Unformatted text preview: continued) Exercise 3.37
Subject: Determination of diffusivity of H2S (A) in water (B) from experimental data for
absorption into a laminar jet, using the penetration theory.
Given: Laminar jet of D = 1 cm and L = 7 cm. Temperature, T = 20oC (293 K). Solubility of
H2S in water = cA i = 100 mol/m3 (100 x 106 mol/cm3). Absorption rates, nA, as a function of jet
volumetric flow rate, Q.
Assumptions: Jet is contacted by pure H2S. Therefore, all masstransfer resistance is in the
liquid. Negligible bulk flow effect. Concentration of H2S in bulk water cA b = 0.
Find: Diffusivity by penetration theory.
Analysis: Use cgs units. From a rearrangement of Eq. (3192),
DAB = ( nA 2A cAi − cAi 2 π tc ) (1) Surface area of jet = A = πDL = (3.14)(1)(7) = 22.0 cm2
Assume the contact time is the time of exposure of the jet = jet volume/jet flow rate = V/Q
Jet volume = V = πD2L/4 = (3.14)(1)2(7)/4 = 5.5 cm3. Therefore, tc = 5.5/Q.
Therefore, Eq. (1) becomes, DAB
Jet rate, cm3/s
0.143
0.568
1.278
2.372
3.571
5.142 nA
=
2(22) 100 × 10 −6 − 0 Absorption rate, mol/s x 106
1.5
3.0
4.25
6.15
7.20
8.75 2 2
(314)(5.5)
.
4 nA
= 89.2 × 10
Q
Q Diffusivity, DAB , cm2/s
1.40 x 105
1.41 x 105
1.26 x 105
1.42 x 105
1.29 x 105
1.33 x 105 Average DAB = 1.35 x 105 cm2/s. Experimental value at 25oC is 1.61 x 105 cm2/s.
Note that a material balance shows that the bulk concentration of H2S in the water only reaches a
high of 10.5 x 106 mol/cm3. Thus, the diffusivities would only be slightly higher than above. Exercise 3.38
Subject: Vaporization of water (A) into air (B) in a wettedwall column.
Given: Column inside diameter = D = 1.46 cm and length = L = 82.7 cm. Air volumetric flow
rate = Q = 720 cm3/s at 24oC and 1 atm (760 torr). Water enters at 25.15oC and exits at 25.35oC.
Partial pressures of water vapor in the air are 6.27 torr entering and 20.1 torr leaving.
Diffusivity of water vapor in air = DAB = 0.22 cm2/s at 0oC and 1 atm.
Assumptions: Negligible bulk flow effect. Ideal gas law. Negligible pressure drop in column.
Find: (a) Rate of mass transfer of water into air.
(b) Overall mass transfer coefficient, KG.
Analysis: Use cgs units. No mass transfer resistance in the liquid phase because pure water.
(a) The rate of mass transfer can be obtained from a material balance on the water
content of the air using the entering and exiting partial pressures of the water vapor.
For the entering air:
Mole fraction of water vapor = yA = pA / P = 6.27/760 = 0.00825
Molecular weight = M = 0.00825(18) + (10.825)(29) = 28.9
Density = ρ =PM/RT = (1)(28.9)/(82.06)(298) = 0.00118 g/cm3
Mole flow rate of gas = n = Qρ/M =(720)(0.00118)/28.9 = 0.02941 mol/s
Mole flow rate of water in entering air = n yA = 0.02941(0.00825) = 2.426 x 104 mol/s
For the exiting air:
Mole fraction of water vapor = yA = pA / P =20.1/760 = 0.02645
1 − 0.00825
= 0.02996 mol/s
By ratio with entering gas, mole flow rate of gas = m = 0.02941
1 − 0.02645
Mole flow rate of water in exiting air = m yA = 0.02...
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 Spring '11
 Levicky
 The Land

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