化工基础学习知识原理习题集规范标准答案英文.doc

!- Problems and Solutions Distillation 1、 A continuous fractionating column is used to separate 4000kg/h of a mixture of 30 percent CSand 70 percent CCl. Bottom product contain 5 percent CSat least, and the rate of recovery of CSin the overhead product is 88% by weight,required. Calculate (a) the moles flow of overhead product per hour .(b) the mole fractions of CSand CClin the overhead product, respectively Solution: Form overall material balance Known by the justice of the problem Take the place of 3 types and enter 2 types The unit converts： （mole fraction） 2、 A liquid containing 40 mole percent methanol and 60 mole percent water is to be separated in a continuous fractional column at 1 atm pressure .Calculate the value of q under the three following conditions (a) the feeding is liquid at 40 C (b) the feeding is saturated liquid. The equilibrium data for methanol-water liquid at 1 atm pressure are given in the attached table. If the column is fed with 100koml/h.The molar fractions of methanol in overhead product and bottom product are 0.95 and 0.04,respectively.A reflux ratio at the top of column is 2.5.Calculate (a) the mole flow of overhead product per hour (b) the mole flow of liquid in rectifying column (c) the mole flow of vapor in stripping column .Assume that the constant molar flow applies to this system . Solution: Form overall material balance Solve the eqution we can have： And These upper values are fixed under the three feed conditions。 The feed at 40℃，q=1.07 Feed saturated liquid ，q=1 3、The feeding at dew point is fed to a continuous fractionating column ,and the equilibrium relations are given by equations: y=0.723x + 0.263 ( in rectifying column); y=1.25x – 0.0187 (in stripping column) .Calculate(a) the compositions of feeding, overhead product and bottom product , respectively(b) the reflux ratio Solution: The slope of the operating line in rectifying column is We can get：R=2.61 The intercept of the rectifying line on y axis is The intersection of the stripping line and the diagonal is Y=X=Xw so Xw=0.0748 Form the intersection of these two operating lines, gives: Because the feed at dew point ,so the feed line is horizon ,and the composition of feed is 6.12 It is desired to produce an overhead product containing 95.7 mole percent A form a ideal mixture of 44 mole A and 50 mole percent B feeding into a continuous fractionating column .The average relative volatility equals to 2.5 ,and a minimum reflux ratio is 1.63.Explain the heat state of feeding and calculate the value of q . Solution: ：Form equilibrium equation: Form operating equation : Substituting in eq.1),gives: X=0.365 Y=0.59 Form definiens of the minimum reflux ratio，the intersection of the two upper equations is also that of the equilibrium curve and the feed line. so the feed is a mixture of liquid and vapor. Form the feed equation ( feed line ),gives: Solve and have: 4、 A liquid of benzene and toluene is fed to continuous distillation in a plate column. Under the total reflux ratio condition , the compositions of liquid on the close together plates are 0.28,0.41,and 0.57,respectively.Calculate the Murphree plate efficiency of relatively two low layers. The equilibrium data for benzene—toluene liquid at the operating condition are given as: x 0.26 0.38 0.51 y 0.45 0.60 0.72 Solution : Under the total reflux ratio condition Form Murphree efficiency: Checked by 5、 A continuous fractionating column is used to separate 2500kg/h of a mixture of 25 percent acetone and 75 percent water . Overhead product containing 99 percent acetone,required. These percentages are by weight ,and 80 mole percent of acetone in feeding is entering to the overhead product. The feeding at 20c ,and the operate reflux ratio is 2.7 times as much as the minimum . there are two condensers at the top of column.Partial condenser is used to control the reflux flow to be saturated liquid . The steam not condensed enter the final condenser to be condensed and cooled, regarded as the products. If the overall efficiency is 60%,then calculate the number of ideal plates. Solution: The compositions of each flow rate Known by the meaning of the question： Calculate the number of actual plates From the attached figure 1，at the point，the corresponding bubble point is 67℃，so the average of the feed temperature and the bubble point is 1/2（67+20）=43.5℃。 The specific heats of each component at 43.5℃ are ： the latent heats of each component at 67℃ are: the slope of the feed line= From fig.2 , the minimum reflux ratio must computed from the slope of operating line ag that is tangent to the equilibrium curve, and measure the slope is 0.475 . Solve and have： the intercept of the rectifying line on y axis the intercept of the stripping line on y axis the operating line of the attached figure 2 is drawn according to the slope 0.281，and this operating line of the attached figure 3 is draw according to the intercept 0.709. 7 ideal plates are needed (except reboiler and partial condenser) hence , the number of actual plates 6、The equilibrium data for the system CS-CClat the 1 atm pressure are given in Example 1-11.A continuous fractionating column is used to separate a feed contains 30 mole percent CSand 70 mole percent CCl.Overhead product containing 95 mole percent CS and a bottom product containing 2.5 mole percent CS,required. The feed is saturated liquid and the mass flow is 400kg/h. A reflux ratio equals to 1.7 times as the minimum. The distillation takes place at 61C and 1 atm . the superficial velocity based on empty tower is 0.8m/s and the distance between two boards is 0.4m.The overall efficiency is 50% .Calculate (a) The number of actual plates needed.(b) the mass flow of two products c) the cross-sectional diameter of tower (d) the effective height of tower. Solution: (a) calculate the number of actual plates Painted x-y picture for the equilibrium data by the question, as figure shows。 Since the feed is saturated liquid, ，from fig .X-Y The intercept of operating line in rectifying column = Calculate the number of ideal plates with The graphic method, and the steps is omitted .The result indicates, besides reboiler, 11 ideal plates are needed and feed should be introduced on the seventh plate from the top Actual plates 1） The mass flow rate of product The average molecular weight of feed is M=0.316 + 0.1154=130.6kg/kmol Form overall material balance D + W=30.6 0.95D + 0.025W=0.330.6 From the two upper equations, gives: W=21.5mol/h and 2） The cross-sectional diameter of tower Since the feed is saturated liquid ,hence: Among them Assume that the rising vapor is ideal gas ，hence 3） The validity height of tower is Absorption 7-1 The vapor of methanol mixed with air is absorbed in water, the temperature is 27c and the pressure is 101.3 kpa .The molar density of methanol in bulk of liquid and gas phase are very weak. Henry’s law applies to this system, H=1.995kmol/m*kpa, , Calculate (1) (2) The percent of gas resistance in the whole resistance Where H=solubility coefficient =Individual mass-transfer coefficient for liquid phase =Individual mass-transfer coefficient for gas phase =Overall mass-transfer coefficient based on gas phase Solution : 1) 2) 7-2 A countercurrent flow tower is used to absorb HS from the air- HS steam fed to it ,using pure water as the absorbing liquid ,(Pure water is used in the countercurrent flow tower to absorb H2S from the air-H2S mixed gas),the tower is operating at 25c and 101.3kpa . The density of HS is changed (reduced)from 2% into 1% (in volume). Henry’s law applies to this system ,and Henry constant E=5.52*10kpa .if the amount of the absorbent is 1.2 times as much as the smallest amount according to theory , (1)calculate the liquid-gas ratio L/V and (outlet liquid) concentrations X1(2)Repeat calculate the liquid-gas ratio L/V and (outlet liquid) concentrations,if the operating pressure is 101.3 kpa Solution: 1)From Eq (7-6), the slope of equilibrium line m is : And : Form Eq (7-54) , calculate the limiting liquid—gas ratio So operating liquid-gas ratio is the terminal concentration X1 2) The slope of equilibrium line So The terminal concentration X1 7-3 The butane mixed with the air is absorbed in a sieve-plate tower containing eight ideal plates .The absorbing liquid is a non-volatilization oil having a molecular weight of 250 and a density of 900kg/m.The absorption takes place at 101.3kpa and 15c. 5 percent of butane in the entering gas .The butane is to be recovered to the extent of 95% ,the vapor pressure of butane at 15C is 194.5kpa,and liquid butane has a density of 580kg/m.Assume that Raoult’s and Dalton’s laws apply .(1)calculate the cubic meters of fresh absorbing oil per cubic meter of butane recovered .(Caculate the amout of absorbing oil is required (in volume) when per cubic meter of butane is recovered) (2)Repeat caculate (1), on the assumption that the operating pressure is 3034.4kpa and that all other factors remain constant.(conditions reain the same) Solution: 1）calculate the mass flow rate of water 2） Calculate ideal plates (a) Schematicallyad Plot the operating line BE and the equilibrium line OE (Y=26.7X) in the X-Y right-angle coordinate .Draw the rectangular steps between the operating line and the equilibrium line from the point B, and we can get N So : ideal plates (b) Using Kremser.A. map The absorption coefficient is 95%，and X2=0，so the absorption coefficient of relativity，under the limiting amount of water according to theory condition ，。Looking into fig7-21 according to these data, we can get: The absorption factor in the state of operation Looking into fig 7-21（or calculate according to equation 7-77）,we can know when A=1.43，,the ideal plates。 So the result is the same as that of schematicallyad 3) This question should be estimate by Kremser.A. map absorption coefficient，ideal plates Form fig 7-21， reading So: 7-4 An absorption tower is to recover 99 percent of ammonia in an air stream ,using pure water as the absorbing liquid .(Pure water is used in an absorption tower to absorb ammonia in an air stream ,the absorptivity is 99 percent),the height of the packed section is 3m ,the absorption takes place at 101.3kpa and 20c. The mass flow rate of gas V is 580kg/(m.h),and 6 percent of ammonia in the gas in volume . The mass flow rate of water L is 770kg/( m.h). The tower is countercurrent operated under isothermal temperature, the equilibrium equation Y=0.9X, ka,but has nothing to do with L, try to make out how the height of the packed section to change in order to keep the absorption coefficient unchanged when the conditions of operation have be changed as following (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 original3) the mass flow rate of gas is two times as much as the original Solution: The average molecular weight of the mixed gas M=290.94+170.06=28.28 1） Form Eq(7-6), So Form Eq(7-62) and Eq(7-43a) So: is changed with the operating pressure So So the height of the packed section reduce 1.802m against the original 2） when the mass flow rate of gas increasing, has not remarkable effect the height of the packed section reduce 0.605m against the original 3） when mass flow rate of gas increasing , corresponding grow，according to Poblem: So the height of the packed section increase 4.92m against the original Drying 11-1 The total pressure and humidity of the humid air are 50kPa and 60 C , respectively, and the relative humidity is 40 percent. Calculate (a) the partial pressure of aqueous vapor in the humid air (b) humidity (c) the density of humid air. Solution: a）The partial pressure of aqueous vapor in the humid air From aqueous vapor appendix table, the vapor pressure of aqueous vapor at 60℃ is 149.4mmHg。 b） Humidity c） The humidity volume is： =2.24(mhumid air)/(kg dry air ) So the density of humid air 11-2 A dryer is used to dry material from 5 percent moisture to 0.5 percent moisture (wet basis). The production capacity of dryer is 1.5 kg dry material/s. Hot air enters at 127℃ and 0.007kg water/kg dry air and leaves at 82℃.The temperature of the material in the inlet and outlet are 21℃ and 66℃, respectively. The specific heat of wet material is 1.8KJ/（kg℃）。If the heat loss of the dyer can be ignored, then calculate a) the consumption of the wet air. b) the temperature of the air leaving the dryer. Solution：The moisture content of the material (dry basis) is From the material balance, we have: From the enthalpy balance, we have: Where： kg water/kg dry air （3） From these, the ，L=6.61kg wet air/s 11-3 A humid material is dried at a dryer.55 hours are required to reduce moisture content from 35 percent to 10 percent. The critical moisture content was found to be 15 percent and the equilibrium moisture 4 percent. If under the same drying conditions, it is required that the moisture content of the material drops to 0. 05 from 0.35.Assuming that the rate of drying is proportional to the free-moisture content (X—X*), try to calculate how long will it take to dry the humid material. Solution: The drying time at constant rate stage is The drying time at falling rate stage is So the required drying time under the new condition is 11-4 A atmosphere parallel-counterflow dryer is to be designed to dry a humid material from 1.0kg moisture /kg dry material to 0.1kg moisture /kg dry material using hot air. The air enters at 135℃ and 0.01kg water/kg dry air, and leaves at 60℃. Assuming the air experienced equal enthalpy drying process. According to experiment, the rate of drying under the constant-rate period can be represented by equation -dx/dτ=30ΔH kg water /(kg dry material. H);the rate of drying can be represented by equation -dx/dτ=1.2x kg water/(kg dry material*h). Try to calculate how long should it takes to dry this material? Solution：Because the air ex