--- title: "Properties of moist air numerical" subject: "Applied Thermodynamics" chapter: 4 key: "semester_3/applied_thermodynamics:psychrometric-properties-numerical" source: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/psychrometric-properties-numerical site: BCT Engineering --- # Properties of moist air numerical Applied Thermodynamics · Chapter 4: Air-Conditioning - Asked in: year not given, 6 marks - Video: [NUMERICALS ON PSYCHROMETRY| CALCULATION OF Specific Humidity, Relative Humidity, DPT, Enthalpy](https://www.youtube.com/watch?v=mKQp0PkrI38) — KP Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/psychrometric-properties-numerical - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_thermodynamics/chapter-4 ## Question Atmospheric air at 101.325 kPa has a dry bulb temperature of 32 °C and a relative humidity of 55%. Without using the chart, determine (a) the partial pressure of water vapour, (b) the specific humidity, (c) the dew point temperature, (d) the degree of saturation, (e) the specific enthalpy, and (f) the specific volume. The saturation pressure of water at 32 °C is 4.759 kPa. ## Short answer Compute p_v = φ·p_s = 2.62 kPa; ω = 0.622 p_v/(p-p_v) = 0.01649 kg/kg dry air; dew point from steam tables ≈ 21.8°C; ω_s = 0.03066, degree of saturation μ = 0.538; h = 1.005t + ω(2501+1.88t) = 74.4 kJ/kg dry air; specific volume v = R_a T/(p-p_v) = 0.887 m³/kg dry air. ## Answer #### Given $t = 32$ °C, $\phi = 0.55$, $p = 101.325$ kPa, $p_s = 4.759$ kPa. #### (a) Partial pressure of vapour $$ p_v = \phi\,p_s = 0.55 \times 4.759 = 2.618\ \text{kPa} $$ #### (b) Specific humidity $$ \omega = 0.622\,\frac{p_v}{p - p_v} = 0.622 \times \frac{2.618}{101.325 - 2.618} = 0.01649\ \text{kg/kg dry air} $$ #### (c) Dew point The dew point is the saturation temperature at $p_v = 2.618$ kPa. From steam tables (interpolating between 20 °C, 2.339 kPa and 22 °C, 2.645 kPa): $$ t_{dp} \approx 21.8\ ^\circ\text{C} $$ #### (d) Degree of saturation $$ \omega_s = 0.622\,\frac{4.759}{101.325 - 4.759} = 0.03066,\qquad \mu = \frac{\omega}{\omega_s} = \frac{0.01649}{0.03066} = 0.538 $$ #### (e) Enthalpy $$ \begin{aligned} h &= 1.005\,t + \omega\,(2501 + 1.88\,t)\\ &= 1.005 \times 32 + 0.01649\,(2501 + 1.88 \times 32) = 74.4\ \text{kJ/kg dry air} \end{aligned} $$ #### (f) Specific volume $$ v = \frac{R_a T}{p - p_v} = \frac{0.287 \times 305.15}{101.325 - 2.618} = 0.887\ \text{m}^3/\text{kg dry air} $$ (The wet bulb temperature is about 24.6 °C from the adiabatic saturation equation.) **Answer:** (a) 2.62 kPa; (b) 0.01649 kg/kg; (c) about 21.8 °C; (d) 0.538 (53.8 %); (e) 74.4 kJ/kg; (f) 0.887 m³/kg. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is p_v found from DBT and RH? A. p_v = p_s / phi B. p_v = phi x p_s = 0.55 x 4.759 = 2.618 kPa C. p_v = phi x p D. p_v = p - p_s Answer: B. Relative humidity is p_v/p_s, so p_v = 0.55 x 4.759 = 2.618 kPa. (Key point: Partial pressure of vapour) 2. Which expression gives specific humidity here (about 0.0165 kg/kg)? A. 0.622 x 2.618/101.325 B. 0.622 x 2.618/(101.325 - 2.618) C. 0.622 x 4.759/(101.325 - 2.618) D. 2.618/(101.325 - 2.618) Answer: B. omega = 0.622 p_v/(p - p_v), using the vapour pressure, not the saturation pressure. (Key point: Specific humidity) 3. How is the dew point found? A. It is the saturation temperature of water at p_v = 2.618 kPa (about 21.8 °C) B. It is the saturation temperature at 4.759 kPa C. It equals the wet bulb temperature D. It is 32 x 0.55 Answer: A. Dew point is the saturation temperature at the actual vapour pressure, found by interpolating the steam table. (Key point: Dew point) 4. Degree of saturation mu here equals: A. 0.55 (it equals RH) B. omega / omega_s = 0.01649/0.03066 = 0.538 C. omega_s / omega = 1.86 D. p_v / p = 0.026 Answer: B. mu = omega/omega_s with omega_s computed at p_s = 4.759 kPa; it is close to, but not exactly equal to, RH. 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