1、1,Heat TransferP57. 2-9 A steel tube having K=46W/m has an inside diameter of 3.0cm and a tube wall thickness of 2mm. A fluid flows on the inside of the tube producing a convection coefficient of 1500W/m2 on the inside surface, while a second fluid flows across the outside of the tube producing a co
2、nvection coefficient of 197W/m2 on the outside tube surface. The inside fluid temperature is 223 while the outside fluid temperature is 57. Calculate the heat lost by the tube per meter of length.,解：d1=30mm, d2=34mm,2,Heat Transfer P60. 2-39 The temperature distribution in a certain plane wall is,Wh
3、ere T1 and T2 are the temperatures on each side of the wall. If the thermal conductivity of the wall is constant and the wall thickness is L, derive an expression for the heat generation per unit volume as a function of x ,the distance from the plane where T=T1.Let the heat-generation rate be q0 at
4、x=0,解：,3,4,5,中文教材P51. 9某教室的墙壁是由一层厚度为240mm的砖层和一层厚度为20mm的灰泥构成。现在拟安装空调设备，并在内表面加贴一层硬泡沫塑料，使导入室内的热量比原来减少80。已知砖的导热系数为=0.7 W/(mK)，灰泥的=0.58 W/(mK)，硬泡沫塑料的=0.06 W/(mK)，试求加贴硬泡沫塑料层的厚度。 解：已知：,由,得：,6,中文教材P52. 19一外径为100mm，内径为85mm的蒸汽管道，管材的导热系数=40 W/(mK)，其内表面温度为180，若采用=0.053 W/(mK)的保温材料进行保温，并要求保温层外表面温度不高于40，蒸汽管允许的热损失
5、ql=52.3W/m。问保温材料层厚度应为多少？,解：已知：,设保温材料层厚度为,，则,7,中文教材P51. 27一肋片厚度为3mm，长度为16mm，试计算等截面直肋的肋片效率。（1）铝材料肋片，其导热系数为140 W/(mK)，表面传热系数 h=80W/(m2K)；（2）钢材料肋片，其导热系数为40 W/(mK)，表面传热系数h=125W/(m2K)。,解：(1),ml=19.518(16+3/2)10-3=0.3416查表得 th(ml)=0.3275,8,(2),ml=45.64(16+3/2)10-3=0.7988查表得 th(ml)=0.664,9,Heat TransferP186
6、. 4-16 A 12-mm-diameter aluminum sphere is heated to a uniform temperature of 400 and then suddenly subjected to room air at 20 with a convection heat-transfer coefficient of 10W/m2. Calculate the time for the center temperature of the sphere to reach 200.,解： 0.1,用集总参数法：,10,又有：,11,Heat TransferP188.
7、 4-36 A horizontal copper plate 10mm thick is initially uniform in temperature at 250. The bottom surface of the plate is insulated. The top surface is suddenly exposed to a fluid stream at 80. After 6 min the surface temperature has dropped to 150. Calculate the convection heat-transfer coefficient
8、 which causes this droop.,解：=8954kg/m3, =398J/(msK), C=0.384kJ/(kgK)假设Bi0.1，用集总参数法计算,12,h=84.7W/m2K检验：,可以用集总参数法进行计算。,0.1,13,中文教材P82. 4用不锈钢做底板的家用电熨斗初始时处于室温tf 。当开关接通后，电热器在底板内以的强度qvW/m3发热。不锈钢的热物性参数，c和均已知，不锈钢的体积为V，暴露于空气中的表面面积为A，该表面与空气之间的表面传热系数为h，试用集总参数法(lumped-heat-capacity method)分析电熨斗底板温度变化T()。,解：由热平衡
9、有： (注：having inner heat source),(1),14,设 ，则,(1)式变为：,15,t=tf 时，,16,中文教材P83. 16一细长钢棒，直径为0.04m，均匀加热到400，然后放入温度为30的油浴中，已知钢材=8000/ m3，c=0.46KJ/(.K)，=45W/(m.K)，钢棒表面在油浴中的表面传热系数h=500W/(.K)。试求10min后，棒中心和表面的温度；若为了控制棒中心温度为180，则需冷却多少时间？此期间每米长钢棒的散热量为多少？,解：由题意可知：R=d/2=0.02m, t0=400, tf=30,=1.2210-5/s,h=500W/(.K),
10、17,（1）10min后，棒中心和表面的温度根据Bi数和Fo数查教材图3-13，可得：,，,则10min后，棒中心的温度几乎等于油的温度。,，,表面温度：,则10min后，棒表面的温度也等于油的温度。,（2）控制棒中心温度为180，则有,，Bi=0.22，查图3-13可得Fo=2.21,由,可得,18,（3）每米长钢棒的散热量,，查图3-15得,=0.57，则有,=0.571710.2=974.8kJ,棒中心温度为180时，每米长钢棒的散热量为974.8kJ,19,中文教材P83. 19用常功率平面热源法进行材料的热扩散率和导热系数的测定，试材可以认为是一半无限大物体。平面热源的加热热流密度为50W/。试验开始前试材温度均匀一致为10，通电加热后，经过315s,试材与平面热源接触表面处温度测得为19.5，经过358s，测得距平面热源0.015m处试材的温度为11.5。试求材料的导热系数和热扩散率。,解：常热流密度半无限大物体表面温度分布为：,常热流密度条件下半无限大物体内温度场的表达式：,20,其中q=50W/,t0=10,则有,=19.5-10=9.5；,=11.5-10=1.5,由,，,可得：,，代入数值计算:,21,查附录14可得,，则,=2.2810-7/s,由,可得,则有,=0.05W/（m.K）,
