problems-and-solutions.pdf
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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
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