Unformatted text preview: n steel. Ideal gas law.
Find:. (a) Initial rate of mass transfer of hydrogen.
(b) Initial rate of pressure decrease inside the vessel.
(c) Time in hours for the pressure inside the vessel to decrease to 50 psia at constant
temperature of 80oF.
Analysis: As the hydrogen diffuses through the wall, the pressure in the vessel will decrease.
(a) Assume finitedifference form of Fick's law with no bulk flow effect.
For a spherical shell, Eq. (362) gives, nA = DA,B AGM DA,B = 3 × 10 −9 cA1 − cA 2
r2 − r1 (1) (3600)
.
= 116 × 10−9 ft 2 / h
2
(2.54 × 12) r2 = (2 + 0.125)/12 = 0.1771 ft,
r1 = 2/12 = 0.1667 ft
1/2
2
AGM = (A2A1) , A2 = 4π(r2) = 4(3.14)(0.1771)2 = 0.394 ft2 ,
A1 = 4π(r1)2 = 4(3.14)(0.1667)2 = 0.349 ft2 , AGM = (0.394 x 0.349)1/2 = 0.371 ft2
cA 1 = 0.094 lbmol / ft 3 , cA 2 = 0
Substituting into Eq. (1), nA = 11.6 × 10 −9 ( 0.371)
(b) ( 0.094 − 0 ) ( 0.1771 − 0.1667 ) = 3.89 ×10 −8 lbmol/h Let m = number of moles of hydrogen in the vessel at any time.
By the ideal gas law, m = PV/RT (2) T = 540oR , V = πD3/6 = 3.14(4/12)3/6 = 0.233 ft3
Initially, P = 150 psia and m = (150)(0.233)/(10.73)(540) = 0.00603 mol Exercise 3.20 (continued)
Analysis: (b) (continued)
Differentiating Eq. (2) with respect to time to determine the rate of change of
pressure due to the decrease in the number of moles of hydrogen as it leaves the
vessel by diffusion through the walls,
dm
V dP
=n=−
dt
RT dt (3) Solving Eq. (3) for the rate of change of pressure, the initial rate of change is,
dP
nRT
(3.89 × 10−8 )(10.73)(540)
= −0.00097 psia/h
=−
=−
dt
V
0.233 (c) As the pressure in the vessel decreases, the solubility of hydrogen in steel and,
therefore, the rate of diffusion of hydrogen through the steel will decrease. Assume Henry's law
for the solubility. Then,
cA 1 = 0.094 P
150 (4) From Eqs. (1), (3), and (4), the rate of diffusion of hydrogen as a function of pressure is,
n = 3.89 × 10 −8 P
V dP
(0.233) dP
dP
=−
=−
= −0.000040
150
RT dt
(10.73)(540) dt
dt Integrating Eq. (5) from P = 150 psia to 50 psia,
t
3.89 × 10 −8
dt = −
150(0.000040) 0 50
150 dP
150
= 10986
= ln
.
P
50 Therefore, t = 154,000 (1.0986) = 169,500 h , a very long time! (5) Exercise 3.21
Subject: Mass transfer of gases through a dense polymer membrane
Given: A 0.8micron thick polyisoprene membrane. Partial pressures of methane and hydrogen
on the two sides of the membrane. Membrane properties in Table 14.9.
Assumptions: Applicability of the solutiondiffusion model.
Find:. Masstransfer fluxes for methane and hydrogen
Analysis: The masstransfer flux of a gas species is given by Eq. (352),
Ni = Hi Di
∆pi
∆z (1) Membrane thickness = 0.8 micron = 0.8 x 106 m
Values of Di and Hi = Si are given in Table 14.9 and are used with the given partial
pressures in Eq. (1) to give the following results:
Species
Methane
Hydrogen ∆p, Pa
2.45 x 106
1.80 x 106 H, mol/m3Pa
1.14 x 104
0.17 x 104 D, m2/s
8.0 x 1011
109 x 1011 Flux, N, mol/m2s
0.028
0.042 Exercise 3.22
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 Spring '11
 Levicky
 The Land

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