Question

Moist air at 20°C, 1 atm, and 43% relative humidity and a volumetric flow rate of 900 m²/hrs. enters a control volume at steady state and flow along a surface maintained at 65°C, through which heat transfer occurs. Liquid water at 20°C is injected at a rate of 5 kg/hr and evaporates into flowing stream. For the control volume, Wev=0, and kinetic and potential energy effects are negligible. Moist air exits at 32°C, 1 atm. (a) draw the system schematic with given data, (b) determine the rate of heat transfer in kW and (c) calculate the relative humidity at exit in %(d) use the attached psychrometric chart to compare your answers in (b) and (c) by drawing the process on the chart.

Answer 1

Liquid water en stg/hr T2 = 20c T3 =32c T1=20°c Pz = 1 atom d. = 43% Ocv (Av.), 200 Milne P = 1 atom To=330k at T = 280 from Psychrometry chart and d. = 43% volume 2 = 0.83 milks of dry air 0.0062 kg/kg of dry air flow rate humidity on = ratio

PAVO, m = cima, mz- ms Mz Wz Non mas mars flow rate of dry air ma goo V Ord 1084.33kg / hy - 0.3012 kg/see = 6.722 Kelhar m, + 111722 kg/hr 6.olod 1 kg/kg of (n mä =ma, abyair at Tz=32° and Cazzo o wlodsky log of da from Psychrometry chant Relative of = 37% hemching at by energy conservation hi + (wzow, shfa & Ocv = hz Ora hachi - Cis tanghya Cp CT3-71) + Wer (250071.88 Ty] - W, [2500+ 1.88 T.] – CW2 why, by steam table at T = 26C hf = 94.2043 kJ leg. Ocv = 12.06.+ 27.67-15.73 -0.388 = 23.61 kJ/kg of day air der = ra Oci Zoll1gKW exit