传质与分离习题含答案

1、Problems for Mass Transfer and Separation ProcessAbsorption1 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 ph

2、ases are contact, 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 operation.0_2 When the temperature is 10 c and the overall pressure is 101.3KPa , the solubility of oxygen in water can be represented by equation p=3.27 104x , w

3、here 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:

4、the mole fraction of oxygen in air is 0.21,hence:p = P y =1x0.21=0.21amtSosolubilityx 3.27*1040.213.27*1046.24*10 6Because the x is very small , it can be approximately equal to molar ratio X , that isX x 6.42*10 66.42*10 6(kmolO2)*32(kgO2/kmolO2)1(kmolH 2O) *18(kgH 2O/ kmolH 2O)11.4g(O2)/m3(H2。)3 A

5、n 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 oper

6、ating condition, the equilibrium relationship is y*=1.75x. V olume total absorption coefficient is K ya=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: yayb 10.0002xa=0minyb ya 0.02 0.991.733xbxa0.02 1.751.42.43 L 2.43 0.03 0

7、.0729 kmol. m2 sminmV 1.750.720L 2.43Number of mass transfer units N oy=(y i-y2)/ y=12(yb-ya)=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)= ya=0.0002Or Ney-ln(yb mxa)(1 S) S =121 SyamxbV/S 0.03Hoy 1.36

8、4mKya 0.022H Hoy Noy 1.364 12 16.37m4 The mixed gas from an oil distillation tower contains H 2s=0.04(molar fraction). Triethanolamine (absorbent) is used as the solvent to absorb 99% H 2s in the packing tower, the equilibrium relationship is y*=1.95x, the molar flux rate of the mixed gas is 0.02kmo

9、l m-2 s-1, overall volume absorption coefficient is Kya=0.05 kmol - s 1 m-2y-1, The solvent free of H 2s enters the tower and it contains 70% of the H 2s saturation concentration when leaving the tower. Try to calculate: (a) the number of mass transfer units N oy, and (b) the height of packing layer

10、 needed, 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 Ammonia is removed from ammonia-air mixture by countercurrent scrubbing with water i

11、n apacked 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 NH 3 concentration of liquid out of the tower is 80% of i

12、ts 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 transfer units.(3) if the molar fraction of the ammonia out of the tower will be reduced to 0.002 and the other operating co

13、nditions keep unchanged, is the tower suitable?0.03 0.003/ cl1.350.8 0.03 1.2Lsolution ： (1)G-1 2(2) S 0.891.3510.03NoyIn 0.110.896.261 0.890.003H OYN OY66.260.958mN OY-In0.11 0.031 0.890.0020.898.47Since Z' Hoy Noy0.958 8.47 8.1 6.0m it is not suitable6 Pure water is used in an absorption tower

14、 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 20 ° c, respectively. The flux of gas V is 580kg/(m 2 .h), and 6 percent (volume %) of ammonia is contained in the gas mixture. The f

15、lux of water L is 770kg/( m 2 .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 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 needed to k

16、eep 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 originalSolution: Z 3m, p 1atm,T 293K0

17、.061 0.060.0638Y2 Y(1 0.99) 0.000638The average molecular weight of the mixed gas M=29 0.94+17 0.06=28.285802(1 0.06) 19.28kmol/(m2 h)28.28L 77018_242.78kmol/(m h)mV 0.9 19.2842.780.4056Nog1v'n(, mV1 丫 mX2Y2 mX2)(1 mX)mVH OG1) p 2 Pm' pm p,mVSoNogOGKra1 0.40566.884ln(0.0638 00.000638)(10.405

18、6) 0.4056mpNog0.93 0.4358m6.8842 0.450.45 19.28 0.202842.781Y mX2mXln(-2)(1)1 mVY， mX2 LmV v1 0.20285.496ln(100)(1 0.2028) 0.2028KGaPSo: H OG changes with the operating pressureH OG _H OG 1Hog Hog 0.43580.21792So ZNog Hog 5.496 0.2179 1.198mSo the height of the packed section reduce 1.802m vs the or

19、iginal2) L2L10.4056 0.20282mVmV1mV(L2L2LNog5.496when the mass flow rate of liquid increases, KGa has not remarkable effectHog Hog 0.4358mZ Nog Hog 5.496 0.4358 2.395mthe height of the packed section reduce 0.605m against the originalmVNog3) V 2Vm(2V) 2(mV) 2 0.4056 0.8116 L L1ln(100)(1 0.8116) 0.811

20、6 15.811 0.8116when mass flow rate of gas increaes, KGa also will increase. Since it is gas filmcontrol for absorption, we have as follows:Kg3V°.8Kg aH ogV、0.8KGa(V)V20.8Kg3Kg3P2V20.8KGaP20.2Hog_0 2_ _20.4358 0.501mZ Nog Hog 15.81 0.501m 7.92mSo the height of the packed section increase 4.92m a

21、gainst the originalDistillation1 Certain binary mixed liquid containing mole fraction of easy volatilization componentxF 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 i

22、s x b =0.025. If the mole overflow rates are constant in the column, 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 sections solution :xf =0.35; xb=0.025; xd=0.96; R=3.2。x x(1) d/ F= -= 0.

23、3476 = 34.76% oxdxb(2) the operating line for rectifying section.V X Xd =0.762 X+0.229。R 1 R 1the operating line for stripping sectionL=RD D=0.347FL'=L+FB=F-DL FBy x xB = 1.45X 0.0112。L F B L F B2 A continuous distillation column is to be designed to separate an ideal binary material system ,The

24、 feed which contains 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 operating line for rectifying section is y=0.833x+0.15, the vapor generated in the reboiler enters the col

25、umn through the bottom plate, a complete condenser is used on the top of column and reflux temperature is bubbling point. Find:(1) Mole fraction x d of overhead product and the mole fraction x b of bottom product ?(2) Vapor amount condensed in the complete condenser, in mol/h?(3)The operating line f

26、or stripping section.(4) If the average relative volatility of the column is 3 and the first plate 's Murphree efficiency from the tower top is E m,L=0.6, find the constituent of gas phase leaving the second plate from tower top.Xdhandled by complete condenser V = ( R +1)D=300Solution: (1)the op

27、erating line for rectifying section isand y=0.833x+0.15RXd 0.833 and =0.15,R 1R 1R = 5, xd=0.9, xw=0.1(2) Amount of the condensate vapor (kmol/h )。(3) The operating line for stripping sectionL qF W L qF WXw = 1.25 X 0.025。(4) the constituent of gas phase leaving from the second plate of tower top y2

28、EML= XdXl =0.6, since x1 =y1y1y1(1 y1)3 2y13 2xdXd =0.75So x1= 0.81LRy2=y1 一( xdX1) = 0.9 (0.90.81) = 0.825。VR 13 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 100

29、0kmol/h, and the flow rate of overhead product and the flow rate of bottom product are also 500kmol/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 temp

30、erature 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 stripping section.(L ' mol/h,.(3) The operating

31、line for stripping section .(4)if relative volatilitySolution:(1) reflux ratio R、=2.4, find R /Rminthe mole fraction in overhead product x d and the mole fraction in bottomproduct xBRXD = 0.86 x + 0.12,R= 6.14, xd= 0.857, xw=0.143。(2) upward flow rate in the rectifying section V kmol/h and down ward

32、 flow rate in the stripping section.(L kmbl/hL'=L=RD=3070 kmol/h , V= (R+1) D = 3570 kmol/h V =V-F=3570-1000-2570kmol/h(3) The operating line for stripping sectionL'By x xB = 1.19 x 0.02V'V'reflux ratio and min reflux ratio R / Rmin1xd1 yF1-xD1 = 1.7341 yF= 3.54。4 There is a continuo

33、us rectifying operation column, whose the operation line equation is as follows:Rectifying section: 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) reflux ratio R.solution:R_= 0

34、.723, R=2.61R 1xD = 0.263,xD=0.95。R 1Since y = xW = 1.25 xW-0.0187So xW =0.0748。the feed enters the column at a dew point, q operation line equation as : y= xF . From q line andrectifying operation line, the following is solved.x=0.535, y=0.65 ,xF =0.655 A column is to be designed to separate a liqu

35、id 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 percent A. given: liquid average relative volatility =2.5, min reflux ratio Rmin =1.63, try to illustrate the thermal condition of the feed an

36、d to calculate the value of q. solution :1 (1)xq1 1可Since Rmin/(Rmin+1)=xdyqXdxqMin reflux ratio : RminXdyq0.957 yq=1.63yqxqyqxqFrom the equilibrium equation as : yq=q point:xq=0.365 , yq= 0.59xq=0.365< xF = 0.44, yq= 0.59> xF , feed is a mixture of gas and liquid.q operation line equation as

37、y=qx/(q-1)- xF /(q-1) , let x=xq=0.365, y=yq= 0.59 we haveq= 0.667。6 A continuous rectifying column operated at atmospheric pressure is used to separate benzene toluene mixed liquid. The feed is saturation liquid containing 50 mole percent benzene. Given: the overhead product must contain 90 mole pe

38、rcent benzene and the bottom product contains 10 mole percent benzene. 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 methylbenzene are as followstoC80.1859095100105110.6x1.0000.78

39、00.5810.4110.2580.1300y1.0000.9000.7770.6320.4560.2620Solution: Based on M-T, we can obtain the theoretical plate numbers.Interception of operating line for rectifying sectionXD0.9=0.163。R 14.52 1N= 16。Feed plate is the third one from the column top.7 There is a rectifying column, given : mole fract

40、ion of distillation liquid from tower top x d=0.97, reflux ratio R=2, the gas-liquid equilibrium relationship y=2.4x/(1+1.4x); find: the constituent x 1 of the down liquid leaving from the first plate and the constituent y 2 of the up gas leaving from the second plate in the rectifying section. Supp

41、ose the total condenser is used at the top of column. solution :2.4xFrom the equilibrium relation y ,1 1.4xSince y1 = xd =0.97 , x1 = 0.93. Based on the operating line equation, we haveR xdy2 =x1=0.94R 1 R 18. A liquid of benzene and toluene is continuously fed to a plate column. Under the total ref

42、lux 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 two lower plates. The equilibrium data for benzene toluene liquid and vapor phases under the operating condition are given as follows:x0.260.380.51y0.

43、450.600.72Solution : Under the total reflux ratio conditiony n 1xnFrom the Murphree efficiency:Em,vY3y2Yn y(n 1 *ynyn 1x20.41x1 0.57*Calculate Y20.628*from X20.41, Y30.475 from X3=0.28 by interception.Em2丫2丫3*丫2丫30.57 0.410.628 0.410.73 73%Em3丫3 丫4丫3* 丫40.41 0.280.475 0.2867%Drying1 A wet solid with

44、 1000 kg/h is dried by air from 40% to 5% moisture content (wet basis) under the convective drying conditions. The air primary humidity H 1 is 0.001(kg water/kg dry air), and the humidity of air leaving dryer H 2 is 0.039 (kg water /kg dry air), suppose that the material loss in the drying process c

45、an 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 rate of moist air, L', in kg fresh air/h.(3) rate of moist material out of dryer, G 2, in kg moist solid/h.Solution :G1=1000kg/h, w 1=40%, w 2=5% , H 1=0.001, H 2=0.

46、039Gc =G1(1-w 1)=1000(1-0.4)=600kg/hxi=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)=WW368.6L 9700 kg dry air/hH1 H20.039 0.001L' = L(1+H)=12286.7(1+0.039)=12765.8kg fresh air/h(4) Gc =G2(1-w 2)- Gc 600G2c631.6kg/h1 w21 0.052 The wet solid is to be dried from wa

47、ter 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 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 being preheated, air enter

48、s the dryer. The dry and wet bulb temperatures of air leaving dryer are respectively 60 C and 40 c . If heat loss is negligible and drying is considered as constant 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(

49、Given: Vaporization latent 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) Solution:wi=0.2, w 2=0.05, G1=1000kg/h, 1=40TC , t0=20 C , tw0=16.5 C , t2=60 C , tw2=40 CQ=1.01L(t 2-t0)+W(2501+1.88t 2)+Gc - Cm 2-=)+QlFor c

50、onstant enthalpy process, I i=I 2From the Fig, we have H0=0.01(based on t0=20 C , tw0=16.5 ), H2=0.045(based on t2=60 C , tw2=40C)Ii=(1.01+1.88H 0)t1+25010H0,I2=(1.01+1.88H 2)t2+2501H 2=(1.01+1.88 0>045) 60+2501 >0.045= 177.7Ii= (1.01 + 1.88 0.01)t1+2501 >0.01=1.03t1+24.9= 12=177.7t1177.7 2

51、4.91.03148.4 CGc=G1(1-w 1)=1000(1-0.2)=800xi=0.2/0.8=0.25,x2=5/95=0.053W=Gc(xi-X2)=800(0.25-0.053)=157.6''' H0=HiWH2 Hi157.60.045 0.014502.9 kg/hL' =L(1+H)=4502.9(1+0.01)=4547.9 kg/h 3 The wet solid material is to be dried from water content 42% to 4% (wet basis) in an adiabatic drye

52、r. The solid product 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) Determ

53、ining the air humidity (H i and H2) from the given air state in t-H diagram.(2) If H 0=0.008(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)?Solution:(1) wi=0.42,w2=0.04, G2=0

54、.126kg/s, 10=21,Hp=0.4, 11=93, Hp=0.6, I i=l2From t-H diagram, we have H 0=0.008, H2=0.03(2) G2(1-w2)=G1(1-wi)= GcGiG21 w2i wi0.1261 0.041 0.420.209. W=G1- G2= 0.209-0.126=0.0826Or W= Gc (X 1-X2)H2 Hi0.08260.03 0.0083.752kg/sQp=L(I i-l0)=L(1.01 + 1.88H i)(ti-t0)=3.752(1.01+1.88 0.008)(93-21) =30i.2k

55、J/s= i.08 106kJ/h 4 The 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

56、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 under constant enthalpy, heat loss in the drying system can be negligible.(1) 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, wi=12%, w2=3%,0i=15, 2=28,t0=25 C, H0=0.01, ti=70, t2=45, I i=I2 Gc=1000(1-0.12)=880,x1=12/88=0.136,x2=3/97