在并流换热器中, 用水冷却油。水的进、出口温度分别为15℃和40℃, 油的进、出口温度分别为150℃和100℃。现因生产任务要求油的出口温度降至80℃, 设油和水的流量、进口温度及物性均不变, 若原换热器的管长为1m, 求将此换热器的管长增加多少米才能满足要求。换热器的热损失可忽略。

解:原平均温度差 Δt1 = 150-15 = 135℃, Δt2 = 100-40 = 60℃Δtm =(135-60)/ln(135/60)= 92.5℃由热量衡算 Q = WhCph(T1 - T2)= WcCpc(t2 - t1)WhCph/WcCpc =(t2 - t1)/(T1 - T2)=(40-15)/(150-100)= 0.5当油的温度降至80℃时, 由热量衡算得Q = WhCph(150-80)= WcCpc(t2' -15)或 WhCph/WcCpc =(t2'-15)/(150-80)= 0.5 解得 t2' = 50℃新的平均温度差,Δt1 = 150-15 = 135℃, Δt2 = 80-50 = 30℃,Δtm' =(135-30)/ln(135/30)= 69.8℃由传热速率方程可分别得原换热器 WhCph(150-100)= KSΔtm = KnπdL(92.5)新换热器 WhCph(150-80)= KS'Δtm = KnπdL'(69.8)∴ L' =(70/50)×(92.5/69.8)×1 = 1.855 mΔL = 1.855 – 1 = 0.855 m