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锅炉原理作业题

1. 已知某种煤的组成成分如下:M ar =5.0%, A d =20%, C daf =90.8%, H daf =3.8%, O daf =3.1%, N daf =1.3%, S daf =1.0%。

求煤的收到基组成。

解:根据X=KX 0
100100520%19%100100
10010051990.8%0.7690.8%69.008%100100
1000.76 3.8% 2.888%1001000.76 3.1% 2.356%100
1000.76 1.100
ar ar d ar ar ar daf ar ar ar daf ar ar ar daf ar ar ar daf M A A M A C C M A H H M A O O M A N N --=
=⨯=----==⨯=⨯=--==⨯=--==⨯=--==⨯3%0.988%1000.761%0.76%100
ar ar ar daf M A S S =--==⨯= 验算:100%ar ar ar ar ar ar ar C H O N S M A ++++++=
2. 某锅炉燃用某种无烟煤,煤种特性如下:C daf =94%, H daf =1.4%, O daf =
3.7%, N daf =0.6%, S daf =0.3%, M ar =
4.0%, A ar =24%, 当过量空气系数α=1.1时,求实际空气量V 。

解:根据X=KX 0,100100424, 0.72100100ar ar ar daf M A X X K ----=== 0.7294%67.68%
0.72 1.4% 1.008%
0.72 3.7% 2.664%0.720.6%0.432%
0.720.3%0.216%
ar daf ar daf ar daf ar daf ar daf C KC H KH O KO N KN S KS ==⨯===⨯===⨯===⨯===⨯=
理论空气量:
()03V 0.0889(0.375)0.2650.0333
0.088967.680.3750.2160.265 1.0080.0333 2.664 6.202 m /kg
ar ar ar ar C S H O =++-=⨯+⨯+⨯-⨯= 实际空气量:03V=αV 1.1 6.202 6.823 m /kg =⨯=
3. 某锅炉燃料特性系数β=0.11,运行中测得省煤器前后RO 2值分别为1
4.3%和14%,求该省煤器处的漏风系数Δα。

解:运行中过量空气系数的计算公式:2max
27979RO RO βαβ+=+, max 2212118.9189110.11
RO β===++ 省煤器前过量空气系数:2max 279790.11'14.3' 1.314779790.1118.9189
RO RO βαβ++===++ 省煤器后过量空气系数:2max 279790.11''14'' 1.342379790.1118.9189
RO RO βαβ++===++ 省煤器处的漏风系数:''' 1.3423 1.31470.0276ααα∆=-=-= 解:运行中过量空气系数的计算公式:()max 222
211RO RO RO αβ==+ 省煤器前过量空气系数:()()22121' 1.3231'10.1114.3RO αβ=
==++⨯ 省煤器后过量空气系数:()()22121'' 1.35141''10.1114RO αβ===++⨯ 省煤器处的漏风系数:''' 1.3514 1.3230.0284ααα∆=-=-=
gl 4ar,net 03410t /''3475/963/'1340/2%90.5% 1.5%Q 20935/ 5.5m /kg 1.2gr gs D h h kJ kg h kJ kg h kJ kg q kJ kg
V ηα=========4. 已知锅炉蒸发量,过热蒸汽焓,给水焓,饱和水焓,排污率为,锅炉热效率,其中,求:(1)锅炉的计算煤耗量和标准煤耗量;
(2)若理论空气量,过量空气系数,计算每小时送入锅炉的空气量。

解:过热蒸汽流量:410t/h 1000113.89 kg/s 3600
gr D ⨯== 排污水流量:2%113.89 2.278 kg/s pw gr D pD ==⨯= 有效利用热:()11''(')gr gr gs pw gs Q Q D h h D h h B B ⎡⎤=
-+-=⎣
⎦ 总有效利用热:
()()()''(')
113.893475963 2.2781340963 286950.5 kJ/s gr gr gs pw gs Q D h h D h h =-+-=⨯-+⨯-= 实际煤耗量:1,100100100286950.5=====15.143 kg/s 90.520935
gl r gl ar net Q Q Q B Q Q Q ηη⨯⨯ 计算煤耗量:4q 1.5=1=15.1431=14.916 kg/s 100100j B B ⎛⎫⎛⎫-⨯- ⎪ ⎪⎝
⎭⎝⎭ 标准煤耗量:,20935==15.143=10.816 kg/s 2931029310ar net b Q B B ⨯ 实际空气量:03V=αV 1.2 5.5 6.6 m /kg =⨯= 每小时送入锅炉的空气量:
3353j V'=B V 14.916 6.698.45 m /s=98.453600 m /h=3.5410 m /h =⨯=⨯⨯。

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