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Gas-Liquid Separation - Diagram of Two Film Theory showing...

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Diagram of Two Film Theory showing gas adsorption (solid arrows) and gas stripping (dashed arrows). Dissolved gas is brought to interfacial film surface by mixing (convection) and diffuses to interface where a chemical reaction from liquid (C i ) to gas (p i ) phase. Then, the gas diffuses to film surface and convects to bulk atmosphere. Reverse occurs for stripping. Bulk liquid concentration is convected to surface, diffuses to interface, chemical reaction from liquid to gas phase occurs, gas is convected to bulk and stripping is complete. B u l k G a s Interface Mixing Mixing p a p 1 C i GAS STRIPPING GAS ADSORPTION C b Film Theory Two Theories - Penetration and Film Penetration Theory - Eddies originate in the bulk and migrate to the g/l interface and brief exposure results in gas absorbance before other eddies move the absorbance eddies from the surface due to turbulence. Gas transfer is a function of diffusivity, ******* gradient and surface renewal (eddy transfer). Two Film Theory - two films exist at surface, one gas/one liquid. They are stagnant but furnish the resistance to g/l transfer. Three phases - 1) transport to surface of film (usually controlled by mixing) 2)diffusion through the gas and then the liquid film and 3)transport to the liquid bulk. Transport is slow by diffusion ****** to mixing so that (transport) controls gas transfer. Molecular Diffusion moles gas A through quiescent gas B: where dNa/dt = moles gas A diffusing dN dt DAp RT S P- P) av a a1 a2 = ( P b t = time (hr) D v = vol. diff. (ft 2 /hr) A = area for diffusion p = pressure (atm) p a1 - p a2 = partial pressure diffusing gas (atm) P b = log mean pressure B S = diff. Distance R = 0.729 ft 3 *atm/lb*mol* o R T = o R Gas Film Liquid Film Bulk Gas Diffusion
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Molecular Diffusion Through Liquids dN dt DAC SC C-C) av a + b bm a1 a2 = ( Ca 1 /Ca 2 = molar Conc. Diff. gas C a+b = molar A+B / vol Cbm = log mean concentration Mass Diffusivity NH 3 , CO 2 , + O 2 in Air 25 o C, in Water 20 o C AIR WATER Dv, (ft2/hr) Ns c (u/pDv) Dv (ft 2 /hr) Ns c (u/pDv) NH 3 0.89 0.67 6.83*10- 5 570 CO 2 0.636 0.94 6.87*10- 6 670 O 2 0.8 0.75 7.00*10- 7 558 Assuming a steady state with no interface accumulation N a = dNa/dt = k g A( ρ - ρ i )=k i A(C i -C b ) N a = mol/time k g , k l = mass transfer coeff. (mol/hr*ft 2 *atm) for k g and (ft/hr) for k l ρ , ρ I = partial pressure (atm) C 1 ,C 2 = conc. interface, bulk We have derived Henry’s Law from G = G o + RT ln Prod/React CO 2 g + H 2 0 CO 2 + H 2 0 G = 0; ln P R G RT k x o = = CO 2 gas is expressed as pressure K = CO 2(L) /CO 2(g) = 10 -1.47 (20 o C, 1atm) CO 2(g) = 10 -3.5 atm CO 2(l) = KCO (g) = K(P CO2(g) ) = 10 -3.5 10 -1.45 = 10 -5 CO 2(l) = 10 -5 M/l (44,000mg/mol) = 0.44 mg/l note CO 2 (l) is independent of pH as predicted by Henry’s Law P = (1/K)C = HC
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Example 1: Predict sat. concentration of O2 (mg/l) in H20, at STP H = 4.01*10-4 atm/mol frac ρ O 2 = 0.21 atm. O 2(g) + H 2 O O 2(l) CO 2(l) = k ρ = ρ /N = 0.21 / 4.01*10 4 CO 2(l) = 5.24*10 -6 mol frac.
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Gas-Liquid Separation - Diagram of Two Film Theory showing...

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