第七章 气态污染物控制技术基础.doc

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，5002%=1.88105x，x=5.32105。 由y*=mx，m=y*/x=0.02/5.32105=376。因x=5.32105很小，故CCO2=2.96mol/m3。100g与气体平衡的水中约含441005.32105/18=0.013g。7.2 解：在1atm下O2在空气中含量约0.21。0.21=4.01104x解得O2在水中摩尔分数为x=5.24106。7.3 解：20C时H2S E=0.489105kPa，分压20atm0.1%=2.03kPa。P*=Ex，x=P*/E=4.15105，故C*H2S=2.31mol/m3。H=C/P*=2.3/（2.03103）=1.14103mol/（m3.Pa）=115mol/（m3.atm）由。7.4 解：GB=50000.95=4750m3N/h。Y1=0.053，；。因此用水量Ls=25.4GB1.5=1.81105m3N/h。由图解法可解得传质单元数为5.6。7.5 解：GB=100.89=8.9m3/min，Y1=0.124，Y2=0.02。作出最小用水时的操作线，xmax=0.068。故，Ls=1.531.758.9=23.8m3/min。图解法可解得传质单元数为3.1。Hy=2.393.1=7.4m。7.6解：利用公式，将已知数据代入，解得因此。7.7解： ，。7.8解：XT cm3/gP atmlgXTlgPP/V3011.47700.0335121.7080.3010.0396731.8260.4770.0458141.9090.6020.0499351.9690.6990.05410462.0170.7780.058依据公式，对lgXTlgP进行直线拟合：，即K=30，n=1.43；依据公式，对P P/V进行直线拟合：，即Vm=200，B=0.173。7.9 解：三氯乙烯的吸收量V=2.541040.0299.5%=505.46m3/h，M=131.5。由理想气体方程得因此活性炭用量；体积。7.10 解：Y1=0.025kg苯/kg干空气，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=11.280.0025=0.0282kg苯/kg硅胶。Y*=0.1670.02821.5=0.0008kg苯/kg干空气。，由此可求得近似值；同时， 由此求得f的近似值，列表如下：YY*Yb=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，故HOG=；因此吸附区高度为H2=HOG.NOG=0.070415.948=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（0.0250）2.376090320（kg苯/m2）而吸附床饱和区吸附苯蒸汽量吸附床未饱和区吸附苯蒸汽量因此总吸附量解得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（1xA）mol/（kg催化剂.min）。根据单管物料平衡可列出如下方程：其中，Q单位为mol/min。数据代入并整理得，对等式两边积分，即，解得Q=0.447mol/min。反应管数目：250/0.447560个。7.12 解：由得。Chapter 71. Solution：If the equilibrium line can be assumed to be straight, then its slope is 0.03/0.0027=11.1. The entering mole fraction of SO2 is 0.03 and that leaving is 0.003. A fresh absorbent implies that there is no dissolved SO2 in the entering liquid , i.e., x1=0. The value of (L/ G)min is obtained from the slope of the line between (x1,y1)=(0,0.003) and (0.0027,0.03), namely2. Solution：The flue gas contains 1000 ppm of NO, a volume fraction of 0.001. At a rate of 1000 m3 s-1, this is 1 m3 NO s-1 . The number of moles of NO in 1m3 s-1 at 573K and 1 atm is For 75% removal, 15.95 g-moles NO s-1 are to be removed. The stoichiometric reaction for selective c