发布时间 : 星期三 文章第八章 热量传递的基本方式 - 图文更新完毕开始阅读8b2b935984254b35eefd34ab
第八章 热量传递的基本方式
英文习题
1. Heating of a copper ball
A 10-cm-diameter copper is to be heated from 100℃ to an average temperature of 150℃ in 30 minutes. Taking the average density and specific heat of copper in this temperature range to be ρ=8950 kg/m and Cp=0.395 kJ/kg.℃, respectively. Determine (a) the total amount of heat transfer to the copper ball; (b) the average rate of heat transfer to the ball; (c) the average heat flux.
3
2. Heating of water in an electric teapot
1.2 kg of liquid water initially at 15℃ to be heated to 95℃ in a equipped
with
a
1200
W
electric
heating
element
heat of 0.7 will
take
inside(Figure.8-2).The teapot is 0.5 kg and has an average specific kJ/kg.℃. Taking the specific heat of water to be 4.18 kJ/kg.℃ and disregarding any heat loss from the teapot, determine how long it for the water to be heated.
FIGURE 8-1teapot
3. Heat loss from heating ducts in a basement
A 5-m-long section of an air heating system of a house passes through an unheated space in the basement (Fig.8-2). The cross-section of the rectangular duct of the heating system is 20 cm×25 cm. Hot air enters the duct at 100 kPa and 60℃ at an average velocity of 5m/s. The temperature of the air in the duct drops to 54℃ as a result of heat loss to the cool space in the basement. Determine the rate of heat loss from the air in the duct to the basement under conditions. Also, determine the cost of this heat loss per hour house is heated by a natural gas furnace that has an efficiency of 80 percent, and the cost of the natural gas in that $0.60/therm (1therm=105.500 kJ).
FIGURE 8-2steady if
the
area is
4. The cost of heat loss through the roof
The roof of an electrically heated home is 6 m long, 8 m wide, and 0.25 m thick, and is made of a flat layer of concrete whose thermal conductivity is λ=0.8 W/m.℃ On a certain winter night, the temperature of the inner and surfaces of the roof are measured to be about 15℃ and 4℃, respectively, for a period of 10 hours. Determine (a) the rate loss through the roof that night and (b) the cost of that heat the home owner if the cost of electricity is $0.08./kWh.
FIGURE 8-3(Fig. 8-3). the outer of loss
heat to
5. Measuring the thermal conductivity of a material
A common way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical samples of the material, as shown in Fig. 8-4. The thickness of the resistance heater, including its cover, which is made of thin silicon rubber, is usually less than 0.5mm. A circulating fluid such as tap water keeps the exposed ends of the samples at constant temperature. The lateral surfaces of the samples are well insulated to ensure that heat transfer through the samples is one-dimensional. Two thermocouples are embedded into each sample some distance L apart, and a differential thermometer reads the temperature drop ΔT across this distance along each sample. When
steady operating conditions are reached, the total rate of heat transfer through both samples becomes equal to the electric power drawn by the heater, which is determined by multiplying the electric current by the voltage.
In a certain experiment, cylindrical samples of diameter 5cm and length 10cm are used. The two thermocouples in each sample are placed 3cm apart. After initial transients, the electric heater is observed to draw 0.4A at 110V, and both differential thermometers read a temperature difference of 15℃. Determine the thermal conductivity of the sample.
6. Measuring convection heat transfer coefficient
A 2-m-long, 0.3-cm-diameter electrical wire extends across a room at 15℃, as shown in Fig 9-5. Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be 152℃ in steady operation. Also, the drop and electric current through the wire are measured to and 1.5A, respectively. Disregarding any heat transfer by determine the convection heat transfer coefficient for heat between the outer surface of the wire and the air in the
FIGURE 8-4voltage be
60V radiation, transfer room.
7. Radiation effect on thermal comfort
It is a common experience to feel “chilly” in winter and “warm” in summer in our homes even when the thermostat setting is kept the same. This is due to the so called “radiation effect” resulting from radiation heat exchange between our bodies and the surrounding surfaces of the walls and the ceiling. Consider a person standing in a room maintained at 22℃ at The inner surfaces of the wall, floors, and the ceiling of the house observed to be at an average temperature of 10℃ in winter and summer. Determine the rate of radiation heat transfer between person and the surrounding surfaces if the exposed surface area average outer surface temperature of the person are 1.4m and respectively (Fig.8-5).
FIGURE 8-52
all times.
are 25℃
in this and the 30℃,
8. Heat transfer between two isothermal plates
Consider steady heat transfer between two large parallel plates at constant temperatures of T1=300K and T2=200K that are L=1cm apart, as shown in Fig 8-5. Assuming the surfaces to be black (emissivity ε=1), determine the rate of heat transfer between the unit surface area assuming the gap between the plates is with atmospheric, (b) evacuated, (c) filled with urethane and (d) filled with superinsulation that has an apparent conductivity of 0.00002 W/m.℃.
FIGURE 8-6plates per
(a)filled insulation, thermal
工程热力学与传热学
第八章 热量传递的基本方式 习题 习 题
1. 2. 3. 4. 5.
试说明热传导，热对流和热辐射三种热量传递基本方式之间的联系与区别。 试说明热对流与对流换热之间的联系与区别。
导热系数和表面传热系数是物性参数吗？请写出他们的定义式，并说明其物理意义。 平壁的导热热阻与哪些因素有关，请写出其表达式。
在有空调的房间内，夏天和冬天的室温均控制在20℃，夏天只穿衬衫，但冬天穿衬衫会感到冷，
这是为什么？ 6.
在计算机主机箱中为什么在CPU处理器上和电源旁要加风扇？
习 题 解 答 1.
答；热传导，热对流，热辐射是热量传递的三种基本方式。热传导是指物体的各部分之间不发
生相对位移时，依靠物体内部的微观粒子，分子，原子和自由电子的饿热运动而进行的热量传递现象。无论是固体，液体还是气体，它们之间存在温差时，在相互接触的时候，就会发生导热现象，然而，单纯的导热只能发生在密实的固体中。热对流是指由于流体的宏观运动使温度不同的流体相对位移而产生的热量传递现象，显然，热对流只能发生在流体之中，而且必然伴随有微观粒子热运动产生的导热。由于物体内部微观粒子的热运动使物体向外发射辐射能的现象称为热辐射。热辐射可以不依靠中间媒介，可以在真空中传播，热辐射总是伴随着热能与辐射能这两种能量形式之间的相互转化，物体间以热辐射的方式进行的热量传递是双向的。 2.
答：热对流是指由于流体的宏观运动使温度不同的流体相对位移而产生的热量传递现象，显然，
热对流只能发生在流体之中。流体与固体表面之间的热量传递是热对流和导热两种基本传热方式共同作用的结果，这种传热现象称为对流换热。 3.
答：导热系数是物性参数，表示该物体导热能力的大小，其定义为：??q。
gradt表面传热系数不是物体的物性参数，表面传热系数的大小反映对流换热的强弱，它取决于物体的物性，流动的形态，物体表面的形状和尺寸，换热时流体有无相变等因素。表面传热系数可以根据牛顿冷却公式定义，即：q?h?t,Q?hA?t。 4.
答；平壁导热热阻的计算公式为：Rt?l，热阻与物体的导热系数，导热面积成反比，与平?A壁的厚度成正比。 5.
答：答：首先，冬天和夏天的最大区别是室外温度的不同。夏季室外温度比室内气温高，因此
通过墙壁的热量传递方向是由室外传向室内。而冬季室外气温比室内低，通过墙壁的热量传递方向是由室内传向室外。因此冬季和夏季墙壁内表面温度不同，夏季高而冬季低。因此尽管冬季室内温度比夏季略高，但人体在冬季通过辐射与墙壁的散热要比夏季高很多，根据人体对冷的感受主要是
散热量的原理，在冬季散热量大，因此要穿厚一些。 6.
答；增强换热，降低CPU的温度。