理论烟气组成 63.1224 CO,=m×100%=12100×100%=14.72% 8.00 4.511.7、224 HO= 100% 218100 ×100%=8.12% 8.00 0.8224 100=32100×100%=007% 公×100=2101154M>9 1.7224 21×100%=7426% O2=×100%≈(.15-1)×137 合计:∑(CO2+H2O+SO2+N2+O)=974% 6.解 (1)无遮热板时 q 乌*人A567×10-(1232-693):323×0wm2 1-0.3511-0.75 0.3510.75 (2)有遮热板时: B-Eb2567×10(123-693) =1978v/n 1-0.35 1-0.75 923 035++2x1-004 0.0410.75 1978567×103(11232-t (3)q 1-0.3511-004 0.35 E3t=984K 即t=771℃6 理论烟气组成： 2 2 63.1 22.4 12 100 100% 100% 14.72% 8.00 CO f V CO V × =× = × = 2 2 4.5 11.7 22.4 ( ) 2 18 100 100% 100% 8.12% 8.00 H O f V H O V + × =× = × = 2 2 0.8 22.4 32 100 100% 100% 0.07% 8.00 SO f V SO V × =× = × = 2 2 1.7 22.4 79 1.15 1.37 28 100 21 100% 100% 74.26% 8.00 N f V N V × +×× =× = × = 2 2 (1.15 1) 1.37 100% 100% 2.57% 8.00 O f V O V − × =× = × = 合计： 2 2 2 22 ∑( ) 99.74% CO H O SO N O + + ++ = 6．解： （1）无遮热板时： 84 4 1 2 4 2 1 2 1 12 2 5.67 10 (1123 693 ) 3.23 10 / 1 1 1 0.35 1 1 0.75 1 0.35 1 0.75 Eb Eb q wm ε ε εϕ ε − − × − = = =× − −− − + + + + （2）有遮热板时： 84 4 1 2 2 1 2 3 1 13 3 23 2 5.67 10 (1123 693 ) 1978 / 1 1 1 1 1 1 0.35 1 1 0.04 1 1 0.75 2 2 0.35 1 0.04 1 0.75 Eb Eb q w m ε ε ε εϕ εϕ ε − − × − == = − − − − −− ++ ++ ++× ++ （3） 1 3 1 3 1 13 3 1 1 1 Eb Eb q ε ε ε ϕ ε − = − − + + ⇒ 8 44 5.67 10 (1123 ) 1978 1 0.35 1 1 0.04 0.35 1 0.04 984 t t K − × − = − − + + = 即t = 771℃