T=778℃→%=2311J/kg T=63.5℃→y4=2347KJ/kg 2168 W,=D =0.974D W2=wr(FC.-wc-W,Cpr-W.)(t1-t2) .2216212×377(W+2+W)×4193113-109 277 0.9781+440-0.0302(W1+W3+W4) 0.9781+440-0.0302(8106-W2) 0.97W,=0978,+195 2277 W3=W23=W2 0085W W3Cm)(4-4) Y4 =x2311406212×37×419)×(815-676 2347 2347 =0.96W+362 又W1+W2+W3=8106 解得:D1=1857kg/hW=1809g/hW2=2025kg/h W=1995kg/hW4=2277kg/h Dy11857×2168/3.6 =85.8m K.△t 810×16.1 A1=171=1809×225 86.1m 1160 2025×2277/3.6 A3 986m2 K 1160×112 1995×2311/3.6 A 4 =98.1m K△t 去平均值 85.8+86.1+986+98.1T3  = 77.8 ℃ → 3  = 2311kJ / kg T4  = 63.5 ℃ → 4  = 2347kJ / kg 2 2 4 3 3 2 2 1 1 2 1 1 3 4 1 2 4 1 2 1 3 4 1 2 2 1 2 1 2 1 1 1 0.985 2311 2277 0.97 0.978 195 0.978 440 0.0302(8106 ) 0.978 440 0.0302( ) 2277 [16212 3.77 ( ) 4.193](117.3 100.9) 2277 2227 ( )( ) 0.974 2227 2168 0 W W W W W W W W W W W W W W W W FC W C W C W C t t W W W D D D p p w p w p w = =   = = + = + − − = + − + +  − + +  − =  +  − − − − +   = =  =   =        0.96 362 2347 (16212 3.77 4.19) (81.5 67.6) 2347 2311 ( )( ) 3 3 3 4 3 3 4 4 3 4 3 0 = +  −   − =  +  − − +   = W W W FC W C t t W W p p w    又 W1 +W2 +W3 = 8106 解得： D1 =1857kg / h W1 =1809kg / h W2 = 2025kg / h W3 =1995kg / h W4 = 2277kg / h ⑤ 2 1 1 1 1 1 85.8 810 16.1 1857 2168/ 3.6 m K t D A =   =  =  2 3 4 3 3 4 2 3 3 2 2 3 2 2 2 1 1 2 98.1 1280 10.2 1995 2311/ 3.6 98.6 1160 11.2 2025 2277/ 3.6 86.1 1160 11.2 1809 2227/ 3.6 m K t W A m K t W A m K t W A =   =   = =   =   = =   =   =    去平均值 2 92 4 85.8 86.1 98.6 98.1 A = m + + + =