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

7x assumptions no stripping of water no absorption of

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Unformatted text preview: .247) 1/ 2 0.247 1 = 5.23 ft 33.9 2.02 uL 0.247 = = 0.0234 s-1 H L ( aPh / a ) 5.23 ( 2.02 ) This is 35% of the suggested value of 0.067 s-1 Coefficient kGa is obtained from the gas-phase analog of (1), kG a = uV H G ( aPh / a ) (2) with HG obtained from (6-133), using dimensionless groups from (6-134) and (6-135). From Exercise 6.28, uV = 1.59 ft/s, ρV = 0.148 lb/ft3, µV = 0.018 cP = 1.21 x 10-5 lb/ft-s From Eq. (6-134), N ReV = uV ρV (1.59)(0.148) = = 574 (33.9)(1.21× 10−5 ) aµV From Exercise 6.23, the diffusivity of benzene in air = DV = 0.086 cm2/s at 70oF and 15 psia. Using Eq. (3-36) to correct for temperature and pressure, DV = 0.086 15 (2)(14.7) From Eq. (6-135), NScV = 77 + 460 70 + 460 1.75 = 0.045 cm2/s = 4.84 x 10-5 ft2/s µV 1.21× 10−5 = = 1.69 ρV DV (0.148)(4.84 × 10 −5 ) Exercise 6.29 (continued) Analysis: (continued) From Eq. (6-133) 1 1/ 2 4ε HG = ( ε − hL ) CV a4 1 4(0.919) ( 0.919 − 0.177 )1/ 2 0.368 33.94 From (2), kG a = 1/ 2 1/ 2 (N ) (N ) −3/ 4 ReV ScV ( 574 )−3 / 4 (1.69 )−1/ 3 −1/ 3 uV DV a aPh 1.59 4.84 ×10−5 1.59 = 1.73 s-1 0.455 ( 2.02 ) This is approximately twice the suggested value of 0.80 s-1. = 1 = 0.455 ft 2.02 Exercise 6.30 Subject: Absorption of NH3 from air into water in a packed tower. Given: Column operation at 68oF and 1 atm. Inlet gas mass velocity = 240 lb/h-ft2. Inlet water mass velocity = 2,400 lb/h-ft2. Tower packed with 1.5-inch ceramic Berl saddles. Henry's law applies for NH3 with p (atm) = 2.7x. Assumptions: No stripping of water. No absorption of air. Dilute system. Find: (a) Packed height for 90% absorption of NH3. (b) Minimum water mass velocity for 98% absorption of NH3. (c) KGa, pressure drop, maximum liquid rate, KLa, packed height, column diameter, HOG, and NOG for 1.5-inch ceramic Hiflow rings. Analysis: (a) Assuming Dalton's law, the equilibrium equation for 1 atm is, p = yP = 2.7x atm or with P = 1 atm, y = Kx = 2.7x (1) Because the system is dilute, Eq. (6-89) can be used to compute NOG . Then the packed height is given by Eq. (6-85), lT = NOGHOG (2) The absorption factor from Eq. (6-15) is, A = L/KV = (2,400/18)/2.7(240/29) = 6.0 Entering and exiting ammonia mole fractions are: yin = 0.02, yout = 0.002, xin = 0.0 The ammonia mole fraction in the exiting liquid is obtained by an overall ammonia balance: 240 2,400 240 2,400 yin + xin = yout + xout 29 18 29 18 (3) Solving Eq. (3), xout = 0.00112. From Eq. (6-89), N OG = ln ( A − 1) / A ( yin − Kxin ) / ( yout − Kxin ) + (1 / A) ( A − 1) / A = ln (6 − 1) / 6 0.02 / 0.002 + (1 / 6) (6 − 1) / 6 = 2.57 To obtain an estimate of HOG, the correlations of Billet and Schultes in Chapter 6 or literature data can be used. The latter is found on page 16-38 of Perry's Chemical Engineers' Handbook, 6th edition or on page 663 of "Equilibrium-Stage Separation Operations in Chemical Engineering" by Henley and Seader, as shown below. Analysis: (a) (continued) Exercise 6.30 (continued) Air-Ammonia-Water system with...
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