1、4-1 IntroductionSteady-State: T = f(x, y, z) T = f(x) T = ax+bUnsteady-State: T = f(, x, y, z) T = f(, x) T = ?If a solid body is suddenly subjected to a change in environment, some time must elapse (流逝) before equilibrium temperature condition will prevail (建立) in the body. We refer to the equilibr

2、ium condition as the steady state and calculate the temperature distribution and heat transfer by methods described in Chaps. 2. Chapter 4 Unsteady-State Conduction4-1 IntroductionChapter 4 Unsteady-State ConductionHTiT1A B CGEThe upper figure shows the temperature change in a plane while the left s

3、ide is suddenly kept at temperature T1, and the right side is exposed to a fluid of temperature Ti, which is the same as the initial temperature.The lower figure shows the temperature change for a plane initially at a uniform temperature Ti , at time zero the surfaces are suddenly lowered to T = T1.

4、T1Ti2LConsider an infinite plane of thickness 2L. The plane is initially at a uniform temperature Ti and at time zero the surfaces are suddenly lowered to T = T1.Differential equation:Initial condition:2LxT1T1oTiBoundary conditions:Let:The 4 equations e:Assuming a product solution2LxT1T1oTiSubstitut

5、ing into the origin Eq.:Separating variables:（a）Integrating Eq.(a):（b） must have finite value when , which requires:D 0; let D = - 2, Eq. (a) and (b) e2LxT1T1oTiThis produces two ordinary differential equations:The general solution is:From the condition for x =0 we haveA = 0From the condition for x

6、=2L:The series form of solution is:The initial condition yields:From Fourier expansionThe final solution is:The solution for the third kind of boundary condition is:4-2 Lumped-Heat-Capacity System(集總參數(shù)系統(tǒng))The analysis of transient heat conduction might be much simpler for systems which could be consi

7、dered uniform in temperature. This type of analysis is called the lumped-heat capacity method (集總參數(shù)法).1. IntroductionConsider a plate of thickness 2L. The plane is initially at a uniform temperature Ti and at time zero the surfaces are suddenly exposed to T = T. The boundary condition is:There are t

8、wo thermal resistances:1) Conduction resistance: L/k;2) Convection resistance: 1/hThere will be three different unsteady temperature distributions for different relative values of the two resistances:1) 1/h L/k: The thermal resistance is mainly inside the plate;2) L/k 1/h: inside: T almost uniform ;

9、3) 1/h L/k.TTi1/hL/k2LTTiL/k L/k: Bi 1That is when Bi1, the temperature distribution inside the plate is almost uniform.In practice, the criterion is usually taken as: Bi 0.1Under this condition, the lumped-heat capacity method is applicable.2. Solution:Consider a lumped-heat capacity system, e.g.,

10、a cold metal ball heated by hot air, energy equilibrium yields:VAh, TLet:Initial condition:Integration:The exponent is:Where V/A has a dimension of length, denoted as s; and hs/k is the Biot number, BiV; /s2 is called the Fourier number, (付里葉數(shù)) denoted by FoV; The subscript V means the characteristi

11、c length is V/A. So the solution is:3. Time ConstantThe termhas a dimension of 1/, whenSois called the time constant, cV is the heat capacity, andhA is the surface heat transfer capability.denoted by c./01 c100.368 c2 c3is the ratio of the two factors.It is shown in the figure, the smaller the time

12、constant, the faster the temperature change, and the faster the body reaches the ambient temperature.This means time constant reflects the response of a body to the change of its environment. Smaller time constant corresponds to faster response.cV is the heat capacity, andhA is the surface heat tran

13、sfer capability.When a thermocouple is used to measure the temperature of a fluid, its joint is usually very small to reduce the time constant, so as to get a better measurement within a limit time./01 c100.368 c2 c34. Applicability of Lumped-Capacity AnalysisIt is found that for the geometries of p

14、lane, cylinder and sphere, ifthe Lumped-heat-capacity method can be used to yield reasonable estimates within about 5%, where M is a parameter depending on the shape of the material:For infinite planes: M=1;For infinite long cylinders: M=1/2;For Spheres:M=1/3.As V/A=s, thenFor large planes of thickn

15、ess 2L:For long cylinders of radius R:For spheres of radius R:For the above geometries, if L and R are respectively the characteristic length, a common criterion can be used:Bi 0.1。Example: A cylindrical mercury thermometer of 20mm long and 4mm in diameter is used to measure gas temperature, where h

16、=11.63W/(m2K). Find the temperature difference ratio after 5 min of measurement. k = 10.36 W/(m K), = 13110 kg/m3, c = 0.138 kJ/(kgK)Solution:This means the lumped-capacity method is applicable.The smaller the ratio, the better the measurement, and the ideal value is 0%. 4-3 Semi-Infinite Solid Semi

17、-infinite solid: a solid with sides that can be infinitely extended in the y, z, and +x directions.Consider a semi-infinite solid with uniform initial temperature Ti . The left surface is suddenly lowered and maintained at a temperature T0 .Mathematical description:T0TiWhere erf() is the error funct

18、ion. erf() is the error function. is a dummy variable.When = 2, erf() = 0.9953or: /i = 0.9953Which means: T = Ti So: wheneitheror x2/(16), T is unchanged.The heat flow at x = 0 is:Heat flow:4-4 Heisler Diagram (p125-140) Analytical solution can be found for simple geometries with the third kind boun

19、dary condition: h, T.1. Large PlatesThickness: 2L，Initial temperature: TiMathematical description:T, h-LT, hLxoLet:Then we have：（2）（4）（1）（3）T, h-LT, hLxoUsing the separation of variable method, assuming:Substituting into Eq. (1):Separating variables:（a）Integrating Eq.(a):（b） must have finite value w

20、hen , which requires:D 0.2, One only needs to calculate the first term of the series, and the uncertainty is less than 1%. So when Fo 0.2:1L is the first eigenvalue, and is a function of Bi. (Re1, Return 2)Bi0.010.050.10.51.05.010501001L0.09980.22170.31110.65330.86031.31381.42891.54001.55521.5708Whe

21、n Fo 0.2,For x = 0, the temperature at the center, 0/i, can be calculated form the above equation, which is presented in figure 4-7 (p128). 0/i = f(Fo, 1/Bi)Also from the above equation, the ratio of the temperature at any position to that at the center is:This is shown in figure 4-10 (p133). /0 = f

22、(x/L, Bi)To find the temperature at any position, the two figures must be used together.Also from the above equation, the ratio of the temperature at any position to that at the center is:The ratio of the temperature at the surface to that at the center can be calculated when x = L .From the table,

23、when Bi 0.1, 1L 0.95。which means when Bi 0.1, the difference between the temperature at the surface and that at the center is less than 5%, or the whole plane is almost at a uniform temperature. This is why the criterion to use the lumped-heat-capacity method is set as Bi 0.2,The result of the above

24、 equation is presented in figure 4-14 (p136).Other figures are for cylinders and spheres.3. Discussion(1) As symmetry condition is the same as heat insulation condition, so the result is the same as for a plate of thickness L, where one side of it is heat insulated, and the other side is subject to

25、a convection condition. (2) If h, the plate surface temperature will be the same as that of the fluid. Therefore, when Bi, the solution can be used for the first kind boundary condition.T, h-LT, hLxoLxoT, hExample: A steel plate has a thickness of 100mm and initial temperature Ti = 20. One side of i

26、t is suddenly exposed to a 1000 stove with h = 174 W/(m2K). The other side is heat insulated. Calculate how long it will take for the heated side to reach a temperature of 500 and the center temperature then. (k = 34.8 W/(mK), a=0.55510-5 m2/s)Solution 1: The problem can be considered as a 200mm thi

27、ck plate with convection boundary condition.0/i can be solved first, then from 0/i and Bi find Fo. As:To find w/0 from figure 4-10 (p133) we need Bi first.(h = 174 W/(m2K), k = 34.8 W/(mK), a=0.55510-5 m2/s)From x/L=1 and Bi = 0.5, w/0 can be found from figure 4-10 (p133):w/0 =0.8Now:The center temp

28、erature is:Finally with 0/i and 1/Bi=2, the Fo , and thus the time needed, can be found from figure 4-7(b) (p129): Fo = 1.2Solution 2: Direct calculation:Calculate Bi:(h = 174 W/(m2K), k = 34.8 W/(mK), a=0.55510-5 m2/s)Find from the table: L1 = 0.6533. From the above Eq.:Then: Fo = 1.196.4-5 Multidi

29、mensional Systems2L12L2xyzFor an infinite rectangular bar with cross section 2L12L2, the initial temperature is Ti. The bar is suddenly exposed to a fluid with temperature T and convection coefficient h; The differential equation is:Let:Than:2L1xy2L2Initial condition:Boundary condition:2L12L2xyzInit

30、ial condition:Boundary condition:It can be proved that the product of the above two functions will be the solution for that of the infinite bar:Now consider the un-steady state heat conduction of two infinite plates with thicknesses 2L1 and 2L 2, respectively. If the solutions of the dimensionless t

31、emperature for the two plates are x(x, ) and y(y, ), then they should satisfy the differential equations:For x： For y： First, the differential equation is satisfied:Then the initial condition is satisfied:and the boundary condition is satisfied:Left sideSimilarly, it can be proved that:So thatis the

32、 solution of the 2-D system.Note:This production method is only applicable to the third type of boundary and the first type boundary condition where the boundary temperatures are constants.and the boundary condition is satisfied:Left sideEarlier, we have dimensionless solutions for infinite cylinder

33、s, infinite plate, and semi-infinite solids, we now denote them as:C(r, ) = the dimensionless solution for infinite cylinderP(x, ) = the dimensionless solution for infinite plateS(x, ) = dimensionless solution for semi-infinite solidFor multiple dimensional problem:For infinite rectangular bar, the

34、solution is:2L12L2xyThe results for other geometries are listed below.(a) For semi-infinite plate, the solution is:(b) For infinite rectangular bar, the solution is:2L1xz2L12L2xy(c) For semi-infinite rectangular bar, the solution is:(d) For rectangular parallelepiped(六面體), the solution is:2L12L2xyz2L12L22L3xyz(e) For semi-infinite cylinder, the solution is:(f) For short cylinder, the solution is:2r0rz2r02LrzA short steel cyl