锅炉课后计算题答案

1、1.已知煤的空气干燥基成分：Cad=60.5%,Had=4.2%,Sad=0.8%,Aad=25.5%,Mad=2.1%和风干水分M：=3.5%,试计算上述各种成分的收到基含量。解：(100-Mafr)=0.965100Car=Cad(100Mar)30.965=58.325%100Har=Had(100-Mafr)(如=4.20.965=4.053%100&r=Sad(100-Mar)=0.80.965=0.772%100=Sad(100Mar)=25.50.965=24.6075%100Mar=MfMadaiaiad(100Mar)=3.5.2.10.965=5.5625%1002.已知煤

2、的空气干燥基成分:Cad=68.6%,Had=3.66%,Sad=4.84%,Oad=3.22%,Nad=0.83%,Aad=17.35%,Mad=1.5%,Vad=8.75%,空气干燥基发热量Qnet,ad=27528kJ/kg和收到基水分Mar=2.67%,煤的焦渣特性为3类，求煤的收到基其他成分、干燥无灰基挥发物及收到基的低位发热量，并用门捷列夫经验公式进行校核。解:100-M由Mar=MarMadCarHarNarAr100也可得Mf117ar-一=98.5Cad(100-Ma：)能6。0。11798.5100100-=67.79%Had(100-M；)3.66(100-11798.5

3、)-=3.62%100100Sad(100-乂；)4.84(10011798.5100100)一=4.78%Nad(100-M；r)OS3(曲100100Oad(100-M；r)3.22(100100100AJd(100-Mjr)17.35(10010010011798.511798.5)-=0.82%)=3.18%11798.5)一二17.14%-，-、100Mar一Qnet,ar-(Qnet,ad25Mad)25Mar100-Mad100-2.67二(27528251.5)-252.67100-1.5=27171kJ/kg门捷列夫经验公式校核Qnet,ar=339Car+1030Har-1

4、09(Oar-Sar)-25Mar=33967.7910303.62-109(3.18-4.78)-252.67=26817kJ/kg3 .下雨前煤的收到基成分为：Car1=34.2%,小.1=3.4%,Sar1=0.5%,。刎=5.7%,NaM=0.8%,Aar1=46.8%,Mar1=8.6%,Qnet,ar1=14151kJ/kg。下雨后煤的收到基水分变为Mar2=14.3%,57091.4求雨后收到基其他成分的含量及收到基低位发热量，并用门捷列夫经验公式进行校核。解：由Mar2=M：1Mar11001M可得Ma:100Car2=Car1100-MM=34.2-100570就UN07%H

5、ar2=Har1Mi=33.4吧570100Sar2=Sar1100-Mafr1100100570100-=0.5914=0.47%100Oar2=Oa1=5.79100570益产的34%Nar2=3,1世必=0.8-100570益产.0.75%Aar2二A,”理工=46.8-100570益*43.88%100-Mar2Qnet,ar2=(Qnet,ar125Mar1)醛-25Mar2100-Mar1100-14.3二(14151258.6)-2514.3100-8.6=13113kJ/kg门捷列夫经验公式校核Qnet,ar2=339Ca.21030Har2一109(Oar2一Sa)25Mar

6、2=33932.0710303.19-109(5.34-0.47)-2514.3=13269.1kJ/kg4 .某工厂贮存有收到基水分Mar1=11.34%及收到基低位发热量Qnet,ar1=20097kJ/kg的煤100t,由于存放时间较长，收到基水分减少到Mar2=7.18%,问100t煤的质量变为多少？煤的收到基低位发热量将变为多大?解：由Mar=M：MadMa/yCMad100-Mad100-Mf黑安可得10010011.34-1007.18一=4.48t100-7.18煤的质量变为：100-4.48=95.52t,100-Mar由Qnet,ar=(Qnet,ad25Mad)25Mar

7、可得100-MadQnet,ad=(Qnet,ar25Mar)-25M,d100-Mar100-7.18=(200972511.34257.18100-11.34=21157kJ/kg5.已知煤的成分：Cdaf=85.00%,Hdaf=4.64%,Sdaf=3.93%,Odaf=5.11%,Ndaf=1.32%,人=30.05%,Mar=10.33%,求煤的收到基成分，并用门捷列夫经验公式计算煤的收到基低位发热量。解：Aar=Ad(100-Mar)=30.5(100-10.33)-26.95%Car=Cdaf(100AarMar)=85(100一26.95一10.33)=53.31%Har=H

8、daf(100-Aar-Mar)-4.64(100-26.95-10.33)=2.91%Sar=Sdaf(100-Aar-Mar)=3.93(100-26.95-10.33)=2.46%Oar=Odaf(100AarMar)=5.11(10026.9510.33)=3.21%Nar=Ndaf(100AarMar)=1.32(10026.9510.33)=0.83%Qnet,ar=339Car+1030Har-109(Oar-Sar)-25Mar=33953.3110302.91-109(3.21-2.46)-2510.3320729kJ/kg6用氧弹测热计测得某烟煤的弹筒发热量为26578kJ

9、/kg,并知Mar=5.3%,Har=2.6%,M；=3.5%,Sad=1.8%,试求其收到基低位发热量。解：由Marf100-Mafr=Mar-Mad*可得100f100100Mad=(Mar-Mar)r=(5.3-3.5)=1.865%100-Mar100-3.5Had=Har100f-100-Ma；100=2.6=2.69%100-3.5Qgr,ad=Qb,ad-94.1Sb,ad-：Qb,ad=26578-94.11.8-0.00126578=26382kJ/kgQnet,ad=Qgr,ad226Had25l/lad=26382-2262.69-251.86525727kJ/kg-_、

10、100Mar一Qnet,ar=(Qnet,ad25Mad)25Mar100-Mad100-5.3一(25727251.865)-255.3100-1.86524739kJ/kg7.一台4t/h的链条炉，运行中用奥氏烟气分析仪测得炉膛出口处RO2=13.8%,。2=5.9%,CO=0;省煤器出口处RO2=10.0%,。2=9.8%,CO=0。如燃料特性系数P=0.1,试校核该烟气分析结果是否准确？炉膛和省煤器出口处的过量空气系数及这一段烟道的漏风系数有多大？解：将炉膛出口和省煤器出口的各气体含量带入公式CO=一R。2）（RO2*。2）看是否成立。经计算可知烟气分析结果准确。0.605:，、一21

11、21炉膛出口=二1.3921-O221-5.9省煤器出口a=1.87521-O221-9.8漏风系数.二=：-：l=1.875-1.39=0.4858.SZL101.3WL型锅炉所用燃料成分为Car=59.6%,Har=2.0%,Sar=0.5%,Oar=0.8%,Nar=0.8%,Aar=26.3%,Mar=10.0%,Vdaf=8.2%,Qnet,ar=22190kJ/kg。求燃料的理论空气量V：、理论烟气量Vy0以及在过量空气系数分别为1.45和1.55时的实际烟气量Vy,并计算口=1.45时300c和400c烟气的次含和a=1.55时200c及300c烟气的燃。解：V：=0.0889(

12、Car0.375Sar)0.265Har-0.0333Oar=0.0889(59.60.3750.5)0.2652-0.03330.8=5.818Nm3/kg0.0124Mar0.0161乂Vy0-Vn2VH2。=0.01866(Car0.375Sar)0.79Vk00.008Nar0.111Har=0.01866(59.60.3750.5)0.795.8180.0080.80.11120.124100.1615.818=6.158Nm3/kg当二=1.45时，Vy=V：1.0161(11)Vk=6.161.0161(1.45-1)5.82=8.82Nm3/kg当二=1.55时，Vy=V：1.

13、0161(11)Vk=6.161.0161(1.55-1)5.82=9.41Nm3/kg因为VRO2=0.01866(Car0.375Sar)=1.1156Nm3/kgVN02=0.79Vk=4.6Nm3/kgV；2O=0.111Har+0.0124Mar+0.01617/=0.44Nm3/kg所以有(1)当二=1.45,t=300C时，Iy=VRO2。)ro2W)”V；2oG)h2o=1.11565594.63920.44463=2631kJ/kg过量空气的始：=Ik=(：-1)Vk(cu)k=0.455.82403=1055kJ/kg1y=I0Ik=26311055=3686kJ/kg当：=1.45,t=400C时，Iy=VRO2oro2V)“V；2o(cu)h2o=1.11567724.65270.44626=3561kJ/kg过量空气的始：-Ik-(：-1)Vk0(c)k-0.455.82542-1419kJ/kg1y=I；1k=35611419=4980kJ/kg(2)当:=1.55,t=200C时，1y=VRO2(C7I)RO2VN2(CTI)N2VH2O(C7I)H2O=1.11563574.62600.44304=1728kJ/kg过