Question

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11-43·Atmospheric air with a dry bulb temperature of 30°C and a relative humidity f 40 percent enters the cooling section of an air-conditioning system at a rate of 10 kg/s and is cooled to 18°C. Determine the volumetric flow rate of the air leaving the cooling section, the heat-transfer rate required for this process, and the entropy change of the air per unit mass of dry air. If the velocity of the airstream is not to exceed 15 m/s in the cooling section, what should be the dimensions of the cooling section if a rectangular duct with an aspect ratio of 1.5:1 is used?

Answer 1

DBT = 30 degree C, RH = straightphi = 40% Atmospheric pressure = 100 k pa = P_b straightphi = P_v/P_vl doubleheadarrow 0.4 = P_v/P_vl (P_vl at 30 degree C = 4.25 k pa) doublerightarrow 0.4 = P_v/425 doublerightarrow P_v = 1.7 k pa rightarrow Vapour pressure Humidity ratio, w = 0.622 P_v/P_t - P_v = 0.622 times 1.7/100-1.7 w = 0.01076 kg/kg of dry air using psychometric - chart corresponding to DBT = 40 degree C, straightphi = 40% specific volume = 0.874 m^3/kg therefore Volumetric flow rate = sp. Volume times mass flow rate = 0.874 times 10 = 8.74 m^3/sec Heat transfer = (h_- h_) q = Cpa(t_alpha_1 - t_alpha_2) + W Cpa (t_d_2 - t_d_1) q = 1.005(30 - 18) + 0.01076 times 1.8(30 - 18) q = 12.29 kJ/kg therefore Entropy change during the process: - S_2 - S_1 = q/delta T = 12.29/(30 - 18) = 1.04 kJ/kg - k \r\n volume flow rate = A middot v therefore Q = A middot v doublerightarrow 8.74 = (1.5 times 1)x^2 middot 15 doublerightarrow x^2 = 8.74/1.5 times 15 doublerightarrow x = 0.509 m therefore l = 76.33 cm b = 50.9 cm}As