1.已知煤的空气干燥基成分：ad C =60.5%，ad H =4.2%，ad S =0.8%，ad A =25.5%，ad M =2.1%和风干水分f ar M =3.5%，试计算上述各种成分的收到基含量。

%5625.5965.01.25.3100)100(%6075.24965.05.25100)100(%772.0965.08.0100)100(%053.4965.02.4100)100(%325.58965.05.60100)100(965.0100)100(=⨯+=-⨯+==⨯=-⨯==⨯=-⨯==⨯=-⨯==⨯=-⨯==-f ar ad f ar ar f ar ad ar f ar ad ar f ar ad ar f ar ad ar f ar M M M M M S A M S S M H H M C C M 解： 2.已知煤的空气干燥基成分：ad C =68.6%，ad H =3.66%，ad S =4.84%，ad O =3.22%，ad N =0.83%，ad A =17.35%，ad M =1.5%，ad V =8.75%，空气干燥基发热量ad net Q ,=27528kJ/kg 和收到基水分ar M =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。

%14.17100)5.98117100(35.17100)100(%18.3100)5.98117100(22.3100)100(%82.0100)5.98117100(83.0100)100(%78.4100)5.98117100(84.4100)100(%62.3100)5.98117100(66.3100)100(%79.67100)5.98117100(6.68100)100(5.98117100100=-⨯=-==-⨯=-==-⨯=-==-⨯=-==-⨯=-==-⨯=-==-⨯+=f ar ad ar f ar ad ar f ar ad ar f ar ad ar f ar ad ar f ar ad ar f ar f ar ad f arar M A A M O O M N N M S S M H H M C C M M M M M 可得解：由kgkJ M S O H C Q kg kJ M M M M Q Q arar ar ar ar ar net ar ad ar ad ad net ar net /2681767.225)78.418.3(10962.3103079.6733925)(1091030339/2717167.2255.110067.2100)5.12527528(25100100)25(,,,=⨯--⨯-⨯+⨯=⨯---+==⨯---⨯⨯+=---+=：门捷列夫经验公式校核3.下雨前煤的收到基成分为：1ar C =34.2%，1ar H =3.4%，1ar S =0.5%，1ar O =5.7%，1ar N =0.8%，1ar A =46.8%，1ar M =8.6%，1,ar net Q =14151kJ/kg 。

下雨后煤的收到基水分变为2ar M =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。

222222,21211,2,1121121121121121121111225)(1091030339/131133.14256.81003.14100)6.82514151(25100100)25(%88.431004.915701008.46100100%75.01004.915701008.0100100%34.51004.915701007.5100100%47.01004.915701005.0100100%19.31004.915701004.33100100%07.321004.915701002.341001004.91570100100ar ar ar ar ar ar net ar ar ar ar ar net ar net f ar ar ar f ar ar ar f ar ar ar f ar ar ar f ar ar ar f ar ar ar f ar f ar ar f ar ar M S O H C Q kg kJ M M M M Q Q M A A M N N M O O M S S M H H M C C M M M M M ---+==⨯---⨯⨯+=⨯---⨯⨯+==-⨯=-⨯==-⨯=-⨯==-⨯=-⨯==-⨯=-⨯==-⨯=-⨯==-⨯=-⨯==-⨯+=：门捷列夫经验公式校核可得解：由kgkJ /1.132693.1425)47.034.5(10919.3103007.32339=⨯---⨯+⨯= 4.某工厂贮存有收到基水分1ar M =11.34%及收到基低位发热量1,ar net Q =20097kJ/kg 的煤100t ，由于存放时间较长，收到基水分减少到2ar M =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？kgkJ M M M M Q Q M M M M Q Q tt M M M M M M M M ad arad ar ar net ad net ar ad ar ad ad net ar net ad ad ar f ar f ar ad f arar /2115718.72534.1110018.710034.112520097251001002525100100)25(52.9548.410048.418.710018.710034.11100100100100100100,,,,=⨯---⨯⨯+=---+=---+==-≈-⨯-⨯=--=-⨯+=）（）（可得由煤的质量变为：可得解：由5.已知煤的成分：daf C =85.00%，daf H =4.64%,daf S =3.93%,daf O =5.11%,daf N =1.32%，d A =30.05%，ar M =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。

kgkJ M S O H C Q M A N N M A O O M A S S M A H H M A C C M A A arar ar ar ar ar net ar ar daf ar ar ar daf ar ar ar daf ar ar ar daf ar ar ar daf ar ar d ar /2072933.1025)46.221.3(10991.2103031.5333925)(1091030339%83.0)33.1095.26100(32.1)100(%21.3)33.1095.26100(11.5)100(%46.2)33.1095.26100(93.3)100(%91.2)33.1095.26100(64.4)100(%31.53)33.1095.26100(85)100(%95.26)33.10100(5.30)100(,⨯--⨯-⨯+⨯=---+==--⨯=--==--⨯=--==--⨯=--==--⨯=--==--⨯=--==-⨯=-=解：6用氧弹测热计测得某烟煤的弹筒发热量为26578kJ/kg ，并知ar M =5.3%，ar H =2.6%，f arM =3.5%，ad S =1.8%，试求其收到基低位发热量。

可得解：由100100f ar ad farar M M M M -⨯+= kgkJ M M M M Q Q kgkJ M H Q Q kgkJ Q S Q Q M H H M M M M ar ad ar ad ad net ar net adad ad gr ad net adb ad b ad b ad gr f ar ar ad f ar f ar ar ad /247393.525865.11003.5100)865.12525727(25100100)25(/25727865.12569.22262638225226/2638226578001.08.11.94265781.94%69.25.31001006.2100100%865.15.3100100)5.33.5(100100)(,,,,,,,,=⨯---⨯⨯+=---+=⨯-⨯-=--==⨯-⨯-=--==-⨯=-⨯==-⨯-=-⨯-=α 7.一台4t/h 的链条炉，运行中用奥氏烟气分析仪测得炉膛出口处2RO =13.8%，2O =5.9%，CO =0；省煤器出口处2RO =10.0%，2O =9.8%，CO =0。

如燃料特性系数β=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？485.039.1875.1875.18.92121212139.19.521212121605.0)()21(22222=-=''-''=∆=-=-=''=-=-=''++--=l l O O O RO RO CO αααααββ漏风系数省煤器出口炉膛出口知烟气分析结果准确。

看是否成立。

经计算可入公式器出口的各气体含量带解：将炉膛出口和省煤8.∏--W SZL 3.110型锅炉所用燃料成分为ar C =59.6%，ar H =2.0%，ar S =0.5%，ar O =0.8%，ar N =0.8%，ar A =26.3%，ar M =10.0%，daf V =8.2%，ar net Q ,=22190kJ/kg 。

求燃料的理论空气量0k V 、理论烟气量0y V 以及在过量空气系数分别为1.45和1.55时的实际烟气量y V ，并计算α=1.45时300℃和400℃烟气的焓和α=1.55时200℃及300℃烟气的焓。

kgkJ c V c V c V I t kgkJ I I I kgkJ c V I kgkJ c V c V c V I t kgkJ I I I kgkJ c V I kgkJ c V c V c V I t kgkJ I I I kgkJ c V I kgkJ c V c V c V I t kg Nm V M H V kg Nm V V kgNm S C V kg Nm V V V kg Nm V V V kgNm V M H N V S C V V V V kgNm O H S C V O H OH N N RO RO y k y y k k k O H OH N N RO RO y k y y k k k O H OH N N RO RO y k y y k k k O H OH N N RO RO y k ar ar O H k N ar ar RO k y y k y y k ar ar ar k ar ar OH N RO y arar ar ar k /263146344.03926.45591156.1)()()(300,55.1/25798511728/85126682.555.0)()1(/172830444.02606.43571156.1)()()(200,55.1)2(/498014193561/141954282.545.0)()1(/356162644.05276.47721156.1)()()(400,45.1/368610552631/105540382.545.0)()1(/263146344.03926.45591156.1)()()(300,45.11/44.00161.00124.0111.0/6.479.0/1156.1)375.0(01866.0/41.982.5)155.1(0161.116.6)1(0161.155.1/82.882.5)145.1(0161.116.6)1(0161.145.1/158.6818.5161.010124.02111.08.0008.0818.579.0)5.0375.06.59(01866.00161.00124.0111.0008.079.0)375.0(01866.0/818.58.00333.02265.0)5.0375.06.59(0889.00333.0265.0)375.0(0889.02222222222222222222222222222220000000000000000003003003300300300030=⨯+⨯+⨯=++====+=∆+=∴=⨯⨯=-=∆=⨯+⨯+⨯=++====+=∆+=∴=⨯⨯=-=∆=⨯+⨯+⨯=++====+=∆+=∴=⨯⨯=-=∆=⨯+⨯+⨯=++====++====+==⨯-⨯+=-+===⨯-⨯+=-+===⨯+⨯+⨯+⨯+⨯+⨯+⨯=++++++=++==⨯-⨯+⨯+=-++=θθθαθαθθθαθαθθθαθαθθθααααα℃时，当过量空气的焓：℃时，当过量空气的焓：℃时，当过量空气的焓：℃时，）当（所以有因为时，当时，当解：kg kJ I I I kgkJ c V I k y y k k k /392112902631/129040382.555.0)()1(00=+=∆+=∴=⨯⨯=-=∆θα过量空气焓：。