传质与分离习题(含答案).doc

Problems for Mass Transfer and Separation Process Absorption 1 The ammonia–air mixture containing 9% ammonia(molar fraction) is contact with the ammonia-water liquid containing 5% ammonia (molar fraction). Under this operating condition, the equilibrium relationship is y*=0.97x. When the above two phases are contact, what will happen, absorption or stripping? Solution： It is an absorption operation. 2 When the temperature is 10 and the overall pressure is 101.3KPa , the solubility of oxygen in water can be represented by equation p=3.27(104x , where p (atm) and x refer to the partial pressure of oxygen in the vapor phase and the mole fraction of oxygen in the liquid phase, respectively. Assume that water is fully contact with the air under that condition, calculate how much oxygen can be dissolved in the per cubic meter of water? Solution: the mole fraction of oxygen in air is 0.21,hence: p = P y =1x0.21=0.21amt Because the x is very small , it can be approximately equal to molar ratio X , that is So 3 An acetone-air mixture containing 0.02 molar fraction of acetone is absorbed by water in a packed tower in countercurrent flow. And 99％ of acetone 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. Under the operating condition, the equilibrium relationship is y*=1.75x. Volume total absorption coefficient is Kya=0.022 kmol·s—1m-2y-1.. What is the molar flow rate of the absorbent and what height of packing will be required? solution： xa=0 Number of mass transfer units Noy=(y1-y2)/(y=12 (yb-ya)=0.02-0.0002 (y=[(yb-y*b)- (ya-y*a)]/ln[(yb-y*b)/ (ya-y*a )] (yb-y*b )=0.02-1.75xb=0.0057 Xb=V/L (yb-ya)= (0.02-0.0002)/2.43=0.00815 (ya-y*a)= ya=0.0002 Or =12 4 The mixed gas from an oil distillation tower c