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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 105 cm2/s = 2.5 x 109 m2/s. Solubility of O2
in water = a mole fraction of 2.3 x 104 = 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. (3103) 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. (392), 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 (393). For a water viscosity, µL= 0.89 cP =
0.00089 kg/ms 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. (3103) and (3113), 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 masstransfer 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 104 (55.5) = 0.0128 kmol/m3.
i The average masstransfer coefficient is given by (3110), provided that the factor η, given by
Eq. (3100), is greater than 0.1. This equation requires the Schmidt number. From Eq. (3101),
N Sc = µ
ρDO2 ,H 2 O = 0.00089
= 356
(1,000)(2.5 × 10−9 ) From Eq. (3100),
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. (3110),
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 105 cm2/s, u y = 0.0486 m/s, () δ = 1.15 x 102 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.
 Spring '11
 Levicky
 The Land

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