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时间:2020-06-18
《《燃料与燃烧》部分习题答案.pdf》由会员上传分享,免费在线阅读,更多相关内容在行业资料-天天文库。
1、《燃料与燃烧》习题解答第一篇燃料概论1.某种煤的工业分析为:Mar=3.84,Ad=10.35,Vdaf=41.02,试计算它的收到基、干燥基、干燥无灰基的工业分析组成。解:干燥无灰基的计算:V41.02dafFc100V58.98;dafdaf收到基的计算FC100AMVarararar100MAararVV35.36ardaf100Aar=9.95FCar=50.85干燥基的计算:A10.35dVd=36.77;FC100VA52.88ddd2.某种烟煤成分为:Cdaf=83.21Hdaf=5.87Od
2、af=5.22Ndaf=1.90Ad=8.68Mar=4.0;试计算各基准下的化学组成。解:干燥无灰基:S100CHON3.80dafdafdafdafdaf100Mar收到基:AA8.33ard100100AMararCC72.95ardaf100Har=5.15Oar=4.58Nar=1.67Sar=3.33Mar=4.0干燥基:A8.68d100AdCC75.99ddaf100HH0.9135.36ddafOO0.9134.77ddafNd=Ndaf×0.913=1.74SS0.9
3、133.47ddaf干燥无灰基:Cdaf=83.21Hdaf=5.87Odaf=5.22Ndaf=1.90Sdaf=3.80一3.人工煤气收到基组成如下:H2COCH4O2N2CO2H2O48.019.313.00.812.04.52.4计算干煤气的组成、密度、高热值和低热值;解:干煤气中:H2,d=48.0×[100/(100-2.4)]=49.18CO,d=19.3×1.025=19.77CH4,d=13.31O2,d=0.82N2,d=12.30CO2,d=4.61ρ=M干/22.4=(2×49.18%+28×19.77%+16×13
4、.31%+32×0.82%+28×12.30%+44×4.61%)/22.43=0.643kg/mQ高=4.187×(3020×0.1977+3050×0.4918+9500×0.1331)333=14.07×10kJ/m=14.07MJ/mQ低=4.187×(3020×0.1977+2570×0.4918+8530×0.1331)333=12.55×10kJ/m=12.55MJ/m第二篇燃烧反应计算第四章空气需要量和燃烧产物生成量5.已知某烟煤成分为(%):Cdaf—83.21,Hdaf—5.87,Odaf—5.22,Ndaf—1.90,S
5、daf—3.8,Ad—8.68,War—4.0,试求:3(1)理论空气需要量L0(m/kg);3(2)理论燃烧产物生成量V0(m/kg);3(3)如某加热炉用该煤加热,热负荷为17×10kW,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:100War1004AA%8.68%8.33%ard100100100AW1008.334ararCC%83.21%72.95%ardaf100100HH%0.87675.870.8767%5.15%ard
6、afOO%0.87675.220.8767%4.58%ardafNN%0.87671.90.8767%1.66%ardafSS%0.87673.800.8767%3.33%ardaf二W4%ar计算理论空气需要量L0:181L0C8HSO1.4290.2131001872.9585.153.334.580.011.4290.21337.81m/kg(2)计算理论燃烧产物生成量V0:CSHWN22.479V0L01232218
7、2810010072.953.335.1541.660.2240.797.81100322182838.19m/kg(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]3=29.80(MJ/m)每小时所需烟煤为:3317103600171036003m2.05310kg/hQ29809每小时烟气生成量:43VmV20538.190
8、.357.812.2410(m/h)toln每小时供风量:43LmnL20531.357.812.1610m/htol0计算烟气成分:C22.4
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