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

# 39 subject absorption of ammonia a from air b into 2

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Unformatted text preview: 996(0.02645) = 7.924 x 10-4 mol/s Increase in water content of the gas = rate of mass transfer of water = nA = 7.924 x 10-4 - 2.426 x 10-4 = 5.5 x 10-4 mol/s (b) From Eqs. (3-211) and (3-218), noting that there is no resistance in the liquid phase, KG = k p = ( NA pAi − pAb ) = LM nA AP yAi − yAb ( ) (1) LM Exercise 3.38 (continued) Analysis: (b) (continued) A = πDL = 3.14(1.46)(82.7) = 379 cm2 At 25oC, the interface mole fraction of water vapor = yA i = y A i − yA b LM = yA i − yA in − yA i − yA out ln yA i − yA in yA i − yA out = PAs 0.46 psi = = 0.0313 P 14.7 psi 0.0313 − 0.00825 − 0.0313 − 0.02645 = 0.0117 0.0313 − 0.00825 ln 0.0313 − 0.02645 From Eq. (1), KG = 5.5 × 10−4 = 1.24 × 10−4 mol/s-cm2-atm (379)(1)(0.0117) Note that the correction for bulk flow amounts to only about 2.5% because the average mole fraction of air between the interface and the bulk is 0.975. Exercise 3.39 Subject: Absorption of ammonia (A) from air (B) into 2 N H2SO4 in a wetted-wall column, with countercurrent flow. Given: Column inside diameter = D = 0.575 inch (0.0479 ft) and length = L = 32.5 in. (2.71 ft) Air flow rate = n = 0.260 lbmol/h at 1 atm. Partial pressures of ammonia in the air are 0.0807 atm entering and 0.0205 leaving. Air enters at 77oF and exits at 84oF. Aqueous acid enters at 76oF and exits at 81oF. Change in acid strength is negligible. Assumptions: Reaction of the basic ammonia with the sulfuric acid is instantaneous such that the partial pressure, pA i , of ammonia at the gas-liquid interface is zero. Therefore, the only mass transfer resistance is in the gas phase. Air flow rate refers to the total entering gas. Find: Mass-transfer coefficient, kp , for the gas phase. Negligible bulk flow effect. Analysis: From a rearrangement of Eq. (3-207), kp = nA A pAb − pAi ( ) = LM nA A pA b − 0 ( ) = LM nA A pAb () (1) LM The rate of mass transfer is obtained by material balance for the decrease in partial pressure of the ammonia in the air. First compute the flow rate of just the air in the gas. P − pA in 10 − 0.0807 . = 0.239 lbmol/h P 1.0 pA in 0.0807 Ammonia in entering gas = 0.239 = 0.021 lbmol/h nB = 10 − 0.0807 . pBin nB = n Ammonia in exiting gas = pA out pBout nB = = 0.260 0.0205 0.239 = 0.005 lbmol/h 10 − 0.0205 . Rate of mass transfer of ammonia = nA = 0.021 - 0.005 = 0.016 lbmol/h Area for mass transfer = A = πDL = 3.14(0.0479)(2.71) = 0.408 ft2 pA b LM = () From Eq. (1), k p = nA / A pAb pA in − pA out 0.0807 − 0.0205 = = 0.0439 atm pA in 0.0807 ln ln 0.0205 pA out LM = 0.016 /(0.408)(0.0439) = 0.893 lbmol/h-ft2-atm The bulk flow effect amounts to a correction of about 5%. Exercise 3.40 Subject: Overall mass-transfer coefficient for a packed cooling tower, where water (A) is evaporated into air (B). Given: Cross-sectional area for air flow = S = 0.5 ft2. Height over which data were taken = H = 7 ft. Air rate = n = 0.401 lbmol/h. Water rate = 20 lbmol/h. At the bottom, P = 14.1 psia, T = 120oF, PA...
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