化工原理英文教材传质分离过程Mass Transfer and Separation Processes

传质过程英语Mass Transfer ProcessMass transfer is a process by which a substance moves from one location to another. The substance can be in solid, liquid or gas phase. The process is important in several areas of chemical engineering, including separation, purification, and reaction engineering. The following are the key aspects of mass transfer process.1. Mechanisms of mass transferThe mass transfer process can occur due to diffusion, convection, or a combination of both. Diffusion occurs when molecules move from high concentration to low concentration. It is driven by the concentration gradient and is a slow process. Convection, on the other hand, involves the movement of substances due to fluid flow, such as mixing or stirring. It is faster than diffusion but requires energy input.2. Mass transfer coefficientsMass transfer coefficients are parameters that indicate the rate of mass transfer under specific conditions. They are dependent on the properties of both the substance being transferred and the medium through which it moves. The coefficients can be determined experimentally or through theoretical equations. A high mass transfer coefficient indicates a fast transfer rate.3. Types of mass transferThere are three types of mass transfer: gas-phase mass transfer, liquid-phase mass transfer, and solid-phase mass transfer. In gas-phase mass transfer, the substance movesbetween gas phases, such as from the air to water. In liquid-phase mass transfer, the substance moves between different liquid phases, such as from water to oil. In solid-phase mass transfer, the substance moves through a solid, such as a catalyst.4. Applications of mass transferMass transfer is used in several areas of chemical engineering, such as separation and purification processes. Distillation, extraction, and adsorption are some examples of separation processes that utilize mass transfer. In reaction engineering, mass transfer plays a critical role in controlling the rate of reactant delivery to the reaction site.In conclusion, mass transfer is an essential process in chemical engineering. Understanding its mechanisms, coefficients, and types is crucial for designing effective separation, purification, and reaction systems.。

单元操作Unit operation单元操作是化学工业和其他过程工业中进行的物料粉碎、输送、加热、冷却、混合和分离等一系列使物料发生预期的物理变化的基本操作的总称。

对这些操作的研究，是化学工程的一个重要分支。

各种单元操作依据不同的物理化学原理，应用相应的设备，达到各自的工艺目的。

如蒸馏根据液体混合物中各组分挥发能力的差异，可以实现液体混合物中各组分分离或某组分提纯的目的。

对单元操作的研究，以物理化学、传递过程和化工热力学为理论基础，着重研究实现各单元操作的过程和设备，故单元操作又称为化工过程及设备。

单元操作的应用遍及化工、冶金、能源、食品、轻工、核能和环境保护等部门，对这些部门生产的大型化和现代化起着重要作用。

Unit operation is a general term for a series of material handling, transportation, heating, cooling, mixing and separation of materials in the chemical industry and other process industries. The study of these operations is an important branch of chemical engineering. Various unit operations according to different physical and chemical principles, the application of the corresponding equipment, to achieve the purpose of their respective processes. Such as distillation according to the difference of the volatile capacity of the liquid mixture, can achieve the purpose of separation of components in liquid mixture or a group of purification. Based on the theory of physical chemistry, transfer process and chemicalthermodynamics, the research on the operation of the unit has focused on the process and equipment of realizing the operation of each unit, so the unit operation is also called the chemical process and equipment. Application of unit operation in chemical industry, metallurgy, energy, food, light industry, nuclear energy and environmental protection departments, the production of these departments and the modernization of large-scale play an important role.单元操作沿革Unit operation evolution单元操作在化学工业的发展过程中，人们最初以具体产品为对象，分别进行各种产品的生产过程和设备的研究。

Problem for Mass Transfer and Separation ProcessesAbsorption1 At 20℃, the ammonia solubility in water is l0kgNH3／1000kgH2O, gas phase equilibrium molar fraction y* is 0.008, try to calculate following coefficients, Henry coefficient E, equilibrium constant m 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);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?3 When the temperature is 10 o C and the overall pressure is 101.3KPa , the solubility of oxygen in water can be represented by equation p=3.27 104x, where p (atm) and x refer to the partial pressure of oxygen in the vapor 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 dissolved in per cubic meter of water?4 An acetone-air mixture containing 0.02 molar fraction of acetone is absorbed by water in a packed tower in countercurrent flow. 99％of acetone is removed when mixed gas molar flow rate is 0.03kmol·s-1m-2 and practice absorbent flow rate L is 1.4 times as much as the min amount required. Under the operating condition, the equilibrium relationship is y*=1.75x, volume total absorption coefficient is K y a=0.022 kmol·s-1m-2y-1.. What is the molar flow rate of the absorbent and how height of packing will be required?5 The mixed gas from an oil distillation tower contains H2S=0.04(molar fraction). Triethanolamine (三乙醇胺) is used as the absorbent to remove 99% H2S in the packing tower, the equilibrium relationship is y*=1.95x, the molar flux rate of the mixed gas is 0.02kmol·m-2·s-1，volume total absorption coefficient is Kya=0.05 kmol·s—1m-2, The solvent free of H2S enters the tower and it contains 70% of the H2S saturation concentration when leaving the tower. Try to calculate: (a) the number of mass transfer units N oy, and (b) the height of packing layer needed, Z.6 Ammonia is removed from ammonia–air mixture by countercurrent scrubbing with water in a packed tower at an atmospheric pressure. Given: the height of the packing layer Z 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, and the equilibrium relation is y*=1.2x. Find:(1)the practical liquid—gas ratio.(2) the number of overall mass transfer units.(3) if the molar fraction of the ammonia out of the tower will be reduced to 0.002 and the other operating conditions keep unchanged, is the tower suitable?7 Pure water is used in an absorption tower with the height of the packed layer 3m to absorb ammonia in an air stream. The absorptivity is 99 percent. The operating conditions of absorber are 101.3kpa and 20o C, 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, andgas phase mass transfer coefficient ka is proportional to V0.8, but it has nothing to do with L.GWhat is the height of the packed layer needed to keep the same absorptivity when the conditions of operation change 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 originalDistillation1 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 more volatile component’s recovery percent of the overhead product is η=0.92, Calculate: Mole fraction in the bottom product x B=?x0.35, 2 Certain binary mixed liquid containing mole fraction of easy volatilization componentF feeding at bubbling point, is separated through a sequence rectify column. The mole fraction in the overhead product is x D=0.96, and the mole fraction in the bottom product is x B =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，x0.5, feed rate 100kmol/h, is saturated The feed which contains more volatile componentFvapor,the flow rate of overhead 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 x D of overhead product and the mole fraction x B of bottom product ?(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 efficiencyfrom the column top is E m,L=0.6(represented by liquid mole fraction), find the constituent of gas phase leaving the second plate from the column top.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 also500kmol/h. Given: the operating line for rectifying section is y=0.86x+0.12, indirect vapor is used in the reboiler for heating and the total 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?(2) upward flow rate of vapor in the rectifying section(V mol/h,) and down ward flow rate of liquid in the stripping section.(L’ mol/h,) (3) The operating line for stripping section .(4)if relative volatility α=2.4, find R／Rmin5 Separate component A and B of mixed liquid in a continuous distillation column, Given: raw liquid flow rate F is 4000 kg·h-1 , mass fraction of A is 0.3. It is required that the A mass fraction in the column bottom liquid can not be beyond 0.05, recovery ratio of A is 88% on the top. Try to calculate the flow rate of the distillated liquid D and its constituent x D at the top of column.( in the terms of molar flow rate and molar fraction) Given: molecular weight of components A and B are 76 and 154, respectively.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 contains 95.7 mole percent A. Given: liquid average relative volatility α=2.5, min reflux ratio Rmin =1.63, 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 (甲苯) 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. In this situation the equilibrium data of benzene—methylbenzene are as followst 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=2.4x/(1+1.4x); 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. Suppose the total condenser is used at the top of column.10 A continuous fractionating column is used to separate 4000kg/h of a mixture of 30 percentCS2and 70 percent CCl4. Bottom product contains 5 percent CS2, and the rate of recovery ofCS2in the overhead product is 88% by weight. Calculate (a) the moles flow of overhead productper hour. (b) the mole fraction of CS2in the overhead product, respectively.11. A liquid mixture of benzene and toluene is continuously fed to 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 two lower plates. The equilibrium data for benzene—toluene liquid and vapor phases under the operating condition are given as follows:x 0.26 0.38 0.51y 0.45 0.60 0.72Drying1 A wet solid with 1000 kg/h is dried by air from 40% to 5% moisture content (wet basis) under the convective drying conditions. The air primary humidity H1 is 0.001(kg water/kg dry air), and the humidity of air leaving dryer H2 is 0.039 (kg water /kg dry air), suppose that the material loss in the drying process can be negligible. Find:(1) rate of water vaporization W, in kg water/ h.(2) rate of dry air G required, in kg bone dry air/h, flow rate of moist air, V, in kg fresh air/h.(3) rate of moist material out of dryer, L2, in kg moist solid/h.2 The wet solid is to be dried from water content 20% to 5% (wet basis) in a convective dryer at an atmospheric pressure. The feed of wet solid into the dryer is 1000 kg/h at a temperature of 40℃. The dry and wet bulb temperatures of air are respectively 20℃ and 16.5℃ before air enters the preheater. After being preheated, air enters the dryer. The dry and wet bulb temperatures of air leaving dryer are respectively 60℃ and 40℃. If heat loss is negligible and drying is considered as constant enthalpy process:(1) What is the fresh (wet) air required per unit time (in kg/h)?(2) The air temperature T G1 before entering the dryer?(Given: Vaporization latent heat of water at 0℃ is 2501kJ/kg, specific heat of dry air C pB is 1.005 kJ/kg·K, specific heat of water vapor C pA 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 drying is under constant enthalpy process.(1) Determining the air humidity H1.(2) If H0=0.008(kg water/kg dry air), H2=0.03( kg water /kg dry air. Find:(a) Dry air flow rate G required, in kg dry air/s.(b)How much heat is supplied to air by the preheater (in k J/h)?4 The wet solid containing 12%(wet basis) moisture is fed to a convective dryer at a temperature of 15℃ and is withdrawn at 28℃, which contains 3% moisture (wet basis). The flow rate of final moist solid (product) is 1000kg/h. 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 constant enthalpy, heat loss in the drying system can be negligible.(1) Drawing the operation process covering various air states in T-H.(2) What is the fresh air required per unit time (in kg/h)?5 Certain wet material is dried under an ordinary pressure in a convective dryer. Given: air temperature and humidity before entering a preheater are T G0=15℃and H0=0.0073 (kg water)/ (kg dry air); air temperature before entering the dryer T G1=90℃; air temperature and humidity leaving the dryer T G2=50℃and H2=0.023(kg water)/ (kg dry air); water content contained by material entering the dryer X1=0.15 (kg water)/(kg bone dry material ); water content contained by material leaving the dryer X2=0.01 kg water/(kg bone dry material); the production capability of dryer is 273kg/h (based on the product leaving the dryer). Find:(1) flow rate of the dry air G ,in kg bone dry air/h:(2) flow rate of wet air before entering the preheater, in fresh air m3/s;(3) heat duty (quantity) provided to the preheater, in kw if heat loss can be negligible.6 Wet material is dried in a dryer at an atmosphere pressure. The operating conditions are as follows:Air conditions: air temperature before entering the preheater T G0=20ºC, air humidity before entering the preheater H0=0.01(kg water/kg dry air), air temperature before entering the dryerT G1=120℃, air temperature leaving the dryer T G2=70ºC, air humidity leaving the dryerH2=0.05(kg water/kg dry air).Material conditions: material temperature before entering the dryer T s1=30ºC, wet content before entering the dryer w1=20% based on wet basis, material temperature leaving the dryer T s2=50ºC, wet content leaving the dryer w2=5% based on wet basis, specific heat of bone-dry materialCs=1.5kJ/(kg bone-dry solids . ºC). Production capacity of dryer is 53.5 kg/h based on product leaving the dryer. Suppose heat loss in the drying system can be negligible. Find:(1) The mass of bone-dry air required per unit time.(2) How much heat is obtained by air when passing through the preheater, in kJ/h?(3) How much supplementary heat transfer rate Q D is obtained?。