11-1燃烧学计算题

5.已知某烟煤成分为（%）：Cdaf—83.21,Hdaf—5.87,Odaf—5.22,Ndaf—1.90,Sdaf—3.8,Ad—8.68,War—4.0,试求：(1)理论空气需要量L0（m3/kg）；(2)理论燃烧产物生成量V0（m3/kg）;(3)如某加热炉用该煤加热，热负荷为17×103kW，要求空气消耗系数n=1.35，求每小时供风量，烟气生成量及烟气成分。解：（1）将该煤的各成分换算成应用成分：%33.81004100%68.8100100%ardarWAA%95.72100433.8100%21.83100100%arardafarWACC%15.5%8767.087.58767.0%dafarHH%58.4%8767.022.58767.0%dafarOO%66.1%8767.09.18767.0%dafarNN%33.3%8767.080.38767.0%dafarSS%4arW计算理论空气需要量L0：kgmOSHCL/81.701.058.433.315.5895.723821.0429.11100183821.0429.1130（2）计算理论燃烧产物生成量V0：kgmLNWHSCV/19.881.779.0224.02866.1184215.53233.310095.72100791004.22281823212300（3）采用门捷列夫公式计算煤的低发热值：Q低=4.187×[81×C+246×H－26×（O－S）－6×W]]=4.187×[81×72.95+246×5.15－26×（4.58－3.33）－6×4]=29.80（MJ/m3）每小时所需烟煤为：hkgQm/10053.2298093600101736001017333每小时烟气生成量：)/(1024.281.735.019.8205334hmVmVntol每小时供风量：hmmnLLtol/1016.281.735.12053340计算烟气成分：)/(1080.220531004.221295.721004.2212332hmmCVco)/(8.461004.223232hmmSVso)/(10296.1)1004.22182(332hmmWHVoH)/(10714.179.01004.2228342hmLmNVnN)/(10188.1)(100213302hmmLLVno计算烟气百分比组成：CO2＇=12.45%SO2＇=0.21%H2O＇=5.73%N2＇=76.36%O2＇=5.25%6.某焦炉干煤气%成分为：CO—9.1；H2—57.3；CH4—26.0；C2H4—2.5；CO2—3.0；O2—0.5；N2—1.6；煤气温度为20℃。用含氧量为30%的富氧空气燃烧，n=1.15，试求：(1)富氧空气消耗量Ln（m3/m3）(2)燃烧产物成分及密度解：应换成湿基（即收到基）成分计算：将煤气干燥基成分换算成湿基成分：当煤气温度为20℃，查附表5，知：gd，H2O=18.9（g/m3）H2O湿=（0.00124×18.9）×100/（1+0.00124×18.9）=2.29CO湿=CO干%×（100－H2O湿）/100=8.89同理：H2湿=55.99CH4湿=25.40O2湿=0.49CO2湿=2.93N2湿=1.57C2H4湿=2.44（1）计算富氧燃烧空气消耗量：332220/0.323421213.01mmOSHHCmnHCOLmn330/45.30.315.1mmLnLn（2）计算燃烧产物成分：332/421.001.093.244.224.2589.810012mmCOHCnCOVmnco3322/14.1100122mmOHHCmHVmnOH332/43.245.37.001.057.11007010012mmLNVnN330/135.045.03.03.02mmLLVnO烟气量为：Vn=4.125（m3/m3）烟气成分百分比：CO2＇=10.20H2O＇=27.60N2＇=58.93O2＇=3.27(3)计算烟气密度：3'2'2'2'2/205.14.2210027.33293.582860.271820.10444.2210032281844mkgONOHCO7.某焦炉煤气，成分同上题，燃烧时空气消耗系数n=0.8，产物温度为1200℃，设产物中O2，=0，并忽略CH4，不计，试计算不完全燃烧产物的成分及生成量。解：（1）碳平衡公式：42.0.01.044.224.2593.289.810012222COCOCOCOCOCOmnVVVVVVCOHCnCO（2）氢平衡方程式：14.1.01.029.244.224.25299.55100122222220.22OHHOHHHHmnVVVVVVOHHCmH（3）氧平衡方程式：81.05.05.05.05.09.08.001.029.25.049.093.289.8212121100121212222222.0222OHCOCOOHCOCOOHCOCOOVVVVVVVVVnLOHOCOCO（4）氮平衡方程式：22.0276.31001NOVnLN2272.29.08.076.301.057.1NNVV（5）水煤气反应平衡常数：OHCONCOOHCONCOVVVVPPPPK222222查附表6当t=1200℃时，K=0.387387.0222OHCONCOVVVV各式联立求解得：算得百分比为：VH2=0.207(m3/m3)H2，=4.84VCO=0.153(m3/m3)CO，=3.57VCO2=0.267(m3/m3)CO2，=6.24VH2O=0.933(m3/m3)H2O，=21.80VN2=2.72(m3/m3)N2，=63.55Vn=4.28(m3/m3)4.某烟煤，收到基成分为：CarHarOarNarSarAarWar76.324.083.641.613.807.553.0其燃烧产物分析结果为：RO2’CO’H2’O2’N2’14.02.01.04.079.0试计算：(1)该燃烧的RO2.max;β；K；P；(2)验算产物气体分析的误差；(3)空气消耗系数n和化学不完全燃烧损失q化；解：（1）)/(88.7)64.333.38.333.308.467.2632.7689.8(01.001.0)33.333.367.2689.8(30kgmOSHCL)/(655.121.030.02kgmLLO)/(46.11004.22)328.31232.76(1004.22)3212(32kgmSCVro)/(5.01004.22)182(32kgmWHVoH)/(25.679.01004.2228302kgmLNVNV0干=1.46+6.25=7.71（m3/kg）Qd=30.03(MJ/kg)所以13.146.1665.122.0ROVLKO1971.710046.11000'max.22干完）（）（VVRORO11.019192121max.2'.2'ROROmaz)/(1090.371.71003.303330mkJVQPd（2）验证分析误差：将各烟气成分及β值代入烟气分析方程：成分误差：%1212179.20.（3）计算空气消耗系数及不完全燃烧损失14.111614.1015.025.041)2'5.05.0(4''2'4'2''2CHCOROKCHHCOOn%11%10021411082126389519%100'108126''22'max.2'COROHCOPROq化79.200.58.011.01185.04211.0605.01411.01)58.0(185.0)605.0()1(4'2'2'''2CHHOCORO