Unformatted text preview: 7(42.3) 50,000
From Eq. (2), k c =
2/3
0.948
2.41 = 0121 m/s
. Using properties in Example 3.14,
PAs
10
cA s = yA s c =
c=
(0.0327) = 4.3 × 10−4 kmol / m3 and cA ∞ = 0
P
760 ( ) From Eq. (1), N A = kc cA s − cA∞ = (0.121)(4.3 × 10−4 − 0) = 5.20 × 10−5 kmol/sm2
From Eq. (3171), packed bed correlation can't used with NRe = 50,000. Exercise 3.35
Subject: Experimental data for stripping of CO2 (A) from water by air (B) in a wettedwall
tube.
Given: Location in the tube where P = 10 atm, T =25oC (298 K), CO2 flux from water into air
= NA = 1.62 lbmol/hft2, partial pressure of CO2 at gasliquid interface = pA i = 8.2 atm, and partial pressure of CO2 in bulk air = pA b = 0.1 atm. Diffusivity of CO2 in air = DAB = 1.6 x 102
cm2/s. Assumptions: Neglect stripping of water, but account for the bulk flow effect. Masstransfer
resistance of the water film does not need to be considered. Ideal gas law.
Find: Using the film theory, find the masstransfer coefficient, kc , for the gas phase and the
film thickness, δ.
Analysis: Solve the problem in cgs units. From Eq. (3105) and a rearrangement of Eq. (3189), kc = N A 1 − yA LM c yA i − yA b = DAB
δ (1) NA = 1.62 lbmol/hft2 = 0.00022 mol/scm2
From the ideal gas law, c =P/RT = (10)/(82.06)(298) = 0.000409 mol/cm3
yA i = pA i (1 − yA ) LM = From Eq. (1), kc = P = 8.2
= 0.82
10 yA b = pA b
P = 0.1
= 0.01
10 (1 − yA b ) − (1 − yA i ) (1 − 0.01) − (1 − 0.82)
=
= 0.475
(1 − yA b )
(1 − 0.01)
ln
ln
(1 − 0.82)
(1 − yA i ) (0.00022)(0.475)
= 0.315 cm/s
(0.000409) ( 0.82 − 0.01) From Eq. (1), δ = DAB / kc = 1.6 x 102/0.315 = 0.051 cm Exercise 3.36
Subject: Experimental data for absorption of CO2 (A) from air into water (B) using a packed
column of Pall rings.
Given: Location in a packed column where pA i = 150 psia, cA b = 0 , and NA = 0.017 lbmol/hft2
(2.31 x 106 mol/scm2). Diffusivity of CO2 in water = DAB = 2.0 x 105 cm2/s. Henry's law for
CO2 between air and water is p, psia = 9,000 x.
Assumptions: Negligible bulk flow effect. Only the masstransfer resistance in the water need
be considered. Neglect stripping of water.
Find: (a) Liquidphase masstransfer coefficient, kc , and film thickness, δ, for the film theory.
(b) Contact time for penetration theory.
(c) Average eddy residence time and probability distribution for surface renewal theory.
Analysis: Solve the problem in cgs units.
(a) From Eq. (3105) and a rearrangement of Eq. (3189), kc = NA
D
= AB
δ
c xA i − xA b (1) Total liquidphase concentration, since almost pure water, = c = ρ/MB = 1.0/18 = 0.0556 mol/cm3
From given Henry's law, xA i = pA i /9,000 = 150/9,000 = 0.0167 and xA b = 0.0.
From Eq. (1), kc = 2.31× 10−6
= 0.0025 cm/s
(0.0556) ( 0.0167 − 0 ) From Eq. (1), δ = DAB /kc = 2 x 105/0.0025 = 0.0080 cm
(b) From a rearrangement of Eq. (3194),
tc = 4 DAB
4(2 × 10 −5 )
=
= 4.08 s
(3.14)(0.0025) 2
πkc2 1 DAB
2 × 105
(c) From Eq. (3201), t = = 2 =
= 3.2 s
s
kc
(0.0025) 2
From Eq. (3196), the probability distribution is:
φ{t} = s exp(− st ) = 0.313 exp(−0.313t ) Exercise 3.36 (...
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 Spring '11
 Levicky
 The Land

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