The intermediate product of a chemical process is a chemical paste that is saturated with water....

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Chemistry

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

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

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

C.) What is the maximum concentration of water vapor allowable inthe air (far from the paste surface) if water must be removed at arate 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 thepaste?

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4.1 Ratings (651 Votes)

  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 )*

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