传质与分离作业题(英文)新

吸收部分模拟试题及答案一、填空1气体吸收计算中，表示设备（填料）效能高低的一个量是 传质单元高度 ，而表示传质任务难易程度的一个量是 传质单元数 。

2 在传质理论中有代表性的三个模型分别为 双膜理论 、 溶质渗透理论 、表面更新理论。

3如果板式塔设计不合理或操作不当，可能产生 严重漏液 、 严重泡沫夹带及 液泛 等不正常现象，使塔无法工作。

4在吸收塔某处，气相主体浓度y=0.025，液相主体浓度x=0.01，气相传质分系数k y =2kmol/m 2·h ，气相传质总K y =1.5kmol/m 2·h ，则该处气液界面上气相浓度y i 应为⎽⎽0.01⎽⎽⎽。

平衡关系y=0.5x 。

6单向扩散中飘流因子 A>1 。

漂流因数可表示为 ，它反映 由于总体流动使传质速率比单纯分子扩散增加的比率。

7在填料塔中用清水吸收混合气中HCl ，当水量减少时气相总传质单元数N OG 增加 。

8一般来说，两组份的等分子反相扩散体现在 精馏 单元操作中，而A 组份通过B 组份的单相扩散体现在 吸收 操作中。

9 板式塔的类型有 泡罩塔 、 浮阀塔 、 筛板塔 （说出三种）；板式塔从总体上看汽液两相呈 逆流 接触，在板上汽液两相呈 错流 接触。

10分子扩散中菲克定律的表达式为⎽⎽⎽⎽⎽dzdC D J A AB A -= ，气相中的分子扩散系数D 随温度升高而⎽⎽⎽增大⎽⎽⎽（增大、减小），随压力增加而⎽⎽⎽减小⎽⎽⎽（增大、减小）。

12易溶气体溶液上方的分压 小 ，难溶气体溶液上方的分压 大 ，只要组份在气相中的分压 大于 液相中该组分的平衡分压，吸收就会继续进行。

13压力 减小 ,温度 升高 ,将有利于解吸的进行 ；吸收因素A= L/mV ,当 A>1 时,对逆流操作的吸收塔,若填料层为无穷高时,气液两相将在塔 顶 达到平衡。

14某低浓度气体吸收过程， 已知相平衡常数m=1 ，气膜和液膜体积吸收系数分别为k ya =2×10-4kmol/m 3.s, k xa =0.4 kmol/m 3.s, 则该吸收过程及气膜阻力占总阻力的百分数分别为 气膜控制，约100% ；该气体为 易 溶气体。

传质分离过程试卷一、选择题（共10题，每题2分，共20分）1.以下不属于传质分离过程的是：– A. 蒸馏– B. 气体吸附– C. 曝气– D. 结晶2.传质分离过程中，分馏是利用物质的什么性质实现的？– A. 密度差异– B. 温度差异– C. 压力差异– D. 溶解度差异3.以下哪种传质分离过程利用了膜的选择性通透性？– A. 萃取– B. 吸附– C. 渗透– D. 结晶4.下列哪种传质分离过程主要利用了溶剂的不同挥发性？– A. 蒸馏– B. 萃取– C. 气体吸附– D. 结晶5.反渗透是一种什么类型的传质分离过程？– A. 物理传质分离过程– B. 化学传质分离过程– C. 生物传质分离过程– D. 不确定6.以下哪种传质分离过程是基于物质在溶液和固体表面之间的吸附作用？– A. 吸附– B. 渗透– C. 萃取– D. 结晶7.结晶是通过什么方式实现物质之间的分离？– A. 溶解度差异– B. 密度差异– C. 温度差异– D. 压力差异8.下列哪个条件对于蒸馏过程的实现是必要的？– A. 压力大于饱和蒸汽压力– B. 温度高于沸点– C. 设备具备分离精馏的结构– D. 所有选项都对9.萃取是一种利用分散相在连续相中的亲和性实现物质分离的过程，其中分散相也称为：– A. 溶液– B. 固相– C. 气相– D. 透析10.以下哪个选项不属于传质分离过程的应用？– A. 生活中的水的净化– B. 石油炼制过程中的裂化– C. 水果的蒸馏提取– D. 医药领域中的药物合成二、简答题（共4题，每题10分，共40分）1.请简要描述传质分离过程的定义及目的。

传质分离过程是指通过运用不同物质在不同条件下的传质特性，利用物质之间的差异来实现分离纯化目标物质的过程。

其目的是根据不同物质的传质特性，使混合物中的目标物质与其他物质进行分离，以达到提纯、浓缩、分级等目的。

2.传质分离过程的分类及其基本原理有哪些？传质分离过程可以分为物理传质分离和化学传质分离两大类。

传质与分离课后练习题一、填空题1. 传质过程主要包括________、________和________三种基本方式。

2. 在气体吸收过程中，根据溶质与溶剂的接触方式，可分为________和________两种类型。

3. 蒸馏操作中，将混合液加热至沸腾，产生的蒸汽通过________冷却后，可得到纯净的液体。

4. 萃取过程中，常用的萃取剂应具备________、________和________等特点。

5. 吸附分离技术中，根据吸附剂与吸附质之间的作用力，可分为________和________两种类型。

二、选择题1. 下列哪种传质方式属于质量传递？（）A. 动量传递B. 能量传递C. 质量传递D. 热量传递2. 在下列吸收操作中，属于物理吸收的是（）。

A. 氨气吸收B. 二氧化硫吸收C. 丙酮吸收D. 氯气吸收3. 下列哪种蒸馏方法适用于分离沸点相近的液体混合物？（）A. 简单蒸馏B. 蒸馏C. 蒸馏D. 分子蒸馏A. 萃取剂的性质B. 混合液的温度C. 萃取剂的浓度5. 下列哪种吸附剂属于物理吸附剂？（）A. 活性炭B. 离子交换树脂C. 氢氧化钠D. 氧化铝三、判断题1. 传质过程中，质量传递速率与浓度梯度成正比。

（）2. 在气体吸收过程中，气膜控制表示溶质在气相中的扩散速率较慢。

（）3. 蒸馏过程中，塔板数越多，分离效果越好。

（）4. 萃取操作中，萃取剂的选择对萃取效果具有重要影响。

（）5. 吸附分离过程中，吸附剂的选择与吸附质的性质无关。

（）四、简答题1. 简述传质过程的基本原理。

2. 请列举三种常见的气体吸收设备，并简要说明其工作原理。

3. 蒸馏操作中，如何提高塔板的效率？4. 萃取过程中，影响萃取效果的因素有哪些？5. 简述吸附分离技术的应用领域。

五、计算题1. 某混合液中含有甲、乙两种组分，其摩尔分数分别为0.4和0.6。

现将该混合液进行蒸馏分离，求在塔顶和塔底得到的馏分中甲、乙组分的摩尔分数。

Problems for Mass Transfer and Separation ProcessAbsorption1 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?Solution ：09.0=y 05.0=x x y 97.0=*09.00485.005.097.0=<=⨯=*y y It is an absorption operation.2 When the temperature is 10 c 0 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 the per cubic meter of water?Solution: the mole fraction of oxygen in air is 0.21,hence:p = P y =1x0.21=0.21amt64410*24.610*27.321.010*27.3-===p x Because the x is very small , it can be approximately equal to molar ratio X , that is 610*42.6-=≈x XSo[])(/)(4.11)/(18*)(1)/(32*)(10*42.6lub 2322222226O H m O g O kmolH O kgH O kmolH kmolO kgO kmolO ility so ==-3 An acetone-air mixture containing 0.02 molar fraction of acetone is absorbed by water in a packed tower in countercurrent flow. And 99％ of acetone is removed, mixed gas molar flow fluxis 0.03kmol ·s —1m -2 , practice absorbent flow rate L is 1.4 times as much as the min amountrequired. Under the operating condition, the equilibrium relationship is y*=1.75x. V olume totalabsorption coefficient is K y a=0.022 kmol ·s —1m -2y -1.. What is the molar flow rate of the absorbentand what height of packing will be required?solution ：()0002.01=-=ηb a y y x a =0733.175.102.099.002.0*min =⨯=--=⎪⎭⎫ ⎝⎛ab a b x x y y V L 43.24.1min=⎪⎭⎫ ⎝⎛=V L V L s m kmol L 20729.003.043.2=⨯=720.043.275.1===L mV S Number of mass transfer units N oy =(y 1-y 2)/∆y=12(y b -y a )=0.02-0.0002∆y=[(y b -y*b )- (y a -y*a )]/ln[(y b -y*b )/ (y a -y*a )](y b -y*b )=0.02-1.75x b =0.0057X b =V/L (y b -y a )= (0.02-0.0002)/2.43=0.00815(y a -y*a )= y a =0.0002Or ])1)(ln[(11S S mx y mx y S N ba ab Oy +----==12 m Kya S V H OY 364.1022.003.0/=== m N H H OY OY 37.1612364.1=⨯==4 The mixed gas from an oil distillation tower contains H 2S=0.04(molar fraction). Triethanolamine (absorbent) is used as the solvent to absorb 99% H 2S 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，overall volumeabsorption coefficient is Kya=0.05 kmol ·s —1m -2y -1, The solvent free of H 2S enters the tower andit contains 70% of the H 2S 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.solution ：ya=yb(1-0.99)=0.04*1%=0.0004xb*=yb/m=0.04/1.95= 0.0205 xb=0.7xb*=0.0144yb*=1.95*0.0144=0.028yb-yb*=0.04-0.028=0.012△ym=0.0034Z=HoyNoyNoy=(yb-ya)/ △ym=11.6m a K G H y m oy 4.005.0/02.0/===Z=11.6*0.4=4.64m5 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 is6 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 NH 3 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 and the min liquid —gas ratio L/V=?. (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?solution ：(1) 35.12.103.08.0003.003.0=⨯-=G L (2) 89.035.12.1==S 26.689.0003.003.011.089.011=⎥⎦⎤⎢⎣⎡+-=In N OY (3) m N Z H OY OY 958.026.66===47.889.0002.003.011.089.011=⎥⎦⎤⎢⎣⎡+⨯-='In N OY Since m N H Z OYOY 0.61.847.8958.0'>=⨯='= it is not suitable6 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 200c, respectively. The flux of gas V is 580kg/(m 2.h), and 6 percent (volume %) of ammonia is contained in the gas mixture. The flux of water L is 770kg/( m 2.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 k G a is proportional to V 0.8, but it has nothing to do with L. What 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 original Solution: 3,1,293Z m p atm T K ===1210.060.063810.06(10.99)0.000638Y Y Y ==-=-= The average molecular weight of the mixed gas M=29×0.94+17×0.06=28.2822580(10.06)19.28/()28.2877042.78/()180.919.280.405642.78V kmol m h L kmol m h mV L =-=⋅Ω==⋅Ω⨯==12221ln[()(1)]110.06380ln[()(10.4056)0.4056]10.40560.0006386.88430.43586.884OG OG OG N mV LH Y mX mX mV Y mX L L Z m N =-+-=----+-==== 1） 2p p '=''p p m m = So 1222ln[()(1)]10.90.4520.4519.280.202842.78111ln[(100)(10.2028)0.2028]10.20285.496OG mp m p m V L Y mX m X m V N mV Y mX L L L-+='==⨯=''⨯==''-'=---+-= OG r G V V H K a K aP ==ΩΩSo:OG H changes with the operating pressure10.43580.21792OG OG OG OG H H H H ρρρρ'=''=⋅=⨯='So 5.4960.2179 1.198OGOG Z N H m '''=⋅=⨯= So the height of the packed section reduce 1.802m vs the original2） 2L L '=11()0.40560.20282225.496OGmV mV mV L L L N ===⨯=''=when the mass flow rate of liquid increases,G K a has not remarkable effect0.43585.4960.4358 2.395OG OG OG OG H H m Z N H m'=='''=⋅=⨯= the height of the packed section reduce 0.605m against the original3） 2V V '=(2)2()20.40560.81161ln[(100)(10.8116)0.8116]15.8110.8116OG mV m V mV L L L N '===⨯='=-+=- when mass flow rate of gas increaes,G K a also will increase. Since it is gas film control for absorption, we have as follows:0.80.80.80.20.80.2()222220.43580.50115.810.5017.92G G G G OG OG G G OGOG K a V V K a K a K a V V V H H K aP K aP mZ N H m m ∝''==''===Ω'Ω=⨯='''==⨯= So the height of the packed section increase 4.92m against the originalDistillation1 Certain binary mixed liquid containing mole fraction of easy volatilization component F x 0.35, 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 sections solution ：F x ＝0.35；x B =0.025；x D =0.96；R=3.2。

《传质与分离》课程考试试卷(B卷)姓名，考号，班级。

注意事项：1.请首先按要求在试卷上填写您的姓名，考号和所在单位的名称。

2.请仔细阅读各种题目的回答要求，在规定的位置填写您的答案。

3.一、选择题（选择正确的答案，将相应的字母填入（）内，每题1分，共80分）1.下列属于现象自然蒸发的方式是。

（）A.水放杯子里慢慢变少B.加热硫酸铜溶液C.碘升华D.加热食盐水2.蒸发可以用于。

（）A.溶有不挥发性溶质的溶液B.溶有挥发性溶质的溶液C.溶有不挥发性溶质和溶有挥发性溶质的溶液D.挥发度相同的溶液3.工业三效蒸发并流加料装置温度最高。

( )A.第一效B.第二效C.第三效D.第二效和第三效4.工业三效蒸发三种加料流程中有输送泵。

( )A.并流加料流程B.逆流加料流程C.平流加料流程D.三种加料流程中均有5.有结晶析出的蒸发过程，适宜用的流程是。

（）A.并流加料B.平流加料C.逆流加料D.错流加料6.有关蒸发，下列项中说法错误的是。

（）A.蒸发器的效数并非越多越好B.其他条件确定时，原料液进料温度降低，单位蒸汽消耗量增加C.多效蒸发流程降低了加热蒸汽的经济性D.其他条件确定时，原料液进料温度提高，单位蒸汽消耗量会减少7．影响蒸发器生产能力因素在。

( )A. 传热系数KB.传热面积AC.传热推动力△t mD.以上都是8.下列蒸发器中，属于自然循环类型的是 。

( )A.中央循环管式(或标准式)蒸发器B.悬筐式蒸发器C.外热式蒸发器D.以上都是9.下列除沫器中， 对1~10μm 的雾滴捕获效果最好。

（ ）A.折流板式除沫器B.离心式除沫器C.丝网除沫器D.球形除沫器10.提高蒸发装置的真空度，一定能取得的效果为 。

（ ）A.将增大加热器的传热温差B.将增大冷凝器的传热温差C.将提高水蒸汽的总传热系数D.会降低二次蒸汽流动的阻力损失11.天津蒸发实训装置中，在装置的钢架上放置两个探头，测量是 。

( )A.温度B.压力C.液位D.电导率12.下列 属于蒸发装置运行中异常现象。

,考试作弊将带来严重后果！华南理工大学期末考试2007《传质与分离工程英文》试卷A （含答案）1. 考前请将密封线内填写清楚；所有答案请直接答在试卷上(或答题纸上)；．考试形式：开（闭）卷；S in the air is absorbed by NaOH solution is ( A ).2B. liquid film “controls”;C. two film “controls”.AB=1. WhenC ).>x A; B. y A<x A; C. y A=x A. D. uncertainAD ) is true.>t d; B. t>t w=t d; C. t=t w>t d; D. t=t w=t dwo C, ( C ) of reading is o C B. 77 o C C. 77.01 o C; D. 77.010 o CB ).When the water content of some material is close to its equilibrium water content X*, its drying rate will ___ C _.B. decrease;C. be close to zero;D. be uncertainFor the desorption (stripping) process, when ( B ) increases and ( A )A ) increases, it is good for the operation.D ) decreases, it is good for the operation.A.dry-bulb temperature of air;B. wet-bulb temperature of air ;C. dew point temperature of air;D. size of material to be dried(9) For absorption process, when the coefficients k x a and k y a are of the same order of magnitude and equilibrium constant m is much greater than 1, ( B ) is said to be controlling. In order to increase mass transfer rate, it is better to increase ( E ).A. gas film;B. liquid film;C. both gas film and liquid film;D. gas phase flow rate;E. liquid phase flow rate;F. both gas phase flow rate and liquid phase flow rate;(10) For absorption process, if it is the liquid film control, ( D ) is feasible to increase mass transfer rate of absorption.A to decrease gas flow rate.B to decrease liquid flow rateC to increase gas flow rate.D to increase liquid flow rate.(11)When the reflux ratio R increases and other conditions keep the same, overflow of vapor V will (increase ); the concentration of overhead product x D will (increase ); steam consumption of reboiler will (increase ), the theoretic plate numbers required will (decrease ) for the same recovery percentage of overhead product.(12) The factors to influence mass transfer coefficients of gas phase include (T, P, gas flow rate).(13) When some moist air is preheated in a preheater, its humidity will (be the same ), its relative humidity will (decrease ), its enthalpy will (increase), its dew point will (be the same ), its wet-bulb temperature will (increase ).(14) When the liquid flooding at a packed tower occurs, (pressure drop ) increases remarkably and the entire tower may fill with (liquid )..(15) Mathematic expression of Fick’s law is ( J A=-DdC A/dz ).(16) The requirements to satisfy constant molal overflow are (vaporization heat of different components will be the same, sensible heat exchange of liquid and gas phases on the plates can be ignored, heat loss of column can be igored. )(17)Some wet solids are dried by air with temperature t and humidity H.（1)When air velocity increases and air temperature t and humidity H keep constant, the drying rate at constant-rate period will (increase); If air velocity and air conditions keep unchanged, but thickness of materials increases, the critical moisture content of wet material Xc will (decrease ).2. (20 points) At a constant pressure operation a continuous distillation column with reflux is used for the separation of a binary mixture with feed rate F=1000kmol/h and concentration x F=0.36 (mol fraction of more volatile component). Feed is saturated vapor, the relative volatility ofmixture α is 3, indirect steam heating is used at a rebolier, and total condenser and bubble point reflux at the top of column are employed and the reflux ratio R is 3.2, respectively. It is required that the recovery percent of more volatile component at the top of column is 92%, and that the concentration of more volatile component at the top x D is 0.9. Calculate:（1）Operating line equation for rectifying section;(2) Overhead product flow rate D, vapor overflow V for rectifying section and vapor overflow V ’ for stripping section, in kmol/h.(3）If the practical liquid concentration leaving the first plate (x 1) is 0.825, what is this plate efficiency of plate 1L M E ,?Solution (1) y=0.762x+0.214(2) D x D /Fx F =0.92, D=0.92(1000*0.36)/0.9=368kmol/hV=(R+1)D=4.2*368=1545.6kmol/hV’=V -F=1545.6-1000=545.6kmol/h(3) E mL =(x D -x 1)/(x D -x 1*)=(0.9-0.825)/(0.9-0.75)=0.5y 1=x D =αx 1*/([α-1)x 1*+1]=0.9=3x 1*/[(3-1)x 1*+1]x 1*=0.753.(20 points) Some gas mixture contains acetone vapor with 3％(molar fraction) which is absorbed by pure water at a countercurrent operation absorber. 98% acetone of outlet gas is removed. Ratio of liquid to gas flow rate is 2 and the equilibrium relation is =*y 1.05x Find(1）what is the acetone concentration of outlet water at the tower bottom?(2) what is the overall mass transfer unit number of gas phase?(3) if the inlet compositions of gas and liquid phases keep the same and the ratio of liquid to gas flow rate is 1.04, where (in what position of tower) is the equilibrium of gas and liquid phases when the packing layer height is infinite? What is the maximum recovery percentage of acetone? Solution: (1) x 1=V(y 1-y 2)/L=0.03*0.98/2=0.0147(2) N oy =(y 1-y 1)/∆y m =0.03*0.98/0.00438=6.71∆y m =[(0.03-1.05*0.0147)-0.0006]/ln[(0.03-1.05*0.0147)/0.0006]=0.00438(3) since104.105.104.1 05.1 ='=='=VL m S VL m So the equilibrium will be reached at the bottom of tower. Max removal efficiency of acetone is121max y y y '-=η0286.005.103.005.11*1===y x 0297.00286.004.1*121=⨯=='-x VL y y %9903.00297.0121max =='-=y y y η4. (15 points) A moist material is to be dried from water content 25% to 5% (wet basis) in an adiabatic dryer (constant enthalpy drying process) under atmospheric pressure. The feed of moist material solid into the dryer is 1000 kg/h. After some fresh air with a dry-bulb temperature of 25ºC and a humidity of 0.013 kg water/kg dry air is preheated to 100ºC, it is sent to the dryer. And the air temperature leaving the dryer is 65ºC.(1) What are the bore-dry air (in kg/h) and the volume of fresh air required per unit time (in m 3/h)?(2) How much heat is obtained by air when it passes through the preheater (in kJ/h)? [Hint: 273273)244.1772.0(00+⨯+=t H v H , enthalpy of moist air=(1.01+1.88H)t+2492H] Solution: (1) W=Gc(X1-X2)=1000*0.75(1/3-5/95)=210.5kg/hI 1=I 2(1.01+1.88H 1)t 1+2492H 1=(1.01+1.88H 2)t 2+2492H 2H 1=0.013, t 1=100,t 2=65, H 2=0.0268L=W/(H 2-H 1)=210.5/(0.0268-0.013)=15253.6kgdry air/hv H =(0.772+1.244*0.013)(298/273)=0.86m 3/kgdry airV=Lv H =15253.6*0.86=13118.1 m 3/h(2)Qp=L(I 1-I 0)=15253.6(1.01+1.88*0.013)(100-25)=1183420 kJ/h5. Question (5 points)Draw the capacity performance chart of plate column and indicate the meaning of every line, describe the satisfactory operation zone. When the liquid flow rate keeps constant and vapor flow rate increases remarkably, please explain what may happen to the column.Solution: When the liquid flow rate keeps constant and vapor flow rate increases remarkably, more liquid will be taken upward the plate. So make separation efficiency of column decrease.。

试卷A/ B一、填空题：（共计25）1.按所依据的物理化学原理，传质分离过程可以分为平衡分离过程和速率分离过程，常见的平衡分离过程有精馏、吸收、闪蒸。

（5分）2.多组分精馏的FUG简捷计算法中，F代表Fenske 方程，用于计算N m ，U代表Underwood 公式，用于计算R m ，G代表Gilliland 关联，用于确定实际理论板数N 。

（6分）2.在多组分精馏的FUG简捷计算法中，用Fenske 方程计算N m，用Underwood 公式计算R m，用Gilliland晰分割法的基本假定是：釜液中除了 B 外没有其他 D 。

A. heavy key component;B. light key component;C. heavy non-key components;D. light non-key components.2.多组分精馏过程最多只能有b) 个关键组分；多组分吸收过程最多只能有a) 个关键组分（2分）。

a) 1 b) 2 c) 3 d) 43.The extent of separation achieved between or among the product phases for each of the chemical species present in the feed depends on the exploitation of differences in molecular分子性质、thermodynamic热力学性质and transport传递性质properties of the species in the different phases present.（4分）A. molecular;B. thermodynamic;C. transport;D. dielectric;E. diffusive;F. pH value.4.有效能定义为: B= . （2分）A. H-T0S;B. H-TS三、判断题（共10分，每题2分）1.相对挥发度相对于汽-液平衡常数而言，随温度和压力的变化更敏感。

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。

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?。

(8)In order to increase the rate of drying of wet solids mainly containing bound water, must the air velocity be increased ? A. Yes ； B. No.(9)In the air with C 025, and relative humidity 60%, the equilibrium water content of the wood is *1X . If the air relative humidity is changed into 40% under the same temperature, the equilibriumwater content of the wood becomes *2X . The relationship of the equilibrium water content is ( ).A. *2*1X X =;B. *2*1X X >;C. *2*1X X <; D. uncertain.(10)Generally speaking, the diffusivities of gases in liquid water are than in gas phases.2. Questions ( 7 points )What is the critical water content (X C ) in drying process? Describe the factors that influence X C .3. (15 points) A continuous distillation column with reflux is used for the separation of a mixture of heptane and octane (Heptane is the more volatile component). It is required that the distillate product is to contain 0.95 (mole fraction of heptane) and the bottoms product not more than 0.02 (mole fraction of heptane). The feed is liquid at its boiling point and contains 0.45 mole fraction of heptane. The vapor and liquid equilibrium data are given in the attached table. Determine: (1) The minimum number of ideal plates under total reflux.(2) The operating line equation for rectifying section. (3) The number of ideal plates if the reflux ratio is 2.5.y x10.90.80.70.60.50.40.30.20.1010.90.80.70.60.50.40.30.20.1Solution.(1) Draw the equilibrium line and the operating line (diagonal line), then draw steps between these two lines. As shown in the following figure, N min=8.3 ideal plates plus a reboiler0.020.95(2)If RD=2.5,(3) RD=2.5,0.950.450.0216.316151413121110987654321N=15.3 ideal plates plus a reboiler)2(4. (15 points) A porous solid is dried in a batch dryer under constant drying conditions. Seven hours are required to reduce the moisture content from 35 to 10 percent. The critical moisture contents are on the dry basis. Assuming that the rate of drying during the falling-rate period is proportional to the free-moisture content, how long should it take to dry a sample of the same solid from 35 to 5 percent under the same drying conditions?5. (15 points) Experimental data have been obtained for air containing 1.6% by volume SO2 being scrubbed with pure water in a packed column of 1.5 m2 in cross-sectional area and 3.5 m in packed height. Entering gas and liquid flow rates are 0.062 and 2.2 kmol/h, respectively. If the outlet mole fraction of SO2 in the gas is 0.004 and column temperature is near ambient with m=35.48 (that is, the equilibrium relation is y*=35.48x), calculate from the data:(a) The overall number of transfer units based on gas phase N oy for absorption of SO2(b) The overall height of a transfer unit based on gas phase H oy in meters(c) The volumetric overall mass transfer coefficient, K y a for SO2 in kmol / (m3 s).6. (18 points) Some kind of wet solids is dried in a continuous dryer under atmospheric pressure. The operation conditions are given as follows: States of air:At the inlet of the pre-heater, Air temperature and humidity are: C t 0020=, andair dry kg vapor water kg H /01.00=, respectively.At the inlet of the dryer, C t 01120=.At the outlet of the dryer, air dry kg vapor water kg H C t /05.0,70202==. States of wet solids: At the inlet of the dryer, %20,30101==w C θ(wet basis).At the outlet of the dryer, %5,50202==w C θ(wet basis).Specific heat of bone-dry solids, C kg kJ C s 0/5.1⋅=. The mass flow rate of final moist solids (product) is 53.5kg/h. Calculate: (1)Mass flow rate of dry air L , kg dry air/h ;(2)Heat transfer rate input into the preheater Q P , kJ/h ;(3)Supplementary heat transfer rate into the dryer Q D , kJ/h . Assume that heat losses of the drying equipment can be omitted.。

试卷A/ B 一、填空题：（共计25）1.按所依据的物理化学原理，传质分离过程可以分为平衡分离过程和速率分离过程，常见的平衡分离过程有精馏、吸收、闪蒸。

（5分）2.多组分精馏的FUG简捷计算法中，F代表 Fenske 方程，用于计算N m ，U代表Underwood 公式，用于计算R m ，G代表 Gilliland 关联，用于确定实际理论板数 N 。

（6分）2.在多组分精馏的FUG简捷计算法中，用Fenske 方程计算N m，用Underwood 公式计算R m，用Gilliland 关联确定N 。

（6分）3.The driving force for gas absorption is 气相中溶质的实际分压与溶液中溶质的平衡蒸气压力之差。

（2分）4．泡点计算是分离过程设计中最基本的汽液平衡计算。

泡点计算可分两种：泡点温度计算和泡点压力计算。

（2分）4.露点计算是分离过程设计中最基本的汽液平衡计算。

露点计算可分两种：露点温度计算和露点压力计算。

（2分）5.多组分吸收简捷计算中所用到的Horton-Franklin方程关联了吸收因子、吸收率、和理论板数。

（3分）6.多组分多级分离过程严格计算中围绕平衡级所建立的MESH方程分别是指：物料衡算方程，相平衡关系，组分摩尔分率加和方程和热量衡算方程（4分）7共沸精馏是：原溶液加新组分后形成最低共沸物，塔顶采出。

（3分）7萃取精馏是：原溶液加新组分后不形成共沸物且新组分沸点最高，从塔釜采出。

（3分）二、单项或多项选择题（共计10）1.晰分割法的基本假定是：馏出液中除了 A 外没有其他 C 。

（2分）A. heavy key component;B. light key component;C. heavy non-key components;D. light non-key components.1.晰分割法的基本假定是：釜液中除了 B 外没有其他 D 。

Problems for Mass Transfer and Separation ProcessAbsorption1 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?Solution ：09.0=y 05.0=x x y 97.0=*09.00485.005.097.0=<=⨯=*y y It is an absorption operation.2 When the temperature is 10 c 0and 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 the per cubic meter of water?Solution: the mole fraction of oxygen in air is 0.21,hence: p = P y =1x0.21=0.21amt64410*24.610*27.321.010*27.3-===p xBecause the x is very small , it can be approximately equal to molar ratio X , that is 610*42.6-=≈x XSo[])(/)(4.11)/(18*)(1)/(32*)(10*42.6lub 2322222226O H m O g O kmolH O kgH O kmolH kmolO kgO kmolO ility so ==-3 An acetone-air mixture containing 0.02 molar fraction of acetone is absorbed by water in apacked tower in countercurrent flow. And 99％ of acetone is removed, mixed gas molar flow fluxis 0.03kmol ·s —1m -2 , 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. V olume totalabsorption coefficient is K y a=0.022 kmol ·s —1m -2y -1.. What is the molar flow rate of the absorbent and what height of packing will be required?solution ：()0002.01=-=ηb a y y x a =0 733.175.102.099.002.0*min =⨯=--=⎪⎭⎫⎝⎛ab a b x x y y V L43.24.1min=⎪⎭⎫⎝⎛=V L V L s m k m o l L 20729.003.043.2=⨯=720.043.275.1===L mV S Number of mass transfer units N oy =(y 1-y 2)/∆y=12(y b -y a )=0.02-0.0002∆y=[(y b -y*b )- (y a -y*a )]/ln[(y b -y*b )/ (y a -y*a )] (y b -y*b )=0.02-1.75x b =0.0057X b =V/L (y b -y a )= (0.02-0.0002)/2.43=0.00815 (y a -y*a )= y a =0.0002 Or ])1)(ln[(11S S mx y mx y S N ba ab Oy +----==12 m Kya S V H OY 364.1022.003.0/===m N H H OY OY 37.1612364.1=⨯==4 The mixed gas from an oil distillation tower contains H 2S=0.04(molar fraction). Triethanolamine (absorbent) is used as the solvent to absorb 99% H 2S 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，overall volumeabsorption coefficient is Kya=0.05 kmol ·s —1m -2y -1, The solvent free of H 2S enters the tower and it contains 70% of the H 2S 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.solution ：ya=yb(1-0.99)=0.04*1%=0.0004xb*=yb/m=0.04/1.95= 0.0205 xb=0.7xb*=0.0144 yb*=1.95*0.0144=0.028yb-yb*=0.04-0.028=0.012 △ym=0.0034 Z=HoyNoyNoy=(yb-ya)/ △ym=11.6m a K G H y m oy 4.005.0/02.0/=== Z=11.6*0.4=4.64m5 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 is6 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 NH 3 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 and the min liquid —gas ratio L/V=?. (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?solution ：(1)35.12.103.08.0003.003.0=⨯-=GL(2) 89.035.12.1==S26.689.0003.003.011.089.011=⎥⎦⎤⎢⎣⎡+-=In N OY(3) m N Z H OYOY 958.026.66===47.889.0002.003.011.089.011=⎥⎦⎤⎢⎣⎡+⨯-='In N OYSince m N H Z OYOY 0.61.847.8958.0'>=⨯='= it is not suitable6 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 200c, respectively. The flux of gas V is 580kg/(m 2.h), and 6 percent (volume %) of ammonia is contained in the gas mixture. The flux of water L is 770kg/( m 2.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 k G a is proportional to V 0.8, but it has nothing to do with L. What 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 original Solution: 3,1,293Z m p atm T K ===1210.060.063810.06(10.99)0.000638Y Y Y ==-=-= The average molecular weight of the mixed gas M=29×0.94+17×0.06=28.2822580(10.06)19.28/()28.2877042.78/()180.919.280.405642.78V kmol m h Lkmol m h mV L =-=⋅Ω==⋅Ω⨯==12221ln[()(1)]110.06380ln[()(10.4056)0.4056]10.40560.0006386.88430.43586.884OG OG OGN mV LH Y mX mX mVY mX L L Z mN =-+-=----+-==== 1） 2p p '=''p p mm =So 1222ln[()(1)]10.90.4520.4519.280.202842.78111ln[(100)(10.2028)0.2028]10.20285.496OG mp m p m V LY mX m X m VN mV Y mX L L L -+='==⨯=''⨯==''-'=---+-= OG r G V V H K a K aP ==ΩΩSo:OG H changes with the operating pressure10.43580.21792OG OGOG OG H H H H ρρρρ'=''=⋅=⨯='So 5.4960.21791.1O G OG Z N H m '''=⋅=⨯= So the height of the packed section reduce 1.802m vs the original2） 2L L '=11()0.40560.20282225.496OGmV mV mV L L L N ===⨯=''=when the mass flow rate of liquid increases,G K a has not remarkable effect0.43585.4960.4358 2.395OGOG OGOG H H m Z N H m '=='''=⋅=⨯=the height of the packed section reduce 0.605m against the original3） 2V V '=(2)2()20.40560.81161ln[(100)(10.8116)0.8116]15.8110.8116OGmV m V mV LL LN '===⨯='=-+=-when mass flow rate of gas increaes,G K a also will increase. Since it is gas film control for absorption, we have as follows:0.80.80.80.20.80.2()222220.43580.50115.810.5017.92G G G G OG OG G G OGOG K a V V K a K a K a VV V H H K aP K aP mZ N H m m ∝''==''===Ω'Ω=⨯='''==⨯= So the height of the packed section increase 4.92m against the originalDistillation1 Certain binary mixed liquid containing mole fraction of easy volatilization component F x 0.35, 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 sections solution ：F x ＝0.35；x B =0.025；x D =0.96；R=3.2。

3月16日作业：一、请写出下列2个课题的中英文检索式1、拟对二氧化钛（TiO2）进行分子动力学研究，请写出检索表达式（中英文）。

提示：关键词1——二氧化钛（TiO2、Titanium、Titanate ）关键词2——分子动力学（Molecular Simulation、Molecular Dynamics）答：二氧化钛AND 分子动力学( TiO2 OR Titanium OR Titanate) AND ( Molecular Simulation OR Molecular Dynamics ）2.查找形状记忆合金作为密封元件在法兰连接中应用的相关文献，编写检索式。

（注意寻找近义词、相关词）提示：关键词1——形状记忆合金、形状记忆效应（shape memory alloy、SMA、shape momory effect、SME）关键词2——密封、垫圈、垫片（seal，seal*，washer，gasket）关键词3——法兰连接、螺栓连接（flanged connection , flange*, bolted flanged connection, bolt* , BFC）答：（形状记忆合金OR形状记忆效应) AND (密封OR垫圈OR垫片) AND (法兰连接OR 螺栓连接)（“shape memory alloy” OR SMA OR “shape momory effect” OR SME）AND （seal OR seal* OR washer OR gasket）AND（flanged connection OR flange* OR”bolted flanged connection” OR bolt* OR BFC）二、请根据以下文献的信息，将它们以参考文献的标准著录格式列出。

（1）期刊文章题名: 原油常减压蒸馏塔的流程模拟Author-作者: 殷卫兵;罗雄麟;史伟;Organ-单位: 中国石油大学自动化研究所;中国石油锦西石化分公司;Source-文献来源: 化工自动化及仪表Keyword-关键词: 原油;;常减压蒸馏;;流程模拟PubTime-发表时间: 2010-05-10Year-年: 2010Period-期: 05PageCount-页码: 88-93答：[1]殷卫兵,罗雄麟,史伟.化工自动化及仪表[J].原油常减压蒸馏塔的流程模拟,2010(05):88-93.（2）期刊文章题名: 无冷凝器及再沸器的热集成蒸馏塔技术进展Author-作者: 李娟娟;陆恩锡;张翼;Organ-单位: 华南理工大学传热强化与过程节能教育部重点实验室;华南理工大学传热强化与过程节能教育部重点实验室;华南理工大学传热强化与过程节能教育部重点实验室广东广州510640;广东广州510640;广东广州510640Source-文献来源: 化学工程Keyword-关键词: 蒸馏;;内部热集成;;热力学效率PubTime-发表时间: 2006-09-30Year-年: 2006Period-期: 09PageCount-页码: 1-4答：[1]李娟娟,陆恩锡,张翼.化学工程[J].无冷凝器及再沸器的热集成蒸馏塔技术进展,2006(09):1-4.（3）学位论文题名: 松香蒸馏塔控制系统的设计研究与仿真Author-作者: 蒋晶晶Source-文献来源: 昆明理工大学Keyword-关键词: 神经网络,PID参数自适应,MA TLAB仿真,解耦控制PubTime-发表时间: 2004-03-30FirstDuty-第一责任人.: 蒋晶晶答：[1]蒋晶晶.松香蒸馏塔控制系统的设计研究与仿真[D]. 昆明:昆明理工大学,2004.（4）学位论文题名: 渗流型催化蒸馏塔内件结构及流体力学性能研究Author-作者: 颜亚盟Source-文献来源: 天津大学Keyword-关键词: 催化蒸馏塔内件;;积液高度;;压降;;混合装填PubTime-发表时间: 2009-06-01FirstDuty-第一责任人: 颜亚盟答：[1]颜亚盟.渗流型催化蒸馏塔内件结构及流体力学性能研究[D].天津:天津大学,2009.（5）图书ISBN: 7-5025-8720-9 CNY29.00题名/责任: 传热、蒸发与冷冻操作实训/ 潘学行主编出版发行项: 北京: 化学工业出版社, 2006载体形态项: 249页: 图; 24cm中图法分类号: TQ021.3 4中图法分类号: TQ028.6 4个人著者等同: 潘学行主编答：[1]潘学行.传热、蒸发与冷冻操作实训.北京: 化学工业出版社, 2006,0-249（6）图书ISBN: 7-5025-8441-2 CNY27.00题名/责任: 传质与分离操作实训/ 潘文群主编出版发行项: 北京: 化学工业出版社, 2006载体形态项: 221页: 图; 24cm中图法分类号: TQ021.4 4中图法分类号: TQ028 4个人著者等同: 潘文群主编答：[1]潘文群.传质与分离操作实训.北京: 化学工业出版社, 2006,0-221（7）中国专利答：[1]北京工业大学.外部/内部热耦合蒸馏塔的控制与优化方法:中国,CN201110190641.4[P].2011-11-16.（8）中国专利（9）期刊论文题名: Characteristics of dynamic membrane filtration:structure;operation mechanisms;and cost analysisAuthor-作者: Yalei Zhang;Yangying Zhao;Huaqiang Chu;Bingzhi Dong;Xuefei Zhou;Organ-单位: State Key Laboratory of Pollution Control and Resources Reuse;Key Laboratory of Yangtze Aquatic Environment;School of Environmental Science and Engineering;Tongji University;Source-文献来源: Chinese Science BulletinPubTime-发表时间: 2014-01-25Year-年: 2014Period-期: 03PageCount-页码: 247-260答：[1] Yalei Zhang , Yangying , Zhao.Huaqiang Chu , Bingzhi Dong , Xuefei Zhou.Characteristics of dynamic membrane filtration:structure;operation mechanisms;and cost analysis[J].Chinese Science Bulletin,2014,(03):247-260（10）期刊论文题名: Reduction of DOM fractions and their trihalomethane formation potential in surface river water by in-line coagulation with ceramic membrane filtration Author-作者: Pharkphum Rakruam;Suraphong Wattanachira;Organ-单位: International Postgraduate Programs in Environmental Management; Graduate School; Chulalongkorn University;Center of Excellence on Hazardous Substance Management;Chulalongkorn University;Department of Environmental Engineering/Center of Excellence on Hazardous Substance Management; Faculty of Engin...Source-文献来源: Journal of Environmental SciencesPubTime-发表时间: 2014-03-01Year-年: 2014Period-期: 03PageCount-页码: 529-536答:[1]Pharkphum , Rakruam , Suraphong Wattanachira.Reduction of DOM fractions and their trihalomethane formation potential in surface river water by in-line coagulation with ceramic membrane filtration[J].Journal of Environmental Sciences,2014,(03):529-536.。