(完整版)传质与分离习题(含答案)

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1、Problems for Mass Transfer and Separation Process Absorption 1 The ammoniaair 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 ph

2、ases are contact, what will happen, absorption or stripping? Solution：09. 0y 05. 0x xy97. 0 09. 00485. 005. 097. 0yy It is an absorption operation. 2 When the temperature is 10 c0 and the overall pressure is 101.3KPa , the solubility of oxygen in water can be represented by equation p=3.27104x , whe

3、re 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: t

4、he mole fraction of oxygen in air is 0.21,hence: p = P y =1x0.21=0.21amt 64410*24. 610*27. 321. 010*27. 3px Because the x is very small , it can be approximately equal to molar ratio X , that is 610*42. 6 xX So )(/ )(4 .11)/(18*)( 1)/(32*)(10*42. 6lub2322222226OHmOgOkmolHOkgHOkmolHkmolOkgOkmolOility

5、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 0.03kmols1m-2 , practice absorbent flow rate L is 1.4 times as much as the min amount required. Under the op

6、erating condition, the equilibrium relationship is y*=1.75x. Volume total absorption coefficient is Kya=0.022 kmols1m-2y-1. What is the molar flow rate of the absorbent and what height of packing will be required? solution：0002. 01bayy xa=0 733. 175. 102. 099. 002. 0*minababxxyyVL 43. 24 . 1minVLVL

7、smkmolL20729. 003. 043. 2 720. 043. 275. 1LmVS 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 )1)(ln(11SSmxymxySNbaabOy=12 mKyaSVHOY364. 1022.

8、 003. 0/ mNHHOYOY37.1612364. 1 4 The mixed gas from an oil distillation tower contains H2S=0.04(molar fraction). Triethanolamine (absorbent) is used as the solvent to absorb 99% H2S in the packing tower, the equilibrium relationship is y*=1.95x, the molar flux rate of the mixed gas is 0.02kmol m-2 s

9、-1， overall volume absorption coefficient is Kya=0.05 kmols1m-2y-1, The solvent free of H2S enters the tower and it contains 70% of the H2S saturation concentration when leaving the tower. Try to calculate: (a) the number of mass transfer units Noy, and (b) the height of packing layer needed, Z. sol

10、ution：ya=yb(1-0.99)=0.04*1%=0.0004 xb*=yb/m=0.04/1.95= 0.0205 xb=0.7xb*=0.0144 yb*=1.95*0.0144=0.028 yb-yb*=0.04-0.028=0.012 ym=0.0034 Z=HoyNoy Noy=(yb-ya)/ ym=11.6 maKGHymoy4 . 005. 0/02. 0/ Z=11.6*0.4=4.64m 5 Ammonia is removed from ammoniaair mixture by countercurrent scrubbing with water in a pa

11、cked tower at an atmospheric pressure. Given: the height of the packing layer Z is 6 m, the mixed gas entering the tower contains 0.03 ammonia (molar fraction, all are the same below), the gas out of the tower contains ammonia 0.003; the NH3 concentration of liquid out of the tower is 80% of its sat

12、uration concentration, and the equilibrium relation is y*=1.2x. Find: (1)the practical liquidgas ratio and the min liquidgas ratio L/V=?. (2) the number of overall mass transfer units. (3) if the molar fraction of the ammonia out of the tower will be reduced to 0.002 and the other operating conditio

13、ns keep unchanged, is the tower suitable? solution：(1) 35. 12 . 103. 08 . 0003. 003. 0GL (2) 89. 035. 12 . 1S 26. 689. 0003. 003. 011. 089. 011InNOY (3) mNZHOYOY958. 026. 66 47. 889. 0002. 003. 011. 089. 011InNOY Since mNHZOYOY0 . 61 . 847. 8958. 0 it is not suitable 6 Pure water is used in an absor

14、ption 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 absorber are 101.3kpa and 200c, respectively. The flux of gas V is 580kg/(m2.h), and 6 percent (volume %) of ammonia is contained in the gas mixture. The

15、 flux of water L is 770kg/( m2.h). The gas and liquid is countercurrent in the tower at isothermal temperature. The equilibrium equation y*=0.9x, and gas phase mass transfer coefficient kGa is proportional to V0.8, but it has nothing to do with L. What is the height of the packed layer needed to kee

16、p the same absorptivity when the conditions of operation change as follows: (1)the operating pressure is 2 times as much as the original.(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 original Solution: 3 ,1,293Zm patm TK 1

17、210.060.063810.06(10.99)0.000638YYY The average molecular weight of the mixed gas M=290.94+170.06=28.28 22580(10.06)19.28/()28.2877042.78/()180.9 19.280.405642.78VkmolmhLkmolmhmVL 12221ln()(1)110.06380ln()(10.4056)0.405610.40560.0006386.88430.43586.884OGOGOGNmVLHYmXmXmVYmXLLZmN 1） 2pp ppmm So 1222ln

18、()(1)10.90.4520.45 19.280.202842.78111ln(100)(10.2028)0.202810.20285.496OGmpmpm VLYmXm Xm VNmVYmXLLL OGrGVVHK aK aP So:OGH changes with the operating pressure 10.43580.21792OGOGOGOGHHHH So 5.4960.21791.198OGOGZNHm So the height of the packed section reduce 1.802m vs the original 2） 2LL 11()0.40560.2

19、0282225.496OGmVmVmVLLLN when the mass flow rate of liquid increases,GK a has not remarkable effect 0.43585.4960.43582.395OGOGOGOGHHmZNHm the height of the packed section reduce 0.605m against the original 3） 2VV (2 )2()20.40560.81161ln(100)(10.8116)0.811615.8110.8116OGmVm VmVLLLN when mass flow rate

20、 of gas increaes,GK a also will increase. Since it is gas film control for absorption, we have as follows: 0.80.80.80.20.80.2()222220.43580.50115.81 0.5017.92GGGGOGOGGGOGOGK aVVK aK aK aVVVHHK aPK aPmZNHmm So the height of the packed section increase 4.92m against the original Distillation 1 Certain

21、 binary mixed liquid containing mole fraction of easy volatilization component Fx 0.35, feeding at bubbling point, is separated through a sequence rectify column. The mole fraction in the overhead product is xD=0.96, and the mole fraction in the bottom product is xB =0.025. If the mole overflow rate

22、s are constant in the column, try to calculate (a)the flow rate ratio of overhead product to feed(DF)? (b)If the reflux ratio R=3.2, write the operating lines for rectifying and stripping sections solution： Fx0.35；xB=0.025；xD=0.96；R=3.2。 （1） D/ FBDBFxxxx0.347634.76。 （2）the operating line for rectify

23、ing section. DxRxRRy1110.762x0.229。 the operating line for stripping section L=RD D=0.347F L=L+F B=F-D BxBFLBxBFLFLy1.45x0.0112。 2 A continuous distillation column is to be designed to separate an ideal binary material system，The feed which contains more volatile component Fx 0.5, feed rate 100kmol/

24、h, is saturated vapor, the flow rate of overhead product and the flow rate of bottom product are also 50kmol/h. Suppose the operating line for rectifying section is y=0.833x+0.15, the vapor generated in the reboiler enters the column through the bottom plate, a complete condenser is used on the top

25、of column and reflux temperature is bubbling point. Find: (1) Mole fraction xD of overhead product and the mole fraction xB of bottom product ? (2) Vapor amount condensed in the complete condenser, in mol/h? (3)The operating line for stripping section. (4) If the average relative volatility of the c

26、olumn is 3 and the first plates Murphree efficiency from the tower top is Em,L=0.6, find the constituent of gas phase leaving the second plate from tower top. Solution: (1)the operating line for rectifying section isDxRxRRy111 and y=0.833x+0.15 833. 01RR and 1RxD0.15， R5，xD0.9，xW=0.1 （2）Amount of th

27、e condensate vapor handled by complete condenser V(R+1)D=300（kmol/h） 。 （3）The operating line for stripping section WxWqFLWxWqFLqFLy1.25x0.025。 （4）the constituent of gas phase leaving from the second plate of tower top2y EML=11xxxxDD0.6，since 1x)1 (111yyy1123yyDDxx23=0.75 So 1x0.81 2y1yVL( xD1x)0.91R

28、R（0.90.81）0.825。 3 An ideal binary solution of a volatile component A containing 50% mole percent A is to be separated in a continuous distillation column. The feed is saturated vapor, the feed rate is 1000kmol/h, and the flow rate of overhead product and the flow rate of bottom product are also 500

29、kmol/h. Given: the operating line for rectifying section is y=0.86x+0.12, indirect vapor is used in the reboiler for heating and the total condenser is used on the top of tower. Assume that the reflux temperature is at its bubble point. Find: (1) reflux ratio R, the mole fraction of overhead product

30、 xD and the mole fraction of bottom product xB? (2) upward flow rate of vapor in the rectifying section(V mol/h,) and down ward flow rate of liquid in the stripping section.(L mol/h,) . (3) The operating line for stripping section . (4)if relative volatility =2.4, find RRmin Solution: （1）reflux rati

31、o R、the mole fraction in overhead product xD and the mole fraction in bottom product xB DxRxRRy1110.86x0.12， R6.14，xD0.857，xW=0.143。 （2）upward flow rate in the rectifying section V kmol/h and down ward flow rate in the stripping section.(L kmol/h LLRD=3070 kmol/h， V（R1）D3570 kmol/h V=V-F=3570-1000-2

32、570kmol/h （3）The operating line for stripping section BxVBxVLy1.19x0.02 （4）reflux ratio and min reflux ratio RRmin minR11111FDFDyxyx1.734 minRR3.54。 4 There is a continuous rectifying operation column, whose the operation line equation is as follows: Rectifying section: y=0.723x+0.263 Stripping sect

33、ion: y=1.25x-0.0187 if the feed enters the column at a dew point, find (a)the molar fraction of feed、overhead product and bottom product. (b) reflux ratio R. solution: 1RR0.723， R=2.61 1RxD0.263， Dx=0.95。 Since yWx1.25Wx-0.0187 So Wx0.0748。 the feed enters the column at a dew point, q operation line

34、 equation as：y=Fx. From q line and rectifying operation line, the following is solved. x=0.535, y=0.65， Fx0.65 5 A column is to be designed to separate a liquid mixture containing 44 mole percent A and 56 mole percent B, the system to be separated can be taken as ideal. The overhead product is to co

35、ntain 95.7 mole percent A. given: liquid average relative volatility =2.5, min reflux ratio Rmin =1.63, try to illustrate the thermal condition of the feed and to calculate the value of q. solution： From the equilibrium equation as： yqqqxx) 1(1qqxx5 . 115 . 2。 Since Rmin/(Rmin+1)= qDqDxxyx Min reflu

36、x ratio：qqqqqqDxyyxyyxR957. 0min=1.63 q point： xq=0.365， yq0.59 xq=0.365Fx，feed is a mixture of gas and liquid. q operation line equation as y=qx/(q-1)-Fx/(q-1)，let x=xq=0.365， y=yq0.59 we have q0.667。 6 A continuous rectifying column operated at atmospheric pressure is used to separate benzenetolue

37、ne mixed liquid. The feed is saturation liquid containing 50 mole percent benzene. Given: the overhead product must contain 90 mole percent benzene and the bottom product contains 10 mole percent benzene. If reflux ratio is 4.52, try to calculate how many ideal plates are needed? And determine the p

38、osition of feed plate. In this situation the equilibrium data of benzenemethylbenzene are as follows toC 80.1 85 90 95 100 105 110.6 x 1.000 0.780 0.581 0.411 0.258 0.130 0 y 1.000 0.900 0.777 0.632 0.456 0.262 0 Solution: Based on M-T, we can obtain the theoretical plate numbers. Interception of op

39、erating line for rectifying section1RxD152. 49 . 00.163。 N16。 Feed plate is the third one from the column top. 7 There is a rectifying column, given : mole fraction of distillation liquid from tower top xD=0.97, reflux ratio R=2, the gas-liquid equilibrium relationship y=2.4x/(1+1.4x); find: the con

40、stituent x1 of the down liquid leaving from the first plate and the constituent y2 of the up gas leaving from the second plate in the rectifying section. Suppose the total condenser is used at the top of column. solution： From the equilibrium relation xxy4 . 114 . 2, Since 1yxD =0.97， x10.93. Based

41、on the operating line equation, we have 2y111RxxRRD=0.94 8. A liquid of benzene and toluene is continuously fed to a plate column. Under the total reflux ratio condition, the compositions of liquid on the close plates are 0.28, 0.41 and 0.57, respectively. Calculate the Murphree plate efficiency of

42、two lower plates. The equilibrium data for benzenetoluene liquid and vapor phases under the operating condition are given as follows: x 0.26 0.38 0.51 y 0.45 0.60 0.72 Solution : Under the total reflux ratio condition nnxy1 From the Murphree efficiency: 57. 041. 012231*1(,xyxyyyyyEnnnnvm Calculate 6

43、28. 0*2Y from 41. 02X, 475. 0*3Y from X3=0.28 by interception. %6728. 0475. 028. 041. 0*E %7373. 041. 0628. 041. 057. 04343m33*2322YYYYYYYYEm Drying 1 A wet solid with 1000 kg/h is dried by air from 40% to 5% moisture content (wet basis) under the convective drying conditions. The air primary humidi

44、ty H1 is 0.001(kg water/kg dry air), and the humidity of air leaving dryer H2 is 0.039 (kg water /kg dry air), suppose that the material loss in the drying process can be negligible. Find: (1) rate of water vaporization W, in kg water/ h. (2) rate of dry air L required, in kg bone dry air/h, flow ra

45、te of moist air, L, in kg fresh air/h. (3) rate of moist material out of dryer, G2, in kg moist solid/h. Solution : G1=1000kg/h, w1=40%, w2=5% , H1=0.001, H2=0.039 Gc =G1(1-w1)=1000(1-0.4)=600kg/h x1=0.4/0.6=0.67, x2=5/95=0.053 (1)W=Gc(x1-x2)=600(0.67-0.053)=368.6kg/h (2)L(H1-H2)=W 9700001. 0039. 06

46、 .368HHW21L kg dry air/h L= L(1+H1)=12286.7(1+0.039)=12765.8kg fresh air/h (3) Gc =G2(1-w2) h/6kg.63105. 01600w122cGG 2 The wet solid is to be dried from water content 20% to 5% (wet basis) in a convective dryer at an atmospheric pressure. The feed of wet solid into the dryer is 1000 kg/h at a tempe

47、rature of 40. The dry and wet bulb temperatures of air are respectively 20 and 16.5before air enters the preheater. After being preheated, air enters the dryer. The dry and wet bulb temperatures of air leaving dryer are respectively 60 and 40. If heat loss is negligible and drying is considered as c

48、onstant enthalpy process: (1) What is the fresh (wet) air required per unit time (in kg/h)? (2) The air temperature t1 before entering the dryer (Given: Vaporization latent heat of water at 0 is 2501kJ/kg, specific heat of dry air Cg is 1.01 kJ/kgK, specific heat of water vapor Cv is 1.88 kJ/kgK) So

49、lution: w1=0.2, w2=0.05, G1=1000kg/h, 1=40, t0=20, tw0=16.5, t2=60, tw2=40 Q=1.01L(t2-t0)+W(2501+1.88t2)+GcCm (2-1)+QL For constant enthalpy process, I1=I2 From the Fig, we have H0=0.01(based on t0=20, tw0=16.5), H2=0.045(based on t2=60, tw2=40) I1=(1.01+1.88H0)t1+25010H0, I2=(1.01+1.88H2)t2+2501H2=

50、(1.01+1.880.045)60+25010.045177.7 I1= (1.01+1.880.01)t1+25010.01=1.03t1+24.9= I2=177.7 4 .14803. 19 .247 .1771t Gc=G1(1-w1)=1000(1-0.2)=800 x1=0.2/0.8=0.25, x2=5/95=0.053 W=Gc(x1-x2)=800(0.25-0.053)=157.6 H0=H1 9 .450201. 0045. 06 .15712HHWL kg/h L=L(1+H0)=4502.9(1+0.01)=4547.9 kg/h 3 The wet solid

51、material is to be dried from water content 42% to 4% (wet basis) in an adiabatic dryer. The solid product out of the dryer is 0.126kg/s. After the fresh air at a dry-bulb temperature of 21C and a percentage humidity of 40 is preheated to 93C, it is sent to the dryer, and leaves the dryer at percenta

52、ge humidity of 60. If the drying is under constant enthalpy process. (1) Determining the air humidity (H1 and H2) from the given air state in t-H diagram. (2) If H0=0.008(kg water/kg dry air), H2=0.03( kg water /kg dry air. Find: (a) Dry air flow rate L required, in kg dry air/s. (b)How much heat is

53、 supplied to air by the preheater (in k J/h)? Solution: (1) w1=0.42,w2=0.04, G2=0.126kg/s, t0=21,Hp=0.4, t1=93, Hp=0.6, I1=I2 From t-H diagram, we have H0=0.008, H2=0.03 (2) G2(1-w2)=G1(1-w1)= Gc 209. 042. 0104. 01126. 0111221wwGG W=G1- G2= 0.209-0.126=0.0826 Or W= Gc (X1-X2) skgHHWL/752. 3008. 003.

54、 00826. 012 Qp=L(I1-I0)=L(1.01+1.88H1)(t1-t0)=3.752(1.01+1.880.008)(93-21) =301.2kJ/s1.08106kJ/h 4 The wet solid containing 12%(wet basis) moisture is fed to a convective dryer at a temperature of 15 and is withdrawn at 28, which contains 3% moisture (wet basis). The flow rate of final moist solid (

55、product) is 1000kg/h. After the fresh air at a dry-bulb temperature of 25 and a humidity of 0.01 kg water/kg dry air is preheated to 70, it is sent to the dryer, and leaves the dryer at 45C. Suppose the drying process is under constant enthalpy, heat loss in the drying system can be negligible. (1)

56、Drawing the operation process covering various air states in t-H diagram. (2) What is the fresh air required per unit time (in kg/h)? Solution: G2=1000, w1=12%, w2=3%, 1=15, 2=28, t0=25, H0=0.01, t1=70, t2=45, I1=I2 Gc=1000(1-0.12)=880, x1=12/88=0.136, x2=3/97=0.0309 W=880(0.136-0.0309)=92.5 I1=(1.0

57、1+1.88H0)t1+2501H0=(1.01+1.880.01)70+25010.0196.9 I2=(1.01+1.88H2)45+2501H2=45.5+2574.6H2=96.9 Since I1= I2 H2=(96.9-45.5)/2574.6=0.02 H0=H1 hkgHHWL/925001. 002. 05 .9212 L=9250(1+0.01)=9343 5 Certain wet material is dried under an ordinary pressure in a convective dryer. Given: air temperature and

58、humidity before entering a preheater are t0=15 and H0=0.0073 (kg water)/ (kg dry air); air temperature before entering the dryer tl= 90; air temperature and humidity leaving the dryer t2=50 and H2=0.023(kg water)/ (kg dry air); water content contained by material entering the dryer x1=0.15 (kg water

59、)/(kg bone dry material ); water content contained by material leaving the dryer x2=0.01 kg water/(kg bone dry material); the capability of production of dryer is 273kg/h (based on the product leaving the dryer). Find: (1) flow rate of the dry air L ,in kg bone dry air/h: (2) flow rate of wet air be

60、fore entering the preheater, in fresh air m3/s; (3) heat duty (quantity) provided to the preheater, in kw if heat loss can be negligible. Solution: t1=15, H0=0.0073, I0=34, t1=90, I1=110, t2=50, H2=0.023, x1=0.15, x2=0.01, G2=273kg/h x2=w2/(1-w2), w2=x2/(1+x2)=0.01/1.01=0.01 Gc=G2(1-w2)=273(1-0.02)=

61、270.3kg dry-bone material/h W=Gc(x1-x2)=270.3(0.15-0.01)=37.8kg water /h L=W/(H2-H1)=37.8/(0.023-0.0073)=2407.6kg dry air/h Lv=LvH 824. 0273273150073. 0244. 1772. 0273273244. 1772. 000tHH Lv=2407.60.8241983.9m3/h=1983.9m3/3600s=0.551 m3/s Qp=L(1.01+1.88H0)(t1-t0)=2407.6(1.01+1.880.0073)（90-15）183951

62、kJ/h=183951kJ/3600s=51.1kW 6 Wet material is dried in a dryer at an atmosphere pressure. The operating conditions are as follows: Air conditions: air temperature before entering the preheater t0=20C, air humidity before entering the preheater H0=0.01(kg water/kg dry air), air temperature before ente

63、ring the dryer t1=120, air temperature leaving the dryer t2=70C, air humidity leaving the dryer H2=0.05(kg water/kg dry air). Material conditions: material temperature before entering the dryer 1=30C, wet content before entering the dryer w1=20% based on wet basis, material temperature leaving the d

64、ryer 2=50C, wet content leaving the dryer w2=5% based on wet basis, specific heat of bone-dry material Cs=1.5kJ/(kg bone-dry solids . C). Production capacity of dryer is 53.5 kg/h based on product leaving the dryer. Suppose heat loss in the drying system can be negligible. Find: (1) The mass of bone

65、-dry air required per unit time. (2) How much heat is obtained by air when passing through the preheater, in kJ/h? t0=20, H0=0.01, t1=120, t2=70, H2=0.05, 1=30, w1=0.2, 2=50, w2=5%, Cs=1.5, qm2G2=53.5 x1=0.2/0.8=0.25, x2=5/95=0.053 Gc=G2(1-w2)=53.5(1-0.05)=50.8 kg/h W=Gc(x1-x2)=50.8(0.25-0.053)=10 kg/h L=W/(H2-H0)=10/(0.05-0.01)=250 kg/h Qp=L(1.01+1.88H0)(t1-t0)=250(1.01+1.880.01)(120-20)=25720 kJ/h