质量流量为7200kg/h的常压空气,要求将其温度由20℃加热到80℃,选用108℃的饱和水蒸气作加热介质。若水蒸气的冷凝传热膜系数为1×104 W/(m2·℃),且已知空气在平均温度下的物性数据如下:比热容为1kJ/(kg·℃),导热系数为2.85×10-2 W/(m·℃),粘度为1.98×10-5Pa·s,普兰特准数为0.7。现有一单程列管式换热器,装有Ф25mm×2.5mm钢管200根,管长为2m,核算此换热器能否完成上述传热任务?计算中可忽略管壁及两侧污垢的热阻,不计热损失解:空气需要吸收的热量是已知的,蒸汽冷凝放出热量能否通过该换热器的传递为空气所获得,就与列管换热器的传热速率密切相关。核算现有的列管换热器是否合用,就是用工艺本身的要求与现有换热器相比较,最直接的方法就是比较两者的Q或S(1)核算空气所需的热负荷应小于换热器的传热速率,即Q需要换热器(2)核算空气所需的传热面积应小于换热器提供的传热面积,即S需要换热器解题时,首先应确定列管换热器中流体的流径,因蒸汽安排在壳程易排出冷凝水,故蒸汽走壳程,空气走管程。空气热负荷为_(UND)=(U)_(m2{C)_(m2)}((V)_(2)-(l)_(1))=dfrac (7200)(3600)times (10)^3(80-20)=1.2times (10)^5换热器的传热速率 _(UND)=(U)_(m2{C)_(m2)}((V)_(2)-(l)_(1))=dfrac (7200)(3600)times (10)^3(80-20)=1.2times (10)^5管内空气的对流传热系数计算:_(UND)=(U)_(m2{C)_(m2)}((V)_(2)-(l)_(1))=dfrac (7200)(3600)times (10)^3(80-20)=1.2times (10)^5所以_(UND)=(U)_(m2{C)_(m2)}((V)_(2)-(l)_(1))=dfrac (7200)(3600)times (10)^3(80-20)=1.2times (10)^5因忽略壁阻及污垢两侧的热阻,则_(UND)=(U)_(m2{C)_(m2)}((V)_(2)-(l)_(1))=dfrac (7200)(3600)times (10)^3(80-20)=1.2times (10)^5所以总传热系数 _(UND)=(U)_(m2{C)_(m2)}((V)_(2)-(l)_(1))=dfrac (7200)(3600)times (10)^3(80-20)=1.2times (10)^5平均温度差为_(UND)=(U)_(m2{C)_(m2)}((V)_(2)-(l)_(1))=dfrac (7200)(3600)times (10)^3(80-20)=1.2times (10)^5换热器传热面积为 _(UND)=(U)_(m2{C)_(m2)}((V)_(2)-(l)_(1))=dfrac (7200)(3600)times (10)^3(80-20)=1.2times (10)^5换热器的传热速率为_(UND)=(U)_(m2{C)_(m2)}((V)_(2)-(l)_(1))=dfrac (7200)(3600)times (10)^3(80-20)=1.2times (10)^5则Q换热器>Q需要,说明该换热器能完成上述传热任务。

换热器的传热速率

因忽略壁阻及污垢两侧的热阻,则

所以总传热系数

换热器传热面积为

换热器的传热速率为