第八章 热量传递的基本方式

第八章 热量传递的基本方式英文习题1. Heating of a copper ballA 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 3 and C p =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.2. Heating of water in an electric teapotteapot1.2 kg of liquid water initially at 15℃ to be heated to 95℃ in aequipped with a 1200 W electric heating elementinside(Figure.8-2).The teapot is 0.5 kg and has an average specific heat of 0.7kJ/kg.℃. Taking the specific heat of water to be 4.18 kJ/kg.℃ and disregarding any heat loss from the teapot, determine how long itwill take for the water to be heated. 3. Heat loss from heating ducts in a basementA 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 understeady conditions. Also, determine the cost of this heat loss per hourif the house is heated by a natural gas furnace that has anefficiency of 80 percent, and the cost of the natural gas in thatarea is$0.60/therm (1therm=105.500 kJ). 4. The cost of heat loss through the roofThe roof of an electrically heated home is 6 m long, 8 m wide, and 0.25 m thick, and is made of a flat (Fig. 8-3). layer of concrete whose thermal conductivity is λ=0.8 W/m.℃On a certain winter night, the temperature of the inner andthe outer surfaces of the roof are measured to be about 15℃ and 4℃,respectively, for a period of 10 hours. Determine (a) the rateof heat loss through the roof that night and (b) the cost of that heatloss to the home owner if the cost of electricity is $0.08./kWh.5. Measuring the thermal conductivity of amaterialA 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 ther mometer reads the temperature drop ΔT across this distance along each sample. WhenFIGURE 8-1FIGURE 8-2FIGURE 8-3steady 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 coefficientA 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 voltage measured to be 152℃ in steady operation. Also, thedrop and electric current through the wire are measured tobe 60V and 1.5A, respectively. Disregarding any heat transfer byradiation, determine the convection heat transfer coefficient for heattransfer between the outer surface of the wire and the air in the room.7. Radiation effect on thermal comfortIt is a common experience to feel “chilly” in winter and “warm” in summer in our homes even whe n 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.all times. Consider a person standing in a room maintained at 22℃ atThe inner surfaces of the wall, floors, and the ceiling of the houseare observed to be at an average temperature of 10℃ in winter and25℃ in summer. Determine the rate of radiation heat transfer betweenthis person and the surrounding surfaces if the exposed surface areaand the average outer surface temperature of the person are 1.4m 2 and30℃, respectively (Fig.8-5).8. Heat transfer between two isothermal platesConsider steady heat transfer between two large parallel plates at constant temperatures of T 1=300K and T 2=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 theplates per unit surface area assuming the gap between the plates is(a)filled with atmospheric, (b) evacuated, (c) filled with urethaneinsulation, and (d) filled with superinsulation that has an apparentthermalconductivity of 0.00002 W/m.℃.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 FIGURE 8-4FIGURE 8-5FIGURE 8-61.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。