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

3 62 gives na dab agm dab 3 10 9 ca1 ca 2 r2

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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 finite-difference form of Fick's law with no bulk flow effect. For a spherical shell, Eq. (3-62) 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 0.8-micron 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 solution-diffusion model. Find:. Mass-transfer fluxes for methane and hydrogen Analysis: The mass-transfer flux of a gas species is given by Eq. (3-52), Ni = Hi Di ∆pi ∆z (1) Membrane thickness = 0.8 micron = 0.8 x 10-6 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/m3-Pa 1.14 x 10-4 0.17 x 10-4 D, m2/s 8.0 x 10-11 109 x 10-11 Flux, N, mol/m2-s 0.028 0.042 Exercise 3.22 S...
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