problems-and-solutions2.docx
《problems-and-solutions2.docx》由会员分享,可在线阅读,更多相关《problems-and-solutions2.docx(48页珍藏版)》请在得力文库 - 分享文档赚钱的网站上搜索。
1、problems-and-solutions2 Chapter 66 .1. Carbon dioxide is diffusing through nitrogen in one direction at atmospheric pressure and 0C. The mole fraction of CO 2 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 cm 2 /s. The gas phase as a whole
2、 is stationary; that is, nitrogen is diffusing at the same rate as the carbon dioxide, but in the opposite direction. ( a ) What is the molal flux of CO 2 , 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 m
3、eters per second, would an observer have to move from one 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
4、 dioxide relative to the observer under condition ( d )?solution : ( a ) from (6.1-8)from equation (6.1- 19 )( b ) net mass fluxfor carbon dioxide (molecular weight=44)mass flux of CO 2 = 441.38810 -4 kg/m 2 hfor nitrogen (molecular weight=28)mass flux of N 2 = 281.38810 -4 kg/m 2 hso the net mass f
5、lux in the direction of CO 2 diffusionm =(44-28) 1.38810 -4 =2.221 10 -3 kg/m 2 h( c ) Here J A = N A = N B , 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)
6、 and (6.1-3b). To select two points, y A =0.2 and 0.02, respectively, to calculate the positions of observerfor y A =0.2, C A = C m y A = from equation (6.1-3a)the diffusing velocity of A: for B: C B = C m (1- y A )= the diffusing velocity of B: let u o is the velocity of the observer moving in the
7、direction of CO 2 diffusion, then net velocity ( u A - u o ) gives a mass transfer rate m A equal to that in opposite direction m B corresponding to ( u B + u o ) from equations (6.1-4) and (6.1-5)for m A = m B , and rearranging two equations above givesat y A =0.2,It is similar to calculating u o a
8、t the point of y A =0.02( d ) When the velocity of observer moving is equal to that of nitrogen diffusing, the nitrogen is stationaryu o = ( e ) When the velocity of observer moving is equal to that of nitrogen diffusing, the molal flux of carbon dioxide diffusing is indicated byChapter77.1 solution
9、:The data from third column in the table are used as calculating Mole fraction x of ammonia in water:molar ratio of ammonia to water X in liquidmolar ratio of ammonia to inert gas the results of calculation are list in the tablep /kPa0 0.4 0.8 1.2 1.6 2.0 2.43 3.32 4.23 6.67 9.28x0 0.00527 0.01048 0
10、.01563 0.02074 0.0258 0.03079 0.04063 0.05028 0.07357 0.09574X0 0.00530 0.01059 0.01588 0.02118 0.0265 0.03177 0.04235 0.05294 0.07940 0.1059Y0 0.00396 0.00796 0.01198 0.01600 0.0201 0.02453 0.03387 0.04353 0.07040 0.1008The data in the table are used to plot the mole fraction x versus partial press
11、ure p diagram and molar ratio X-Y diagram7.3 Vapor-pressure data for a mixture of pentane (C 5 H 12 ) and hexane (C 6 H 14 ) are given by the table. Calculate the vapor and liquid composition in equilibrium for the pentane-hexane at 13.3kpa pressure on assuming that the vapor of mixture approaches i
12、deal behavior and liquid follows Raoults low. t,K260.6265270275280285289P A , kPa13.317.321.926.534.542.548.9P B , kPa2.833.54.265.08.5311.213.3Solution: 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 substance
13、srearranging equation above 1and rearranging equation above gives 2 substituting the data for the table into the equations 1 and 2 gives the results in the following table t,K260.6265270275280285289P A , kPa13.317.321.926.534.542.548.9P B , kPa , 2.833.54.265.08.5311.213.3x1 .00.71 00.5130.3860.1840
14、. 0 670y1 .00.9240.8450.7690.4770.21407.4 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
15、 problem 7.3.Solution : (a) calculate the value of by The results are given in the table :t,K260.6265270275280285289P A , kPa13.317.321.926.534.542.548.9P B , kPa , 2.833.54.265.08.5311.213.34.704.945.145.34.043.793.68Average relative volatilityand equilibrium relation with relative volatility for a
16、 mixture of pentane (C 5 H 12 ) and hexane (C 6 H 14 ) is expressed by(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,KxyThe results of problem 7.3The calculating result by using 260.62652702752802852894
17、.511.00.71 00.5130.3860.1840.06701.00.9240.8450.7690.4770.21401.00.91 70.8260.7390.5040.24507.5 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 y A and x A using Raoults law at total pressure of 101.325kPa . If a mi
18、xture has a composition of x A = 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 pressure P A =204.2kPa, for toluene, vapor pr
19、essure P B =86.0kPa, substituting these data into Eq(7.1-11) and solving for x A (mole fraction of benzene)substituting into Eq(7.1-12) F rom figure7.1-2, i f a mixture has a composition of x A = 0.40 and is at 358.2 K and 101.32 kPa pressure, it will not boil , and will boil at temperature of 95 o
20、C(368.2K);At 95 o C(368.2K), from Table 7.1-1 for benzene, vapor pressure P A =155.7kPa, and substituting into Eq(7.1-12)7.7 What are the effects on the concentrations of the exit gas and liquid streams of the following changes in the operating conditions of the column of Example 7.5 ? ( a ) A drop
21、in the operating temperature that changes the equilibrium relationship to y = 0.6 x . Unchanged from the original design: N , L / V , y b , and x a . ( b ) A reduction in the L / V ratio from 1.5 to 1.25. Unchanged from original design: temperature, N , y b , and x a . ( c ) An increase in the numbe
22、r of ideal stages from 5.02 to 8. Unchanged from original design: temperature, L / V , y b , and x a .solution:( a ) For a dilute solution and a dilute gas, L and V are assumed constant, and the stripping factor is= 0 . 6 1 . 5 = 0.9All concentrations can be expressed in terms of x a , the mole frac
- 配套讲稿:
如PPT文件的首页显示word图标,表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
- 特殊限制:
部分文档作品中含有的国旗、国徽等图片,仅作为作品整体效果示例展示,禁止商用。设计者仅对作品中独创性部分享有著作权。
- 关 键 词:
- problems and solutions2
限制150内