Separation Process Principles- 2n - Seader & Henley - Solutions Manual

Find rate of absorption analysis the exercise is

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Unformatted text preview: t 2 a 2.54 Solving Eq. (1), gives t = 1.03 x 105 s = 28.7 h Exercise 3.26 Subject: Absorption of pure oxygen into a downward flowing laminar water film Given: Film, at 10 atm and 25oC, flows down a wall, with L = 1 m high and W = 6 cm (0.06 m) wide. Flow is laminar, without ripples, at a Reynolds number of 50. Assumptions: Diffusivity of O2 in water = 2.5 x 10-5 cm2/s = 2.5 x 10-9 m2/s. Solubility of O2 in water = a mole fraction of 2.3 x 10-4 = xO 2 . Negligible vaporization of water. Find: Rate of absorption Analysis: The exercise is similar to Example 3.13. By material balance on the O2 in the water film, Eq. (3-103) gives: nO2 = u y δW cO2 where, W = 0.06 m and cO2 o L − cO2 (1) o = oxygen concentration in the water at the top of the film = 0. From Eq. (3-92), the bulk average film velocity is, ρgδ 2 uy = 3µ 3µΓ where from Eq. (3 - 92), film thickness = δ = 2 ρg (2) 1/ 3 (3) The Reynolds number for film flow is given by (3-93). For a water viscosity, µL= 0.89 cP = 0.00089 kg/m-s at 25oC, the film flow rate per unit width of film is, µ 0.00089 = 50 = 0.01112 kg / m - s 4 4 From Eq. (3), the film thickness for ρ = 1,000 kg/m3 and g = 9.807 m/s2 is, Γ = N Re 3(0.00089)(0.01112) δ= (1,000) 2 (9.807) 1/ 3 = 1.45 × 10 −4 m (1,000)(9.807)(1.45 × 10 −4 ) 2 From Eq. (2), u y = = 0.0772 m / s 3(0.00089) Combining Eqs. (3-103) and (3-113), we obtain an equation for the O2 concentration in the water at the bottom of the film, noting that with pure oxygen gas, all of the mass-transfer resistance resides in the liquid water film: A k cavg W cO2 = cO2 − cO2 − cO2 exp − (4) L i i o uyδ Exercise 3.26 (continued) Analysis: (continued) The total concentration in the liquid is approximately that of water = c = ρ/Mwater = 1,000/18.02 = 55.5 kmol/m3. The concentration of oxygen in the water at the interface = cO2 = xO2 c = 2.3 x 10-4 (55.5) = 0.0128 kmol/m3. i The average mass-transfer coefficient is given by (3-110), provided that the factor η, given by Eq. (3-100), is greater than 0.1. This equation requires the Schmidt number. From Eq. (3-101), N Sc = µ ρDO2 ,H 2 O = 0.00089 = 356 (1,000)(2.5 × 10−9 ) From Eq. (3-100), η= k cavg 8/3 8/3 = = 1.033 (> 0.1) N Re N Sc δ / L (50)(356) 145 × 10−4 / 1 . From Eq. (3-110), uyδ 0.0772(1.45 × 10 −4 ) = 0.241 + 51213η = . 0.241 + 51213(1.033) = 6.2 × 10−5 m / s . L 1 Area for mass transfer = A = LW = (1)(0.06) = 0.06 m2. From Eq. (4), 0.06 0.06 = 0.0128 − 0.0128 − 0 exp − = 0.01275 kmol / m3 (0.0772)(145 × 10−4 ) . 6.2 × 10 −5 cO 2 L Thus, the water is better than 99% saturated with oxygen. From Eq. (1), nO2 = (0.0772) (1.45 × 10 −4 ) ( 0.06 ) ( 0.01275 − 0 ) = 8.6 × 10−9 kmol/s Exercise 3.27 Subject: Absorption of pure carbon dioxide into a downward flowing laminar water film. Given: Results in Example 3.13, including: DAB = 1.96 x 10-5 cm2/s, u y = 0.0486 m/s, () δ = 1.15 x 10-2 cm, and cCO2 o =0 Find: Height, y, from the top of the film where the water is 50% saturated with CO2. Analysis: At height...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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