Nur=0.023Rer°Pr 6.54×10 =0.023×(7728098×(431)4×(3147×10 0.11 363.8 0.635 Nu-=363.8× d 00259420w/m2°C mCp(te-te 1×4147×(65-15) a(t-t+)d9240×50×314×0255:75m 校核L/d=5.75/0.025=230>50 E=1与假设相符则L=575m 退出‹#› ) 363.8 3.147 10 6.54 10 0.023 (77280) (4.31) ( N u 0.023Re Pr ( ) 0.1 1 4 4 0.8 0.4 0.1 1 w 0.4 f f 0.8 f f =   =      = − − 9420w / m C 0.025 0.635 363.8 d Nu 2 f  =  =   = 5.75m 9240 50 3.14 0.025 1 4147 (65 15) (t t ) d mCp(t t ) L w f f 2 f1 =      − =  −  − = 校核 L/d=5.75/0.025=230>50 ξl=1与假设相符 则L=5.75m 第二章 热 量 传 输