Thermodynamics and Psychrometry

What to remember

• First law: Q − W = ΔU for a closed system (energy is conserved). Second law: no cycle can convert all heat to work (Kelvin-Planck) and heat cannot flow from cold to hot by itself (Clausius). Carnot efficiency η = 1 − T_L/T_H is the upper limit.
• Ideal gas: pV = mRT, c_p − c_v = R, γ = c_p/c_v (air: 1.4). Reversible adiabatic: pV^γ = constant; polytropic: pV^n = constant.
• Psychrometry: specific humidity ω = 0.622 p_v/(p − p_v); relative humidity φ = p_v/p_g; for saturated air DBT = WBT = DPT.

1. Basic concepts

• System: closed (mass fixed, energy may cross), open (control volume; mass and energy cross), isolated (neither crosses).
• Property: point function (pressure, volume, temperature, internal energy, enthalpy, entropy). Heat and work are path functions.
• Intensive properties do not depend on mass (p, T, specific volume). Extensive ones do (V, U, H, S).
• Zeroth law: two bodies in thermal equilibrium with a third are in equilibrium with each other. It is the basis of temperature measurement.
• Quasi-static (reversible) process: passes through a series of equilibrium states. Cycle: system returns to its initial state.
• Triple point of water: 273.16 K (0.01 °C). Absolute temperature T = t + 273.15.

2. First law

• Closed system: Q − W = ΔU. Example: Q = 100 kJ, W = 40 kJ gives ΔU = 60 kJ.
• Cycle: ∮δQ = ∮δW.
• Enthalpy h = u + pv. Steady-flow energy equation (SFEE): q − w = Δh + ΔV²/2 + gΔz. Applies to turbines (adiabatic, w = h₁ − h₂), compressors, nozzles, boilers, condensers and throttling valves.
• Throttling: constant enthalpy (h₁ = h₂), pressure falls, entropy rises. Nozzle: velocity V₂ = √(2(h₁ − h₂)) from zero inlet velocity (h in J/kg).
• Perpetual motion machine of the first kind (PMM1) violates the first law.

3. Ideal gas and processes

• pV = mRT; R = R_u/M with R_u = 8.314 kJ/kmol·K. For air R = 0.287 kJ/kg·K.
• c_p − c_v = R; γ = c_p/c_v; c_v = R/(γ − 1); c_p = γR/(γ − 1). Air: c_p about 1.005, c_v about 0.718 kJ/kg·K.
• ΔU = m c_v ΔT; ΔH = m c_p ΔT for ideal gases in any process.

• Adiabatic relations: T₂/T₁ = (p₂/p₁)^((γ−1)/γ) = (V₁/V₂)^(γ−1).
• Worked example: 1 kg air at 300 K expands isothermally from 1 MPa to 0.5 MPa: W = 0.287 × 300 × ln 2 = 59.7 kJ.
• Constant-volume heating: 2 kg air by 50 K needs 2 × 0.718 × 50 = 71.8 kJ.

4. Second law, entropy, Carnot

• Kelvin-Planck: no heat engine can work in a cycle exchanging heat with only one reservoir. Clausius: no device can transfer heat from cold to hot body without work input. PMM2 violates the second law.
• Carnot cycle: two reversible isothermal and two reversible adiabatic processes. η = 1 − T_L/T_H (Kelvin). Example: 600 K and 300 K gives 50%.
• Reversible heat engines between the same two reservoirs have the same efficiency (Carnot's theorem), independent of the working fluid.
• COP of refrigerator = T_L/(T_H − T_L); COP of heat pump = T_H/(T_H − T_L) = COP_ref + 1.
• Clausius inequality ∮δQ/T ≤ 0. Entropy dS = δQ_rev/T. Entropy of an isolated system never decreases. Reversible adiabatic means isentropic.
• Sources of irreversibility: friction, free expansion, heat transfer across a finite temperature difference, mixing.

5. Pure substances and steam

• Phases: sub-cooled (compressed) liquid, saturated liquid, wet mixture, saturated vapour, superheated vapour. Critical point of water about 374 °C and 22.1 MPa; beyond it no liquid-vapour distinction.
• Dryness fraction x = mass of vapour / total mass. Properties of wet steam: v = v_f + x v_fg; h = h_f + x h_fg; s = s_f + x s_fg. Latent heat h_fg falls to zero at the critical point.
• Degree of superheat = superheat temperature − saturation temperature at that pressure.
• Mollier (h–s) chart is used for turbine and nozzle expansion. In the wet region constant-pressure lines are also constant-temperature lines.

6. Power cycles

• Otto example: r = 8, γ = 1.4 gives η = 1 − 8^(−0.4) = 1 − 0.435 = 0.565. Brayton example: r_p = 10 gives η = 1 − 10^(−0.2857) = 0.482.
• For the same compression ratio Otto is more efficient than Diesel. For the same maximum pressure and temperature Diesel is more efficient.
• Rankine improvements: higher boiler pressure and superheat temperature; lower condenser pressure; reheat (reduces moisture at turbine exit); regenerative feed heating (raises mean temperature of heat addition). Open and closed feed heaters; the bleed steam is extracted from the turbine.
• Combined gas-steam cycle plants raise efficiency by using gas-turbine exhaust to raise steam.

7. Psychrometry

Moist air is a mixture of dry air and water vapour (Dalton's law: total pressure = p_a + p_v).

• Dry bulb temperature (DBT): ordinary thermometer. Wet bulb temperature (WBT): thermometer with wetted wick. Dew point temperature (DPT): temperature at which vapour begins to condense on cooling at constant pressure.
• Specific (absolute) humidity ω = 0.622 p_v/(p − p_v). Example: p_v = 2 kPa, p = 100 kPa gives ω = 0.622 × 2/98 = 0.0127 kg/kg dry air.
• Relative humidity φ = p_v/p_g (p_g = saturation pressure at DBT). Degree of saturation μ = ω/ω_s.
• Enthalpy of moist air: h = 1.005 t + ω(2500 + 1.88 t) kJ/kg dry air.
• Saturated air: DBT = WBT = DPT; φ = 100%. Unsaturated air: DBT > WBT > DPT.
• Psychrometric chart processes: sensible heating or cooling (horizontal line; ω constant); humidification; cooling and dehumidification (below the dew point); adiabatic mixing; evaporative cooling (constant WBT approx., DBT falls). Sensible heat factor SHF = sensible heat / total heat.
• Bypass factor of a coil: fraction of air that passes without contact.

8. Mixtures, availability and Maxwell relations (short)

• Gas mixtures (Dalton): each gas fills the whole volume at its own partial pressure; mole fraction = pressure fraction = volume fraction. Mixture molecular mass M = sum of (mole fraction × M of each gas).
• Availability (exergy) is the maximum useful work obtainable from a system as it comes into equilibrium with the surroundings. Irreversibility = T₀ × entropy generated.
• Gibbs function g = h − Ts and Helmholtz function a = u − Ts. For a pure substance, T ds = du + p dv = dh − v dp (the two T-ds relations).
• Joule-Thomson coefficient μ = (∂T/∂p) at constant h. It is zero for an ideal gas, so an ideal gas does not change temperature on throttling.
• Steam generators: fire-tube boilers (hot gases inside tubes) suit low pressure; water-tube boilers suit high pressure and large capacity, as in thermal power plants. Boiler accessories: economiser, air pre-heater, superheater, reheater.

Exam traps

• Heat and work are path functions; internal energy, enthalpy and entropy are point functions.
• Carnot efficiency uses Kelvin, not Celsius.
• Throttling is constant enthalpy, not constant entropy.
• Otto is constant-volume heat addition; Diesel is constant-pressure heat addition.
• Isothermal heat equals work only for an ideal gas.
• Polytropic index n = 0 is isobaric, n = 1 isothermal, n = γ adiabatic, n = ∞ isochoric.
• COP of a heat pump = COP of a refrigerator + 1 (same temperatures).
• In sensible heating the specific humidity does not change but relative humidity falls.

One-liners

• 1. Zeroth law defines temperature.
• 2. R for air is 0.287 kJ/kg·K.
• 3. γ for air is 1.4.
• 4. Enthalpy h = u + pv.
• 5. Entropy of an isolated system never decreases.
• 6. Carnot efficiency = 1 − T_L/T_H.
• 7. Dryness fraction is 1 for saturated vapour.
• 8. Rankine cycle is the ideal steam power cycle.
• 9. Reheat reduces moisture in the last turbine stages.
• 10. Triple point of water is 273.16 K.
• 11. Dew point is the temperature of vapour condensation on cooling.
• 12. Specific humidity ω = 0.622 p_v/(p − p_v).

Practice questions

1. The zeroth law of thermodynamics provides the basis for the measurement of

   1. entropy
   2. pressure
   3. temperature
   4. enthalpy
   Answer

   C. temperature

   Bodies in equilibrium with a third are in equilibrium with each other, so thermometers work.

2. Which of the following is a path function?

   1. Internal energy
   2. Heat transfer
   3. Enthalpy
   4. Entropy
   Answer

   B. Heat transfer

   Heat and work depend on the process; u, h, s are point functions.

3. A closed system receives 100 kJ of heat and does 40 kJ of work. The change in internal energy is

   1. 60 kJ
   2. −60 kJ
   3. 40 kJ
   4. 140 kJ
   Answer

   A. 60 kJ

   ΔU = Q − W = 100 − 40 = 60 kJ.

4. In an ideal throttling process the quantity that remains constant is

   1. enthalpy
   2. pressure
   3. temperature of every fluid
   4. entropy
   Answer

   A. enthalpy

   Adiabatic, no work, negligible velocity change: h₁ = h₂.

5. For air with γ = 1.4 and R = 0.287 kJ/kg·K, c_v is about

   1. 0.402 kJ/kg·K
   2. 0.718 kJ/kg·K
   3. 0.287 kJ/kg·K
   4. 1.005 kJ/kg·K
   Answer

   B. 0.718 kJ/kg·K

   c_v = R/(γ − 1) = 0.287/0.4 = 0.7175.

6. One kg of air at 300 K expands isothermally from 1 MPa to 0.5 MPa (R = 0.287 kJ/kg·K). The work done is about

   1. 29.9 kJ
   2. 86.1 kJ
   3. 119.4 kJ
   4. 59.7 kJ
   Answer

   D. 59.7 kJ

   W = RT ln(p₁/p₂) = 0.287 × 300 × 0.693 = 59.7 kJ.

7. Heat required to raise 2 kg of air by 50 K at constant volume (c_v = 0.718 kJ/kg·K) is

   1. 35.9 kJ
   2. 100.5 kJ
   3. 71.8 kJ
   4. 143.6 kJ
   Answer

   C. 71.8 kJ

   Q = m c_v ΔT = 2 × 0.718 × 50.

8. Air at 300 K is compressed reversibly and adiabatically to four times its pressure (γ = 1.4). The final temperature is about

   1. 446 K
   2. 400 K
   3. 1200 K
   4. 520 K
   Answer

   A. 446 K

   T₂ = 300 × 4^(0.2857) = 300 × 1.486 = 446 K.

9. A polytropic process with index n = 1 for an ideal gas is

   1. adiabatic
   2. isothermal
   3. isochoric
   4. isobaric
   Answer

   B. isothermal

   pV¹ = constant is Boyle's law, T constant.

10. One kg of air expands polytropically with n = 1.25 from 500 K to 400 K (R = 0.287 kJ/kg·K). The work done is about

    1. 28.7 kJ
    2. 143.5 kJ
    3. 57.4 kJ
    4. 114.8 kJ
    Answer

    D. 114.8 kJ

    W = R(T₁ − T₂)/(n − 1) = 0.287 × 100/0.25 = 114.8 kJ.

11. A Carnot engine works between 600 K and 300 K. Its efficiency is

    1. 66.7%
    2. 25%
    3. 50%
    4. 100%
    Answer

    C. 50%

    η = 1 − 300/600 = 0.5.

12. A reversible engine operates between reservoirs at 327 °C and 27 °C. Its efficiency is

    1. 25%
    2. 8.3%
    3. 50%
    4. 91.7%
    Answer

    C. 50%

    T_H = 600 K, T_L = 300 K; η = 1 − 0.5.

13. A reversible refrigerator works between 250 K and 300 K. Its COP is

    1. 1.2
    2. 5
    3. 6
    4. 0.2
    Answer

    B. 5

    COP = T_L/(T_H − T_L) = 250/50.

14. A refrigerator has COP 4. The same machine run as a heat pump between the same temperatures would have COP

    1. 3
    2. 5
    3. 4
    4. 0.25
    Answer

    B. 5

    COP_HP = COP_ref + 1.

15. Statements: 1. The Kelvin-Planck statement concerns a heat engine. 2. The Clausius statement concerns heat transfer from a cold body to a hot body. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    Both are statements of the second law.

16. The entropy of an isolated system during any real process

    1. never decreases
    2. is equal to heat divided by temperature
    3. always stays constant
    4. always decreases
    Answer

    A. never decreases

    Entropy increases for irreversible processes, stays constant for reversible ones.

17. A perpetual motion machine of the second kind violates the

    1. second law of thermodynamics
    2. first law of thermodynamics
    3. zeroth law of thermodynamics
    4. law of conservation of mass only
    Answer

    A. second law of thermodynamics

    It would convert heat from a single reservoir wholly into work.

18. A reversible adiabatic process is also

    1. isobaric
    2. isenthalpic
    3. isothermal
    4. isentropic
    Answer

    D. isentropic

    No heat transfer and reversible means ΔS = 0.

19. Wet steam has dryness fraction 0.9, h_f = 500 kJ/kg and h_fg = 2000 kJ/kg. Its enthalpy is

    1. 1800 kJ/kg
    2. 2500 kJ/kg
    3. 2050 kJ/kg
    4. 2300 kJ/kg
    Answer

    D. 2300 kJ/kg

    h = h_f + x h_fg = 500 + 0.9 × 2000.

20. The critical temperature of water is about

    1. 220 °C
    2. 100 °C
    3. 500 °C
    4. 374 °C
    Answer

    D. 374 °C

    Critical point: about 374 °C and 22.1 MPa.

21. At the critical point the latent heat of vaporisation of water is

    1. maximum
    2. zero
    3. about 2257 kJ/kg
    4. equal to sensible heat
    Answer

    B. zero

    Liquid and vapour become identical, so h_fg = 0.

22. A Mollier chart is a plot of

    1. enthalpy against entropy
    2. temperature against volume
    3. pressure against volume
    4. enthalpy against volume
    Answer

    A. enthalpy against entropy

    The h–s chart is used for steam expansion.

23. Which change generally increases the efficiency of a Rankine cycle?

    1. Lowering the superheat temperature
    2. Raising the condenser pressure
    3. Lowering the condenser pressure
    4. Lowering the boiler pressure
    Answer

    C. Lowering the condenser pressure

    It lowers the mean temperature of heat rejection.

24. The main purpose of reheating steam in a Rankine cycle is to

    1. reduce moisture content at turbine exhaust
    2. reduce pump work to zero
    3. increase condenser pressure
    4. reduce boiler pressure
    Answer

    A. reduce moisture content at turbine exhaust

    Reheat keeps the last stages dry and adds some efficiency.

25. Regenerative feed heating improves Rankine efficiency because it

    1. removes the need for a condenser
    2. raises the mean temperature of heat addition
    3. lowers the boiler pressure
    4. increases the turbine back-pressure
    Answer

    B. raises the mean temperature of heat addition

    Feed water enters the boiler hotter, using bled steam.

26. The air-standard Otto efficiency for compression ratio 8 and γ = 1.4 is about

    1. 65%
    2. 30%
    3. 43.5%
    4. 56.5%
    Answer

    D. 56.5%

    η = 1 − 8^(−0.4) = 1 − 0.435.

27. The air-standard efficiency of an Otto cycle depends on

    1. cut-off ratio and γ
    2. pressure ratio only
    3. compression ratio and γ
    4. maximum temperature only
    Answer

    C. compression ratio and γ

    η = 1 − 1/r^(γ−1).

28. Statements: 1. For the same compression ratio the Otto cycle is more efficient than the Diesel cycle. 2. For the same maximum pressure and temperature the Diesel cycle is more efficient than the Otto cycle. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    Both comparisons are standard results.

29. A Brayton cycle with pressure ratio 10 and γ = 1.4 has ideal efficiency about

    1. 28.6%
    2. 48.2%
    3. 35.7%
    4. 64.3%
    Answer

    B. 48.2%

    η = 1 − 10^(−0.2857) = 1 − 0.518.

30. Heat addition in the ideal Brayton cycle takes place at

    1. constant pressure
    2. constant volume
    3. constant entropy
    4. constant temperature
    Answer

    A. constant pressure

    The combustor is a constant-pressure heater.

31. The cut-off ratio is a parameter of the

    1. Brayton cycle
    2. Carnot cycle
    3. Otto cycle
    4. Diesel cycle
    Answer

    D. Diesel cycle

    ρ = V₃/V₂ for constant-pressure heat addition.

32. According to Dalton's law the total pressure of moist air is

    1. the difference of dry air and vapour pressures
    2. the product of partial pressures
    3. equal to vapour pressure only
    4. the sum of the partial pressures of dry air and water vapour
    Answer

    D. the sum of the partial pressures of dry air and water vapour

    p = p_a + p_v.

33. At total pressure 100 kPa and vapour partial pressure 2 kPa, the specific humidity is about

    1. 0.0200 kg/kg dry air
    2. 0.0127 kg/kg dry air
    3. 0.1270 kg/kg dry air
    4. 0.0062 kg/kg dry air
    Answer

    B. 0.0127 kg/kg dry air

    ω = 0.622 × 2/(100 − 2).

34. If the vapour pressure in air is 1.6 kPa and the saturation pressure at the dry bulb temperature is 3.2 kPa, the relative humidity is

    1. 200%
    2. 1.6%
    3. 50%
    4. 25%
    Answer

    C. 50%

    φ = p_v/p_g = 1.6/3.2.

35. The dew point temperature is the temperature at which

    1. water vapour starts to condense when air is cooled at constant pressure
    2. wet bulb equals dry bulb at any humidity
    3. the specific humidity doubles
    4. air is fully dry
    Answer

    A. water vapour starts to condense when air is cooled at constant pressure

    It is the saturation temperature at the vapour partial pressure.

36. For saturated air, which relation holds?

    1. DBT = WBT = DPT
    2. DBT > WBT > DPT
    3. DBT < WBT
    4. WBT = 0
    Answer

    A. DBT = WBT = DPT

    Saturated air has 100% relative humidity.

37. Statements on sensible heating of moist air: 1. The specific humidity remains constant. 2. The relative humidity decreases. Which is/are correct?

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    Moisture content is unchanged but saturation pressure rises with temperature.

38. The sensible heat factor is the ratio of

    1. total heat to latent heat
    2. sensible heat to dry air mass
    3. latent heat to sensible heat
    4. sensible heat to total (sensible plus latent) heat
    Answer

    D. sensible heat to total (sensible plus latent) heat

    SHF = SH/(SH + LH).

39. In an ideal evaporative cooling process the dry bulb temperature

    1. rises while humidity falls
    2. stays constant while humidity rises
    3. falls while the wet bulb temperature stays nearly constant
    4. and wet bulb both rise
    Answer

    C. falls while the wet bulb temperature stays nearly constant

    Adiabatic humidification follows a constant-WBT line.

40. The enthalpy of moist air at 30 °C with ω = 0.01 kg/kg using h = 1.005t + ω(2500 + 1.88t) is about

    1. 25.6 kJ/kg
    2. 55.7 kJ/kg dry air
    3. 30.2 kJ/kg
    4. 85.6 kJ/kg
    Answer

    B. 55.7 kJ/kg dry air

    30.15 + 0.01 × 2556.4 = 55.7.

41. Steam enters an adiabatic turbine with h = 3200 kJ/kg and leaves with h = 2400 kJ/kg (neglect kinetic energy). The work per kg is

    1. 400 kJ
    2. 800 kJ
    3. −800 kJ
    4. 5600 kJ
    Answer

    B. 800 kJ

    w = h₁ − h₂ from the SFEE.

42. A steam nozzle has an enthalpy drop of 50 kJ/kg from negligible inlet velocity. The exit velocity is about

    1. 100 m/s
    2. 1000 m/s
    3. 224 m/s
    4. 316 m/s
    Answer

    D. 316 m/s

    V = √(2 × 50 000) = 316 m/s.

43. The triple point temperature of water is

    1. 273.16 K
    2. 273.00 K
    3. 0 K
    4. 373.15 K
    Answer

    A. 273.16 K

    It is the fixed point of the Kelvin scale.

44. Which of the following is a source of irreversibility?

    1. Reversible heat transfer through infinitesimal temperature difference
    2. Quasi-static expansion
    3. Friction
    4. Frictionless adiabatic expansion
    Answer

    C. Friction

    Friction dissipates work into heat.

45. In a polytropic process pVⁿ = constant, the isobaric process corresponds to n equal to

    1. 0
    2. 1
    3. γ
    4. ∞
    Answer

    A. 0

    pV⁰ = p = constant.