传质与分离作业题(英文)新

Chapter 20\21 Distillation1 A binary material system is to be separated in a continuous distillation column. The feed is liquid phase and mole fraction in feed is x F=0.42, Mole fraction in the overhead product is x D =0.95, Givens: The easy volatilization component’s recovery ratio of the overhead product is η=0.92, Calculate: Mole fraction in the bottom product x B(或x W)=?x=0.35, feeding at 2 Certain bianry mixed liquid containing mole fraction of easy volatilization component isFbubbling point, is separated through a sequence rectify column. The mole fraction in the overhead product isx D=0.96, and the mole fraction in the bottom product is x B(或x W) =0.025. If the mole overflow rates are constant in the column, try to calculate(a)the flow rate ratio of overhead product to feed(D／F)?(b)If the reflux ratio R=3.2, write the operating lines for rectifying and stripping sections3 A continuous distillation column is to be designed to separate an ideal binary material system，The feed whichx=0.5, feed rate 100kmol/h, is saturated vapor,the flow rate of contains easy volatilization componentFoverhead product and the flow rate of bottom product are also 50kmol/h. Suppose the operating line for rectifying section is y=0.833x+0.15, the vapor generated in the reboiler enters the column through the bottom plate, a complete condenser is used on the top of column and reflux temperature is bubbling point. Find:(1) Mole fraction of overhead product x D and the mole fraction of bottom product x B(或x W)?(2) Vapor amount condensed in the complete condenser, in mol/h?(3)The operating line for stripping section.(4) If the average relative volatility of the column is 3 and the first plate’s Murphree efficiency of the top plate is Em,L=0.6, find the constituent of gas phase leaving from the second plate of tower top.( plate’s Murphree efficiency Em,L is the liquid phase single plate efficiency of the duality liquid is rectified in continuum rectify tower)4 An ideal binary solution of a volatile component A containing 50% mole percent A is to be separated in a continuous distillation column. The feed is saturated vapor, the feed rate is 1000kmol/h, and the flow rate of overhead product and the flow rate of bottom product are also 500kmol/h. Given: the operating line for rectifying section is y=0.86x+0.12, the reboiler uses indirect vapor to heat and the complete condenser is used on the top of tower. Assume that the reflux temperature is at its bubble point. Find:(1) reflux ratio R, the mole fraction of overhead product x D and the mole fraction of bottom product x B(或x W)?(2) upward flow rate of vapor in the rectifying section(V mol/h,) and down ward flow rate of vapor in the stripping section.(L’ mol/h,) .(3) The operating line for stripping section .(4)if relative volatility a=2.4, find reflux ratio and min reflux ratio R／Rmin5 Separate component A and B of mixed liquid in a continuous distillation column, Given :raw liquid flow rate isF is 4000 kg·h-1 , mass fraction of A is 0.3. Require the mass fraction in the kettle liquid shouldn’t be beyond 0.05, recovery ratio of A is 88% on the top. Try to calculate the flow rate of the distillation liquid D and its constituent x D on the column top.( in the terms of molar flow rate and molar fraction)6 There is a continuous rectifying operation column, whose the operation line equation is as follows:Rectifying section: y=0.723x+0.263Stripping section: y=1.25x-0.0187if the feed enters the column at a dew point, find (a)the molar fraction of feed、overhead product and bottom product. (b) reflux ratio R.7 A column is to be designed to separate a liquid mixture containing 44 mole percent A and 56 mole percent B, the system to be separated can be taken as ideal. The overhead product is to contain 95.7 mole percent. given: liquid average relative volatility a=2.5, reflux ratio R=4.52, try to illustrate the thermal condition of the feed, and to calculate the value of q.8 A continuous rectifying column operated at atmospheric pressure is used to separate benzene—methylbenzene mixed liquid. The feed is saturation liquid containing 50 mole percent benzene. Given: the overhead product must contain 90 mole percent benzene and the bottom product contains 10 mole percent benzene. If reflux ratio is 4.52, try to calculate how many ideal plates are need? and locate the feed plate. Materials system equilibrium refers to Case 2—6In this situation the equilibrium materials of benzene—methylbenzenet o C 80.1 85 90 95 100 105 110.6x 1.000 0.780 0.581 0.411 0.258 0.130 0y 1.000 0.900 0.777 0.632 0.456 0.262 09 There is a rectifying column, given : mole fraction of distillation liquid from tower top x D=0.97, reflux ratio R=2, the gas-liquid equilibrium relationship y=0.86x+0.12；find: the constituent x1 of the down liquid leaving from the first plate and the constituent y2 of the up gas leaving from the second plate in the rectifying section.10.A continuous fractionating column is used to separate 4000kg/h of a mixture of 30 percent CS2and 70percent CCl4. Bottom product contains 5 percent CS2at least, and the rate of recovery of CS2in the overheadproduct is 88% by weight. Calculate (a) the moles flow of overhead product per hour. (b) the mole fractions ofCS2and CCl4in the overhead product, respectively.11. 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 plates are 0.28, 0.41 and 0.57, respectively. Calculate the Murphree plate efficiency of relatively two low plates. The equilibrium data for benzene—toluene liquid under the operating condition are given as:x 0.26 0.38 0.51y 0.45 0.60 0.7212 Problem 20.1 , 20.2 21.8 21.1Chapter 17/18 Absorption1 At 20℃, the ammonia solubility in water is l0kgNH3／1000kgH2O, gas phase equilibrium molar fraction is 0.008, try to calculate following coefficients, solubility coefficient, Henry coefficient, equilibrium constant and equilibrium concentration when the following conditions are(1) Total pressure above the ammonia is 101.3kPa (absolute atmosphere);(2) Total pressure above the ammonia is 301.9kPa (absolute atmosphere);(3) Total pressure above the ammonia remains 301.9kPa (absolute atmosphere) but the ammonia’s temperature rises to 50℃and the equilibrium partial pressure raises 5.9 kPa above ammonia.2.The ammonia–air mixture containing 9% ammonia(molar fraction) is contact with the ammonia-water liquid containing 5% ammonia (molar fraction). Under this operating condition, the equilibrium relationship is y*=0.97x. When the above two phases are contact, what will happen, absorption or stripping? What are the max( or min) values of gas and liquid phases?3.When the temperature is 10℃and the pressure is 101.3KPa , the solubility of oxygen in water can be10x, where p (kPa) and x refer to the partial pressure of oxygen in the represented by equation p=3.31×6vapor phase and the mole fraction of oxygen in the liquid phase, respectively.Assume that water is fully contact with the air under that condition, calculate how much oxygen can be solute in the per cubic meter（立方米）of water?4．There is a packing tower whose diameter is 250mm, which contains 16 mm Rachig ring（拉西环）with the height of packing layer 2.6m. It is used to absorb the carbon dioxide in CO2-air mixture at atmospheric pressure by 2.5mol/L sodium hydroxide lye（NaOH碱液）. Given: the molar flow flux of gas mixture V=0.0117kmol/m2.s, the molar flow flux of NaOH is L=0.208k kmol/m2.s. The mixed gas of inlet tower contains CO2 =3.15×10—4 (molar fraction), and outlet gas contains CO2=3.1×10—5.(molar fraction).Try to calculate the gas phase total transfer mass coefficient Kya.5 An acetone-air mixture（丙酮一空气混合物）containing 0.02 molar fraction of acetone 丙酮is absorbed by water in a packed tower in countercurrent flow. And 99％of acetone is removed, mixed gas molar flow rate is 0.03kmol·s—1m-2 , practice absorbent flow rate L is the 1.4 times more than the min amount required. Under the operating condition, the equilibrium relationship is y*=1.75x. Volume total absorption coefficient is Kya=0.022 kmol·s—1m-2y-1.. What is the molar flow rate of the absorbent and what height of packing will be required? 6．The mixed gas from petrol(石油) distillation tower contains H2S=0.04(molar fraction). Triethanolamine 三乙醇胺(absorbent) is used as the water solvent to absorb 99% H2S in the packing tower, the equilibrium relationship is y*=195x, the molar flux rate of the mixed gas is 0.02kmol·m-2·s-1，volume total absorption coefficient is Kya=0.005 kmol·s—1m-2y-1, The solvent enters the tower free of H2S and it contains 70% of the H2S saturation concentration when leaving the tower. Try to calculate: (a) the number of mass transfer units required , and (b) the height of the of packing layer needed.7.Ammonia氨is removed from ammonia氨–air mixture by countercurrent scrubbing with water in a packed tower at an atmospheric pressure. Given: the height of the packing layer is 6 m, the mixed gas entering the tower contains 0.03 ammonia (molar fraction, all are the same below), the gas out of the tower contains ammonia 0.003; the NH3 concentration of liquid out of the tower is 80% of its saturation concentration. Find:(1)the practical liquid—gas ratio and the min liquid—gas ratio.(2) the number of overall mass transfer units.(3) if the molar fraction of the ammoniac out of the tower will be reduced to 0.002 and the other operating conditions keep unchanged, is the tower suitable?8.Pure water is used in an absorption tower with the height of the packed section 3m to absorb ammonia氨in an air stream. The absorptivity is 99 percent. The operating conditions of absorber are 101.3kpa and 20℃, respectively. The flux of gas V is 580kg/(m2.h), and 6 percent (volume %) of ammonia is contained in the gas mixture. The flux of water L is 770kg/( m2.h). The gas and liquid is countercurrent in the tower at isothermal temperature. The equilibrium equation Y*=0.9X, and gas phase mass transfer coefficient ka is proportional toGV0.8, but it has nothing to do with L. Try to find what is the height of the packed tower to change in order to keepthe absorption coefficient unchanged when the conditions of operation have be changed as follows (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 original. 3) the mass flow rate of gas is two times as much as the origina9. problem 17.2，17.4，17.17，18.3，18.9,Chapter 19/24 Drying1 A wet solid is dried by air from 40% to 5% moisture content (wet basis) under the convective drying conditions in 1000 kg/h. The air primary humidity H1 is 0.001(kg water/kg dry gas), and the humidity of the air leaving dryer H2 is 0.039 (kg water /kg dry gas), suppose that the materials loss in the drying process can be negligible. Find:(1) Mass rate of water vaporization W, in kg water/ h.(2) Mass rate of dry air required L, in kg dry air/h, volume flow rate of moist air, V, in kg primary air/h.(3) Mass rate of moist solids out of dryer, G2, in kg moist solids/h.2 The wet solid is to be dried from water content 20% to 5% (wet basis) in a convective dryer at atmospheric pressure. The feed of wet solid into the dryer is 1000 kg/h at a temperature of 40℃. Suppose there is no significant temperature change in dryer, Given: dry bulb temperature of air is 20℃, and wet bulb temperature is 16.5℃. After being preheated, air enters the drier. The temperature leaving dryer is 60℃, wet bulb temperature is 40℃, and heat loss is negligible, Find:(1) What is the volume of fresh air required per unit time (in m3/h)? (Based on the preheated state)(2) The temperature of the air into the dryer.(Given: Vaporization latent heat of water at 0℃is 2491.27 kJ/kg, specific heat of dry air Cg is 1.011 kJ/kg·K, specific heat of water vapor Cv is 1.88 kJ/kg·K)3 The wet solid material is to be dried from water content 42% to 4% (wet basis) in an adiabatic dryer. The solid product out of the dryer is 0.126kg/s. After the fresh air at a dry-bulb temperature of 21ºC and a relative humidity of 40％is preheated to 93ºC, it is sent to the dryer, and leaves the dryer at relative humidity of 60％. If the air is equal enthalpy in the drying process, find:(1) Determining the air state parameters from the given air state in H-I diagram.(2) If Ho=0.008(kg water/kg dry air), H2=0.03( kg water /kg dry air. Find:(a) Mass rate of dry air required L [kg dry air/s](b)How much heat is supplied to air by the preheater (in kJ/h)?4 The wet solid湿物料containing 12%(wet basis湿基) moisture is fed to a convective dryer at a temperature of 15℃and withdrawn at 28℃containing 3% moisture (wet basis). The flow rate of final moist solid (product) is 1000kg/h.(according to the dry production). After the fresh air at a dry-bulb temperature of 25℃and a humidity of 0.01 kg water/kg dry air is preheated to 70℃, it is sent to the dryer, and leaves the dryer at 45ºC. Suppose the drying process is under the constant enthalpy, heat loss in the drying system can be negligible. Find:(1) Drawing the operation process covering various air states in H-I diagram.(2) What is the volume of fresh air required per unit time (in m3/h)?(3) In order to keep the enthalpy unchanged, how much supplementary heat Q D is needed?5 Drying wet materials by hot air, primary temperature of air is t。