第七章气态污染物控制技术基础

作业习题解答第七章气态污染物控制技术基础7.1解：由亨利定律P*=Ex，500×2%=1.88×105x，x=5.32×10－5。由y*=mx，m=y*/x=0.02/5.32×10－5=376。因x=5.32×10－5很小，故CCO2=2.96mol/m3。)/(1096.210%250096.2343*PammolPCH100g与气体平衡的水中约含44×100×5.32×10－5/18=0.013g。7.2解：在1atm下O2在空气中含量约0.21。0.21=4.01×104x解得O2在水中摩尔分数为x=5.24×10－6。7.3解：20》C时H2SE=0.489×105kPa，分压20atm×0.1%=2.03kPa。P*=Ex，x=P*/E=4.15×10－5，故C*H2S=2.31mol/m3。H=C/P*=2.3/（2.03×103）=1.14×10－3mol/（m3.Pa）=115mol/（m3.atm）由185.1,542.0108121611511hKkkHKAllgAl。)/(3.431.285.1)(32*2hmmolCCKNSHSHAlA。7.4解：GB=5000×0.95=4750m3N/h。Y1=0.053，321063.2%54750)47505000(Y；4.257.26/053.000263.0053.00)(max21minXYYGLBS。因此用水量Ls=25.4GB×1.5=1.81×105m3N/h。由图解法可解得传质单元数为5.6。7.5解：GB=10×0.89=8.9m3/min，Y1=0.124，Y2=0.02。作出最小用水时的操作线，xmax=0.068。故53.1068.002.0124.0)(minBsGL，Ls=1.53×1.75×8.9=23.8m3/min。图解法可解得传质单元数为3.1。mLaHy39.2)(3.333.0。Hy=2.39×3.1=7.4m。7.6解：利用公式0KL，将已知数据代入002.05051.0220KK，解得min65min/28500mK因此min27856512850max。7.7解：mVaKbmin/8.95103020230)0129.02629.0(30min8.95'KL，kgaSLxb2.452301141)0129.02629.0(2。7.8解：XTcm3/gPatmlgXTlgPP/V3011.47700.0335121.7080.3010.0396731.8260.4770.0458141.9090.6020.0499351.9690.6990.05410462.0170.7780.058依据公式nTkPX1，对lgXT~lgP进行直线拟合：7.030PXT，即K=30，n=1.43；依据公式mmVPBVVP1，对P~P/V进行直线拟合：PVP005.00289.0，即Vm=200，B=0.173。7.9解：三氯乙烯的吸收量V=2.54×104×0.02×99.5%=505.46m3/h，M=131.5。由理想气体方程RTMmPV得hkgRTPVMm/1075.329431.85.13146.5051038.135因此活性炭用量kgm4301036.541075.328100；体积3409.925771036.5mmV。7.10解：Y1=0.025kg苯/kg干空气，硅胶苯kgkgYX/282.0)167.0(5.1111，Y2=0，X2=0。故操作线方程为X=11.28Y。当Y=Yb=0.0025kg苯/kg干空气时，X=11.28×0.0025=0.0282kg苯/kg硅胶。Y*=0.167×0.02821.5=0.0008kg苯/kg干空气。08.5881*YY，由此可求得YYbYYdY*近似值；同时，WeWbababWeWbAAb))(1()1(000WeWbYeYbYYbYeYbdYYYbYYdYYYdYYYdYYYdYYY))(1(****0由此求得f的近似值，列表如下：YY**1YY*YYdYYYbYYdY*ab01YYAWdwYY)1(0wewbAWdwYY)1(0Yb=0.00250.0008588.080000.9000.00500.0022361.901.1841.1840.19900.80.16920.16920.00750.0041294.930.8212.0050.33710.70.10350.27270.01000.0063272.240.7092.7140.45630.60.07750.35020.01250.0088273.370.6823.3960.57090.50.06310.41330.01500.0116296.120.7124.1080.69060.40.05390.46710.01750.0146350.460.8084.9160.82650.30.04760.5147Ye=0.02000.0179475.001.0325.9481.00000.20.04340.5580NOG=5.948，f=0.5580；2atm，298K时，=2.37kg/m3，因此)/(37.22smkgvG，故HOG=mGDap07041.0)10835.131.21060.0(60042.1)(42.151.05251.0；因此吸附区高度为H2=HOG.NOG=0.07041×5.948=0.419m。对单位横截面积的床层，在保护作用时间内吸附的苯蒸汽量为（0.025－0）×2.37×60×90＝320（kg苯/m2）而吸附床饱和区吸附苯蒸汽量)0()(2TbxHH吸附床未饱和区吸附苯蒸汽量)1)(0(2fxHTb因此总吸附量320442.0282.0625419.0282.0625)419.0(H解得H=2.05m，此即所需要的最小床高。7.11解：反应管转化率为xA时，反应速度为RA=－0.15（1－xA）mol/（kg催化剂.min）。根据单管物料平衡可列出如下方程：AAQdxAdxx)1(15.0其中2322101.1)108.3(4mA，Q单位为mol/min。数据代入并整理得AAxdxQdx1098668.0，对等式两边积分，即74.001.601098668.0AAxdxQdx，解得Q=0.447mol/min。反应管数目：250/0.447＝560个。7.12解：kJQ531056.438.171106475.56227由TcmQ得KcmQT31472642.01056.45。Chapter71.Solution：Iftheequilibriumlinecanbeassumedtobestraight,thenitsslopeis0.03/0.0027=11.1.TheenteringmolefractionofSO2is0.03andthatleavingis0.003.AfreshabsorbentimpliesthatthereisnodissolvedSO2intheenteringliquid,i.e.,x1=0.Thevalueof(L´/G´)minisobtainedfromtheslopeofthelinebetween(x1,y1)=(0,0.003)and(0.0027,0.03),namely1000027.0003.003.02.Solution：Thefluegascontains1000ppmofNO,avolumefractionof0.001.Atarateof1000m3s-1,thisis1m3NOs-1.ThenumberofmolesofNOin1m3s-1at573Kand1atmis61(1)(10)21.27(82.05)(573)PVngmolesNOsRTFor75%removal,15.95g-molesNOs-1aretoberemoved.Thestoichiometricreactionforselectivecatalyticreductionis4NO+4NH3+O2→4N2+6H2OThusaNH3feedrateof15.95g-moless-1,271.2gs-1,or976.3kgh-1,isrequired.