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

# 332 094 10 5 25 105 05 1 2 241 1 3 000417 m

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Unformatted text preview: From Eq. (3-105), the average rate of sublimation per unit area is, nA,B / A = k cavg cA i − cA o (1) From Eqs. (3-137) and (3-138), the average mass-transfer coefficient is given by, k cavg = 0.664 DA,B N Re L L 1/ 2 N Sc 1/ 3 = 0.664 0.94 × 10 −5 5 × 105 1 From Example (3.14), cA i =4.3 x 10-2 kmkol/m3 and 1/ 2 2.41 1/ 3 = 0.0059 m / s cA o = 0. From Eq. (1), nA,B / A = 0.0059 4.3 × 10 −2 − 0 = 2.54 × 10 −4 kmol / s - m2 (b) From Eqs. (3-132) and (3-133), using physical properties in Example (3.14), with a Reynolds number of one-half of 5 x 105, since x = 0.5 m, k cx = 0.332 DA,B N Re x x 1/ 2 N Sc 1/ 3 = 0.332 0.94 × 10 −5 2.5 × 105 0.5 1/ 2 2.41 1/ 3 = 0.00417 m / s From Eq. (3-105), the local rate of sublimation per unit area is, nA,B / A = kcx ( cA )i − ( cA )o = 0.00417 ( 4.3 × 10−2 − 0 ) = 1.79 × 10 −4 kmol/s-m 2 Exercise 3.30 Subject: Sublimation of a straight, circular tube of naphthalene (A) into air (B) at 100oC (373 K) and 1 atm that flows through it in laminar flow. Given: Tube diameter, D, is 5 cm (0.05 m). Reynolds number = NRe = 1,500. Physical properties in Example 3.14. Assumptions: Fully established laminar-flow velocity profile for air entering the tube. Negligible pressure drop. Equilibrium at naphthalene-air interface. Find: Tube length, x, for average naphthalene mole fraction in exiting air, yA L , to equal 0.005. Analysis: From Example 3.14, yA i = PAs / P = 10 / 760 = 0.0132 By material balance for sublimation of naphthalene into air, the rate of mass transfer over the tube length must be: nA = ux S cA x − cA 0 (1) This rate of mass transfer is given by (3-105), which with a log-mean driving force is, nA = k cavg A cA i − cA ln 0 − cA i − cA cA i − cA (2) 0 cA i − cA x x For the entering air, cA 0 = 0. Therefore, combining Eqs. (1) and (2), and solving for k cavg , k cavg = cA i − cA ux S ln A cA i − cA 0 (3) x From the tube geometry, S = πD2/4 = (3.14)(0.05)2/4 = 0.00196 m2 A = πDx = (3.14)(0.05)x = 0.157 x m2 For laminar flow, Eq. (3-150) applies, which contains the Peclet number, Eq. (3-102): N Pe M = N Re N Sc = (1,500)(2.41) = 3,620 From Eq. (3-150), using the definition of the Sherwood number in Eq. (3-148), but for average conditions, Exercise 3.30 (continued) Analysis: (continued) k cavg = 0.0668 N Pe M / x / D DA,B D N Sh avg = A,B 3.66 + D D 1 + 0.04 N / x/D 2/3 = Pe M = 0.000688 + N Re = 0.00227 x 1 + 1.28 / x 2 / 3 0.0668 3620 / x / 0.05 0.94 × 10−5 3.66 + = 2/3 0.05 1 + 0.04 3620 / x / 0.05 (4) Dux ρ N µ (1,500)(0.000215) , therefore, u x = Re = = 68 cm / s = 0.68 m / s 0.0327(29) µ Dρ (5) 1,000 ln cA i − cA 0 cA i − cA x = ln yA i − yA 0 yA i − yA x = ln 0.0132 − 0.0 = 0.476 0.0132 − 0.005 Combining Eqs. (3) and (4), 0.000688 + 0.00227 (0.68)(0.00196) 0.00404 = (0.476) = 0.157 x x x 1 + 128 / x 2 / 3 . Rearranging Eq. (5), f x = 0.000688 + Solving the nonlinear Eq. (6), x = 3.73 m 0.00227 0.00404 − =0 x x 1 + 1.28 / x 2 / 3 (6) (5) Exercise 3.3...
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