Question: C A i = ( 1 . 0 3 ) ( 0 . 1 1 3 6 ) = 0 . 1 1 7 kmo

CAi=(1.03)(0.1136)=0.117kmolm2
CAb=0(because of the small contact time, the depth of penetration of the solute in the jet will be small and the bulk liquid free from H2S)
The given rate of absorption of H2S=4.4210-4gs=(4.4234)10-410-3, i.e.1.310-8kmols.
Using Eqs. (i) and (ii) and the given rate of absorption [note that r=(Qv)12],
1.310-8=2lQv2*2DABvl2(0.117-0)=4DAB(1.3210-5)(0.05)2(0.117)
=>,DAB=
[Diffusivity of H2S in water at 25C reported in the literature =1.2110-9m2s]
EXAMPLE 3.6(Gas absorption from bubbles) A mixture of 50%CO2 and 50%N2 is bubbling through water in a laboratory column at 30C and 1atm. The depth of water in the column is 30cm. A single-nozzle gas distributor is used. The gas flow rate is 15cm3 per minute and the bubbles are of 1cm diameter on the average. The bubble rise velocity is 20cms. Calculate the rate of absorption of carbon dioxide. The diffusivity of CO2 in water is 2.1910-5cm2s. Henry's law can be used to calculate the solubility of CO2 in water at the given temperature, p=1860x**(p= partial pressure of CO2, in atm; x** is its mole fraction in water at equilibrium).
Hints: Contact time of a liquid element with a gas bubble, tc=dbvb=1cm20cms=0.05s Mass transfer coefficient, kL=2(DABtc)12=0.0236cms. Residence time of a single bubble in the liquid =(liquid depth
 CAi=(1.03)(0.1136)=0.117kmolm2 CAb=0(because of the small contact time, the depth of penetration

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