Question

Use: Air at a temperature of 293 K, with 50% saturation pressure for water in the air far

above the pool and a velocity of 0.5 m/sec. The vapor pressure of water at this temperature

is 2.3388 kPa. The length is 10 m and the width is 4 m. How fast will the level drop in a

day? The kinematic viscosity of air is 1.51 x 10-5m2/sec. The gas constant is 8.314 Pa.m3/(mol.K). The diffusion coefficient is 0.25 cm2/sec.

Show all work and intermediate calculations WITH UNITS. Answers should be boxed with units.

Answer 1

Air at temperature T = 293 K and relative humidity R. H = 50%.

This air is flowing over a pool with a velocity v = 0.5 m/s

Vapor pressure of water at this temperature, Pa1= Pv = 2.3388 kPa

Length of Pool L = 10 m

Width of pool W = 4 m

R. H = Partial pressure of water vapor into air /vapor pressure

0.5 = Pa2/Pv

Partial pressure of water vapor into air Pa2 = 0.5*2.3388 = 1.1694 kPa.

Kinematic viscosity of air, nu = 1.51*10^-5 m2/s

Diffusion coefficient Dab = 0.25 cm2/s = 0.25*10^-4 m2/s

Gas constant R = 8.314 Pa. m3/mol.k

Calculations:

Reynolds number ReL = v*L/nu

ReL =( 0.5m/s)10m/(1.51*10^-5m2/s) = 331125

Schmidt number Sc = nu/Dab = (1.51*10^-5/0.25*10^-4) = 0.604

Here we can say that flow is over pool which having lenght L = 10 m it means flow over falt plate so that correlation for flat plate:

Re < 5*10^5 and Sc> 0.6 then it is laminar flow then,

Sherwood number Sh = 0.664*ReL^0.5 *Sc^0.33

Sh = 0.664*(331125)^0.5 *(0.604)^0.33

Sh = 323.52

Kc*L/Dab = 323.52

Kc = (323.52*0.25*10^-4 m2/s)/10m = 8*10^-4 m/s

mass transfer coefficient Kc = 8*10^-4 m/s

molar flux of Water evaporated Na = Kc/RT * (Pa1 - Pa2)

Na =[( 8*10^-4 m/s)/(8.314Pa.m3/mol.k)*293k][2.3388 - 1.1694]*10^3 Pa

Na = 3.84*10^-4 mol/m2.s

Pool area where water evaporated A = 10m*4m = 40 m2

Molar evaporation of water n = Na*A =( 3.84*10^-4 mol/m2.s)*40 m2 = 0.01536 mol/s

Molar evaporation per day n = 0.01536 mol/s * (24*3600s/1day) = 1327.23 mol/day

Water evaporation per day = 1327.23 mol/day * (18 gm/mol) = 23890 gm/day = 23.890 kg/day .

Mass flow rate = 23.890 kg/day.

dm/dt = 23.890

d(rho*V)/dt = 23.890

density of water rho = 1000 kg/m3 which is constant throughout the process.

Volume V = A*h ( liquid height into the pool)

Where A = 40 m2 is constant

Rho*A*dh/dt = 23.89

(1000 kg/m3)*(40m2) * dh/dt = 23.89 kg/day

dh/dt = 5.9725*10^-4 m/day

liquid level drop dh/dt = 0.59725 mm/day