传热-2013-2.ppt

4.3 对流传热 4.3.1 对流传热过程分析 4.3.2 对流传热速率 4.3.3 影响对流传热系数的因素 4.3.4 对流传热系数关联式的建立 4.3.5 无相变时的对流传热系数关联式 4.3.6 有相变时的对流传热系数关联式 4.3.1 对流传热过程分析 对流传热机理 通过流体内分子的定向流动和混合而导致热量的传递。 指流体与固体壁面直接接触时的传热，是流体的对流与导热两者共同作用的结果。 传热速率与流动状况有密切关系。 湍动程度越高，对流的传热速率越大 对流传热的分类 热边界层的形成过程 牛顿冷却定律 A layer of insulation is installed around the outside of a cylinder whose radius r1 is fixed and with a length L. The cylinder has a high thermal conductivity and the inner temperature T1 at point r1 To determine the effect of the thickness of insulation on Q, we take the derivative of Q with respect to r2, equate this result to zero, and obtain the following for maximum heat flow: (r2)cr is the value of the critical radius when the heat-transfer rate is a maximum. r2 (r2)cr, the increasing of the thickness of insulation increase Q. r2 (r2)cr,, the increasing of r2 will decrease Q. adding insulation on small electrical wires could increase the heat loss. Adding insulation to large pipes decreases Q. EXAMPLE Insulating an Electrical Wire and Critical Radius An electric wire having a diameter of 1.5 mm and covered with a plastic insulation (thickness = 2.5 mm) is exposed to air at 300 K and h0 = 20 W/m2 · K. The insulation has a λ of 0.4 W/m · K. It is assumed that the wire surface temperature is constant at 400 K and is not affected by the covering. (a).Calculate the value of the critical radius. (b).Calculate the heat loss per m of wire length with no insulation. ( C) Repeat (b) for insulation being present. 重要结论 管路中常常要保温，以减少热损失，通常热损失随着保温层厚度的增加而减少 在小直径管外包扎性能不良的保温材料，随保温层厚度的增加，可能反而使热损失增大： 习题： 有ф25×2的蒸汽管道,管内饱和蒸汽温度为130℃,管外包一层 导热系数为0.8 w/m2.℃的保温层,保温层外面的大气温度为30℃ 空气的传热膜系数为10w/m2℃ , 蒸气的传热膜系数为10000w/m2.℃ 试问: 1. 是否在任何条件下都是保温层越厚,热损失减少 ? 2. 最佳的保温层的r为多少? 问答题？ 在无相变的对流传热过程中，热阻主要集中在？，减少热阻的最有效措施是 ？ 提高流体湍动程度 。 【例】质量流量为8.32kg/s的苯通过由38根φ25×2.5mm钢管组成的列管换热器。20oC的苯通过管内并被加热到80oC。求管壁对苯的传热系数。当苯的流量提高50%时，传热系数有何变化？ 忽略温度变化对物性的影响，当苯流量增 加50%时，传热系数为α′ 壁面冷凝的存在形态 ①　膜状冷凝