锅炉第二章课后计算地训练题目答案详解

1、实用标准文案1. 已知煤的空气干燥基成分：Cad =60.5%， H ad =4.2%， Sad =0.8%， Aad =25.5%， M ad =2.1%和风干水分 M arf =3.5%，试计算上述各种成分的收到基含量。(100M arf )0.965解：100CarCad(100M arf)60.50.96558.325%100H arH ad(100M arf )4.20.9654.053%100SarSad(100M arf)0.80.9650.772%100AarSad(100M arf)25.50.96524.6075%100M arM arfM ad(100M arf)3.52

2、.1 0.9655.5625%1002. 已 知 煤 的 空气 干 燥基 成分 ： Cad =68.6% ， H ad=3.66% ， Sad =4.84% ， Oad =3.22% ，N ad =0.83%， Aad =17.35%， M ad =1.5%，Vad=8.75%，空气干燥基发热量 Qnet ,ad =27528kJ/kg和收到基水分 M ar =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：由 M ar M arfM ad100 M arf可得 M arf 11710098.5Cad (100f)

3、68.6 (100117 )CarM ar98.567.79%100100H ad (100f)3.66 (100 117 )H arM ar98.53.62%100100117 )Sad (100f)4.84 (100SarM ar98.54.78%100100117NarNad (100M arf)0.83 (10098.5)1001000.82%117Oad (100M arf)3.22 (100)Oar98.53.18%100100117Aad (100M arf)17.35 (100)Aar98.510010017.14%精彩文档实用标准文案Qnet , ar(Qnet ,ad25M

4、 ad ) 100M ar25 M ar100M ad1002.67( 27528251.5)252.6727171kJ / kg门捷列夫经验公式校核：Qnet , ar339Car1030H ar109(OarSar )25M ar339 67.79 1030 3.62 109 (3.18 4.78) 25 2.67 26817 kJ / kg3. 下雨前煤的收到基成分为：Car 1 =34.2%， H ar 1 =3.4%，Sar 1 =0.5%，Oar 1 =5.7%， N ar 1 =0.8%，Aar 1 =46.8%， M ar1 =8.6%， Qnet ,ar 1 =14151kJ

5、/kg 。下雨后煤的收到基水分变为M ar 2 =14.3%，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 M ar 2 M arf1 M ar 1 100M arf1 可得 M arf157010091.4100M arf100570Car 2191 .432.07%Car 134.2100100100M arf100570H ar 2191.43.19%H ar 133.4100100100f10057091.4Sar 2M ar 10.50.47%Sar1100100570100f100Oar 2M ar 15.791.45.34%Oar 11001

6、00570100f100Nar 2M ar 10.891.40.75%N ar 1100100100f10057091.4Aar 2M ar 146.843.88%Aar1100100Qnet , ar 2(Qnet ,ar125100M ar 225M ar 2M ar 1 )100 M ar1(141512510014.314.38.6)251008.613113 kJ / kg门捷列夫经验公式校核：Qnet , ar 2339Car 2 1030 H ar 2 109(Oar 2 Sar 2 ) 25M ar 2精彩文档实用标准文案33932.0710303.19109(5.340.47

7、)2514.313269 .1kJ / kg4. 某工厂贮存有收到基水分M ar 1 =11.34%及收到基低位发热量Qnet ,ar 1 =20097kJ/kg的煤100t ，由于存放时间较长， 收到基水分减少到M ar 2 =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 M ar M arfM ad100M arf可得100M arf100M ar100M ad10011.34100 7.184.48t100M ad1007.18煤的质量变为：4.4895.52t100由 Qnet ,ar(Qnet , ad100M ar25M ar 可得25M ad

8、 )M ad100Qnet , ad（Qnet , ar100M ad25M ad25M ar ）M ar100（2009725） 1007.18257.1811.3411.3410021157kJ / kg5. 已知煤的成分：Cdaf =85.00%， H daf =4.64%, Sdaf =3.93%, Odaf =5.11%, Ndaf =1.32% ，Ad =30.05%， M ar =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 M ar )30.5(10010.33)26.95%CarCdaf (100AarM ar)8

9、5(10026.9510.33)53.31%H arH daf (100AarM ar )4.64(10026.95 10.33)2.91%SarSdaf (100AarM ar )3.93(10026.9510.33)2.46%OarOdaf (100AarM ar )5.11(10026.95 10.33) 3.21%NarN daf (100AarM ar )1.32(10026.95 10.33) 0.83%Qnet , ar339C ar1030H ar109(OarSar ) 25M ar339 53.31 1030 2.91109(3.212.46)25 10.3320729 k

10、J / kg6 用氧弹测热计测得某烟煤的弹筒发热量为26578kJ/kg，并知 M ar =5.3%， H ar =2.6%，M arf =3.5%， Sad =1.8%，试求其收到基低位发热量。精彩文档实用标准文案解：由 M arM arfM ad100M arf可得100M ad( M arM arf )100(5.3 3.5)1001.865%100M arf100 3.5H adH ar1002.61002.69%100 M arf100 3.5Qgr ,adQb,ad94.1Sb ,adQb,ad2657894.11.80.0012657826382 kJ / kgQnet , ad

11、Q gr ,ad226H ad25 M ad263822262.69251.86525727 kJ / kgQnet , ar(Qnet, ad25M ad ) 100M ar25 M ar100M ad( 25727251.865)1005.325 5.31.86510024739 kJ / 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%， O2 =5.9%，CO =0；省煤器出口处RO2 =10.0%， O2 =9.8%， CO =0。如燃料特性系数=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

12、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(21RO2 ) ( RO2 O2 ) 看是否成立。经计算可 知烟气分析结果准确。0.605炉膛出口21211.39lO22121 5.9省煤器出口212121 O2211.8759.8漏风系数l1.8751.39 0.4858. SZL10 1.3 W型锅炉所用燃料成分为Car =59.6%，H ar =2.0%，Sar =0.5%，Oar =0.8%，N ar=0.8%， Aar =26.3%， M ar =10.0%， Vdaf=8.2%， Qnet ,ar=22190kJ/kg 。求燃料的理论空气量 Vk0 、理论烟气量 Vy0

13、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=1.45时 300和 400烟气的焓和=1.55时 200及 300烟气的焓。精彩文档实用标准文案解： Vk00.0889(Car0.375Sar )0.265H ar0.0333Oar0. 0889(59.6 0.375 0.5) 0.265 2 0.0333 0.85.818Nm3 / kgVy0VRO2VN 2VH 2O0.01866(Car0.375Sar ) 0.79Vk00.008N ar0.111H ar0.0124M ar 0.0161Vk00.01866(59.60.3750.5)0.795.818

14、0.0080.80.11120.124100.1615.8186.158Nm3 / kg当1.45时， VyVy01.0161(1)Vk06.161.0161(1.451)5.828.82 Nm3 / kg当1.55时， VyVy01.0161(1)Vk06.161.0161(1.551)5.829.41Nm3 / kg因为 VRO20.01866(Car0.375Sar )1.1156Nm3 / kgVN020.79Vk04.6Nm3 / kgVH02O0.111H ar0.0124M ar0.0161Vk00.44Nm3 / kg所以有（）当1.45, t时，1300I y0VRO2 (c

15、 ) RO2VN02 ( c ) N2VH02O (c ) H2O1.11565594.63920.444632631kJ / kg过量空气的焓：I k(1)Vk0 (c)k0.455.824031055kJ / kgI yI y0I k263110553686kJ / kg当1.45, t 400时，I y0VRO2 (c ) RO2VN02 ( c ) N2VH02O (c ) H2O1.11567724.65270.446263561kJ / kg过量空气的焓：I k(1)Vk0 (c )k0.455.825421419kJ / kgI yI y0I k3561 14194980kJ / kg当1.55,t时，(2)200I y0VRO(c ) RO2VN0 ( c ) N2VH0O (c ) HO22221.11563574.62600.443041728kJ / kg过量空气的焓：