What Is the Mass Driving Force for Mass Transfer in Gas Phase?

[joe98]

Consider a chemical species A, which is soluble in water. The equilibrium of this system with A in air is described by the following question:

C(air) = HC(water)

where C(air) is mass concentration of the species in air
C(water) is the mass concentration in water, and the Henrys Law constant is H=0.00070(dimensionless)

the film mass transfer coefficient for the air side is 0.07m/s, and the water side is 8x10^-6 m/s

Answer the following, assuming the air stream contains 10 mol% of A at 100KPa and the aqueous solution contains 5 mass% of A at 25C...and use 17g/mol for molar mass of the speciesA) What is the mass driving force for the mass transfer, in mass concentration units, viewed from the gas phse?

I used C=PM/RT=100000*0.017/8.314*(273+25)=0.069 kg/m^3
What is the mass transfer rate( express as mass flux)?

Here i use
N=KC

where K=1/k + H/k2=5.33*10^-3 m/sCould someone assist me in this question...Any suggestions

 

[joe98]

Any suggestions guys on calculating the mass transfer driving force?

 

[joe98]

Heyguys, could you have a look if i am on the right track

Heres my working out to calculate the driving force in mass concentration units from the gas phase

I converted 10 mol% to 0.17 mass%, (which is in mole fraction)

then i calculated the partial pressures of inlet and outlet

PP(waterinlet)=100*10^3* 0=0
PP(gas inlet)= 100*10^3 * 0.005=5000Pa

Mass fraction(water)=0.05*0.0007=3.5*10^-5

Therefore , PP(water out) = 3.5*10^-5 * 100*10*3 = 3.5 Pa

Mass fraction (air) = 1.7*10^-3 * 0.0007=1.19*10^-6

therefore, PP(air out) = 1.19*10^-6 * 100*10^3 = 0.119 Pa

then i calculated C(lmcd) = (5000-3.5) - (0.119-0))/ln(5000-3.5)/0.119 =469 PaTherefore i used C=PM/RT = 469*0.017/8.314*298=3.22*10^-3 Kg/m^3Is this correct guys

 

[Chestermiller]

Heyguys, could you have a look if i am on the right track

Heres my working out to calculate the driving force in mass concentration units from the gas phase

I converted 10 mol% to 0.17 mass%, (which is in mole fraction)

then i calculated the partial pressures of inlet and outlet

PP(waterinlet)=100*10^3* 0=0
PP(gas inlet)= 100*10^3 * 0.005=5000Pa

Mass fraction(water)=0.05*0.0007=3.5*10^-5

Therefore , PP(water out) = 3.5*10^-5 * 100*10*3 = 3.5 Pa

Mass fraction (air) = 1.7*10^-3 * 0.0007=1.19*10^-6

therefore, PP(air out) = 1.19*10^-6 * 100*10^3 = 0.119 Pa

then i calculated C(lmcd) = (5000-3.5) - (0.119-0))/ln(5000-3.5)/0.119 =469 Pa


Therefore i used C=PM/RT = 469*0.017/8.314*298=3.22*10^-3 Kg/m^3


Is this correct guys

Getting the driving force:

1. calculate the concentration of the species in the bulk gas phase, using concentration units for the gas phase

2. calculate the concentration of the species in the bulk liquid phase, using concentration units for the liquid phase

3. take the result in step 2 and multiply by the Henry's Law constant to get the gas phase concentration that would be in equilibrium with the species concentration in the liquid phase

4. subtract the result from step 3 from the result for step 1. This is the mass transfer driving force for species flux from the gas phase to the liquid phase. The mass transfer driving force for species flux from the liquid phase to the gas phase is minus this value.