Water in the bottom of a narrow metal tube is held at a constant temperature of
ID: 1058156 • Letter: W
Question
Water in the bottom of a narrow metal tube is held at a constant temperature of 293K. The total pressure of air (dry basis) is 1 atm and the temperature is 293K. Water evaporates and diffuses through the air in the tube and the diffusion path is kept constant and is 6 inches long and the I.D of the tube is 0.17 cm. Calculate the rate of evaporation at steady state in kgmol/s. Assume the diffusivity of water vapor at 293K and 1 atm pressures 0.250 times 10 m^2/s and the vapor pressure of water at 20 degree C is 17.54mmHg, and the partial pressure of water at the outlet of the metal tube is assumed to be zero.Explanation / Answer
rate of evaporation = DAB*A*(CA1-CA2)/ZRT
DAB= diffusivity = 0.250*10-4 m2/s
Z= diffusion length = 6 inches = 6/12 ft = 0.5ft =0.5*0.3048 m =0.1525
A = sufrace area =- 2PI*0.17/2* Z
CA1= PA!/RT = (17.54/1760 atm/ (0.0821*293 =0.00959 mol/ L =0.000958*10-3 kgmol/10-3m3.s =0.000959 kgmol/m3
CA2=0
rate of evaporation =0.250*10-4*0.000959*2*3.142*0.17/200 =1.280*10-10 kgmol/s
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