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.已知煤的空气干燥基成分Cad=
60.5%Had=
4.2%S.d=
0.8%4ad=
25.5%M.=
2.1%和风干水分例=
3.5%试计算上述各种成分的收到基含量解/°一见二
0.965100C/Ir=C.X100-〃“=
60.5X
0.965=
58.325%“rad100H=H.xJOO-=42x
0.965=
4.053%“d100Sir=S“dXaOOf=o.8x
0.965=
0.772%“ad100=S“dx=
25.5x
0.965=
24.6075%al100v+MX100一见=35+
2.1x
0.965=
5.5625%«*Ur«]
00.己知煤的空气干燥基成分C“=
68.6%Had=
3.66%Sad=
4.84%Oll=
3.22%N”“=
0.83%Art/=
17.35%Mrw/=
1.5%匕4=
8.75%空气干燥基发热量Q〃“d=27528kJ/kg和收到基水分Mg=
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,7।»...r__100—M[―rxp.r117解由+M“x丝可得M=araradjQQ〃
98.5「nnna
68.6x10°-黑==台=
67.79%100„zinna/八
3.66x100--=/川00一此=燹1=
3.62%100117A100-M41735X1Q°-
98.5\714%100100-MQ.m=Qi+25M.d”-25Mar100-MaJ=27528+25xl.5x10°-
2.6725x267100-
1.5=21il\kJ/kg门捷列夫经验公式校核=339Q,+1030Hxi090”-S“-25xMar=339x
67.79+1030x
3.62-109x
3.18-
4.78-25x
2.67=268170/心.下雨前煤的收到基成分为Carl=
34.2%Harl=
3.4%Sarl=
0.5%必=
5.7%N.=
0.8%Aarl=
46.8%Marl=
8.6%Qu/nrl=14151kJ/kgo下雨后煤的收到基水分变为Mg=
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Mg=M*+MMx可得M匕=荒100--x—=
34.2x21^=
32.07%100loo_570Hm=Harix10°-Mg=334x9L4=319%100100sc山100--Sar2=S“riX—=.5X见盘=
0.47%2arl100100A〃100--Oar2=0MX—=
5.7x21^=
5.34%2100100loo_570Ng=N“”x—此”=.8x2=.75%5100100100-143=14151+25x
8.6x--25x
14.3100-
8.6=13H3U/kg门捷列夫经验公式校核=3390陋+1030HT09CU—Sg—25Mg=339x
32.07+1030x
3.19-
1095.34-
0.47-25x
14.3=
13269.1%〃依.某工厂贮存有收到基水分Mar]=
11.34%及收到基低位发热量Qw|=20097kJ/kg的煤100t由于存放时间较长,收到基水分减少到加g=
7.18%问100t煤的质量变为多少?煤的收到基低位发热最将变为多大?吐也可得100100x
11.34-100x
7.18一0«
4.48/煤的质量变为100-
4.48=
95.52/100—M由=2“+25M.125M”可得0=或小+25也;胃一1UU—Mar=20097+25x
11.34x-718100-
11.34=211570/必.已知煤的成分Cm=
85.00%Hdaf=
4.64%=
3.93%Odaf=
5.11%Ndaf=
1.32%4=
30.05%M0=
10.33%求煤的收到基成分,并用门捷列夫经验公式计算燥的收到基低位发热量解4=41-%〃.=
30.5x10-
10.33=
26.95%Car=100-4r-M/r=85x100-
26.95-
10.33=
53.31%Har=#何1-Mar=
4.64X100-
26.95-
10.33=
2.91%Sar=Slaf100-=
3.93X10-
26.95-
1.33=
2.46%0ar=,q100-4r-M/r=
5.11x100-
26.95-
10.33=
3.21%Nar=00-4r-K=
1.32x100-
26.95-
10.33=
0.83%Qi=339C”+1030”“-1090“-S“-25Mar=339x
53.31+l3x
2.91-109x
3.21-
2.46-25x
10.3320729V6用氧弹测热计测得某烟煤的弹筒发热量为26578kJ/kg并知M“=
5.3%”“=
2.6%M,=
3.5%2H=
1.8%试求其收到基低位发热量解由叫“=%;>也以><笔等可得1UUMud=Mg-M》x一丝7=
5.3-
3.5x100=
1.865%“darIOO—M/100-
3.5„„
100.100-/Hd~Hx-=
2.6x=
2.69X°lOO-M4100-
3.5Qgr.ad=Qbad-
94.1S^-=26578-
94.1x
1.8-
0.001x26578=263820/口^=2^-226^-25^^=26382-226x
2.69-25x
1.865257270/依100-M=Qw+25M.——-25M〃100-Ma/=25727+25x
1.865x-——--25x
5.3100-
1.865=24739kJ/kg
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/2=
13.8%02=
5.9%=0;省煤器出口处R02=
10.0%=
9.8%CO=
0.如燃料特性系数夕=
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•段烟道的漏风系数有多大?解将炉膛出口和省耀出口的各气体含量带入公式CO=(21一然°2)-°2+U)看是否成立经计算函烟气分析结果准确
0.605+/
78.SZL10—
1.3—WJ1型锅炉所用燃料成分为C.,=
59.6%Hir=
2.0%S,=
0.5%“=
0.8%Na=
0.8%4=
26.3%Mdr=
10.0%Vdaf=
8.2%“=22190kJ/kg求燃料的理论空气量匕°、理论烟气量乂°以及在过量空气系数分别为
1.45和
1.55时的实际烟气量匕,并计算2=
1.45时300℃和400c烟气的熔和a=l.55时200℃及300℃烟气的焰解匕°=
0.0889C“+
0.375S,+
0.265〃“—
0.0333,=
0.
088959.6+
0.375x
0.5+
0.265x2-
0.0333x
0.8=
5.818M〃3/依W=%—=
0.01866C”.+
0.3750“+
0.79噂+
0.008%+
0.111Har+
0.0124Mar+
0.0161匕°=
0.01866x
59.64-
0.375x
0.5+
0.79x
5.818+
0.008x
0.8+
0.111x2+
0.124x10+
0.161x
5.818=
6.158M//kg当a=
1.45时,Vv=V°+
1.01617-1^0=
6.16+
1.0161x
1.45-1x
5.82=
8.82Mh3/^当a=
1.55时,匕=K°+
1.0161a—1匕°=
6.16+
1.0161x
1.55-lx
5.82=
9.41N//依因为%=
0.01866C“+
0.375S”=
1.1156N而ikgV二.79匕°=
4.6可加3/必V%=0A\\Har+
0.0124M“+
0.016啖=
0.44M//kg所以有1当夕=
1.4=30℃时,I=匕3孙a+%%+喂的%=
1.1156x559+
4.6x392+
0.44x463二2631X7均过量空气的焰:M=a-1匕°c0卜=
0.45x
5.82x403=1055/4+△/«=2631+1055=3686%〃kg当a=
1.45=40C时,I;=%c0r02+V;2响电+C”=
1.1156x772+
4.6x527+
0.44x626二3561/kg过量空气的焰△/«=a-1匕°c0=
0.45x
5.82x542=141%J/kg=3561+1419=4980攵J/依2当a=
1.55/=200℃时=匕0c0RO2M“2=
1.1156x357+
4.6x260+
0.44x304二1728/4过量空气的焰Mk=a-\^cOk=
0.55x
5.82x266=S5\kJ/kg.•.//+Mk=1728+851=257〃kg当a=
1.55f=300℃时,1;=Vrocro2+v;ce$+V;oc/0=
1.1156x559+
4.6x392+
0.44x463=263\kJ/kg过量空气焰:Mk=a-\V^cOk=
0.55x
5.82x403=1290kJ/kg/v=/;+A/=2631+1290=392\kJIkg。