Subtopic: Psychrometry – properties of air-water vapour mixture
Question Statement:
Air at 70°C, 0.015 humidity ratio, is cooled adiabatically by spraying water; final temperature 55°C. c_p(dry air)=1.005, c_p(vapour)=1.88 kJ/kg·K; latent heat at 0°C = 2501.7 kJ/kg. The absolute humidity of the outlet air, in kg water/kg dry air, is:
Options:
(A) 0.017
(B) 0.019
(C) 0.021
(D) 0.023
Official Answer Key:
C
Step-by-Step Solution & Explanation
Adiabatic saturation (constant enthalpy) process: h1 = h2. h = 1.005T + W(2501.7 + 1.88T). At state 1: h1 = 1.005×70 + 0.015(2501.7+1.88×70) = 70.35+0.015×2632.3 = 70.35+39.48=109.83 kJ/kg. At state 2 (T2=55°C), solve for W2: 109.83 = 1.005×55 + W2(2501.7+1.88×55) = 55.28+W2×2604.1. W2 = (109.83-55.28)/2604.1 = 54.55/2604.1 = 0.02095 ≈ 0.021 kg/kg. Answer: C.
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