Mixtures of Ideal Gases: Dalton and Amagat Laws, Mixture Properties and Psychrometry

This chapter answers Section XE4.8 of XE4 Thermodynamics, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam: “Dalton’s and Amagat’s laws, properties of ideal gas mixtures, air-water vapor mixtures and simple thermodynamic processes; Specific and relative humidities; Dew point, dry bulb and wet bulb temperatures, adiabatic saturation temperature, Simple psychrometric processes.” It teaches the two laws and the composition relations, the properties of an ideal-gas mixture including the entropy of mixing, the vocabulary of humid air, and the simple psychrometric processes (sensible heating and cooling, cooling with dehumidification, humidification, adiabatic mixing and evaporative cooling). The numericals are mixture molar mass and partial pressure, humidity ratio, entropy of mixing, cooling-coil load and the humidity of a mixed stream. Every saturation pressure, enthalpy and p_atm is given in the stem. Convention: all gases are ideal, enthalpies are per kg of dry air, and heat removed from the air is quoted as a positive load.

1. Dalton’s and Amagat’s laws and composition

A gas mixture of components i, with n_i moles and mass m_i, is described by the mole fraction y_i = n_i/n, the mass fraction mf_i = m_i/m, and Σy_i = Σmf_i = 1. The two laws say how the mixture pressure and volume are built from the components:

The two laws
LawStatementIdeal-gas consequence
Dalton (additive pressure)p = Σp_i, where p_i is the pressure component i would exert alone at the mixture T and Vp_i = n_iRuT/V = y_i p
Amagat (additive volume)V = ΣV_i, where V_i is the volume component i would occupy alone at the mixture T and pV_i = n_iRuT/p = y_i V

For ideal gases both laws hold exactly, and y_i = p_i/p = V_i/V: mole fraction, pressure fraction and volume fraction are the same number. For real gases the two laws give different answers, and Amagat’s is usually the better one at high pressure. Mass fraction and mole fraction are not the same: y_i = (mf_i/M_i)/Σ(mf_j/M_j), mf_i = y_iM_i/M_m.

Mixture molar mass M_m = m/n = Σy_iM_i = 1/Σ(mf_i/M_i), and R_m = Ru/M_m with Ru = 8.314 kJ/kmol·K. Worked example: 2 kg of N₂ (M = 28 kg/kmol) and 3 kg of CO₂ (M = 44 kg/kmol) at a total pressure of 200 kPa. n = 2/28 + 3/44 = 0.07143 + 0.06818 = 0.13961 kmol; M_m = 5/0.13961 = 35.8 kg/kmol; y_CO₂ = 0.06818/0.13961 = 0.488, so p_CO₂ = 0.488 × 200 = 97.7 kPa and p_N₂ = 102.3 kPa, summing to 200. The trap is to use the mass fraction 0.6 as the pressure fraction, which gives 120 kPa.

2. Properties of an ideal-gas mixture and the entropy of mixing

Under the Gibbs-Dalton model each ideal-gas component behaves as if alone at the mixture temperature, and extensive properties add:

  • U = Σm_iu_i, H = Σm_ih_i, so per kg of mixture u_m = Σmf_iu_i and h_m = Σmf_ih_i
  • c_p,m = Σmf_ic_p,i and c_v,m = Σmf_ic_v,i on a mass basis; on a molar basis C̄_p,m = Σy_iC̄_p,i. Weights are mass fractions for per-kg values and mole fractions for per-kmol values.
  • c_p,m − c_v,m = R_m, and γ_m = c_p,m/c_v,m.
  • s_m = Σmf_is_i(T, p_i): each component’s entropy is evaluated at its partial pressure, so s_i(T, p_i) = s_i(T, p) − R_i ln y_i.

Entropy of mixing. Let different ideal gases, each initially at the same T and p, be allowed to mix in an insulated vessel at constant total pressure and temperature (the partitions are removed). Each gas ends at partial pressure y_ip, so its entropy rises by −n_iRu ln y_i, and

ΔS_mix = −nRu Σy_i ln y_i > 0.

It is positive and independent of the gases chosen as long as they are different. Worked example: 1 kmol N₂ and 1 kmol O₂ mixed (y = 0.5 each, take both as distinct ideal gases, Ru = 8.314 kJ/kmol·K): ΔS = −2 × 8.314 × (0.5 ln 0.5 + 0.5 ln 0.5) = 2 × 8.314 × 0.6931 = 11.53 kJ/K. The process is irreversible: unmixing needs work.

⚠️ Same gas on both sides
If the two gases are the same, removing the partition changes nothing and ΔS = 0. The formula applies only when the components are distinguishable. Also, the entropy of mixing is per mixture moles, so doubling n doubles ΔS.

3. Humid air: humidity, dew point, wet bulb and adiabatic saturation

Treat humid air as dry air plus water vapour, both ideal gases, with p = p_a + p_v. The specific humidity (humidity ratio) ω = m_v/m_a = 0.622 p_v/(p − p_v), where 0.622 ≈ M_v/M_a = 18/28.97. The relative humidity φ = p_v/p_g(T) = m_v/m_g at the same T and p, where p_g is the saturation pressure of water at the dry-bulb temperature. Combining, ω = 0.622φp_g/(p − φp_g). Note that φ and ω are different things: ω counts the mass of vapour, φ says how close the vapour is to saturation at that temperature. Enthalpy per kg of dry air: h = c_pa T + ω h_g, with h_g ≈ 2501 + 1.82T (T in °C) and c_pa = 1.005 kJ/kg·K.

Worked example: T = 30 °C, φ = 0.60, p = 101.325 kPa, p_g(30 °C) = 4.246 kPa. p_v = 0.60 × 4.246 = 2.548 kPa; ω = 0.622 × 2.548/(101.325 − 2.548) = 0.01604, i.e. 16.0 g of vapour per kg of dry air.

The temperatures of humid air
QuantityMeaningOrdering for unsaturated air
Dry-bulb temperature (DBT)ordinary thermometer readinghighest
Wet-bulb temperature (WBT)steady reading of a thermometer whose bulb is covered by a wetted wick in a moving air stream, with evaporation cooling itbetween
Dew-point temperature (DPT)saturation temperature at the vapour pressure p_v: where condensation starts on cooling at constant p and ωlowest

So DPT ≤ WBT ≤ DBT, with all three equal for saturated air. The adiabatic saturation temperature T* is defined thermodynamically: unsaturated air flows over a long pool of water in an insulated duct and leaves saturated at T, with make-up water supplied at T. Energy balance per kg of dry air: h₁ + (ω* − ω₁)h_f* = h*, which gives

ω₁ = [c_pa(T* − T₁) + ωh_fg]/(h_g1 − h_f*),

where ω* is the saturation humidity ratio at T. For the air-water system at atmospheric pressure the wet-bulb temperature is close to T, which is why the wet-bulb reading is used for it. (For other vapour-gas mixtures the two differ.)

Worked example: p = 101.325 kPa, T₁ = 30 °C, T* = 20 °C, with p_g(20 °C) = 2.339 kPa, h_fg* = 2454, h_f* = 84, h_g1 = 2556 kJ/kg. ω* = 0.622 × 2.339/(101.325 − 2.339) = 0.01470. ω₁ = [1.005 × (20 − 30) + 0.01470 × 2454]/(2556 − 84) = (−10.05 + 36.07)/2472 = 0.01053.

4. Simple psychrometric processes

The processes on the psychrometric chart
ProcessWhat changesKey relation (per kg dry air)
Sensible heating or coolingω and DPT constant; DBT changes; φ falls on heating, rises on coolingq = h₂ − h₁ ≈ (c_pa + 1.82ω)(T₂ − T₁)
Cooling with dehumidification (below the DPT)T and ω fall; air leaves nearly saturated; liquid is drainedcondensate = m_a(ω₁ − ω₂); Q = m_a[(h₁ − h₂) − (ω₁ − ω₂)h_f]
Humidification with steamω rises; T rises slightlyh₂ = h₁ + (ω₂ − ω₁)h_steam
Adiabatic mixing of two streamsmixture lies on the straight line joining the two statesω₃ = (m_a1ω₁ + m_a2ω₂)/(m_a1 + m_a2); likewise for h
Evaporative cooling (water spray, adiabatic)T falls, ω rises, φ rises; h nearly constant; follows the constant-WBT lineT₂ can never go below the WBT of the entering air

Worked example, coil load: air at 1 kg/s of dry air enters a cooling coil with h₁ = 75.0 kJ/kg and ω₁ = 0.0180, and leaves with h₂ = 40.0 kJ/kg and ω₂ = 0.0090. The condensate leaves at h_f = 42 kJ/kg. Condensate = 1 × (0.0180 − 0.0090) = 0.0090 kg/s. Q = 1 × (75.0 − 40.0) − 0.0090 × 42 = 35.0 − 0.378 = 34.6 kW. Omitting the condensate term gives 35.0, a small but exam-relevant slip.

Worked example, mixing: 2 kg/s of dry air with ω = 0.010 mixes adiabatically with 1 kg/s of dry air with ω = 0.020. ω₃ = (2 × 0.010 + 1 × 0.020)/3 = 0.01333, i.e. 13.33 g/kg. Always weight by dry-air flows, not by total moist-air flows, and the mixture can become supersaturated (a fog) when the straight line passes through the saturation curve.

⚠️ Sensible cooling is not dehumidification
Cooling humid air from a state with φ = 50 % does not remove any water until the air reaches its dew point. Only below the DPT does ω fall. A coil whose surface is above the entering air’s DPT cools without dehumidifying.

5. Sensible heating, evaporative cooling and moist-air volume in numbers

Sensible heating. Because ω does not change, the heat per kg of dry air is (c_pa + ω c_pv)ΔT, with the vapour specific heat taken as 1.82 kJ/kg·K. Example: 1 kg/s of dry air at ω = 0.010 heated from 15 °C to 30 °C: q = (1.005 + 1.82 × 0.010) × 15 = 1.0232 × 15 = 15.35 kW. The dew point stays where it was, but φ falls sharply because p_g rises with T.

Evaporative cooling. Air at 35 °C with ω = 0.010 is passed through a wetted pad and leaves at 25 °C. The liquid water added carries little enthalpy, so h₂ ≈ h₁. h₁ = 1.005 × 35 + 0.010 × (2501 + 1.82 × 35) = 35.18 + 25.65 = 60.82 kJ/kg. At 25 °C: 25.13 + ω₂(2501 + 1.82 × 25) = 60.82, so ω₂ = 35.69/2546.5 = 0.01402. The cooler has evaporated 14.02 − 10.00 = 4.0 g of water per kg of dry air, and the sensible heat lost by the air has become latent heat in the added vapour. Cooling stops when the air is saturated at the WBT, so no pad can cool below the entering WBT, and the method works best in dry air.

Moist-air volume. The specific volume per kg of dry air is v = R_aT/(p − p_v), since the dry air alone fills the volume at its partial pressure p_a = p − p_v. For the 30 °C, φ = 0.60 state above: v = 0.287 × 303.15/(101.325 − 2.548) = 0.881 m³/kg dry air. Moist air at the same T and p is lighter than dry air, because 18 < 28.97, so a rise in humidity lowers the density.

Key takeaways

  • Dalton: p = Σp_i; Amagat: V = ΣV_i. For ideal gases y_i = p_i/p = V_i/V, and mass fraction differs from mole fraction through y_i = (mf_i/M_i)/Σ(mf_j/M_j).
  • Mixture M_m = Σy_iM_i = 1/Σ(mf_i/M_i); c_p,m = Σmf_ic_p,i per kg; each component’s entropy is taken at its partial pressure.
  • Entropy of mixing of distinct ideal gases at equal T and p: ΔS = −nRu Σy_i ln y_i > 0; it is zero for identical gases.
  • ω = 0.622p_v/(p − p_v) and φ = p_v/p_g; DPT is the saturation temperature at p_v; DPT ≤ WBT ≤ DBT with equality when saturated.
  • Adiabatic saturation: ω₁ = [c_pa(T* − T₁) + ωh_fg]/(h_g1 − h_f); for air-water the wet-bulb temperature is nearly T.
  • Psychrometric processes: sensible change keeps ω fixed; dehumidification needs a surface below DPT; mixing is weighted by dry-air flow; evaporative cooling stays at about constant h and cannot go below WBT.

Practice questions (18)

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.

  1. Dalton’s law of partial pressures defines the partial pressure of a component of a gas mixture as the pressure that

    1. the component alone would exert in the whole mixture volume at the mixture temperature
    2. the component would exert if it alone were at the mixture pressure and temperature
    3. equals its mass fraction multiplied by the total pressure
    4. each component exerts only when the mixture is at the critical point
    Show answer

    Answer: A — the component alone would exert in the whole mixture volume at the mixture temperature

    Dalton’s partial pressure is that of the component alone in the full volume V at the mixture T; for an ideal gas it equals y_ip with the mole fraction (not the mass fraction). Placing the component at the mixture pressure instead defines the Amagat partial volume picture.

  2. A mixture contains 1 kg of hydrogen (M = 2 kg/kmol) and 8 kg of oxygen (M = 32 kg/kmol), both ideal gases. Its molar mass, in kg/kmol, is

    1. 17.0, the simple mean of the two molar masses
    2. 12.0, the total mass divided by the total moles
    3. 28.7, the mass-fraction-weighted sum of the molar masses
    4. 9.0, the total mass divided by the number of components
    Show answer

    Answer: B — 12.0, the total mass divided by the total moles

    n = 1/2 + 8/32 = 0.5 + 0.25 = 0.75 kmol, so M_m = 9/0.75 = 12.0 kg/kmol, which is also Σy_iM_i = (2/3)(2) + (1/3)(32). The mass-fraction-weighted sum (1/9 × 2 + 8/9 × 32 = 28.7) mixes a mass weighting with a per-kmol quantity, and the simple mean ignores the amounts.

  3. A gas mixture of 2 kg of N₂ (M = 28 kg/kmol) and 3 kg of CO₂ (M = 44 kg/kmol) is at a total pressure of 200 kPa, and both gases are ideal. The partial pressure of CO₂ is closest to

    1. 120 kPa, the mass fraction times the total pressure
    2. 102 kPa, the partial pressure of the nitrogen
    3. 80 kPa, the mass fraction of nitrogen times the total pressure
    4. 97.7 kPa, the mole fraction of CO₂ times the total pressure
    Show answer

    Answer: D — 97.7 kPa, the mole fraction of CO₂ times the total pressure

    n_N₂ = 2/28 = 0.07143, n_CO₂ = 3/44 = 0.06818 kmol, y_CO₂ = 0.4884, p_CO₂ = 0.4884 × 200 = 97.7 kPa. The 120 kPa answer uses the mass fraction, which is wrong for pressure; 102 kPa is the nitrogen’s partial pressure, and 80 kPa is the nitrogen mass fraction times the total pressure.

  4. Two different ideal gases, each initially at the same temperature and pressure in separate compartments, are allowed to mix in an insulated rigid container (no heat or work exchange, final state at the same T and total p equal to the initial p). The entropy of the gases

    1. is unchanged, because the temperature and pressure are unchanged and no heat flows
    2. increases by −nRu Σy_i ln y_i as each gas falls to its partial pressure
    3. decreases, because the mixture is more ordered than the separate gases
    4. increases only if the two gases have different specific heats
    Show answer

    Answer: B — increases by −nRu Σy_i ln y_i as each gas falls to its partial pressure

    Each component goes from p to y_ip at the same T, so Δs_i = −R_i ln y_i > 0 and the total is −nRu Σy_i ln y_i; mixing is irreversible. Unchanged T and p of the total do not mean unchanged entropy of each component, and the result does not depend on the specific heats.

  5. Humid air at constant total pressure and constant humidity ratio is heated. Which of the following is correct?

    1. The relative humidity falls and the dew-point temperature is unchanged
    2. The relative humidity rises and the dew-point temperature rises
    3. The relative humidity is unchanged because the humidity ratio is unchanged
    4. The relative humidity falls and the dew-point temperature falls
    Show answer

    Answer: A — The relative humidity falls and the dew-point temperature is unchanged

    At constant ω and p, p_v = ωp/(0.622 + ω) is fixed, so the dew point (the saturation temperature at p_v) is fixed. The saturation pressure p_g rises with T, so φ = p_v/p_g falls. Equal ω does not give equal φ at different temperatures.

  6. Moist air is compressed isothermally from 100 kPa to 200 kPa with no condensation, so the humidity ratio is unchanged. Compared with the initial state, the dew-point temperature of the compressed air

    1. is lower, because the air is drier after compression
    2. is unchanged, because the humidity ratio is unchanged
    3. is higher, because the vapour partial pressure p_v = ωp/(0.622 + ω) doubles
    4. is equal to the new dry-bulb temperature
    Show answer

    Answer: C — is higher, because the vapour partial pressure p_v = ωp/(0.622 + ω) doubles

    With ω fixed, p_v is proportional to p, so it doubles and the dew point (saturation temperature at p_v) rises. This is why compressed air must be dried or drained: after cooling, it can easily reach its dew point. The humidity ratio, not the dew point, is what stays unchanged.

  7. For unsaturated humid air, the dry-bulb temperature (DBT), wet-bulb temperature (WBT) and dew-point temperature (DPT) are related by

    1. DBT < WBT < DPT
    2. WBT < DPT < DBT
    3. DPT < WBT < DBT
    4. DBT = WBT, with DPT lower
    Show answer

    Answer: C — DPT < WBT < DBT

    Evaporation from the wick cools it below the DBT, but not below the DPT, since the air could not take up the water needed for that. All three are equal only when the air is saturated. WBT never exceeds DBT in unsaturated air.

  8. Humid air flows over a cooling coil. Water vapour condenses out of the air only if

    1. the coil surface is hotter than the air stream flowing over it
    2. the relative humidity of the entering air is above 50 %
    3. the entering air has a wet-bulb temperature above 20 °C at the coil
    4. the coil surface is below the dew point of the entering air
    Show answer

    Answer: D — the coil surface is below the dew point of the entering air

    Condensation starts when air touching the surface falls to its dew point. Relative humidity thresholds such as 50 % and fixed wet-bulb values have no meaning without the surface temperature, and a hot surface cannot condense anything.

  9. Air passes through an adiabatic wetted-pad evaporative cooler and the make-up water enters at about the exit temperature. Which statement about the air is correct?

    1. Its dry-bulb temperature falls, its humidity ratio rises, and its enthalpy stays almost constant
    2. Its dry-bulb temperature falls and its humidity ratio is unchanged
    3. Its dry-bulb temperature falls below the entering wet-bulb temperature
    4. Its relative humidity falls and its dew point falls
    Show answer

    Answer: A — Its dry-bulb temperature falls, its humidity ratio rises, and its enthalpy stays almost constant

    The sensible heat lost by the air equals the latent heat of the added vapour, so h is nearly constant while T falls and ω (and the dew point) rises along the line of constant wet-bulb temperature. The exit DBT cannot go below the entering WBT, and the relative humidity rises.

  10. Humid air enters an adiabatic saturator at 30 °C. It leaves saturated at 20 °C (this is the adiabatic saturation temperature). Data: p = 101.325 kPa; p_g(20 °C) = 2.339 kPa; h_fg at 20 °C = 2454 kJ/kg; h_f at 20 °C = 84 kJ/kg; h_g at 30 °C = 2556 kJ/kg; c_pa = 1.005 kJ/kg·K. The humidity ratio of the entering air is closest to

    1. 0.0147 kg/kg
    2. 0.0105 kg/kg
    3. 0.0083 kg/kg
    4. 0.0126 kg/kg
    Show answer

    Answer: B — 0.0105 kg/kg

    ω* = 0.622 × 2.339/(101.325 − 2.339) = 0.01470. ω₁ = [c_pa(T* − T₁) + ω*h_fg]/(h_g1 − h_f) = [1.005 × (−10) + 0.01470 × 2454]/(2556 − 84) = (−10.05 + 36.07)/2472 = 0.0105. The value 0.0147 is the saturated exit humidity ratio, not the entering one.

  11. Which statements are correct for a mixture of ideal gases at a given temperature?

    1. The partial pressure of a component equals its mole fraction times the total pressure
    2. The partial volume of a component (Amagat) equals its mole fraction times the total volume
    3. The mole fraction equals the pressure fraction and the volume fraction
    4. The mass fraction of a component equals its mole fraction
    Show answer

    Answer: A — The partial pressure of a component equals its mole fraction times the total pressure; B — The partial volume of a component (Amagat) equals its mole fraction times the total volume; C — The mole fraction equals the pressure fraction and the volume fraction

    For ideal gases p_i/p = V_i/V = n_i/n, so the first three hold. The mass fraction is y_iM_i/M_m, which equals the mole fraction only if all the molar masses are equal.

  12. Which statements about humid air are correct?

    1. The humidity ratio depends on the vapour partial pressure and on the total pressure
    2. The relative humidity is 100 % when the dry-bulb temperature equals the dew-point temperature
    3. Heating air at constant humidity ratio raises its dew-point temperature
    4. The wet-bulb temperature of unsaturated air is above its dry-bulb temperature
    Show answer

    Answer: A — The humidity ratio depends on the vapour partial pressure and on the total pressure; B — The relative humidity is 100 % when the dry-bulb temperature equals the dew-point temperature

    ω = 0.622p_v/(p − p_v) involves both pressures, and p_v = p_g(DBT) exactly when DBT = DPT, i.e. φ = 1. Heating at fixed ω leaves p_v and so the dew point unchanged, and the WBT of unsaturated air is below the DBT.

  13. Which statements about simple psychrometric processes are correct?

    1. Sensible cooling of air without reaching its dew point leaves the humidity ratio unchanged
    2. Evaporative cooling lowers the dry-bulb temperature and raises the humidity ratio
    3. In adiabatic mixing of two streams the mixture humidity ratio is the dry-air-mass-weighted average of the two
    4. Cooling and dehumidification raises the humidity ratio of the air
    Show answer

    Answer: A — Sensible cooling of air without reaching its dew point leaves the humidity ratio unchanged; B — Evaporative cooling lowers the dry-bulb temperature and raises the humidity ratio; C — In adiabatic mixing of two streams the mixture humidity ratio is the dry-air-mass-weighted average of the two

    ω is fixed while no vapour is added or removed. The evaporative cooler adds vapour while T falls. The mixing balance m_a1ω₁ + m_a2ω₂ = m_a3ω₃ gives the weighted average. Dehumidification removes water, so ω falls.

  14. A mixture consists of 2 kg of nitrogen (M = 28 kg/kmol) and 3 kg of carbon dioxide (M = 44 kg/kmol), both ideal gases. The molar mass of the mixture is ____ kg/kmol (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 35.8

    n = 2/28 + 3/44 = 0.071429 + 0.068182 = 0.139610 kmol; M_m = 5/0.139610 = 35.81 ≈ 35.8 kg/kmol. The mass-weighted average of the molar masses (Σmf M = 0.4 × 28 + 0.6 × 44 = 37.6) is the usual wrong route.

  15. One kmol of nitrogen and one kmol of oxygen, both ideal gases at the same temperature and pressure, are allowed to mix adiabatically at constant total pressure, so that they form a uniform mixture. Take Ru = 8.314 kJ/kmol·K. The entropy of mixing of the gases is ____ kJ/K (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 11.5

    y_N₂ = y_O₂ = 0.5, n = 2 kmol. ΔS = −nRu Σy_i ln y_i = −2 × 8.314 × (2 × 0.5 ln 0.5) = 2 × 8.314 × 0.693147 = 11.526 ≈ 11.5 kJ/K. Omitting n (using one kmol of mixture) would give 5.76.

  16. Humid air is at 30 °C and a total pressure of 101.325 kPa, with relative humidity 60 %. The saturation pressure of water at 30 °C is 4.246 kPa, and take the ratio of the molar masses of water and dry air as 0.622. The humidity ratio of the air is ____ g of vapour per kg of dry air (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 16.0

    p_v = φp_g = 0.60 × 4.246 = 2.5476 kPa. ω = 0.622 p_v/(p − p_v) = 0.622 × 2.5476/(101.325 − 2.5476) = 1.5846/98.777 = 0.016042 kg/kg = 16.0 g/kg. Using p instead of p − p_v gives 15.6, a near miss.

  17. A cooling coil handles 1 kg/s of dry air. The air enters with enthalpy 75.0 kJ/kg of dry air and humidity ratio 0.0180, and leaves with enthalpy 40.0 kJ/kg of dry air and humidity ratio 0.0090. The condensate leaves the coil with an enthalpy of 42 kJ/kg. The heat removed from the air stream, as a positive load, is ____ kW (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 34.6

    Energy balance: Q = m_a[(h₁ − h₂) − (ω₁ − ω₂)h_f] = 1 × [(75.0 − 40.0) − 0.0090 × 42] = 35.0 − 0.378 = 34.62 ≈ 34.6 kW. Neglecting the condensate enthalpy gives 35.0.

  18. Two adiabatic streams of humid air at the same total pressure are mixed. Stream 1 carries 2 kg/s of dry air with humidity ratio 0.010, and stream 2 carries 1 kg/s of dry air with humidity ratio 0.020; no condensation occurs. The humidity ratio of the mixture is ____ g of vapour per kg of dry air (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 13.33

    Dry-air and water balances: ω₃ = (m_a1ω₁ + m_a2ω₂)/(m_a1 + m_a2) = (2 × 0.010 + 1 × 0.020)/3 = 0.040/3 = 0.013333 = 13.33 g/kg. A simple average of 0.010 and 0.020 gives 15.00, which ignores the flow weights.