Question

2- Water in the bottom of a narrow metal tube is held at a constant temperature of 303 K. The total pressure of air (assumed dry) is 1.01325 x 105 Pa (1.0 atm) and the temperature is 303 K. Water evaporates and diffuses through the air in the tube and the diffusion path (z2 - z1) is 0.2 m long. The tube diameter is 10 mm. The vapor pressure of water vapors at 303 K is 4242 Pa. The experimental value of the diffusion coefficient at 298 K is 2.6×10-5 m2/s.

i- Calculate the rate of evaporation (NA) at steady state in ?? ???/?2∙? and ??/?2∙?

ii- Calculate the steady state rate of evaporation (NA) if the air is humid (not dry) and the percentage relative humidity, i.e. 100(??/ ???⁄) =30%, where ?? is the partial pressure of the water vapor in

the air and ??? is the vapor pressure.

Answer 1

solution- Let us here we have given data Temperature = 303K, 1. Pressure of air= 601325 x 105 Pallo atm) Az = o.am d=comm PAO = H2Hapa DAB= 2.6x10 s mals cat 298 K) PA=0 A we can see the gas Phase diffusivity. Now DAB 2 73/2 30 , 303 298 DAB,298

298 DAB, 503 = 2.6x10-5x ( 303 ja DAB = 2.666x10-5 malls Now from the fick's first law of 'A' through non diffusing 1) Now NA = DABP lor P-PA diffusion for B. RTZ P-P No 2,666 x 10x101325 nr 101325-0 8.314 x 303 XD2 101325-4242 so here NA= 2.293 8154 mod mas so finally INA=2.293 x 107 xmol ) mas Now we have now of H2O = 18 kg/kmol. then NAF 2-293x167 x 18 kg so NA = 4.127 X166 kg * so for Now PAL =0.3 mas mas so A = 0.3 -> Po = 1272.6 Pa 4242 so NA = DAB P RTZ -BAD 2.666 x16x101325 P101325-12726 8.314 X 303x02 (101325-4242) so, NM = 1.615 x 104 kmbil mas we get NA= 1-615 X107 kmal más so for Hyo m.W = 18 ,NA = 1.615 x107x18 *9)mis I. NA = 2.908% 166 kg, Thank your mas 1