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2.1已知重油元素分析结果如下:C:85.5%H:11.3%O:2.0%N:0.2%S:1.0%,试计算:1)燃油1kg所需理论空气量和产生的理论烟气量;2)干烟气中SO2的浓度和CO2的最大浓度;3)当空气的过剩量为10%时,所需的空气量及产生的烟气量。解:1kg燃油含:重量(g)摩尔数(g)需氧数(g)C85571.2571.25H113-2.555.2527.625S100.31250.3125H2O22.51.250(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2.2普通煤的元素分析如下:C65.7%;灰分18.1%;S1.7%;H3.2%;水分9.0%;O2.3%。(含N量不计)1)计算燃煤1kg所需要的理论空气量和SO2在烟气中的浓度(以体积分数计);2)假定烟尘的排放因子为80%,计算烟气中灰分的浓度(以mg/m3表示);3)假定用硫化床燃烧技术加石灰石脱硫。石灰石中含Ca35%。当Ca/S为1.7(摩尔比)时,计算燃煤1t需加石灰石的量。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molCg/26.18475.5100。燃烧方程式为:222222013.0010.0584.078.3010.0292.0)78.3(nNSOOHCONOnOSCHn=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%10018644.11495.178.3010.0292.01010.02)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310826.68.144mg/m3=2.12×104mg/m33)需石灰石kg21.103%35407.100.32%7.11000/t煤2.3煤的元素分析结果如下S0.6%;H3.7%;C79.5%;N0.9%;O4.7%;灰分10.6%。在空气过剩20%条件下完全燃烧。计算烟气中SO2的浓度。解: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%10043.4361875.02.4某锅炉燃用煤气的成分如下:H2S0.2%;CO25%;O20.2%;CO28.5%;H213.0%;CH40.7%;N252.4%;空气含湿量为12g/m3N,2.1,试求实际需要的空气量和燃烧时产生的实际烟气量。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2.1,则实际干空气1.2×1.066mol=1.279mol。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=2.125mol2.5干烟道气的组成为:CO211%(体积),O28%,CO2%,SO2120×10-6(体积分数),颗粒物30.0g/m3(在测定状态下),烟道气流流量在700mmHg和443K条件下为5663.37m3/min,水气含量8%(体积)。试计算:1)过量空气百分比;2)SO2的排放浓度(3/mg);3)在标准状态下(1atm和273K),干烟道体积;4)在标准状态下颗粒物的浓度。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(99.78264.025.082)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。2.6煤炭的元素分析按重量百分比表示,结果如下:氢50%;碳75.8%;氮1.5%;硫1.6%;氧7.4%;灰8.7%,燃烧条件为空气过量20%,空气的湿度为0.0116molH2O/mol干空气,并假定完全燃烧,试计算烟气的组成。解:按1kg煤进行计算重量(g)摩尔数(mol)需氧数(mol)C75863.1763.17H40.7520.37510.19S160.50.5H2O83.254.6250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%25.14%10038.44317.63;SO2%11.0%10038.4435.0;H2O%74.6%10038.44391.29;N2%55.75%10038.44361.7079.019.279。O2%33.3%10038.443209.061.70。2.7运用教材图2-7和上题的计算结果,估算煤烟气的酸露点。解:SO2含量为0.11%,估计约1/60的SO2转化为SO3,则SO3含量51083.1601%11.0,即PH2SO4=1.83×10-5,lgPH2SO4=-4.737。查图2-7得煤烟气酸露点约为134摄氏度2.8燃料油的重量组成为:C86%,H14%。在干空气下燃烧,烟气分析结果(基于干烟气)为:O21.5%;CO600×10-6(体积分数)。试计算燃烧过程的空气过剩系数。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%5.1x。总氧量xxxx5.2467.10635)67.71(5.010600%5.16,干空气中N2:O2体积比为3.78:1,则含N23.78×(106.67+24.5x)。根据干烟气量可列出如下方程:6610600)5.2467.106(78.367.7110600%5.1xxx,解得x=0.306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因此氮在油中的最大含量为%52.0%1002012014052.014052.0。Chapter21.Wepassagasstreamthroughafiberfilterthatcollects85percentoftheparticlespresent.Ifweweretousethreesuchfiltersinseriesandifweassumethateachofthemhasan85percentsefficiency,whatwouletheexpectedoverallcollectionefficiencybe?2.FigurebelowshowsthefrequencydistributionofvariousvaluesfortheproductofHtimesuforPeoria,Illinois.Iftheaveragedailyemissionrateforcarbonmonoxideforthiscityis72210/()ghmi醋andthestandardtobemetis103/mgm,whatfractionofthetimemustweimplementintermittentcontroltomeetthisstandard?Assumethatthesimpleboxmodel(Section6.2)applies,thatthesizeofthecityis5miby5mi,andthatthebackgroundconcentrationis23/mgm。3.Airat2200Fand1atmisflowingatarateof000lb/mino
本文标题:大气第二章习题及答案
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