Question

The intermediate product of a chemical process is a chemical paste that is saturated with water. The paste is rolled into a thin layer and placed to dry in a flat area 30m long 15m wide. Drying is enhanced by blowing hot air over the paste, and the mass transfer coefficient is 0.017 m/s. Diffusion of water through the paste is rapid so that the concentration of water vapor at the surface of the paste remains constant at 0.002 gmol/L. Water must be removed from the paste remains at a minimum rate of 9.5 L/min (liquid volume). (Note:MW water=18 and p water=1 g/cm^3.)

a.) What mechanism controls the transfer of water from the paste in the air?

b.) What is the molar rate of water removal that corresponds to a volumetric removal rate of 9.5 L/min?

C.) What is the maximum concentration of water vapor allowable in the air (far from the paste surface) if water must be removed at a rate of at least 9.5 L/min?

D.) Assuming that the mass transfer coefficient remains constant, what can be done to increase the rate of water removal from the paste?

Answer 1

1. a)Steady –state Diffusion when the rate of mass transfer becomes constant
2. b)volumetric removal rate of water=9.5 L/min

Also density of water=1 g/cm^3 =1g/0.001L=1000g/L

Mass removal rate=9.5 L * 1000g/L per min=9500g/ min

Molar removal rate=9500g/molar mass per min=9500g/18g/mol per min=527.78 mol/min

c)Using equation, kc=n/A*C   where,

    kc = mass transfer coefficient in or [ (mol/min*m2)/mol/L] or m/s

    n = mass transfer rate in mol/ min

    A =mass transfer area [ in m2]

    ΔCA =concentration difference [mol/m3]= C(final) –C(initial)

Putting the given values, 0.017 m/s=527.78 mol/min / ( 30m *15m )* [Cf-0.002 gmol/L)

    Or, [Cf-0.002 gmol/L)= 527.78 mol/min / ( 30m *15m )*0.017 m/s *60 s/min

   Or , [Cf-0.002 gmol/L)=1.15 mol/L (gmol=mol)

Cf=1.148 gmol/L

D) If mass transfer coefficient remains constant,then area and concentration difference must be increased to increase mass transfer rate