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

# 2 is solved for substituting this into eq 1 y is

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Unformatted text preview: permeate side = 15 psia. Assumptions: No pressure drop on feed side or on the permeate side. Negligible film resistances on either side of the membrane. Find: Material balance and membrane area in m2 as a function of the cut, θ, for: (a) Perfect mixing on both sides of the membrane. (b) Crossflow Analysis: Ideal separation factor from Eq. (14-40) = α * C3=,C3 = PM C3= PM C3 = 9 = 3.21 2.8 Pressure ratio = PPiPF = 15/300 = 0.05. From the bottom of Eq. (14-1) and the definition of a barrer in Section 14.1, 9 1 × 10 −10 PM PM C3= = C3= = = 9 × 10 −5 cm3 (STP) / cm2 − s − cmHg −4 01 × 10 lM . αx y= (1) , Rearranging Eq. (14-36), 1 + (α − 1) x where, y and x refer to C3= and α = α C3=,C3 . x (α − 1) + 1 − rα x (α − 1) + 1 − 0.05α = 3.21 (2) x (α − 1) + 1 − r x (α − 1) + 1 − 0.05 For a given value of x, Eq. (2) is solved for α. Substituting this α into Eq. (1), y is obtained. From Eq. (14-43), α = α * 3=,C3 C (a) Perfect mixing: On the feed side, xR = x. On the permeate side, yP = y. From a rearrangements of Eqs. (3) and (5) in Example 14.5, x − x 0.6 − x for C3=, θ = F = (3) y−x y−x yθnF and the membrane area , AM = PM C3= xPF − yPP (4) where, nF = 100 lbmol/h or 100(453.6)(22,400)/3600 = 282,000 cm3 (STP)/s Pressures are PF = 300(76/14.7) = 1,551 cmHg, and PP = 15(76/14.7) = 77.6 cmHg Exercise 14.10 (continued) Analysis: (a) (continued) yθ(282,000) 9.0 × 10 x (1,551) − y (76.6) (10,000) The calculations are carried out in a spreadsheet, with the following results: Therefore, A M in m 2 = −5 (b) Crossflow: We have the following relationships between C3= mole fractions in the permeate and retentate as a function of the cut, θ = nP/nF. x − x R (1 − θ) 0.60 − x R (1 − θ) From material balance Eq. (1), Example 14.6, y P = F = (5) θ θ From Eq. (14-49), where the y and x pertain to C3=, yP = x 1 1− α R 1− θ θ 1 − xR α α −1 xF 1 − xF α α −1 − xR α α −1 =x 1 1− α R 1− θ θ 1 − xR α α −1 0.60 1 − 0.60 α α −1 − xR α α −1 Exercise 14.10 (continued) Analysis: (b) (continued) Combine this equation with Eq. (5) to obtain: 0.60 1− θ = xR + x 1 1− α R 1 − xR α α −1 0.60 0.40 (6) α α −1 − xR α α −1 For C3=, xR begins at 0.60 at the feed end and decreases toward the retentate end. For each step in x, Eq. (2) is used to calculate α. Use Eq. (6) to compute values of θ as xR decreases. Use Eq. (5) to compute the corresponding value of yP. Eq. (1) is used to compute the local value of y. For a differential molar transfer rate, dn, the differential membrane area required is given by Eq. (14-50), which is also given in incremental form for numerical calculation. If we base this calculation on C3=, d AM = PM O 2 ydn , xPF − yPP ∆AM , m2 = 9.0 × 10 −5 yθ(282,000) x (1,551) − y (76.6) (10,000) (7) where y is the local value given by Eq. (3) and ∆n is the incremental decrease in the feed flow. Using the equations, the calculations are carried out on...
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