传质与分离习题含答案.docx

1、传质与分离习题含答案Problems for Mass Tran sfer and Separati on ProcessAbsorpti on1The ammonia -air mixture containing 9% ammon ia(molar fraction) is con tact with the ammonia-water liquid containing 5% ammonia (molar fraction). Under this operating condition, the equilibrium relati on ship is y*=0.97x. When

2、the above two phases are con tact, what will happen, absorption or stripping?Solution: y 0.09 x 0.05 y 0.97xy 0.97 0.05 0.0485 y 0.09 It is an absorption operatio n.02When the temperature is 10 c and the overall pressure is 101.3KPa , the solubility of4oxygen in water can be represented by equation

3、p=3.27 10 x , where p (atm) and x refer to the partial pressure of oxyge n in the vapor phase and the mole fraction of oxyge n in the liquid phase, respectively. Assume that water is fully con tact with the air un der that con diti on, calculate how much oxyge n can be dissolved in the per cubic met

4、er of water?Soluti on: the mole fraction of oxyge n in air is 0.21,he nee:p = P y =1x0.21=0.21amtBecause the x is very small , it can be approximately equal to molar ratio X , that is6.42*10So6.42*10 6(kmolO2)*32(kgO2/kmolO2)1(kmolH 2O) *18(kgH 2O/ kmolH 2O)3An acet on e-air mixture containing 0.02

5、molar fraction of acet one is absorbed by water in a packed tower in coun tercurre nt flow. And 99 % of acet one is removed, mixed gas molar flow flux is 0.03kmol s 1m-2 , practice absorbent flow rate L is 1.4 times as much as the min amount required. Un der the operati ng con diti on, the equilibri

6、um relati on ship is y*=1.75x. V olume total1 2 1absorption coefficient is K ya=0.022 kmol s my. What is the molar flow rate of the absorbent and what height of pack ing will be required?L yb ya 0.02 0.99V min x； xa 0.02 1.751.42.43L 2.43 0.03 0.0729 kmoL m2 sminL 2.43Number of mass transfer units N

7、 oy=(y i-y2)/ y=12(y b-y a)=0.02-0.0002y=(y b-y* b)- (y a-y* a)/ln(y b-y* b)/ (y a-y* a )(yb-y* b )=0.02-1.75x b=0.0057Xb=V/L (y b-ya)= (0.02-0.0002)/2.43=0.00815(ya-y* a)= y a=0.0002Or Noy -ln( yb mXa)(1 S) S =121 S ya mXbHoy Noy 1.364 12 16.37m4The mixed gas from an oil distillati on tower contain

8、s H 2S=0.04(molar fraction). Trietha no lam ine (absorbe nt) is used as the solve nt to absorb 99% H2S in the pack ing tower, the equilibrium relationship is y*=1.95x, the molar flux rate of the mixed gas is 0.02kmol -m-2 s-1, overall volume absorpti on coefficie nt is Kya=0.05 kmol - s 1 my The sol

9、ve nt free of H 2S en ters the tower and it contains 70% of the H 2S saturation concentration when leaving the tower. Try to calculate: (a) the nu mber of mass tran sfer un its N oy, and (b) the height of pack ing layer n eeded, Z.solution : ya=yb(1-0.99)=0.04*1%=0.0004xb*=yb/m=0.04/1.95= 0.0205 xb=

10、0.7xb*=0.0144yb*=1.95*0.0144=0.028yb-yb*=0.04-0.028=0.012 ym=0.0034Z=HoyNoyNoy=(yb-ya)/ ym=11.6Hoy Gm/Kya 0.02/0.05 0.4mZ=11.6*0.4=4.64m5Ammon ia is removed from ammo nia -air mixture by coun tercurre nt scrubb ing with water in a packed tower at an atmospheric pressure. Given: the height of the pac

11、king layer Z is 6 m, the mixed gas en teri ng the tower contains 0.03 ammonia (molar fraction, all are the same below), the gas out of the tower contains ammonia 0.003; the NH 3 concentration of liquid out of the tower is 80% of its saturation concentration, and the equilibrium relation is y*=1.2x.

12、Find:(1)the practical liquid gas ratio and the min liquid gas ratio L/V=?. (2) the number of overall mass tran sfer un its.(3) if the molar fraction of the ammonia out of the tower will be reduced to 0.002 and the other operat ing con diti ons keep un cha nged, is the tower suitable?L0.030.003 ,solu

13、tion : (1)1.35G0.80.03 1.21 2(2)S0.891.35N OY10.03In 0.11 0.89 6.2610.890.003Noy0.958 8.47 8.1 6.0m it is not suitable6Pure water is used in an absorption tower with the height of the packed layer 3m to absorb ammonia in an air stream. The absorptivity is 99 percent. The operating conditions of abso

14、rber are0 2101.3kpa and 20 c, respectively. The flux of gas V is 580kg/(m .h), and 6 percent (volume %) ofammonia is contained in the gas mixture. The flux of water L is 770kg/( m 2 .h). The gas and liquid*is coun tercurre nt in the tower at isothermal temperature. The equilibrium equati on y =0.9x,

15、 and gas phase mass transfer coefficient k G a is proportional to V 0.8, but it has nothing to do with L.What is the height of the packed layer n eeded to keep the same absorptivity whe n the con diti ons of operati on cha nge as follows:(1)the operat ing pressure is 2 times as much as the origi nal.(2)the mass flow rate of water is one time more than the original. 3) the mass flow rate of gas is two times as much as the originalSolution: Z 3m, p 1atm,T 293KY2 Y(1 0.99) 0.000638The average molecular weight of the mixed gas M=29 0.94+17 0.06=28.28(1 0.06) 19.28kmol/(m2