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1、Problems for Mass Tran sfer and Separati on ProcessAbsorpti on1 The 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 the above

2、 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 0.09It is an absorption operatio n.02 When the temperature is 10 c and the overall pressure is 101.3KPa , the solubility of4oxygen in water can be represented by equation p=3.27 10

3、 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 meter of wat

4、er?Soluti on:the mole fraction of oxyge n in air is 0.21,he nee:p = P y =1x0.21=0.21amtP3.27* 1040.213.27* 1046.24*10Because the x is very small , it can be approximately equal to molar ratio X , that is6.42*10Sosolubility11.4g(O2)/m3(H2。)6.42*10 6(kmolO2)*32(kgO2/kmolO2)1(kmolH 2O) *18(kgH 2O/ kmol

5、H 2O)3 An acet on e-air mixture containing 0.02 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.

6、Un der the operati ng con diti on, the equilibrium 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?solution :ya yb 10.0002xa=0Lyb ya 0.02 0.99V minx；xa 0.02 1.751.4

7、2.43L 2.43 0.030.0729 kmoL m2 sminmV 1.750.720Hoy尹Kya0.030.0221.364mL 2.43Number of mass transfer units N 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(

8、yb mXa)(1 S) S =121 SyamXbHoy Noy 1.364 1216.37m4 The mixed gas from an oil distillati on tower contains 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 mi

9、xed gas is 0.02kmol -m-2 s-1, overall volume absorpti on coefficie nt is Kya=0.05 kmol - s 1 my The solve 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 he

10、ight of pack ing layer n eeded, Z.solution : ya=yb(1-0.99)=0.04*1%=0.0004xb*=yb/m=0.04/1.95= 0.0205xb=0.7xb*=0.0144yb*=1.95*0.0144=0.028yb-yb*=0.04-0.028=0.012 ym=0.0034Z=HoyNoyNoy=(yb-ya)/ ym=11.6HoyGm/Kya 0.02/0.05 0.4mZ=11.6*0.4=4.64m5 Ammon ia is removed from ammo nia -air mixture by coun tercur

11、re nt scrubb ing with water in a packed tower at an atmospheric pressure. Given: the height of the packing 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 liq

12、uid out of the tower is 80% of its saturation concentration, and the equilibrium relation is y*=1.2x. 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

13、to 0.002 and the other operat ing con diti ons keep un cha nged, is the tower suitable?L0.030.003,solution : (1)1.35G0.80.03 1.21 2(2)S0.891.35N OY10.03In 0.110.896.2610.890.003NoyH OYZN OY66.260.958mIn0.89O1 .30.890.0028.47Since Z H oy N oy0.958 8.478.16.0m it is not suitable6 Pure water is used in

14、 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 absorber 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 mix

15、ture. 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, 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 lay

16、er 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 originalSolutio

17、n:Z 3m, p 1atm,T 293K0.061 0.060.063858028.2877018242.78kmol /(m h)mV 0.9 19.2842.780.4056Y2Y(1 0.99)0.000638The average molecular weight of the mixed gas M=290.94+17 0.06=28.28(10.06)19.28kmol/(m2 h)1)mSoOGNog2pPPmVNogKra1八2ln (-12)(1)1 mV mX2LL1Y mX2mX、 mV0.0638 0“ ln()(11 0.40560.0006386.8840.405

18、6)0.4056Nog蠢 04358mmp0.9p0438 0.202842.781-mV1 -L1YmX2m Xln(YT)(1mVIn (100)(1 0.2028) 0.2028 1 0.20285.496daPSo: H OG changes with the operating pressureH OG H OG1Hog Hog 0.43580.21792So Z Nog Hog 5.496 0.21791.198mSo the height of the packed secti on reduce 1.802m vs the origi nal2)L 2LmVmV 1mV1()-

19、0.40560.2028L2L 2L2Nog5.496whe n the mass flow rate of liquid in creases, KGa has not remarkable effectHog Hog 0.4358mZNog Hog 5.496 0.4358 2.395mthe height of the packed secti on reduce 0.605m aga inst the orig inal3) V 2VmVLNogm(2V)L2 0.40560.8116J0.8116ln (100)(10.8116)0.811615.81whe n mass flow

20、rate of gas in creaes, KGa also will in crease. Since it is gas filmcon trol for absorpti on, we have as follows:Kg3V.8Kg aH ogV 0.80.8心&(廠)2 Kg8V2VKg3P20.8KP0 220.4358 0.501m20.2HogZ Nog Hog 15.81 0.501m7.92mSo the height of the packed secti on in crease 4.92m aga inst the origi nalDistillatio n1 C

21、ertain binary mixed liquid containing mole fraction of easy volatilization component xF 0.35, feedi ng at bubbli ng poin t, is separated through a seque nee rectify colu mn. The mole fracti on in the overhead product is xd=0.96, and the mole fraction in the bottom product is x b =0.025. If the mole

22、overflow rates are con sta nt in the colu mn, try to calculate(a) the flow rate ratio of overhead product to feed(D / F)?(b) If the reflux ratio R=3.2, write the operating lines for rectifying and stripping sectionssolution :xf = 0.35; xb=0.025 ; xd=0.96; R=3.2。x x(1) D/ F = -FB = 0.3476 = 34.76%。Xd

23、Xb(2) the operati ng line for rectifyi ng sect ion.yXXd = 0.762 x + 0.229。R 1 R 1the operati ng line for stripp ing sect ionL=RD D=0.347FL=L+FB=F-DL FByxxB = 1.45 x 0.0112。L F B L F B2 A continuous distillation column is to be designed to separate an ideal binary material system ,The feed which cont

24、ains more volatile component xF 0.5, feed rate 100kmol/h, is saturated vapor, the flow rate of overhead product and the flow rate of bottom product are also 50kmol/h. Suppose the operati ng line for rectify ing sect ion is y=0.833x+0.15, the vapor gen erated in the reboiler en ters the colu mn throu

25、gh the bottom plate, a complete conden ser is used on the top of colu mn and reflux temperature is bubbli ng point. Find:(1) Mole fraction x d of overhead product and the mole fraction x b of bottom product ?(2) Vapor amount conden sed in the complete conden ser, i n mol/h?(3) The operat ing line fo

26、r stripp ing sect ion.(4) If the average relative volatility of the colu mn is 3 and the first plate Murphree efficie ncy from the tower top is E m,l=0.6, find the constituent of gas phase leaving the second plate from tower top.Soluti on:(1)the operat ing line for rectifyi ng sect ion isR 1y RlxXda

27、nd y=0.833x+0.15R 0.833 a nd亠=0.15,R 1R 1han dled by complete con de nser V = (R+1)D=300R = 5, xd= 0.9, xw=0.1(2) Amount of the condensate vapor(kmol/h )。(3) The operating line for stripping sectiony L qF xXw = 1.25 x 0.025 oL qF W L qF Wy1(4) the constituent of gas phase leaving from the second pla

28、te of tower top y2EML=X1 = 0.6, si nee x1 =y1= 址 =0.75Xd X1y1(1 y1)3 2y13 2xdSo X1 = 0.81LRy2 =y1 V( Xd x1)=0.9 Fl(0.90.81)= 0.82503 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, t

29、he feed rate is 1000kmol/h, and the flow rate of overhead product and the flow rate of bottom product are also 500kmol/h. Give n: the operat ing line for rectifyi ng sect ion is y=0.86x+0.12, in direct vapor is used in the reboiler for heat ing and the total conden ser is used on the top of tower. A

30、ssume that the reflux temperature is at its bubble point. Find:(1) reflux ratio R, the mole fraction of overhead product 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 stripp ing sect ion .(L mo

31、l/l),.=2.4, find R /Rmin(3) The operating line for stripping section .(4) if relative volatilitySoluti on:the mole fraction in overhead product x d and the mole fraction in bottom(1) reflux ratio R、 product xBXd = 0.86 x + 0.12,R1y xR 1 R 1R= 6.14, xd = 0.857, xw=0.143。(2) upward flow rate in the re

32、ctifying section V kmol/h and down ward flow rate in the stripping sect ion .(L kml/hL= L = RD=3070 kmol/h ,V=( R+ 1) D = 3570 kmol/hV=V-F=3570-1000-2570kmol/h(3) The operating line for stripping sectionLBy xxB = 1.19x 0.02VV B(4) reflux ratio and min reflux ratio R / RminXd1 yF1Xd1 = 1.7341 yF=3.54

33、 oRmin4 There is a continuous rectifying operation column, whose the operation line equation is as follows:Rectifyi ng sectio n: y=0.723x+0.263Stripping section: y=1.25x-0.0187if the feed enters the column at a dew point, find (a)the molar fraction of feed 、overhead product and bottom product. (b) r

34、eflux ratio R.soluti on:R=0.723, R=2.61R 1XdR 1=0.263,Xd =0.95 oSi nee y = xW = 1.25 xW -0.0187So Xw = 0.0748 othe feed enters the column at a dew point, q operation line equation as : y= xF . From q line and rectifying operation line, the following is solved.x=0.535, y=0.65 ,xF = 0.655 A column is

35、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 contain 95.7 mole perce nt A. give n: liquid average relative volatility =2.5, min reflux ratio Rmin =1.63, try to illustrate t

36、he thermal con diti on of the feed and to calculate the value of q.solution :Xq2.5XqFrom the equilibrium equation as : yq=1(1)Xq1 1.5XqSi nee Rmi n/(Rmi n+1)=Xd yqXd XqXdyqMin reflux ratio : RminyqXq0.957 yq=1.63yqXqq point:xq=0.365 , yq= 0.59xq=0.365 xF ,feed is a mixture of gas and liquid.q operat

37、ion line equation as y=qx/(q-1)- xF /(q-1) , let x=xq=0.365, y=yq= 0.59 we haveq= 0.667 o6 A continuous rectifying column operated at atmospheric pressure is used to separate benzene tolue ne mixed liquid. The feed is saturati on liquid containing 50 mole perce nt benzene. Given: the overhead produc

38、t must contain 90 mole percent benzene and the bottom product contains 10 mole perce nt benzen e. If reflux ratio is 4.52, try to calculate how many ideal plates areneeded? And determine the position of feed plate. In this situation the equilibrium data of benzene methylbe nzene are as followstoC80.

39、1859095100105110.6x1.0000.7800.5810.4110.2580.1300y1.0000.9000.7770.6320.4560.2620Solution: Based on M-T, we can obtain the theoretical plate numbers.Interception of operating line for rectifying sectionXdR 10.9=0.163。4.521N = 16。Feed plate is the third one from the colu mn top.7 There is a rectifyi

40、ng column, given : mole fraction of distillation liquid from tower top x d=0.97, reflux ratio R=2, the gas-liquid equilibrium relati on ship y=2.4x/(1+1.4x): find: the con stitue nt x 1 of the down liquid leaving from the first plate and the constituent y 2 of the up gas leaving from the sec ond pla

41、te in the rectifying secti on. Suppose the total conden ser is used at the top of colu mn.solution :2.4xFrom the equilibrium relati on y1 1.4xSince y1 = xd =0.97 , X1 = 0.93. Based on the operating line equation, we haveR Xd y = X1=0.94R 1 R 18. A liquid of benzene and tolue ne is con ti nu ously fe

42、d to a plate colu mn. Un der the total reflux ratio con diti on, the compositi ons of liquid on the close plates are 0.28, 0.41 and 0.57, respectively. Calculate the Murphree plate efficiency of two lower plates. The equilibrium data for benzene tolue ne liquid and vapor phases un der the operat ing

43、 con diti on are give n as follows:x0.260.380.51y0.450.600.72Solution : Under the total reflux ratio conditionyn 1xnFrom the Murphree efficie ncy:y3y2x20.41x10.57*丫20.628from X20.丫2丫30.570.41*丫2丫30.628 0.41丫3丫40.410.28丫3* 丫40.475 0.28CalculateEm,vEm2E m3yn yn 1yn y(n 1Y30.475 from X3=0.28 by interce

44、ption.0.7373%67%Dryi ng1 A wet solid with 1000 kg/h is dried by air from 40% to 5% moisture conten t (wet basis) un der the con vective drying con diti ons. The air primary humidity H 1 is 0.001(kg water/kg dry air), and the humidity of air leavi ng dryer H 2 is 0.039 (kg water /kg dry air), suppose

45、 that the material loss in the drying process can be n egligible. Find:(1) rate of water vaporizati on W, in kg water/ h.(2) rate of dry air L required, in kg bone dry air/h, flow rate of moist air, L,i n kg fresh air/h.(3) rate of moist material out of dryer, G 2, in kg moist solid/h.Soluti on :G1=

46、1000kg/h, w 1=40%, w 2=5% , H 1=0.001, H 2=0.039Gc =G1(1-w 1)=1000(1-0.4)=600kg/hX1=0.4/0.6=0.67,X2=5/95=0.053(1) W=Gc(x 1-X2)=600(0.67-0.053)=368.6kg/h(2) L (H1-H2)=W0.039 0.0019700 kg dry air/hL = L(1+H)=12286.7(1+0.039)=12765.8kg fresh air/h Gc =G2(1-w 2)G2Gc1 W26001 0.05631.6kg/h2 The wet solid

47、is to be dried from water content 20% to 5% (wet basis) in a con vective dryer at an atmospheric pressure. The feed of wet solid into the dryer is 1000 kg/h at a temperature of 40 C . The dry and wet bulb temperatures of air are respectively 20 C and 16.5 C before air enters the preheater. After bei

48、ng preheated, air enters the dryer. The dry and wet bulb temperatures of air leaving dryer are respectively 60 C and 40C . If heat loss is negligible and drying is considered as con sta nt en thalpy process:(1) What is the fresh (wet) air required per un it time (in kg/h)?(2) The air temperature t1

49、before en teri ng the dryer(Give n: Vaporization late nt heat of water at 0 C is 2501kJ/kg, specific heat of dry air C g is 1.01 kJ/kg K；specific heat of water vapor C v is 1.88 kJ/kg K)Soluti on:w1=0.2, w 2=0.05, G1= 1000kg/h,1=4C , t0=20 C , tw0=16.5 C , t2=60 C , tw2=40CQ=1.01L(t2-t0)+W(2501+1.88

50、t2)+Gc - Cm 2-時(shí))+QlFor con sta nt en thalpy process, 11=12From the Fig, we have H0=0.01(based on t0=20 C , tw0=16.5 C), H2=0.045(based on t2=60 C , tw2=40C)l1=(1.01+1.88H 0)t1+25010H0,12=(1.01+1.88H 2)t2+2501H 2=(1.01+1.88 0.045) 60+2501 0.045= 177.7I1= (1.01 + 1.88 0.01)t1+2501 .01=1.03t1+24.9= 12=

51、177.7t1177.724.91.03148.4 CGc=G1(1-w 1)=1000(1-0.2)=800xi=0.2/0.8=0.25,X2=5/95=O.O53W=Gc(xi-X2)=800(0.25-0.053)=157.6Ho=HlWH2 H!4502.9 kg/h0.045 0.01L =L(1+H)=4502.9(1+0.01)=4547.9 kg/h3 The wet solid material is to be dried from water content 42% to 4% (wet basis) in an adiabatic dryer. The solid p

52、roduct out of the dryer is 0.126kg/s. After the fresh air at a dry-bulb temperature of 21oC and a percentage humidity of 40 % is preheated to 93oC, it is sent to the dryer, and leaves the dryer at percentage humidity of 60 % . If the drying is under constant enthalpy process.(1) Determining the air

53、humidity (H 1 and H2) from the given air state in t-H diagram.(2) If H o=O.OO8(kg water/kg dry air), H 2=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 supplied to air by the preheater (in k J/h)?Soluti on:(1) w1=0.42,w 2=0.04, G2=0.126kg/s, t

54、 o=21,Hp=O.4,11=93, Hp=0.6, I 1=12From t-H diagram, we have H o=0.008, H2=0.03(2) G2(1-W2)=G1(1-W1)= GcG21w21 w10.1261 0.041 0.420.209 W=G1- G2= 0.209-0.126=0.0826Or W= Gc (X 1-X2)W0.0826LH2 H13.752kg/s0.03 0.008Qp=L(I 1-lo)=L(1.01 + 1.88H 1)(t1-to)=3.752(1.01+1.88 0.008)(93-21)=301.2kJ/s= 1.08 10Th

55、e wet solid containing 12%(wet basis) moisture is fed to a convective dryer at a temperature of 15 C and is withdrawn at 28 C , which contains 3% moisture (wet basis). The flow rate of final moist solid (product) is 1000kg/h. After the fresh air at a dry-bulb temperature of 25 C and a humidity of 0.01 kg water/kg dry air is preheated to 70 C , it is sent to the dryer, and leaves the dryer at 45oC. Suppose the drying process is un der con sta nt en thalpy, heat loss in the drying system can be n egligible.(1) Drawi ng the operati