大气污染控制工程第三版课后习题答案(18章全).docx

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1、大气污染控制工程课后作业习题解答第一章概论Ll解*按ImOl干空气计算，空气中各组分摩尔比即体积比，11N2=.781mol,11o2=0209mol,nAr=000934mol,11co2=.00033molo质量百分数为n2% =0.78128.01sc。/ruv0.20932.00CCCC-X100%=75.51%,。2%=X100%=23.08%；28.971228.97x10.93439.94八fw0.03344.01Aroo100%=1.29%，CO2%=X100%=0.05%。28.971228.9711.2?: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：SO

2、2：0.15mgm3,NO2：0.12mgm3,CO：4.0Omg/m。按标准状态下Im3干空气计算，其摩尔数为1Xin3-=44.(A3nolo故三种污染物体积百分数分别为:22.40.15x10-30.12x10-3SO：=0.052Ppm，NO，：=0.058PPm64x44.64346x44.6434.00x10-3CO：=3.20ppm。28x44.643L3解:I)Pl gm3N)1.50 104 15422.4 10 3L031g*1.50 10422.4103=6.70 10 3 mol / 町;。2)每天流经管道的CCI4质量为1.03110X360024103kg=891k

3、gL4解，每小时沉积量200X(5001560106)X0.12图=10.8由LS解，由大气污染控制工程P14(1-1),取M=210=M区=2102x1=0.2369,O2HbPo,19.510-2COHb饱和度pcoCOHhCoHbIo?HbCOHb-O2Hb +COHb IO2Hb 1 + 0.23690.2369 =915%1.6解:含氧总量为480X2=96OmL.不同Co百分含量对应Co的量为:1002%：-2%=19.59nL,7%：7%=72.26/nL98%93%1）最初Co水平为0%时t=-=172.0min；4.2104103r.l.ITAL11.72.2619.591.

4、2）最初CO水平为2%时t=;-=125.4mm,4.2x10TXIo3L7解，由大气污染控制工程P18(1-2),最大能见度为=11581 .Sm o2.6ppdp_2.6x1400x1.4Kp-2.20.2第二章燃烧与大气污染2.1解:Ikg燃油含:重量(g)摩尔数(g)需氧数(g)C85571.2571.25H113-2.555.2527.625S100.312503125H2O22.51.250N元素忽略。1）理论需氧量71.25+27.625+0.3125=99.1875molkg设干空气O2：N2体积比为1：3.7干那么理论空气量99.1875X4.78=474.12molkg重油

5、。即474.12X22.4/1000=10.62m3N/kg重油。烟气组成为C271.25mol,H2O55.25+1.25=56.50mol,SO20.1325mol,N23.7899.1875=374.93molo理论烟气量71.25+56.50+0.3125+374.93=502.99molkg重油。即502.9922.4/1000=1L27m3N/kg重油。2）干烟气量为502.99-56.50=446.49molkg重油。03125SO2百分比浓度为100%=0.07%,446.497125空气燃烧时CO2存在最大浓度100%=l5.96%。446.493）过剩空气为10%时，所需空

6、气量为1.1X10.62=11.68Nkg重油，产生烟气量为U.267+0.1X10.62=12.33m3N/kg重油。2.2?：相对于碳元素作如下计算：%1质量)mol100g煤mol/mol碳C65.75.4751H3.23.20.584S1.70.0530.010O2.30.0720.013灰分18.13.306gmol碳水分9.0L644gmol碳故煤的组成为CH,584S,01O.013燃料的摩尔质量(包括灰分和水分)为=18.26g/MHC。燃烧方程式为5.475n=1+0.584/4+0.010-0.013/2=1.14951)理论空气量495xQ+3.78)XIOooX22.4

7、X10-3而Ikg=6.74/kg；18.26SO2在湿烟气中的浓度为ET100%=0.174%16441+0.292+0.010+3.781.1495-182)产生灰分的量为18.Ix国独X80%=144.8gkg100烟气量(1+0.292+0.010+3.78X1.1495+1.644/18)1000/18.2622.4103=6.826m3kg144R、灰分浓度为-X10mgm3=2.12104mgm36.8263)需石灰石一也理35%2.3解：-=103.2Lt煤按燃烧Ikg煤计算重量(g)摩尔数(mol)需氧数iC79566.2566.25H31.12515.56257.78S60

8、.18750.1875H2O52.8752.940设干空气中N2:。2体积比为3.78：1,(mol)10001.7%小1.740所需理论空气量为4.78X(66.25+7.78+0.1875)=354.76molkg煤。理论烟气量CO266.25mol,SO20.1875mol,H2O15.5625+2.94=18.50molN2 3.78x354.764.78=280.54?。/总计66.25+8.50+0.1875+280.54=365.48molkg煤01875实际烟气量365.48+02X35476=43643mMg煤，S5浓度为曲x1。=0.043%。2.4解:耗氧量0.003mo

9、l00.143mol0.063mol0.014mol取ImoI煤气计算H2S0.002molCO20.05molCO0.285molH2(0.13-0.004)molCH0.007mol三O20.003+0.143+0.063+0.014=0.223mol0设干空气中N2：。2体积比为3.78：1,那么理论干空气量为0.223X(3.78+1)=1.066moL取=1.2,那么实际干空气1.21.066mol=1.279moL空气含湿量为12gnN,即含H2OO67mol/11A1,14.94Lm3No故H2O体积分数为1.493%。故实际I279空气量为一:=1.298机Ho1-1.493%

10、烟气量SO2:0.002mol,CO2：0.285+0.007+0.05=0.342mol,N2:0.223X3.78+0.524=1367mol,H200.0020.126+0.014+1.298 1.493%0.004=0.201 mol 故实际烟气量 0002+0342+1.367+0.201+02X 1.066=2.125mol 2.5 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8-0.520.264 78.99-(8 - 0.5 2) 100% = 50.5%2)在测定状态下，气体的摩尔体积为_

11、 PiVl T2 _ 101325 X 22.4 X 4432) 丁月 一 273x7(X)x133.322=39.46L/mol ；取In?烟气进行计算，那么SCM 20X 10 6n3,排放浓度为120x10639.46x10-3(l-8%)64 = 0.17n29 43) 5663.37XX(1-8%) = 2957喝/min。39 464) 30.0= 52.85g/63按Ikg煤进行计算摩尔数(mol)需氧数(mol)重量(g)C75863.1763.17H40.7520.37510.19S160.50.5H2O83.254.62502.6 解:需氧 63.17+10.19+0.5=

12、73.86mol设干空气中N2：。2体积比为3.78： 1,那么干空气量为73.86X478X1.2=423.66mol, 含水 423.66 0.0116=4.91 mol。烟气中：CO2 63.17mol; SO2 0.5mol; H2O 4.91+4.625+20.375=29.9ImohN2: 73.863.78=279.19mol;过剩干空气 0.2X73.86X4.78=70.6Imo1。实际烟气量为 63.17+0.5+29.91+279.1970.61 =44338mol其中CChH2O63 100% = 14.25%; SO2-100% = 0.11%;443.38443.3

13、829.91 SS皿 Z 279.19 + 0.79x70.61 皿 100% = 6.74%； N2 100% = 75.55% -O2443.3870.610.209 皿 Cc X100% = 3.33% o443.38443.382.7 解*SO?含量为0.11%,估计约1/60的SCh转化为SO3，那么SO3含量0.11%-=1.83105,即Ph2so4=183X105,gpH2s04=4737060查图27得煤烟气酸露点约为134摄氏度。2.8 解：以Ikg油燃烧计算，C860g71.67mol;H140g70mol,耗氧35mol.设生成CoXmoL耗氧0.5XmO1,那么生成CCh(71.67x)mol,耗氧(71.67x)molo烟气中O2量一I%600x10-61 5%总氧量一,+0.5x+(71.67-x)+35=106.6724.5x，干空气中N2：Ch体积比为3.78：1,600106那么含N23.78X(106.67+24.5x)o根据干烟气量可列出如下方程：1 5%YX,+71.67+3.78(106.67+24.5%)=r，解得x=0.306600x10-6600W6故Co2%：7吗段306xl00%=13.99%：0.306600106N2%：3.78(24.50.306+106.67)0.306