(5.2)--Chapter 2 Steady-state heat cond传热学传热学传热学.ppt

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1、Chapter 2 Steady-state heat transfer nBasic law of heat conductionFouriers LawnMathematical description of heat conductionnAnalytical solution to typical one-dimensional steady-state heat conductionnHeat conduction through finsnOne-dimensional heat conduction with heat generationResearch method:With

2、 the assumption of continuum,the relationship between heat rate and temperature distribution of objects and other influencing factors is discussed from a macroscopic point of view.In general,most solids,liquids and gases can be considered as continuum.However,when the mean molecule free walk is not

3、negligible compared to the macroscopic size of the object,such as a lean gas whose pressure reduces to a certain extent,it cannot be considered as a continuum.2.1 Basic law of heat coauction1.Temperature field At time,the temperature distribution of all locations in the object is called temperature

4、field of the object at that moment.For steady state problems,temperature field do not change over time.Temperature field is a function of space coordinates and time.nTwo-dimension steady state:nOne-dimension steady state:nZero-dimension steady state：2.Isothermal surface,isotherm1.Isothermal surface:

5、the same moment,the temperature of each point2.Isotherm:the intersection of the isothermal surface and either sidePhotographs taken by infrared cameras,satellite optical remote sensors,etc.3.A point in an object can only have one temperature value at the same time,so the isothermal surfaces(lines)of

6、 different temperatures cannot intersect at the same time.The object studied is a continuum,and the isothermal surface(line)will not terminate inside the object.It can only terminate at the boundary of the object,or the closed surface(line)formed by itself.3.Temperature gradientnDefinition:The chang

7、e rate of temperature in the normal direction of the isothermal surface,nVector,point to the direction of temperature increaseIn a Cartesian coordinate system,temperature gradient can be expressed as:are partial derivatives in the x,y,and z directions;In the normal direction of the isothermal surfac

8、e,the temperature change rate per unit length is the largest.i,j,and k are unit vectors in the x,y,and z directions.Heat FluxThe magnitude and direction of heat flux can be expressed by the heat flux vector q:The direction of the heat flux vector points to the direction of temperature decrease.ntdAd

9、qIn a Cartesian coordinate system,the heat flux vector can be expressed as:qx,qy,qz represent the component of q in three coordinate directionsnFouriers law was obtained by French physicist Fourier in 1822.nGeneral form(i.e.written as heat flux vector)Negative sign indicates the direction of heat tr

10、ansfer is opposite to the temperature gradient4.Fouriers lawnFor isotropic materialIsotropic1.Thermal conductivity is temperature and direction dependent(anisotropic,e.g.wood,fiber,car tires)2.Value：Carbon steel 36.7，Water at 20 0.599Dry air at 20 0.0259 5.Thermal ConductivityAg 427Cu 399Pure metalT

11、hermalInsulationmaterialBuilding materialsAlloyArtificial materialSuper insulationArtificial heat pipeDry airConductivemedium Insulation materialPorous material,filled by air,3.Many materials,with comprehensive processes including heat conduction,convection and radiation,are uniformly expressed by t

12、he thermal conductivity coefficient.At this time,it is called the apparent(converted)thermal conductivity.Such as:light furnace wall(refractory fiber),cotton tire clothing,space clothing,super insulation material.4.Obtained through experiments:non-steady state measurement(quasi steady state);steady

13、state measurement5.Relationship with temperature:In engineering,linear approximations are usually available.Where:is the temperature;is a constant;is the intercept of the extension of the straight line on the ordinate.2.2 Thermal differential equations and conditions1.Statement of problem1.One-dimen

14、sional problem,can directly integrate Fouriers law to obtain heat flux.2.Multidimensional problem:Fouriers law still appliesFouriers law reveals the relationship between temperature gradient and heat flux at each point in continuous temperature field,so when the temperature distribution in the objec

15、t is obtained,the corresponding heat flux distribution can be obtained.3.The thermal differential equation is to reveal the intrinsic relationship between continuous temperature field and space coordinates and time.2.Ideas and principlesTaking a Cartesian coordinate system as an example,a micro-para

16、llel hexahedron is segmented from the object,using:1.Fouriers law2.Energy conservation law(thermal balance principle)establishes thermal differential equation1.Isotropic,homogeneous2.Uniform density,specific heat,independent of temperature(constant properties)3.Assumptions4.Derivation1.Law of conser

17、vation of energy In any time interval:Analogy:Total heat into elementHeat generation of heat sources in elementsIncrease of element energyTotal heat export elementFund from parents and siblingsMy income(scholarship,other income)Total expenditureDeposit increment in hand(+or-)1.Law of conservation of

18、 energy2.Energy increase in element within unit time Specific heat，(The amount of heat required to increase the unit mass by 1)Amount of heat required to increase the unit mass by 1 Amount of heat(internal energy)required for element to rise t per unit time 3.Heat generation of heat source in elemen

19、t within unit time Heat generated by heat source per unit volume per unit time,4.Import heatFouriers lawOn surface at ,temperature gradient is;Thermal conduction area5.export heatTaylor series expansion,take the first two itemsDue toExpressed in Taylor series,omitting high-order small quantities6.fi

20、nishingSubstituting(c)into equation(a):Substituting(d)into equation(c):(2-7)When is temperature dependent，(2-7)is heat transfer differential equation in three-dimensional Cartesian coordinate system with heat generation,non-steady state,and variable thermal conductivity.Can not mention the different

21、ial symbol outside.5.Equation formnIn three-dimensional rectangular coordinate system,heat conduction differential equation with internal heat generation,non-steady and constant thermal conductivity:nor:nwhere the Laplace calculusThermal diffusivity nHeat conduction differential equation for non-ste

22、ady,with no heat generation,and constant thermal conductivity conditions in three-dimensional rectangular coordinate system:Since,then:oror (Laplace equation)nDifferential equation of heat conduction with internal heat generation,non-steady and variable thermal conductivity in cylindrical and spheri

23、cal coordinate system,please see textbook(2-12),(2-13)on p.44.Cylindrical spherical coordinate（r,z）Spherical coordinate（r,，）,Heat conductivity is large and heat capacity is small,leads to little heat transfer ,Temperature rises faster,leads to faster heat transfer6.Heat diffusion rate(Temperature co

24、nductivity coefficient)(also a property)From express,means transfer rate of temperaturene.g.：Cu:Air:Wood:nPut a copper-rod and a wood-rod in fire,then hold one end of both rods,copper-rod is hot but wood-rod is not so hot.nOnly in unsteady heat conduction,a shows its effect.7.Definite conditions nFo

25、r any differential equation,it can be solved by mathematical method.But for a specific problem,heat conduction differential equation with general solution is not enough,it must satisfy the special solution of some additional conditions.These additional conditions are mathematically called the defini

26、te solution.There are four kinds of definite solutions.nInitial condition:initial temperature distribution(for steady-state heat conduction,there isnt initial condition)nBoundary conditions:temperature or heat change on the boundary of the object.There are three categories:ConditionsEquations(1)Firs

27、t boundary conditionWall temperature UnsteadySteady state(2)Second boundary conditionHeat flux on wall UnsteadySteady state(3)Third boundary conditionHeat transfer coefficient ,Fluid temperaturenGeometry conditions:geometry and sizenPhysical conditions:2.3 Analysis of typical one-dimensional steady-

28、state heat conductionI.Heat conduction in plane WallProblem description:(a)1D；(b)first boundary condition；(c)1.Single layer plane wall(1)1D;(a)In theory,for infinite wall plane,direction y,z direction x.In engineering,direction y,z 10 direction x.(b)Or no heat conduction in direction y,z,only in dir

29、ection x.(2)Governing eqautions Heat conduction differential equation for one-dimensional,non-steady,with no heat generation,and constant thermal conductivity conditions:Boundary conditions(3)SolutionIntegrate equation:By boundary conditions:So,(Temperature distribution)Heat flux,(4)Thermal resistan

30、ce RR2.Multi-layer plane wall(calculated by thermal resistance)Take three layers plane wall for example:Thermal equilibrium principle(steady-state):For n-layer plan wall:Unknown temperature t2,t3 in the intersection of layers can be solved through known q:3.Contact thermal resistanceWhen the surface

31、s of two nominally flat solid are in contact with each other,in fact the solid-to-solid contact only occurs on some discrete contact faces.When the two surfaces are in contact,the void existing in the non-adhering portion will generate a thermal resistance(due to small thermal conductivity of air),w

32、hich is called contact thermal resistance.Temperature distributionContact thermal resistance can only be measured by experiment.Methods to reduce the contact thermal resistance:(a)Smooth surface;(b)Clean surface;(c)Add conductive oil,carbon powder,metal powder Example 2-1nA boiler furnace wall is ma

33、de of cement perrock with density 300kg/m3,wall thickness ,known inner wall temperature t1=500,the outer wall temperature t2=50,try to solve the heat dissipation per hour of the furnace wall.II.Heat conduction in cylinder wall1.1D,heat dissipation at two ends can be ignored:2.Governing equations:Fro

34、m(2-12)in textbook,note the steady state,no heat generation and constant thermal conductivity coefficient:Single layer cylinder wallBoundary conditions:Integrate:Substitute boundary conditions:Combine:By boundary conditions:Substitute c1 and c2:So,when ,temperature distribution in wall of cylinder i

35、s logarithmic.3.Note that,for heat conduction in wall of cylinder，heat flux is not a constant even for steady-state condition,and change with r.However,(steady-state).So its general to use Q when solve such problems.Derivate function of temperature distribution:Then:4.Thermal resistance RMulti-layer

36、 cylinder wall（n layers）Thin cylinder wall described as plane wall Range of application:,Error 4%。III.Heat conduction in spherical shellnTemperature distributionnHeat ratenThermal resistanceIV.One-dimensional heat conduction under second and third boundary conditionsnAn electric iron with power of 1

37、200 W and its bottom plate is vertically placed in a room with an ambient temperature of 25.The metal base is 5 mm thick,with thermal conductivity ,area of A=300 cm2 and heat transfer coefficient considering radiation effect.Find the temperature of both surfaces of the bottom plate under steady-stat

38、e conditions.Heat conduction of plane wall under one-dimensional steady-state conditions,second boundary condition on left side and third boundary condition on right side.Governing equationsBoundary conditionsLeft side,second boundary condition:Right side,third boundary condition:Temperature distrib

39、utionV.one-dimensional heat conduction with variable area or conductivity coefficientTwo steps for solving heat conduction problem:(1)Solve heat conduction differential equation,obtain temperature distribution.(2)Calculate heat rate according to Fouriers Law and obtained temperature fields.For heat

40、condition under one-dimensional,no heat generation and first boundary conditions,the heat rate can be obtained directly regardless temperature fields.Thus,Fouriers Law for one-dimensional：When(t),A=A(x)，Separate variables and integrate,note that heat rate is independent with x,then:When changes line

41、arly with temperature,that is 0(1bt),In fact,whatever changes,if the average conductivity coefficient could be obtained,it could be used in all the heat conduction equations with constant conductivity coefficient by replacing the constant.Analysis:x o t1 t t2 Fouriers Law:If ,whats the temperature d

42、istribution in plane wall?If ,whats the temperature distribution in plane wall?t1 t2 t2 t1 2.4 Heat conduction through finFin wall:Straight fin,Cylinder fin;Uniform cross section,variable cross sectionCylinderRectangleTriangleConeAnnulusPhysical model of straight fin with uniform cross sectionFor st

43、raight fin,practicable,equivalently cross section,infinite direction z.For annulus fin,practicable,equivalently cross section,independent with circumferential angle for symmetric heat conduction.Impracticable for annulus fin with variable area.Extension,some engineering problems described as above m

44、odel,such as:heat dissipation along glass rod of glass-bulb thermometer,heat dissipation by conduction of thermocouple.Simplification of physical model(Features)First boundary condition on one side,third boundary condition on other sides,convection heat transfer,nonuniform wall temperature.,is large

45、r,variation on direction y is negligible compared to direction,can be regard as one-dimensional.If of rod is large,the thermal resistance along the cross section of the rod will not be large.If ,heat conduction along cross section is fast,but heat transfer to fluid at boundary is slow.So only a smal

46、l temperature gradient is required on cross section to match the small heat flux on boundary.Therefore,it can be assumed that the temperature on cross section is uniform and the temperature changes only along the length of rod.Governing equations-deduced by conservation principleThermal balance：make

47、 as extra temperature,Thermal balance:Second order homogeneous nonlinear ordinary differential equationGoverning equations-process boundary conditions as terms,simplified from general heat conduction differential equationthen:Since heat is continuously transferred to the side(convection heat transfe

48、r)in the x direction,the convection heat transfer can be regarded as an internal cooling source,that is,a negative internal heat source.Note:element volume is .Solving:make ,then above formula intoBoundary conditions:The left side of Format(c)represent the heat conduction on the free end of rod,and

49、the right side means the convection heat rate at the same location.General solution of format(a)is:The final expression of temperature distribution along the rod is:Hyperbolic functions:Hyperbolic sine-sinh Hyperbolic cosine-cosh Hyperbolic tangent-tanhCalculation of rod end Note:When end area of ro

50、d is much smaller than side area,heat transferred from end face to surrounding fluid is negligible,equivalently insulated end face.In this case,in the boundary condition(c),so:Then the temperature distribution expression along the rod(Note,when ,hyperbolic cosine function symmetric about y-axis):See