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课后答案网,用心为你服务! 大学答案---中学答案---考研答案---考试答案 最全最多的课后习题参考答案,尽在课后答案网()!Khdaw团队一直秉承用心为大家服务的宗旨,以关注学生的学习生活为出发点,旨在为广大学生朋友的自主学习提供一个分享和交流的平台。 爱校园()课后答案网()淘答案() 1111/29292929第一章概论1.11.11.11.1解:按1mol干空气计算,空气中各组分摩尔比即体积比,故nN2=0.781mol,nO2=0.209mol,nAr=0.00934mol,nCO2=0.00033mol。质量百分数为,;%51.75%100197.2801.28781.0%2=×××=N%08.23%100197.2800.32209.0%2=×××=O,。%29.1%100197.2894.3900934.0%=×××=Ar%05.0%100197.2801.4400033.0%2=×××=CO1.21.21.21.2解:由我国《环境空气质量标准》二级标准查得三种污染物日平均浓度限值如下:SO2:0.15mg/m3,NO2:0.12mg/m3,CO:4.00mg/m3。按标准状态下1m3干空气计算,其摩尔数为。故三种污染物体积百分数分别为:mol643.444.221013=×SO2:,NO2:ppm052.0643.44641015.03=××−ppm058.0643.44461012.03=××−CO:。ppm20.3643.44281000.43=××−1.31.31.31.3解:1)(g/m3N)ρ334/031.1104.221541050.1Nmg=×××=−−c(mol/m3N)。3334/1070.6104.221050.1Nmmol−−−×=××=2)每天流经管道的CCl4质量为1.031×10×3600×24×10-3kg=891kg1.41.41.41.4解:每小时沉积量200×(500×15×60×10-6)×0.12=10.8gµgµ1.51.51.51.5解:由《大气污染控制工程》P14(1-1),取M=210,2369.0105.19102.22102422=×××==−−∝OppMHbOCOHbCOHb饱和度%15.192369.012369.0/1/222=+=+=+=HbOCOHbHbOCOHbHbOCOHbCOHbCOρ1.61.61.61.6解:含氧总量为。不同CO百分含量对应CO的量为:mL960100204800=×2%:,7%:mL59.19%2%98960=×mL26.72%7%93960=×1)最初CO水平为0%时;min0.17210102.426.7234=××=−t2)最初CO水平为2%时。min4.12510102.459.1926.7234=××−=−t1.71.71.71.7解:由《大气污染控制工程》P18(1-2),最大能见度为。mKdLppv8.115812.02.24.114006.26.2=×××==ρρ2222/29292929第二章燃烧与大气污染2.12.12.12.1解:1kg燃油含:重量(g)摩尔数(g)需氧数(g)C85571.2571.25H113-2.555.2527.625S100.31250.3125H2O22.51.250N元素忽略。1)理论需氧量71.25+27.625+0.3125=99.1875mol/kg设干空气O2:N2体积比为1:3.78,则理论空气量99.1875×4.78=474.12mol/kg重油。即474.12×22.4/1000=10.62m3N/kg重油。烟气组成为CO271.25mol,H2O55.25+11.25=56.50mol,SO20.1325mol,N23.78×99.1875=374.93mol。理论烟气量71.25+56.50+0.3125+374.93=502.99mol/kg重油。即502.99×22.4/1000=11.27m3N/kg重油。2)干烟气量为502.99-56.50=446.49mol/kg重油。SO2百分比浓度为,%07.0%10049.4463125.0=×空气燃烧时CO2存在最大浓度。%96.15%10049.44625.71=×3)过剩空气为10%时,所需空气量为1.1×10.62=11.68m3N/kg重油,产生烟气量为11.267+0.1×10.62=12.33m3N/kg重油。2.22.22.22.2解:相对于碳元素作如下计算:%(质量)mol/100g煤mol/mol碳C65.75.4751H3.23.20.584S1.70.0530.010O2.30.0720.013灰分18.13.306g/mol碳水分9.01.644g/mol碳故煤的组成为CH0.584S0.010O0.013,燃料的摩尔质量(包括灰分和水分)为。燃烧方程式为molCg/26.18475.5100=222222013.0010.0584.078.3010.0292.0)78.3(nNSOOHCONOnOSCH+++→++n=1+0.584/4+0.010-0.013/2=1.14951)理论空气量;kgmkgm/74.6/104.22100026.18)78.31(1495.1333=×××+×−SO2在湿烟气中的浓度为%174.0%10018644.11495.178.3010.0292.01010.0=×+×+++2)产生灰分的量为kgg/8.144%8010010001.18=××烟气量(1+0.292+0.010+3.78×1.1495+1.644/18)×1000/18.26×22.4×10-3=6.826m3/kg灰分浓度为mg/m3=2.12×104mg/m3310826.68.144×3)需石灰石/t煤kg21.103%35407.100.32%7.11000=×××2.32.32.32.3解:3333/29292929按燃烧1kg煤计算重量(g)摩尔数(mol)需氧数(mol)C79566.2566.25H31.12515.56257.78S60.18750.1875H2O52.8752.940设干空气中N2:O2体积比为3.78:1,所需理论空气量为4.78×(66.25+7.78+0.1875)=354.76mol/kg煤。理论烟气量CO266.25mol,SO20.1875mol,H2O15.5625+2.94=18.50molN2mol54.28078.476.35478.3=×总计66.25+`8.50+0.1875+280.54=365.48mol/kg煤实际烟气量365.48+0.2×354.76=436.43mol/kg煤,SO2浓度为。%043.0%10043.4361875.0=×2.42.42.42.4解:取1mol煤气计算H2S0.002mol耗氧量0.003molCO20.05mol0CO0.285mol0.143molH2(0.13-0.004)mol0.063molCH40.007mol0.014mol共需O20.003+0.143+0.063+0.014=0.223mol。设干空气中N2:O2体积比为3.78:1,则理论干空气量为0.223×(3.78+1)=1.066mol。取,则实际干空气1.2×1.066mol=1.279mol。2.1=α空气含湿量为12g/m3N,即含H2O0.67mol/m3N,14.94L/m3N。故H2O体积分数为1.493%。故实际空气量为。mol298.1%493.11279.1=−烟气量SO2:0.002mol,CO2:0.285+0.007+0.05=0.342mol,N2:0.223×3.78+0.524=1.367mol,H2O0.002+0.126+0.014+1.298×1.493%+0.004=0.201mol故实际烟气量0.002+0.342+1.367+0.201+0.2×1.066=2.125mol2.52.52.52.5解:1)N2%=1-11%-8%-2%-0.012%=78.99%由《大气污染控制工程》P46(2-11)空气过剩%5.50%100)25.08(99.78264.025.08=××−−××−2)在测定状态下,气体的摩尔体积为;molLPTTVPV/46.39322.1337002734434.22101325221112=××××=⋅=取1m3烟气进行计算,则SO2120×10-6m3,排放浓度为。63312010(18%)640.179/39.4610gm−−××−×=×3)。322.45663.37(18%)2957/min39.46Nm××−=4)。3/85.5222.439.460.03Nmg=×2.62.62.62.6解:按1kg煤进行计算重量(g)摩尔数(mol)需氧数(mol)C75863.1763.17H40.7520.37510.19S160.50.5H2O83.254.62504444/29292929需氧63.17+10.19+0.5=73.86mol设干空气中N2:O2体积比为3.78:1,则干空气量为73.86×4.78×1.2=423.66mol,含水423.66×0.0116=4.91mol。烟气中:CO263.17mol;SO20.5mol;H2O4.91+4.625+20.375=29.91mol;N2:73.86×3.78=279.19mol;过剩干空气0.2×73.86×4.78=70.61mol。实际烟气量为63.17+0.5+29.91+279.19+70.61=443.38mol其中CO2;SO2;%25.14%10038.44317.63=×%11.0%10038.4435.0=×H2O;N2。%74.6%10038.44391.29=×%55.75%10038.44361.7079.019.279=××+O2。%33.3%10038.443209.061.70=××2.72.72.72.7解:SO2含量为0.11%,估计约1/60的SO2转化为SO3,则SO3含量,即PH2SO4=1.83×10-5,lgPH2SO4=-4.737。51083.1601%11.0−×=×查图2-7得煤烟气酸露点约为134摄氏度。2.82.82.82.8解:以1kg油燃烧计算,C860g71.67mol;H140g70mol,耗氧35mol。设生成COxmol,耗氧0.5xmol,则生成CO2(71.67-x)mol,耗氧(71.67-x)mol。烟气中O2量。610600%5.1−×x总氧量,干空气中N2:O2体积比为3.78:1,则xxxx5.2467.10635)67.71(5.010600%5.16+=+−++×−含N23.78×(106.67+24.5x)。根据干烟气量可列出如下方程:,解得x=0.3066610600)5.2467.106(78.367.7110600%5.1−−×=+++×xxx故CO2%:;%99.13%10010600306.0306.067.716=××−−N2%:%62.84%10010600306.0)67.106306.05.24(78.36=××+×−由《大气污染控制工程》P46(2-11)空气过剩系数07.1)06.05.05.1(62.84264.006.05.05.11=×−−××−+=α第三章大气污染气象学3.13.13.13.1解:由气体静力学方程式,大气中气压随高度的变化可用下式描述:(1)dPgdZρ=−⋅将空气视为理想气体,即有可写为(2)mPVRTM=mPMVRTρ==将(2)式带入(1),并整理,得到以下方程:dPgMdZPRT=−5555/29292929假定在一定范围内温度T的变化很小,可以忽略。对上式进行积分得
本文标题:大气污染控制工程课后答案
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