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1、.Chapter 6 6.1.Carbon dioxide is diffusing through nitrogen in one direction at atmospheric pressure and 0C.The mole fraction of CO2 at point A is 0.2;at point B,3 m away,in the direction of diffusion,it is 0.02.Diffusivity D is 0.144 cm2/s.The gas phase as a whole is stationary;that is,nitrogen is
2、diffusing at the same rate as the carbon dioxide,but in the opposite direction.(a)What is the molal flux of CO2,in kilogram moles per square meter per hour?(b)What is the net mass flux,in kilograms per square meter per hour?(c)At what speed,in meters per second,would an observer have to move from on
3、e point to the other so that the net mass flux,relative to him or her,would be zero?(d)At what speed would the observer have to move so that,relative to him or her,the nitrogen is stationary?(e)What would be the molal flux of carbon dioxide relative to the observer under condition(d)?solution:(a)fro
4、m(6.1-8)27308206.01RTPcccMBA from equation(6.1-19)hmkmolyyzDcJNAAiMAA244/10388.102.02.027308206.03360010144.0(b)net mass flux for carbon dioxide(molecular weight=44)mass flux of CO2=441.38810-4 kg/m2h for nitrogen(molecular weight=28)mass flux of N2=281.38810-4 kg/m2h so the net mass flux in the dir
5、ection of CO2 diffusion m=(44-28)1.38810-4=2.22110-3 kg/m2h (c)Here JA=NA=NB,since the diffusion is equimolal.The concentration at any point depends on position due to the concentration profile of the equimolal diffusion,so does velocity based on equations(6.1-3a)and(6.1-3b).To select two points,yA=
6、0.2 and 0.02,respectively,to calculate the positions of observer for yA=0.2,CA=Cm yA=00893.027308206.02.0 from equation(6.1-3a)410388.1AAAAuCNJ the diffusing velocity of A:01555.000893.010388.14Au for B:CB=Cm(1-yA)=03571.027308206.02.01 the diffusing velocity of B:003887.003571.010388.14BABCJu let u
7、o is the velocity of the observer moving in the direction of CO2 diffusion,then net velocity(uA-uo)gives a mass transfer rate mA equal to that in opposite direction mB corresponding to(uB+uo)from equations(6.1-4)and(6.1-5)(44oAAAAAuuCJMm.)(28oBBABBuuCJMm for mA=mB,and rearranging two equations above
8、 gives hmCCNCCCuuCuBAABABBAAo/00159.099988.039292.00022208.003571.02800893.0441610388.12844)2844(284428444 at yA=0.2,It is similar to calculating uo at the point of yA=0.02 (d)When the velocity of observer moving is equal to that of nitrogen diffusing,the nitrogen is stationary uo=m/h003887.003571.0
9、10388.14BABCJu(e)When the velocity of observer moving is equal to that of nitrogen diffusing,the molal flux of carbon dioxide diffusing is indicated by hkmoluuCuuCJAAoAAA/10736.1)003887.001555.0(00893.0)()(4B Chapter7 7.1 solution:The data from third column in the table are used as calculating Mole
10、fraction x of ammonia in water:01048.018100170.1170.1x molar ratio of ammonia to water X in liquid.01059.001048.0101048.01xxX molar ratio of ammonia to inert gas 00796.08.03.1018.0pPpY the results of calculation are list in the table p/kPa 0 0.4 0.8 1.2 1.6 2.0 2.43 3.32 4.23 6.67 9.28 x 0 0.00527 0
11、.01048 0.01563 0.02074 0.0258 0.03079 0.04063 0.05028 0.07357 0.09574 X 0 0.00530 0.01059 0.01588 0.02118 0.0265 0.03177 0.04235 0.05294 0.07940 0.1059 Y 0 0.00396 0.00796 0.01198 0.01600 0.0201 0.02453 0.03387 0.04353 0.07040 0.1008 The data in the table are used to plot the mole fraction x versus
12、partial pressure p diagram and molar ratio X-Y diagram 0.000.020.040.060.080.100246810p/kPax/mole fraction B 虚线范围表示符合 Herrys law.0.000.020.040.060.080.100.120.000.020.040.060.080.10Y/molar ratioX/molar ratio B 7.3 Vapor-pressure data for a mixture of pentane(C5H12)and hexane(C6H14)are given by the t
13、able.Calculate the vapor and liquid composition in equilibrium for the pentane-hexane at 13.3kpa pressure on assuming that the vapor of mixture approaches ideal behavior and liquid follows Raoults low.t,K 260.6 265 270 275 280 285 289 PA,kPa 13.3 17.3 21.9 26.5 34.5 42.5 48.9 PB,kPa 2.83 3.5 4.26 5.
14、0 8.53 11.2 13.3 Solution:From Raoults low（equations(7.1-4)and(7.1-5)）And the total pressure of system is equal to sum of the partial pressures of two substances)1(ABAABBAABAxPxPxPxPppP rearranging equation above BABAPPPPx 1 and AAAAxPpyp rearranging equation above gives PxPyAAA 2 substituting the d
15、ata for the table into the equations 1 and 2 gives the results in the following table t,K 260.6 265 270 275 280 285 289 PA,kPa 13.3 17.3 21.9 26.5 34.5 42.5 48.9 PB,kPa,2.83 3.5 4.26 5.0 8.53 11.2 13.3 x 1.0 0.710 0.513 0.386 0.184 0.067 0 y 1.0 0.924 0.845 0.769 0.477 0.214 0 虚线范围表示符合 Herrys law.7.
16、4 Using the conditions in the problem 7.3 for the pentane-hexane mixture,do as follows:(a)Calculate the relative volatility.(b)Use the relative volatility to calculate the vapor and liquid composition in equilibrium and compare with the result given by the problem 7.3.Solution:(a)calculate the value
17、 of by BAPP The results are given in the table:t,K 260.6 265 270 275 280 285 289 PA,kPa 13.3 17.3 21.9 26.5 34.5 42.5 48.9 PB,kPa,2.83 3.5 4.26 5.0 8.53 11.2 13.3 4.70 4.94 5.14 5.3 4.04 3.79 3.68 Average relative volatility 1(4.704.945.145.304.043.793.68)4.517m and equilibrium relation with relativ
18、e volatility for a mixture of pentane(C5H12)and hexane(C6H14)is expressed by xxxxy51.3151.4)1(1(b)Using the equation above to calculate the vapor and liquid composition in equilibrium and compare with the result given by the problem 7.3.T,K x y The results of problem 7.3 The calculating result by us
19、ingmx 260.6 265 270 275 280 285 289 4.51 1.0 0.710 0.513 0.386 0.184 0.067 0 1.0 0.924 0.845 0.769 0.477 0.214 0 1.0 0.917 0.826 0.739 0.504 0.245 0 7.5 Boiling Point and Raoults Law.For the system benzenetoluene,do as follows,using the data from Table 7.1-1:At 378.2 K,calculate yA and xA using Raou
20、lts law at total pressure of 101.325kPa.If a mixture has a composition of xA=0.40 and is at 358.2 K and 101.32 kPa pressure,will it boil?If not,at what temperature will it boil and what will be the composition of the vapor first coming off?solution:At 378.2K from Table7.1-1 for benzene,vapor pressur
21、e PA=204.2kPa,for toluene,vapor pressure PB=86.0kPa,substituting these data into Eq(7.1-11)and solving for xA(mole fraction of benzene).325.101)1(802.204AAxx 1717.02.124325.21802.20480325.101Ax substituting into Eq(7.1-12)346.01717.0325.1012.204AAAxPPy From figure7.1-2,if a mixture has a composition
22、 of xA=0.40 and is at 358.2 K and 101.32 kPa pressure,it will not boil,and will boil at temperature of 95oC(368.2K);At 95oC(368.2K),from Table 7.1-1 for benzene,vapor pressure PA=155.7kPa,and substituting into Eq(7.1-12)615.040.0325.1017.155AAAxPPy 7.7 What are the effects on the concentrations of t
23、he exit gas and liquid streams of the following changes in the operating conditions of the column of Example 7.5?(a)A drop in the operating temperature that changes the equilibrium relationship to y=0.6x.Unchanged from the original design:N,L/V,yb,and xa.(b)A reduction in the L/V ratio from 1.5 to 1
24、.25.Unchanged from original design:temperature,N,yb,and xa.(c)An increase in the number of ideal stages from 5.02 to 8.Unchanged from original design:temperature,L/V,yb,and xa.solution:(a)For a dilute solution and a dilute gas,L and V are assumed constant,and the stripping factor is LmVAS1=0.6 1.5=0
25、.9 All concentrations can be expressed in terms of xa,the mole fraction of NH3 in the entering solution:0*bx since yb=0 From an ammonia balance,V y=V ya=L(xa-xb)=Lxa.Hence aaxVLy also 9.0*aaaaaxLmVxmVxLmyx From Eq.(7.2-28),02.59.0ln)1(9.0lnln0lnaaabaaxxxSxxxN The separation corresponds to 5.02 ideal
26、 stages,so the stage efficiency is 5.02/7=72 percent.02.59.0ln)1(9.01ln.53.09.0ln02.5)1(9.01ln 589.0)1(9.0153.0e 0.787 V ya=L(xa-xb)=Lxa.decreasing in aaxVLy=1.50.787xa with decreasing in the recovery (b)LmVAS1=0.81.25=1*babbbayyyyyyN 1 abbxmmxy)1(*2 abaaxLxxLVy)(so aaxVLy 3 substituting equations(2
27、)and(3)into equation(1)gives 1)1(*aabaxmxVLyyN=5.02 solving for =0.834 (c)From an ammonia balance,V ya=L x=Lxa.Hence aaaxxVLy5.1)(also aaaaxxmyx833.05.18.0*From Eq.(7.2-28),82.1ln1833.01lnln0)1833.0lnSxxxNaaa（.182.082.1ln81833.01ln=1.456 289.41833.01 solving for=0.95 7.10.A toxic hydrocarbon is stri
28、pped from water with air in a column with eight ideal stages.(a)What stripping factor is needed for 98 percent removal?(b)What percentage removal could be achieved with a stripping factor of 2.0?solution:(a)from equation(7.2-28)NxxxNxxxSaaabaa)1(lnlnln*1 From an ammonia balance,V y=V ya=L(xa-xb)=Lxa
29、.Hence aaxVLy also SxmVxLmyxaaaa*2 substituting equation(2)to(1)gives 898.0198.01ln11ln)1(ln)1(lnln*SNSNxSxxNxxxSaaaaaa the method of trial-and-error is used to solving for stripping factor S (b)for a stripping factor of 2.0,percentage removal could be calculated by NSS11lnln NSln121ln 25621218NS so
30、lving for=0.998 .Chapter 8 8.1 The gas stream from a chemical reactor contains 25 mol%ammonia and the rest inert gases.The total flow is 181.4 kmol/h to an absorption tower at 303 K and 1.013105 Pa pressure,where water containing 0.005 mol frac ammonia is the scrubbing liquid.The outlet gas concentr
31、ation is to be 2.0 mol%ammonia.What is the minimum flow minL?Using 1.5 times the minimum,plot the equilibrium and operating lines.Solution:005.00204.09823333.07525aabXYY hkmolV/05.136100754.181 from Fig9-2 in Chemical Engineeringin Chinese:E=1.85105Pa，83.1013.185.1PEm mass balance on component ammon
32、ia)()(ababXXLYYV 33.240177.057.42005.083.13333.00204.03333.005.136minababXmYYYVLkmol/h 5.36033.2405.1L kmol/h 8.2 A gas stream contains 4.0 mol%NH3 and its ammonia content is reduced to 0.5 mol%in a packed absorption tower.at 293 K and 1.013105 Pa.The inlet pure water flow is 68.0 k mol/h and the to
33、tal inlet gas flow is 57.8 kmol/h.The tower diameter is 0.747 m.The film mass-transfer coefficients are kya=0.0739 kmol/s m3mol frac and kxa=0.169 k mol/s m3mol frac.Using the design methods for dilute gas mixtures,calculate the tower height solution:V=57.8 kmol air/h,yb=0.04 mol frac NH3,ya=0.005,L
34、=68.0 kmol water/h,xa=0.kya=0.0739 kmol/s m3mol frac and kxa=0.169 kmol/s m3mol frac.005.0005.01005.00417.004.0104.0abYY equilibrium relation Y=1.83X Mass balance on ammonia)0()(babXLYYV 00312.0)005.00417.0(688.57)(abbYYLVX overall mass transfer coefficient 36.2483.1053.13169.083.10739.0111akmakaKxy
35、y 225864.0747.0414.34mdS height of transfer unit is maSKVHyoy667.05864.0360036.248.57 number of transfer unit mmaboyYYYYN005.00417.0 0157.0974.1031.0005.0036.0ln031.0005.000312.083.10417.0ln005.000312.083.10417.00ln)0(abbabbmYmXYYmXYY338.20157.0005.00417.0maboyYYYN the height of packing Z=HoyNoy=0.6
36、672.338=1.559m 0774.3ln8.15555.134.85555.11ln5555.111688.5783.1005.00417.0688.5783.11ln688.5783.111001ln11LVmYYLVmLVmNaboy.8.3 In a tower 0.254 m in diameter absorbing acetone from air at 293 K and 101.32 kPa using pure water,the following experimental data were obtained.Height of 25.4-mm Raschig ri
37、ngs=4.88 m,V=3.30 kmol air/h,yb=0.01053 mol frac acetone,ya=0.00072,L=9.03 kmol water/h,xb=0.00363 mol frac acetone.Calculate the experimental value of Kya.solution:00364.000363.0100363.000072.0005.01005.00106.001053.0101053.0babXYY height of packing:Z=4.88m cross sectional area S 2219939.0254.0414.
38、34mdS equilibrium relation Y=2.0X Number of transfer unit mmaboyYYYYN00072.00106.0 0017.0528.10026.000072.000332.0ln0026.000072.000364.00.20106.0ln00072.000364.00.20106.00ln)0(abbabbmYmXYYmXYY812.50017.000072.00106.0maboyYYYN overall mass transfer coefficient 00548.0812.51994.088.43600/3.3oyyNZSVaK
39、kmol/s m3mol frac 75.569.4ln717.3731.0222.22731.01ln731.01103.93.3200072.00106.003.93.321ln03.93.3211001ln11LVmYYLVmLVmNaboy 9.1 A mixture of 100 mol containing 60 mol%n-pentane and 40 mol%n-heptane is vaporized at 101.32 kPa abs pressure until 40 mol of vapor and 60 mol of liquid in equilibrium wit
40、h each other are produced.This occurs in a single-stage system,and the vapor and liquid are kept in contact with each other until vaporization is complete.The equilibrium data are given in Example 9.2.Calculate the composition of the vapor and the liquid.Solution:The equilibrium data are as follows,
41、where x and y are mole fractions of n-pentane:x 1.000 0.867 0.594 0.398 0.254 0.145 0.059 0.y 1.000 0.984 0.925 0.836 0.701 0.521 0.271 0 The given values to be used in Eq.(9.1-2)are F=100 mol,xF=0.60,L=60 mol,and V(moles distilled)=40 mol.Substituting into Eq.(9.1-2),BDFxFVFyFVx so BDxy10060100406.
42、0 1 solving for yD and xB using the Figure and equation(1),q=0.6 for 40 percent of feed to flash.)1(234.06.04.06.011xxqxxqqyf to locate the point where the q line intersects with equilibrium line at yD=0.85,xB=0.43 0.00.20.40.60.81.00.00.20.40.60.81.0composition of n-pentane/mole fraction compositio
43、n of n-pentane/mole fraction 9.2 A binary mixture which contains 60 mole percent more volatile component is subjected to flash distillation and differential distillation at a pressure of 1 atm,respectively.The mole fraction of the feed f that is vaporized is 1/3.what are the compositions of the over
44、head product and the bottoms product.Assume that the equilibrium relation is given by equation y=0.46x+0.549.solution:for flash distillation BDFxFVFyFVx so BDxy32316.0 1 From equilibrium line yD=0.46xB+0.549.2.solving for yD and xB by combining the equations 1 and 2 yD=0.783 and xB=0.508 for differe
45、ntial distillation from equation 9.2-4 121221lnxxLLxydxLLLdL substituting the equilibrium relation into the equation above 6.06.021221254.0549.0549.046.0lnxxxxxdxxxdxxydxLL so 26.02154.0549.06.054.0549.0ln54.0154.0549.0405.03/21lnln2xxdxLLx x2=0.498 from equation 9.2-5 avyLLxLxL)(212211 avy)321(498.
46、0326.0 yav=0.804 9.3 Comparison of Differential and Flash Distillation.A mixture of 100 kg mol which contains 60 mol%n-pentane(A)and 40 mol%n-heptane(B)is vaporized at 101.32 kPa pressure under differential conditions until 40 kg mol are distilled.Use equilibrium data from Example 9.2.What is the av
47、erage composition of the total vapor distilled and the composition of the remaining liquid?If this same vaporization is done in an equilibrium or flash distillation and 40 kg mol are distilled,what is the composition of the vapor distilled and of the remaining liquid?Solution:The equilibrium data ar
48、e as follows,where x and y are mole fractions of n-pentane:.x 1.000 0.867 0.594 0.398 0.254 0.145 0.059 0 y 1.000 0.984 0.925 0.836 0.701 0.521 0.271 0 The given values to be used in Eq.(9.2-4)are L1=100 mol,x1=0.60,L2=60 mol,and V(moles distilled)=40 mol.Substituting into Eq.(9.2-4),6.02510.060100l
49、nxxydx The unknown is x2,the composition of the liquid L2 at the end of the differential distillation.To solve this by numerical integration,equilibrium values of y versus x are plotted so values of y can be obtained from this curve at small intervals of x.Alternatively,instead of plotting,the equil
50、ibrium data can be fitted to a polynomial function.For x=0.594,the equilibrium value of y=0.925.Then f(x)=1/(y-x)=1/(0.925-0.594)=3.02.Other values of f(y)are similarly calculated.The numerical integration of Eq.(9.2-4)is performed from x1=0.6 to x2 such that the integral=0.510 in figure.Hence,x2=0.