07传质与分离工程期末考试题(含答案)

,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.。