Heat Transfer热传去年第二次小考题目

传热考试题及答案一、选择题1. 热传导的三种基本方式是：A. 对流、辐射、传导B. 对流、辐射、对流C. 传导、对流、辐射D. 传导、辐射、对流答案：C2. 以下哪种材料的热导率最高？A. 木头B. 铜C. 橡胶D. 玻璃答案：B3. 热对流的驱动力是：A. 温度差B. 压力差C. 密度差D. 以上都是答案：D二、填空题4. 傅里叶定律表明，单位时间内通过单位面积的热量与温度梯度成正比，其比例常数为______。

答案：热导率5. 热辐射的强度与物体表面温度的四次方成正比，这一定律称为______。

答案：斯特藩-玻尔兹曼定律6. 热交换器中，流体的流动方式可以是______或______。

答案：并流逆流三、简答题7. 简述热传导的基本原理。

答案：热传导是指热量通过物质内部分子振动和碰撞传递的过程，不需要物质的宏观运动。

8. 为什么说对流是流体中热量传递的主要方式？答案：对流是流体中热量传递的主要方式，因为它涉及到流体的宏观运动，能够快速有效地传递热量。

9. 辐射传热的特点是什么？答案：辐射传热不需要介质，可以在真空中进行，且其传递速度与光速相同。

四、计算题10. 已知一金属棒的热导率为200 W/m·K，棒的横截面积为0.01 m²，棒两端的温度差为100°C，求金属棒的热流量。

答案：热流量Q = k * A * ΔT / L = 200 W/m·K * 0.01 m² * 100°C = 2000 W11. 一个房间的表面温度为300K，周围环境的温度为290K，求房间的辐射热损失。

答案：热损失Q = σ * A * (T_room^4 - T_env^4)，其中σ为斯特藩-玻尔兹曼常数，A为房间表面积。

五、论述题12. 论述在不同工况下，热传导、对流和辐射在热交换过程中的作用和影响。

答案：在不同的工况下，热传导、对流和辐射在热交换过程中的作用和影响各不相同。

Chapter 4 Heat transfer4-4. A combustion furnace consists of refractory brick (460 mm thick) and insulation brick (230 mm thick). If the inner surface temperature of refractory brick t1 and the outside surface temperature of t3 of the insulation brick is 1400℃and 100℃respectively, calculate the heat flux and the interface temperature of the two layers. Assume that the two layers contact tightly. Given that the thermal conductivity of the refractory brick is λ1=0.9+0.0007t; and the thermal conductivity of the insulation brick is λ2=0.3+0.0003t, where t can be considered as the average temperature of the layer, the unit of λ is W/(m·℃).4-5. Steel tube with a specification of Φ60 3mm, is packed by cork with 30mm thick, furthermore, is packed by some insulating material with 100mm thick outside as the thermal insulation layer. The outside surface temperature of the steel tube and of the insulation layer is–110℃and 10℃respectively. The thermal conductivity of the cork is 0.037(kcal/m·h·℃) and that of the insulating material is 0.06(kcal/m·h·℃). Calculate the cooling loss per meter tube.4-6. The outside of the steam tube is packed with two insulation layers with different thermal conductivity but the same thickness. Assume that the diameter of the outside layer is twice as that of the inner layer, so does the thermal conductivity. If the position of the two layers is changed with each other, and the other conditions are constant, what’s the change of heat loss per meter tube? Under the above conditions, which material is much appropriate when used as the inner layer?4-7. The inside radius and the outside radius of the hollow sphere is r i and r o respectively. The temperature is t i and t o, respectively, with t i > t o. The thermal conductivity of the hollow sphere is λ. Try to deduce the expression of the thermal resistance through the hollow sphere wall.4-9. In a parallel-current flow heat exchanger, water is used to cool the oil. The temperature of entering water and leaving water is 15℃and 40℃, while that of entering and leaving oil is 150℃and 100℃, respectively. When the production conditions change, the temperature of the leaving oil is required to reduce to 80℃. Suppose that the mass flow, the entering temperature and the physical properties of the two fluids are the same as the former. If the tube length of the original heat exchanger is 1m, how long has the tube to increase to meet the new production requirement? The heat loss of the heat exchanger can be ignored.4-10. Heavy oil and crude oil are flowing parallelly in the single-pass double-pipe heat exchanger, the original temperature of the two oils is 243℃and 128℃, respectively, while the final temperature is 167℃and 157℃, respectively. If the original temperature and the mass flow is the same as the former, while the flow type changes from cocurrent flow to counter flow, calculate the average temperature difference and the final temperatures. Suppose that under the two conditions, the total heat transfer coefficient and the physical properties keep the same, and the heat loss can be neglected.4-11. Air is heated from 20℃to 80℃in a steam heater and flows turbulently in the tube-side of the exchanger. The saturated steam with a vapor pressure of 1.8 kgf/cm²is condensed outside the tube. If the flow rate of air is needed to increase 20%, while the temperature of the air at the inlet and outlet are constant, how can we finish the task? Assume that the heat resistance of the tube wall and the scale can be neglected.4-12. Butyl alcohol at 90℃ is cooled to 50℃. The heat transfer area of the exchanger is 6 m², and the total heat transfer coefficient is 230w/(m²⋅ ℃). If the flow rate of butyl alcohol is 1930kg/h, the cooling medium is water at 18℃, calculate:1) The temperature of the cooling water at the outlet;2) The consumption of the cooling water, in m³/h. （70℃the specific heat of butyl alcohol is: )/(98.2C kg kJ C p ︒⋅= 4-13. In a counter-flow heat exchanger, water at 20℃ is used to cool the liquid from 80℃ to 20℃. The specific heat of the liquid is1.9kJ/(kg ⋅ ℃), and the density is 850kg/m³. The diameter of the tubes in the exchanger is Φ25⨯2.5mm.. Water flows in the tube-side. The convective heat transfer coefficient of water and the liquid is 0.85 and 1.70kW/(m²⋅ ℃), respectively. The resistance of the scale on pipe wall can be ignored. If the temperature of water at the outlet can not be larger than 50℃, calculate the heat transfer area of the exchanger.4-14. In a tubular exchanger, oil is cooled with water, which flows in the tube (Φ19⨯2mm). Given that the convective heat transfer coefficient i αof the water-side is 3490W/(m²⋅K), and that of the oil-side 0α is 258 W/(m²⋅K). After the exchanger is used for some time, scales are formed in both sides of the tube wall. The scale resistance Rsi on the water-side is 0.00026 (㎡⋅K)/W, and that Rso on the oil-side is 0.000176㎡·K/W. T he thermal conductivity λ of the wall is 45 W/(m²⋅K). Calculate: 1) The total heat transfer coefficient K o based on the outside surface of the tube; 2) The increased percentage of the heat resistance after the scale resistance is formed.4-18. Air with a rate of 60m³/h is heated from 20℃ to 80℃ at 2atm in the inner tube of a double-pipe heat exchanger. The diameter of the tube is Φ57⨯3.5mm and the length is 3m. Calculate the convective heat transfer coefficient of air in the tube.4-21. At the normal pressure, methane is flowing in the shell-side of the tubular exchanger at a rate of 10m/s along the axial direction at 120℃. When methane leaves the exchanger, the temperature has changed to 30 ℃. The inner diameter of exchanger shell is 190mm, and the tube bundle consists of 37 steel tubes(Φ19⨯2mm). Try to calculate the convective heat transfer coefficient of methane towards the tube wall.4-24. In the experiment of measuring the total heat transfer coefficient of the tubular exchanger, water flows turbulently in the tube-side of the exchanger, while the saturated vapor is condensed in the shell-side. The tube bundle consists of steel pipes (Φ25⨯2.5mm). When the velocity of water is 1m/s, the total heat transfer coefficient K o based on the outside surface of the tube is 2115 W/(m²⋅℃). If the other conditions are invariable, while the velocity of water changes to 1.5m/s, K o measured would be 2660 W/(m²⋅℃). Try to calculate the heat transfer coefficient of the vapor condensation. The scale resistance is ignored.4-25. Two parallel flat plate are 5mm apart from each other in the air. The blackness of one plate is 0.1, and the temperature is 350K, while the blackness of the other plate is 0.05, the temperature is 300K. If the first plate is coated, reducing the blackness to 0.025. Calculate the change of the heat transfer rate. Assume that the convection heat transfer can be neglected.。

热能的转移和传导练习题一、选择题：1. 热能的传导方式主要有下列几种，其中哪一种是必须通过介质才能进行的？A. 热辐射B. 热对流C. 热传导D. 热传导、热辐射和热对流都需要介质2. 下列哪一项属于热对流的特点？A. 只能在固体和气体之间传导B. 可以在真空中传导C. 传热速度最快D. 传热速度最慢3. 在以下哪一个材料中，热传导的能力最差？A. 金属B. 空气C. 石油D. 水4. 假设一个房间里有一个加热器，窗户是封闭的，有哪种方式是通过热辐射传输热量？A. 加热器向空气传输热能B. 加热器通过墙壁传输热能C. 加热器通过窗户传输热能D. 加热器通过直接接触传输热能5. 热传导的速度与下列哪一项因素无关？A. 材料的导热性能B. 材料的温度差C. 材料的形状D. 材料的密度二、填空题：1. 热传导是指热能通过物体内部的________ 传播。

2. ________ 是靠分子之间的热量传递进行的。

3. 空气和液体传热主要通过________进行。

4. 热辐射可以在真空中传播，因为它不需要通过________。

5. 热对流是通过________介质进行的。

三、解答题：1. 简述热传导的原理及其影响因素。

2. 举例说明热对流的过程。

3. 描述热辐射在日常生活中的应用。

4. 解释导热性能对材料传导热能的影响。

5. 分析热能的转移和传导对人类社会和环境的影响。

参考答案：一、选择题：1. C2. B3. B4. C5. C二、填空题：1.物体内部2.热传导3.热对流4.介质5.流体三、解答题：1. 热传导是指热能通过物体内部的分子传播。

通过分子之间的碰撞和振动，热能会从高温区域传递到低温区域。

热传导的速度受到材料的导热性能、温度差和材料形状等因素的影响。

导热性能越好的材料，传导速度越快；温度差越大，传导速度越快；材料形状越薄，传导速度越快。

2. 热对流是指通过流体介质进行的传热方式。

当流体受热而热胀时，会形成热对流循环。

热传递试题及答案一、选择题1. 热传递的三种基本方式是什么？A. 热传导、热对流、热辐射B. 热传导、热对流、热交换C. 热传导、热交换、热辐射D. 热对流、热交换、热辐射答案：A2. 下列哪种物质的导热性能最好？A. 木头B. 玻璃C. 金属D. 水答案：C3. 热对流主要发生在哪种介质中？A. 固体B. 液体C. 气体D. 真空答案：B4. 热辐射不需要介质，其传播速度是多少？A. 3×10^8 m/sB. 3×10^5 m/sC. 3×10^2 m/sD. 3×10^1 m/s答案：A二、填空题1. 热传递的三种基本方式包括________、________和________。

答案：热传导、热对流、热辐射2. 热传导是通过________进行热量传递的。

答案：物质内部分子的振动和碰撞3. 热对流是依靠________的流动来传递热量的。

答案：流体4. 热辐射是物体通过________向外传递能量的。

答案：电磁波三、简答题1. 请简述热传导的基本原理。

答案：热传导是热量通过直接接触的物体内部分子振动和碰撞传递的过程，从温度高的区域传递到温度低的区域。

2. 热对流与热传导的主要区别是什么？答案：热对流依赖于流体的流动来传递热量，而热传导是通过物体内部分子的振动和碰撞传递热量，不需要流体的流动。

四、计算题1. 假设一个金属棒长2米，横截面积为0.01平方米，热传导系数为200 W/(m·K)，两端温差为100K，求金属棒的热传导率。

答案：Q = k * A * ΔT / L = 200 * 0.01 * 100 / 2 = 100 W2. 一个封闭的容器内装有水，水的比热容为4186 J/(kg·K)，质量为5kg，初始温度为20℃，加热到100℃，求加热过程中需要的热量。

答案：Q = mcΔT = 5 * 4186 * (100 - 20) = 1.8344 * 10^6 J。

PROBLEM 1.22KNOWN: Hot vertical plate suspended in cool, still air. Change in plate temperature with time at the instant when the plate temperature is 245°C.FIND: Convection heat transfer coefficient for this condition.SCHEMATIC:ASSUMPTIONS: (1) Plate is isothermal, (2) Negligible radiation exchange with surroundings, (3)Negligible heat lost through suspension wires.ANALYSIS: As shown in the cooling curve above, the plate temperature decreases with time. The condition of interest is for time t o . For a control surface about the plate, the conservation of energy requirement is()outst in s s pE - E = E dT 2hA T T Mc dt∞−−=&&&where A s is the surface area of one side of the plate. Solving for h, find()ps s Mc -dT h =2A T - T dt ∞⎛⎞⎜⎟⎝⎠()()224.25 kg × 2770 J/kg K h = × 0.028 K/s = 4.7 W/m K 2 × 0.4 × 0.4m 245 - 25K⋅⋅<COMMENTS: (1) Assuming the plate is very highly polished with emissivity of 0.08, determinewhether radiation exchange with the surroundings at 25°C is negligible compared to convection.(2) We will later consider the criterion for determining whether the isothermal plate assumption is reasonable. If the thermal conductivity of the present plate were high (such as aluminum or copper), the criterion would be satisfied.-0.022 K/s-0.028 K/s = 245 °CPlate, 0.3×0.3 m M = 4.25 kgc p = 2770 J/kg·KPlate, 0.4x0.4 mPROBLEM 1.29KNOWN: Air and wall temperatures of a room. Surface temperature, convection coefficient and emissivity of a person in the room.FIND: Basis for difference in comfort level between summer and winter.ASSUMPTIONS: (1) Person may be approximated as a small object in a large enclosure.ANALYSIS: Thermal comfort is linked to heat loss from the human body, and a chilled feeling is associated with excessive heat loss. Because the temperature of the room air is fixed, the different summer and winter comfort levels cannot be attributed to convection heat transfer from the body. In both cases, the heat flux isSummer and Winter : ()22q h T T 2 W/m K 12 C 24 W/m convs ′′=−=⋅×=∞oHowever, the heat flux due to radiation will differ, with values of Summer : ()()448244442q T T 0.9 5.6710W/m K 305300K 28.3 W/m rads sur εσ−′′=−=××⋅−= Winter :()()448244442q T T 0.9 5.6710W/m K 305287K 95.4 W/m rad s surεσ−′′=−=××⋅−=There is a significant difference between winter and summer radiation fluxes, and the chilled condition is attributable to the effect of the colder walls on radiation.COMMENTS: For a representative surface area of A = 1.5 m 2, the heat losses are q conv = 36 W, q rad(summer) = 42.5 W and q rad(winter) = 143.1 W. The winter time radiation loss is significant and if maintained over a 24 h period would amount to 2,950 kcal.PROBLEM 1.44KNOWN: Radial distribution of heat dissipation in a cylindrical container of radioactive wastes. Surface convection conditions.FIND: Total energy generation rate and surface temperature.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible temperature drop across thin container wall.ANALYSIS: The rate of energy generation is()()r 2g o o 022g o o o E qdV=q 1-r/r 2rLdr E 2Lqr /2r /4o ⎡⎤=⎢⎥⎣⎦=−∫∫&&&&&ππor per unit length,&&.′=E q r 2g o o 2π <Performing an energy balance for a control surface about the container yields, at an instant,&&′−′=E E g outand substituting for the convection heat rate per unit length,()()2o o o s qr h 2r T T 2∞=−&ππT T qr 4hs o o =+∞&. <COMMENTS: The temperature within the radioactive wastes increases with decreasing r from T s at r o to a maximum value at the centerline.PROBLEM 1.76KNOWN: Thickness and thermal conductivity, k, of an oven wall. Temperature and emissivity, ε, of front surface. Temperature and convection coefficient, h, of air. Temperature of large surroundings.FIND: (a) Temperature of back surface, (b) Effect of variations in k, h and ε.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Radiation exchange with large surroundings.ANALYSIS: (a) Applying an energy balance, Eq. 1.13 to the front surface and substituting the appropriate rate equations, Eqs. 1.2, 1.3a and 1.7, find()()44122sur 2T T k h T T T T Lεσ∞−=−+−.Substituting numerical values, find()()1220.05m W 448T T 20100K 240.7W/m K m K W 0.8 5.6710400K 300K 200K m K −−=⋅⋅+××−=⋅⎤⎡⎡⎤⎥⎢⎢⎥⎣⎦⎣⎦.Since 2T = 400 K, it follows that 1T = 600 K.<(b) Parametric effects may be evaluated by using the IHT First Law Model for a Nonisothermal Plane Wall. Changes in k strongly influence conditions for k < 20 W/m ⋅K, but have a negligible effect for larger values, as 2T approaches 1T and the heat fluxes approach the corresponding limiting values100200300400Thermal conductivity, k(W/m.K)300400500600T e m p e r a t u r e , T 2(K )100200300400Thermal conductivity, k(W/m.K)0200040006000800010000H e a t f l u x , q ''(W /m ^2)Conduction heat flux, q''cond(W/m^2)Convection heat flux, q''conv(W/m^2)Radiation heat flux, q''rad(W/m^2)Continued…PROBLEM 1.76 (Cont.)The implication is that, for k > 20 W/m ⋅K, heat transfer by conduction in the wall is extremely efficient relative to heat transfer by convection and radiation, which become the limiting heat transfer processes. Larger fluxes could be obtained by increasing ε and h and/or by decreasing T ∞ and sur T .With increasing h, the front surface is cooled more effectively (2T decreases), and although radq ′′ decreases, the reduction is exceeded by the increase in convq ′′. With a reduction in 2T and fixed values of k and L, condq ′′ must also increase.100200Convection coefficient, h(W/m^2.K)400500600T e m p e r a t u r e , T 2(K )100200Convection coefficient, h(W/m^2.K)0100002000030000H e a t f l u x , q ''(W /m ^2)Conduction heat flux, q''cond(W/m^2)Convection heat flux, q''conv(W/m^2)Radiation heat flux, q''rad(W/m^2)The surface temperature also decreases with increasing ε, and the increase in rad q ′′ exceeds the reduction in convq ′′, allowing cond q ′′ to increase with ε. 00.20.40.60.81Emissivity550555560565570575T e m p e r a t u r e , T 2(K )0.20.40.60.81Emissivity0200040006000800010000H e a t f l u x , q ''(W /m ^2)Conduction heat flux, q''cond(W/m^2)Convection heat flux, q''conv(W/m^2)Radiation heat flux, q''rad(W/m^2)COMMENTS: Conservation of energy, of course, dictates that irrespective of the prescribed conditions, cond conv radq q q ′′′′′′=+.PROBLEM 1.77KNOWN: Temperatures at 15 mm and 30 mm from the surface and in the adjoining airflow for a thick stainless steel casting.FIND: Surface convection coefficient, h. SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction in the x-direction, (3) Constant properties, (4) Negligible generation.ANALYSIS: From a surface energy balance, it follows that′′=′′q q cond convwhere the convection rate equation has the form()0convq h T T ,∞′′=−and ′′q cond can be evaluated from the temperatures prescribed at surfaces 1 and 2. That is, from Fourier’s law,()()12cond212condT T q k x x 6050C Wq 1510,000 W/m .m K 30-1510m−′′=−−′′==⋅×oSince the temperature gradient in the solid must be linear for the prescribed conditions, it follows thatT o = 70°C. Hence, the convection coefficient ish =q T T cond 0′′−∞2210,000 W/m h 333 W/m K 30°C==⋅<COMMENTS: The accuracy of this procedure for measuring h depends strongly on the validity of the assumed conditions.T 1= 60°C T 2= 50°C 3015PROBLEM 1.81KNOWN: Conditions associated with surface cooling of plate glass which is initially at 600°C. Maximum allowable temperature gradient in the glass.FIND: Lowest allowable air temperature, T ∞.SCHEMATIC:ASSUMPTIONS: (1) Surface of glass exchanges radiation with large surroundings at T sur = T ∞, (2) One-dimensional conduction in the x-direction.ANALYSIS: The maximum temperature gradient will exist at the surface of the glass and at the instant that cooling is initiated. From the surface energy balance, Eq. 1.13, and the rate equations, Eqs. 1.1, 1.3a and 1.7, it follows that()()44s s sur dT -kh T T T T 0dxεσ∞−−−−=or, with (dT/dx)max = -15°C/mm = -15,000°C/m and T sur = T ∞,()C 2W W1.415,0005873T K m K m m K ∞⎡⎤−−=−⎢⎥⋅⎢⎥⋅⎣⎦o844424W0.8 5.6710873T K .m K −∞⎡⎤+××−⎢⎥⎣⎦⋅T ∞ may be obtained from a trial-and-error solution, from which it follows that, for T ∞ = 618K,21000127519730,,.W m W m Wm222≈+Hence the lowest allowable air temperature isT K =345C.∞≈618o<COMMENTS: (1) Initially, cooling is determined primarily by radiation effects.(2) For fixed T ∞, the surface temperature gradient would decrease with increasing time into the cooling process. Accordingly, T ∞ could be decreasing with increasing time and still keep within themaximum allowable temperature gradient.。

2015/03/12, M3510A, Heat Transfer, Homework III, Due 2015/03/19Problem 2.5 Assume steady-state, one-dimensionalheat conduction through the symmetric shape shown.Assuming that there is no internal heat generation,derive an expression for the thermal conductivity()k x for these conditions: ()()1A x x =-,()()330012T x x x =--, and q = 6000 W, where A is in square meters, T in kelvins, and x in meters.Problem 2.16 Steady-state, one dimensionalconduction occurs in a rod of constant thermalconductivity k and variable cross-sectional area()0ax x A x A e =, where 0A and a are constants. Thelateral surface of the rod is well insulated. (a) Write anexpression for the conduction heat rate, ()x q x . Usethis expression to determine the temperaturedistribution ()T x . (b) Now consider conditions for which thermal energy is generated in therod at a volumetric rate ()0exp q q ax ''''''=-, where 0q ''' is a constant. Obtain an expression for ()x q x when the left face (x = 0) is well insulated.Problem 2.28 Uniform internal heat generation at 73610W m q '''=⨯ is occurring in a cylindrical nuclear reactor fuel rod of 60-mm diameter, and under steady-state conditions the temperature distribution is of the form ()52900 5.2610T r r =-⨯⨯, where T is in degrees Celsius and r is in meters. The fuel rod properties are 30W m K k =⋅, 31100kg m ρ=, and 800J kg K P c =⋅. (a) What is the rate of heat transfer per unit length of the rod at r = 0 (the centerline) and at r = 30 mm (the surface)? (b) If the reactor power level is suddenly increased to 83210W m q '''=, what is the initial time rate of temperature change at r = 0 and r = 30 mm?Problem 2.40 The steady-state temperature distribution in a one-dimensional wall of thermal conductivity k and thickness L is of the form 32T ax bx cx d =+++. Derive expressions for the heat generation rate per unit volume in the wall and the heat fluxes at the two wall faces (x = 0, L).Problem 2.43 The cylindrical systemillustrated has negligible variation oftemperature in the r- and z-directions.Assume that o i r r r ∆=- is small comparedto i r , and denote the length in thez-direction, normal to the page, as L . (a)Beginning with a properly defined controlvolume and considering energy generationand storage effects, derive the differentialequation that prescribes the variation in temperature with the angular coordinate φ. (b) For steady-state conditions with no internal heat generation and constant properties, determine the temperature distribution ()T φ in terms of the constants 1T , 2T , i r , and o r . (c) For the conditions of part (b) write the expression for the heat rate q φ.Problem 2.51 A Spherical shell of inner and outer radii i r and o r , respectively, contains heat-dissipating components, and at a particular instant the temperature distribution in the shell is known to be of the form ()12C T r C r=+. Are conditions steady-state or transient? How do the heat flux and heat rate vary with radius?。

2014/03/18, M3510A, Heat Transfer, Quiz I1. A cylinder of radius o r , length L , and thermal conductivity k is immersed in a fluid of convection coefficient h and unknown temperature T ∞. At a certain instant the temperature distribution in the cylinderis ()2T r a br =+, where a and b are constants. Obtain an expression for the fluid temperature.2. The temperature distribution in a cylinder of thermal conductivity k is of the form432T a r b c d t φ=⨯+⨯++⨯, where a, b, c, and d are constants. Derive expressions for the heat generation rate per unit volume in the cylinder.3. The temperature distribution across a wall 1 m thick at a certain instant of time is given as()290030050T x x x =--, A uniform heat generation, 1000q '''=3W m , is present in the wall of area 102m having the properties 1600ρ=3kg m , 40k =W m K ⋅, and 4c =kJ kg K ⋅. Determine the rate of change of energy storage in the wall.4. An experiment to determine the convection coefficient associated with airflow over the surface of a thick stainless steel casting involves the insertion of thermocouples into the casting at distances of 10 and 20 mm from the surface along a thermal conductivity of 10W m K ⋅. If the thermocouples measure temperatures of 60 and 40℃ in the steel when the air temperature is 200℃, what is the convection coefficient?5. An uninsulated steam pipe passes through a room in which the air and walls are at 25℃. The outsidediameter of the pipe is 70 mm, and its surface temperature and emissivity are 200℃ and 0.8, respectively. If the coefficient associated with free convection heat transfer from the surface to the air is 152W m K ⋅, what is the rate of heat loss from the surface per unit length of pipe?6. The wall of an oven used to cure plastic parts is of thickness 0.1L =m and is exposed to largesurroundings and air at its outer surface. The air and the surroundings are at 300 K. If the temperature of the outer surface is 500 K and its convection coefficient and emissivity are 50h =2W m K ⋅ and 0.6ε=, respectively, what is the temperature of the inner surface if the wall has a thermal conductivity of 1k =W m K ⋅?7. Consider a carton of milk that is refrigerated at a temperature of 5m T =℃. The kitchen temperature on a hot summer day is 30T ∞=℃. If the four side of the carton are of height and width 150L =mm and 80w =mm, respectively, determine the heat transferred to the milk carton as it sits on the kitchen counter for durations of t = 300 s before is it returned to the refrigerator. The convection coefficient associated with natural convection on the sides of the carton is 10h =2W m K ⋅. The surface emissivity is 0.90. Assume the milk carton temperature remains at 5℃ during the process, the top and bottom surfaces are well insulated, and the wall temperature of the room is 30℃.8. An electric resistance heater is embedded in a long cylinder of diameter 10 mm. When water with a temperature of 20℃ and velocity of 1 m/s flows crosswise over the cylinder, the power per unit length required to maintain the surface at a uniform temperature of 100℃ is 10 kW/m. When air, also at 20℃, but with a velocity of 10 m/s is flowing, the power per unit length required to maintain the same surfacetemperature is 400 W/m. Calculate and compare the convection coefficients for the flows of water and air.9. The heat flux that is applied to the left face of a plane wall is 25q ''=2W m . The wall is of thickness 12L =mm and of thermal conductivity 12k =W m K ⋅. If the surface temperatures of the wall aremeasured to be 50℃ on the left side and 30℃ on the right side, determine if steady-state conditions exist?10. Two-dimensional, steady-state conduction occurs in a hollow cylindrical solid of thermal conductivity 22k =W m K ⋅, outer radius 1.5o r =m and overall length 28o z =m, where the origin of the coordinate system is located at the midpoint of the center line. The inner surface of the cylinder is insulated, and the temperature distribution within the cylinder has the form ()22,2015012ln 300T r z r r z =-+--. Determine the inner radius i r of the cylinder.T T T T k k k q c x x y y z z tρ⎛⎫∂∂∂∂∂∂∂⎛⎫⎛⎫'''+++= ⎪ ⎪ ⎪∂∂∂∂∂∂∂⎝⎭⎝⎭⎝⎭ 211T T T T kr k k q c r r r z z tr ρφφ⎛⎫∂∂∂∂∂∂∂⎛⎫⎛⎫'''+++= ⎪ ⎪ ⎪∂∂∂∂∂∂∂⎝⎭⎝⎭⎝⎭ 22222111sin sin sin T T T T kr k k q c r r t r r r θρφφθθθθ⎛⎫∂∂∂∂∂∂∂⎛⎫⎛⎫'''+++= ⎪ ⎪ ⎪∂∂∂∂∂∂∂⎝⎭⎝⎭⎝⎭ sin r T k T k T q kq q r r r θφθθφ∂∂∂''''''=-=-=-∂∂∂ r z T k T T q kq q k r r z φφ∂∂∂''''''=-=-=-∂∂∂ x y z T T T q k q kq k x y z ∂∂∂''''''=-=-=-∂∂∂ ()()()()d f x dx f x f x dx dx +=+⋅。

热传导和热辐射的热能转换练习题热能和热功率一、热传导的热能转换练习题1. 问题描述：一个导热系数为0.2 W/(m·K)的铝棒，长度为1.5 m，截面积为0.02 m²。

其中一端放在温度为100°C的火炉中，另一端放在常温下，求15分钟后该铝棒的另一端温度。

解析：根据热传导的原理，我们可以使用热传导方程进行计算。

热传导方程可以表示为：Q = k·A·(ΔT/Δx)·Δt，其中Q为热能转移量，k 为导热系数，A为截面积，ΔT为温度差，Δx为长度差，Δt为时间。

根据题目给出的数据，我们可以先计算ΔT和Δt：ΔT = 100°C - 25°C = 75°C，Δt = 15分钟 = 900秒。

将数据代入热传导方程，可得：Q = 0.2 W/(m·K) * 0.02 m² * (75°C/1.5 m) * 900 s = 45 W。

由热传导方程我们可以得知热能转移量Q等于铝棒另一端的温度差乘以导热系数k乘以铝棒截面积A除以铝棒长度差Δx乘以时间Δt。

因此，铝棒另一端的温度差为：Q = k·A·(ΔT/Δx)·Δt = ΔT。

将计算得到的热能转移量Q代入上式，可得：ΔT = 45 W，即铝棒另一端的温度差为45°C。

由于另一端的温度为常温25°C，因此铝棒另一端的温度为25°C + 45°C = 70°C。

所以，15分钟后该铝棒的另一端温度为70°C。

二、热辐射的热功率转换练习题1. 问题描述：一个黑体表面积为2 m²，温度为500 K，计算其单位时间内向外部空间辐射的热功率。

解析：根据热辐射的原理，我们可以使用斯特藩-玻尔兹曼定律进行计算。

斯特藩-玻尔兹曼定律可以表示为：P = ε·σ·A·T^4，其中P为热功率，ε为黑体的发射率，σ为斯特藩-玻尔兹曼常数，A为表面积，T 为温度。