problems-and-solutions2.doc

Chapter 66.1. 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 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 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 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 CO2= 441.38810-4 kg/m2hfor nitrogen (molecular weight=28)mass flux of N2= 281.38810-4 kg/m2hso the net mass flux in the direction of CO2 diffusionm=(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=0.2 and 0.02, respectively, to calculate the positions of observerfor yA=0.2, CA=Cm yA=from equation (6.1-3a)the diffusing velocity of A: for B: CB=Cm (1-yA)=the diffusing velocity of B: let uo 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)for mA = mB, and rearranging two equations above gives 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 stationaryuo=(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: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 0.01563 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 pressure p diagram and molar ratio X-Y diagram虚线范围表示符合Herrys law虚线范围表示符合Herrys law7.3 Vapor-pressure data for a mixture of pentane (C5H12) and hexane (C6H14) 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 ideal behavior and liquid follows Raoults low. t,K260.6265270275280285289PA,kPa13.317.321.926.534.542.548.9PB, 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 substancesrearranging equation above 1and rearranging equation above gives 2substituting the data for the table into the equations 1 and 2 gives the results in the following tablet,K260.6265270275280285289PA,kPa13.317.321.926.534.542.548.9PB,kPa, 2.833.54.265.08.5311.213.3x1.00.7100.5130.3860.1840.0670y1.00.9240.8450.7690.4770.21407.4 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 problem 7.3.Solution: (a) calculate the value of by The results are given in the table :t,K260.6265270275280285289PA,kPa13.317.321.926.534.542.548.9PB,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 mixture of pentane (C5H12) and hexane (C6H14) 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 260.62652702752802852894.511.00.7100.5130.3860.1840.06701.00.9240.8450.7690.4770.21401.00.9170.8260.7390.5040.24507.5 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 yA and xA using Raoults 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 pressure 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)substituting into Eq(7.1-12) From figure7.1-2, if a mixture has a composition 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)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 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 1.25. 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= 0.6 1.5 = 0.9All concentrations can be expressed in terms of xa, the mole fraction of NH3 in the entering solution: since yb = 0From an ammonia balance, V y = V ya = L (xa-xb) = Lxa. HencealsoFrom Eq. (7.2-28),The separation corresponds to 5.02 ideal stages, so the stage efficiency is 5.02/7 = 72 percent.0.787V ya = L (xa-xb) = Lxa. decreasing in =1.50.787xa with decreasing in the recovery (b)=0.81.25=1 1 2so 3substituting equations(2) and (3) into equation (1) gives=5.02solving for =0.834(c) From an ammonia balance, V ya = L x = Lxa. HencealsoFrom Eq. (7.2-28),=1.456solving for =0.957.10. A toxic hydrocarbon is stripped 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) 1From an ammonia balance, V y = V ya = L (xa-xb) = Lxa. Hencealso 2substituting equation(2) to (1) givesthe 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 bysolving for =0.998Chapter 88.1 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 concentration is to be 2.0 mol % ammonia. What is the minimum flow ? Using 1.5 times the minimum, plot the equilibrium and operating lines.Solution:from Fig9-2 in Chemical Engineeringin Chinese: E=1.85105Pa，mass balance on component ammoniakmol/h kmol/h8.2 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 total 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 heightsolution:V = 57.8 kmol air/h, yb = 0.04 mol frac NH3, ya = 0.005, L = 68.0 kmol water/h, xa =0 . kya= 0.0739 kmol/s m3mol frac and kxa= 0.169 kmol/s m3mol frac.equilibrium relationY=1.83XMass balance on ammoniaoverall mass transfer coefficientheight of transfer unit isnumber of transfer unitthe height of packingZ=HoyNoy=0.6672.338=1.559m8.3 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 rings = 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: height of packing: Z=4.88mcross sectional area Sequilibrium relationY=2.0XNumber of transfer unitoverall mass transfer coefficient kmol/s m3mol frac9.1 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 with 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, where x and y are mole fractions of n-pentane:x1.000 0.867 0.594 0.398 0.254 0.145 0.059 0y1.000 0.984 0.925 0.836 0.701 0.521 0.271 0The 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),so 1solving for yD and xB using the Figure and equation (1), q=0.6 for 40 percent of feed to flash.to locate the point where the q line intersects with equilibrium line at yD=0.85, xB=0.439.2 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 overhead product and the bottoms product. Assume that the equilibrium relation is given by equation y=0.46x + 0.549.solution:for flash distillation so 1From equilibrium lineyD=0.46xB + 0.549. 2solving for yD and xB by combining the equations 1 and 2yD =0.783 and xB =0.508 for differential distillationfrom equation 9.2-4substituting the equilibrium relation into the equation abovesox2=0.498from equation 9.2-5yav=0.8049.3 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 average 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 are as follows, where x and y are mole fractions of n-pentane:x1.000 0.867 0.594 0.398 0.254 0.145 0.059 0y1.000 0.984 0.925 0.836 0.701 0.521 0.271 0The 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), 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 equilibrium 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 = 0.4057. Substituting into Eq. (9.2-5) and solving for the average composition of the 40 mol distilled,soyav=0.893The 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),so 1using the Figure and equation (1), yD and xB can be calculated by trial and error procedureyD=0.85, xB=0.439.4 Distillation Using McCabeThiele Method. A rectification column is fed 100 kmol/h of a mixture of 50 mol % benzene and 50 mol % toluene at 101.32 kPa abs pressure. The feed is liquid at the boiling point. The distillate is to contain 90 mol % benzene and the bottoms 10 mol % benzene. The reflux ratio is 4.52:1. Calculate the kg mol/h distillate, kmol/h bottoms, and the number of theoretical trays needed using the McCabeThiele method.Solution:F=100kmol/h, xf=0.5, xD=0.9, xB=0.1, R=4.52From equation（9.4-3）soD=50 kmol/h, B=50 kmol/hThe intercept The number of theoretical trays needed is 69.5 A binary mixture is fed to a continuous fractionating column at dew point, and the operating lines are given by equations: y=0.723x + 0.263 (rectifying section); y=1.25x 0.0187 (stripping section). Calculate: (a) the compositions of feed, distillate and bottom product, respectively; (b) the reflux ratio.Solution:From the rectifying line y=0.723x + 0.263 R=2.61 xD=0.95combining equations y=0.723x + 0.263 and y=1.25x 0.01870.723x + 0.263=1.25x 0.0187solving for xx=0.228 and y=0.723x + 0.263=0.7230.228+0.263=0.428so y=xf=0.428 due to feeding at dew pointfrom the stripping liney=1.25x 0.0187 the stripping line intersects with the diagonal at the point y=xx=1.25x 0.0187solving for xx=xB=0.07489.6 A liquid mixture containing 40% mole percent methanol and 60% mole percent water is to be separated in a continuous fractionating column at a pressure of 1atm. Calculate the value of q for the various conditions of feed: (a) the feeding is at 40; (b) the saturated liquid; (c) the saturated vapor.Solution:a）The feed at 40Form the equilibrium data for methanol-water ,the bubble point of feeding is 75.3，so the mass latent heats of methanol and water at 75.3 are 186kcal/kg and 554 kcal/kg, respectively. So the molar latent heats of feed is calculated as followsThe temperature of feed is 40,the bubble point is75.3，The average temperature of feed heated from 40 to 75.3=1/2（40+75.3）=57.7，The specific heat of methanol at 57.7 is 0.64kcal/kg.，and the specific heat of water is 1 kcal/kg.，average specific heat of feed isCp=4.187（0.64320.4+1180.6）=79.5kJ/kmol。The thermal condition is b） Feed at saturated liquid From definition q=1c） Feed at saturated vapor From definitionq=09.7 A rectifying column containing the equivalent of three ideal plates is to be supplied continuously with a feed consisting of 0.4 mol % ammonia and 99.6 mol % water. Before entering the column, the feed is converted wholly to saturated vapor, and it enters between the second and third plates from the top of the column. The vapors from the top plate are totally condensed but not cooled. Per mole of feed, 1.35 mol of condensate is returned to the top plate as reflux, and the remainder of the distillate is removed as overhead product. The liquid from the bottom plate overflows to a reboiler, which is heated by closed steam coils. The vapor generated in the reboiler enters the column below the bottom plate, and bottom product is continuously removed from the reboiler. The vaporization in the reboiler is 0.7 mol per mole of feed. Over the concentration range involved in this problem, the equilibrium relation is given by the equationy = 12.6xL=1.35FD,xDV=0.75F, yB,xBF, xfCalculate the mole fraction of ammonia in (a) the bottom product from the reboiler, (b) the overhead product, (c) the liquid reflux leaving the feed plate.Solution:Feed at dew pointV=V+F=0.7F+F=1.7F 1x1V=L+D=1.35F+D 2Solving for D by substituting equation 1 for V into the equation 2D=0.35F 3For feed at dew pointL=L=1.35F andL=V+B=0.7F+B=1.35F 4Solving for BB=0.65F 5Mass balance on ammonia for reboilerLx1=Vy+BxB