cools the wick until heat gained from the air balances heat used for evaporation. The adiabatic saturation temperature is the temperature air reaches when it is saturated adiabatically by water at that same temperature; it is a thermodynamic property. For air–water mixtures the two are nearly equal because the Lewis number is close to one, so the psychrometric chart uses a single set of wet-bulb lines. Wet-bulb depression, DBT − WBT, is zero for saturated air and grows as the air gets drier. At saturation DBT, WBT and dew point are equal; otherwise DBT > WBT > dew point.
The psychrometric chart plots these properties for a fixed barometric pressure, usually 101.325 kPa (charts for other altitudes exist and should be used at high elevation). Dry-bulb temperature is on the horizontal axis and humidity ratio on the vertical axis. Curved lines are relative humidity, with the saturation (100 percent) curve at the upper left; sloping straight lines are wet-bulb temperature, running nearly parallel to lines of constant enthalpy; steeper lines give specific volume. Any two independent properties locate a state, and all others are read off. The dew point is found by moving horizontally (constant W) to the saturation curve.
Sensible heating or cooling: temperature changes at constant W; heat = ṁa (h2 − h1) ≈ ṁa (1.005 + 1.88W)(t2 − t1). Heating lowers the relative humidity, which is exactly why dryers heat air before passing it through the product. Cooling below the dew point on a coil gives cooling with dehumidification: moisture condenses, W falls, and condensate = ṁa (W1 − W2). Evaporative (adiabatic) cooling passes air over wet pads: it follows a line of nearly constant wet-bulb temperature (and enthalpy), W and φ rise and DBT falls, with the wet-bulb temperature as the lower limit — effective in dry climates and used for produce pre-cooling rooms, greenhouses and animal sheds. Adiabatic mixing of two streams gives a mixed state on the straight line joining them, with W and h equal to the dry-air-mass-weighted means.
Dryer: 1.5 kg/s of dry air at 30 °C with W = 0.012 kg/kg is heated to 60 °C. Heat input = 1.5 × (1.005 + 1.88 × 0.012) × 30 = 46.2 kW. The hot air passes through a bed of grain adiabatically, cooling along a nearly constant wet-bulb (constant enthalpy) line as it picks up water, and leaves with W = 0.020 kg/kg. Water removed = 1.5 × (0.020 − 0.012) × 3600 = 43.2 kg/h.
Cold store and pre-cooler: air cooled on an evaporator coil below its dew point loses moisture, which condenses and freezes on the coil and later has to be defrosted. A large coil-to-room temperature difference dries the air and the produce, so fresh-produce stores use large coils with small temperature differences.
A student should be able to draw a system boundary, pick the right form of the first law (closed or steady-flow), and work any ideal-gas, steam-table or psychrometric problem with units written at every step. The recurring losses of marks are predictable: Celsius used where kelvin is needed, gauge pressure used in the gas law, J/kg of kinetic energy added to kJ/kg of enthalpy, log10 in place of ln, and the sign of work done on the system. Practise reading steam tables and a p–h chart until finding the region and quality of a state takes under a minute, and memorise the Otto, Diesel, Brayton, Carnot and COP formulas with one worked number each. Tie every formula to a tractor engine, a dryer or a cold store so the numbers make physical sense.
An engineer five years into practice uses thermodynamics to size and audit equipment. Typical tasks: calculating the heater and fan duty of a batch or recirculating grain dryer from the psychrometric states of inlet and exhaust air; estimating boiler steam demand for a parboiling or pasteurising line and checking steam quality and trap performance; reading a refrigeration plant's pressures and temperatures, converting them to an actual COP and comparing it with the design value; and interpreting tractor-engine test reports against air-standard limits. The practitioner learns to distrust equipment claims that beat the Carnot limit, to measure before replacing, and to do an energy balance on site before recommending any retrofit.
A senior engineer with 25 years of experience makes energy decisions for whole plants and supply chains: choosing between biomass steam, heat pumps and solar air heaters for drying; specifying refrigerant and plant type for a multi-commodity cold store under local refrigerant regulations; justifying condensate recovery, exhaust-air recirculation and heat recovery with life-cycle costs; and writing specifications and acceptance tests that state capacity, COP or specific energy consumption at defined conditions. Much of the work is judging the gap between the thermodynamic limit and field performance, and deciding which part of that gap is worth closing. The senior engineer also trains juniors and operators, because most energy waste on farms comes from operation rather than design.
A researcher with a PhD and long field experience applies thermodynamics at the frontier: exergy and entropy-generation analysis of dryers, cold chains, biogas and gasifier systems; developing and validating simulation models of grain drying and storage that couple psychrometrics with product sorption behaviour; evaluating low-GWP refrigerants and transcritical CO2 or ammonia systems for rural cold storage; and heat-pump, solar-thermal and thermal-storage integration for post-harvest processing. The work requires defining system boundaries and reference states rigorously, quantifying measurement uncertainty, and publishing performance in comparable second-law terms. Such a researcher also advises standards bodies and funding agencies on test methods and realistic efficiency targets, separating what the laws of thermodynamics allow from what current technology and economics deliver.
This manual does not reproduce steam tables, refrigerant p–h charts or the psychrometric chart, because tables and figures are not yet supported; students must use a textbook or the table supplied in their examination. It does not cover thermodynamic property relations beyond the T ds equations (Maxwell relations, Clapeyron equation), gas mixtures other than moist air, combustion stoichiometry and adiabatic flame temperature, compressible flow and nozzle choking, reheat and regenerative cycle calculations, absorption and air-cycle refrigeration in detail, or real-gas equations of state. Refrigerant property values in the examples are illustrative enthalpies, not data for a named refrigerant. Storage temperatures and humidities for particular crops are deliberately not given here, because they vary by commodity, cultivar and end use; use published storage recommendations. Specific heat of milk and foods is taken as a round value that varies with composition.
Original practice questions written to the syllabus and pattern of the exams named in each question's examTags. They are not past papers and do not reproduce any board's questions. Steam properties were computed from the IAPWS-IF97 formulation; values in printed tables may differ in the last digit.
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