60.25∙e-15.86.2-tfz=346.38+1.83∙φz0.25-tfz第一控制体出口处tfL6=291.69℃,查表可得λ=0.57799W/(m∙℃)φL6=0.48μ=9.1858×10-5kg/(m∙s)Pr=0.8371Δθf1L3=2.25×104×9.1858×10-50.8×0.80.57799×0..83710.4=11.82℃Δθf2L3=346.38+1.82×0.480.25-303.91=56.21℃故tcsL6=tfL6+Δθf1L6=291.69+11.82=303.51℃其余同理由程序计算得出结果如下第二控制体出口处tf2L6=301.38℃,查表可得λ=0.5601W/(m∙℃)φ2L6=1.02μ=8.8×10-5kg/(m∙s)Pr=0.8621Δθf12L6=24.79℃Δθf22L6=46.84℃故tcs2L6=tf2L6+Δθf12L6=301.38+24.79=326.15℃第三控制体出口处tf3L6=314.86℃,查表可得λ=0.5333W/(m∙℃)φL=1.50μ=8.275×10-5kg/(m∙s)Pr=0.9157Δθf1L=35.55℃Δθf2L=33.55℃故tcs3L6=tfL+Δθf2L=348.41℃第四控制体出口处tf4L6=327.70℃,查表可得λ=0.5044W/(m∙℃)φL=1.56μ=7.735×10-5kg/(m∙s)Pr=1.0051Δθf1L=35.67℃Δθf2L=20.73℃故tcs4L6=tfL+Δθf2L=348.43℃第五控制体出口处tf4L6=334.89℃,查表可得λ=0.4868W/(m∙℃)φL=0.96μ=7.401×10-5kg/(m∙s)Pr=1.0873Δθf1L=21.28℃Δθf2L=13.30℃故tcs5L6=tfL+Δθf2L=348.19℃
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