A wet tshirt hung on a hanger has a total surface area of ab

A wet t-shirt hung on a hanger has a total surface area of about 0.6 m^2. It loses water as follows: If the saturation vapor pressure of the water of 20 mm Hg and the room has a relative humidity of 30%, estimate the mass transfer coefficient from the t-shirt.

Solution

Average time:

t = total time / 4 = 863 min/4=215.75 min=12945 s

Average weight:

w= total weight/4 = 608.5 g

Volume of air

V= W/density = 608.5/20 =30.425 m^3 = 30.425×10^3 liters

Now calculated the flux of water awaparated from the shirt

N= [vapor concentration] [air volume]/[liquid area] [time]

= [0.3*20 mmHg*(0.13332 kpa/101 mmHg)*(1mol/22.4 liters)(273/25+273)*30.425×10^3] / [0.6][12945]

= 0.001267 mol/cm^2.s

.............

Mass transfer coefficient:

0.001267=k[20*0.13332*273/101*22.4*10^-3*[25+273]]

k = 1.17*10^-3 cm/s