3.3 Unsteady State Heat Conduction

Transcription

1 3.3 Unsteady State Heat Conduction For many applications, it is necessary to consider the variation of temperature with time. In this case, the energy equation for classical heat conduction, eq. (3.8), should be solved. If the thermal conductivity is independent from the temperature, the energy equation is reduced to eq. (3.).

2 3.3. Lumped Analysis Consider an arbitrarily shaped object with volume V, surface area A s, and a uniform initial temperature of T i as shown in Fig At time t =, the arbitrarily shaped object is exposed to a fluid with temperature of T and the convective heat transfer coefficient between the fluid and the arbitrarily shaped object is h. T(t) V h, T A s Figure 3.5 Lumped capacitance method

3 The convective heat transfer from the surface is q = has ( T T ) Similar to what we did for heat transfer from an extended surface, the heat loss due to surface convection can be treated as an equivalent heat source q has ( T T ) q = = V V Since the temperature is assumed to be uniform, the energy equation becomes ρc p T t has ( T T ) = V (3.88) which is subject to the following initial condition T = T, t = i (3.89) 3

4 Introducing excess temperature, (3.89) become, eqs. (3.88) and Integrating eq. (3.9) and determining the integral constant using eq. (3.9), the solution becomes ϑ t / t = e τ ϑ where dϑ dt = ha ϑ = T T is referred to as the thermal time constant. s ρvc p ϑ = ϑ i, t = i ρvc τ t = ha s p ϑ (3.9) (3.9) (3.9) (3.93) 4

5 The cooling process requires transferring heat from the center of the object to the surface and the heat is further transferred away from the surface by convection. When the lamped capacitance method is employed, it is assumed that the conduction resistance within the object is negligible compared with the convective thermal resistance at the surface, therefore, the validity of the lumped analysis depends on the relative thermal resistances of conduction and convection. The conduction thermal resistance can be expressed as Lc Rcond = ka where L c is the characteristic length and A c is the area of heat conduction. c 5

6 The convection thermal resistance at the surface is R conv = ha Assuming As = Ac, one can define the Biot number, Bi, as the ratio of the conduction and convection thermal resistances R Bi= cond c R conv (3.94) If the characteristic length is chosen as Lc = V / As, the lumped capacitance method is valid when the Biot number is less than., or the conduction thermal resistance is one order of magnitude smaller than the convection thermal resistance at the surface. hl k s 6

7 3.3. One Dimensional Transient Systems For the case that the Biot number is greater than., the temperature distribution can no longer be treated as uniform and the knowledge about the temperature distribution is also of interest. We will now consider the situation where temperature only varies in one spatial dimension. Both homogeneous and nonhomogeneous problem will be considered. 7

8 Homogeneous Problems adiabatic h, T Figure 3.6 shows a finite slab with thickness of L and a uniform initial temperature of T i. At time t =, the left side of the slab is insulated while the right side of the slab is exposed to a fluid with temperature of T. In contrast to the lumped capacitance method that assumes uniform temperature, we will present a more generalized model that takes non-uniform temperature distribution into account. L x Figure 3.6 Transient conduction in a finite slab 8

9 The energy equation for this one-dimensional transient conduction problem is T x subject to the following boundary and initial conditions T =, < x < L, t > α t T =, x = x T k = h( T T ), x = L x T = T, < x < L, t = i (3.95) (3.96) (3.97) (3.98) This is a nonhomogeneous problem because eq. (3.97) is not homogeneous. By introducing the excess temperature, ϑ = T T, the problem can be homogenized, i.e., ϑ ϑ =, < x < L, t > α (3.99) x t 9

10 ϑ =, x = x ϑ k = hϑ, x = L x ϑ = ϑi = Ti T, < x < L, t = (3.) (3.) (3.) To express our solution in a compact form so that it can be used for all similar problems, one can define the following dimensionless variables ϑ x αt θ =, X =, Fo = ϑ L L i (3.3) and eqs. (3.99) (3.) will be nondimensionalized as θ θ =, < X <, Fo > X Fo (3.4) θ =, X = X (3.5)

11 θ = Bi θ, X = X θ =, < X <, Fo = (3.6) (3.7) This problem can be solved using the method of separation of variables. Assuming that the temperature can be expressed as θ ( X,Fo) = Θ( X ) Γ(Fo) (3.8) where Θ and Γ are functions of X and Fo, respectively, eq. (3.4) becomes Θ ( X ) Γ (Fo) = Θ( X ) Γ(Fo) Since the left hand side is a function of X only and the righthand side of the above equation is a function of Fo only, both sides must be equal to a separation constant,, i.e.,

12 Θ ( X ) Γ (Fo) = = µ Θ( X ) Γ(Fo) (3.9) The separation constant µ can be either a real or a complex number. The solution of Γ from eq. (3.9) will be e µfo. If is a positive real number, we will have when Fo, which does not make sense, therefore, µ cannot be a positive real number. If µ is zero, we will have Γ = const, and Θ is a linear function of X only. The final solution for θ will also be a linear function of X only, which does not make sense either. It can also be shown that the separation constant cannot be a complex number, therefore, the separation variable has to be a negative real number. Γ = µ Γ

13 If we represent this negative number by µ = λ, eqs. (3.9) can be rewritten as the following two equations Θ + λ Θ = (3.) The general solutions of eqs. (3.) and (3.) are where C, C, and C 3 are integral constants. λ Γ + Γ = Θ = C cosλ X + C sin λ X (3.) (3.) (3.3) Substituting eq. (3.8) into eqs. (3.5) and (3.6), the following boundary conditions of eq. (3.) are obtained Γ = C e λ 3 Θ () = Fo Θ () = Bi Θ() (3.4) (3.5) 3

14 Substituting eq. (3.) into eq. (3.4) yields Θ () = Cλ sin() + Cλ cos() = Cλ = Since λ cannot be zero, C must be zero and eq. (3.5) becomes Θ = C cosλ X (3.6) Applying the convection boundary condition, eq. (3.5), one obtains λn = cot λn (3.7) where n is an integer. The dimensionless temperature with eigenvalue λ n can be obtained by substituting eqs. (3.6) and (3.3) into eq. (4.8), i.e., n ( ) Fo θn = Cn cos λn X e λ (3.8) Bi 4

15 where C = C C. n 3 Since the one-dimensional transient heat conduction problem under consideration is a linear problem, the sum of different θ n for each value of n also satisfies eqs. (3.4) (3.6). n ( ) Fo θ = Cn cos λn X e λ (3.9) Substituting eq. (3.9) into eq. (3.7) yields C cos( λ X ) n= = n n n= ( ) Multiplying the above equation by cos λ X m and integrating the resulting equation in the interval of (, ), one obtains cos( λ X ) dx = C cos( λ X )cos( λ X ) dx m n m n n= (3.) 5

16 The integral in the right-hand side of eq. (3.33) can be evaluated as cos( λ X )cos( λ X ) dx m n λm sin λm cosλn λn cosλm sin λn, m n λm λn = sin λm λm +, m = n λ m Equation (3.7) can be rewritten as Similarly, for eigenvalue λ m, we have Bi = λ tan λ Combining the above two equations, we have Bi = λ tan λ n m λ tan λ λ tan λ = n m m m n n (3.) or λ sin λ cosλ λ cosλ sin λ = m m n n m n 6

17 therefore, the integral in eq. (3.) is zero for the case that m n, and the right hand side of eq. (3.) becomes m Cn cos( λm X )cos( λn X ) dx λm n= λ m C sin λm = + (3.) Substituting eq. (3.) into eq. (3.) and evaluating the integral at the left-hand side of eq. (3.), we have C sin sin m λm λm = λm λ λ + m m i.e., Changing notation from m to n, we get C m 4sin λm = λ + sin λ C n m n m 4sin λn = λ + sin λ n (3.3) 7

18 The dimensionless temperature, therefore, becomes θ = n= 4sin λn cos λ + sin λ n ( λ ) If the Biot number becomes infinite, the convection boundary condition becomes θ =, X = (3.4) (3.5) which is an isothermal condition at the right-hand side of the wall. Equation (3.7) becomes cosλ n = (3.6) and the eigenvalue is therefore ( π ) n n n X e λ λn = n /, n =,,3,... Fo (3.7) 8

19 The temperature distribution for this case is then 4sin( n π / ) θ = cos [( n π / ) X ] e ( n π / ) + sin ( n π / ) n= (3.8) When Fourier s number is greater than., only the first term in eq. (3.9) is necessary and the solution becomes ( ) θ = θ = C cos λ X e λ Fo ( n π / ) Fo (3.9) 9

20 Example 3.3 A long cylinder with radius of r o and a uniform initial temperature of T i is exposed to a fluid with temperature of T. The convective heat transfer coefficient between the fluid and cylinder is h. Assuming that there is no internal heat generation and constant thermophysical properties, obtain the transient temperature distribution in the cylinder.

21 Solution Since the temperature changes along the r-direction only, the energy equation is r r r t T T T + =, < x < ro, t > α subject to the following boundary and initial conditions T =, r = (axisymmetric) r T k = h( T T ), r = ro r T = T, < r < r, t = i Defining the following dimensionless variables o (3.3) (3.3) (3.3) (3.33) T T r αt hro θ =, R =, Fo =, Bi = T T r r k i o o (3.34)

22 eqs. (3.3) (4.33) will be nondimensionalized as θ θ θ + =, < R <, Fo > Fo R R R θ =, R = R θ = Bi θ, R = R θ =, < R <, Fo = Assuming that the temperature can be expressed as θ ( R,Fo) = Θ( R) Γ(Fo) (3.35) (3.36) (3.37) (3.38) (3.39) and substituting eq. (3.39) into eq. (3.35), one obtains Γ Θ + Θ = = λ Θ R Γ (3.4)

23 which can be rewritten as the following two equations λ R Γ + λ Γ = Θ + Θ + Θ = (3.4) (3.4) Equation (3.) is a Bessel s equation of zero order and has the following general solution Θ ( R) = C J ( λr) + C Y ( λr) (3.43) where J and Y are the Bessel s function of the first and second kind, respectively. The general solution of eq. (3.4) is where C, C, and C 3 are integral constants. Γ = C e λ 3 Fo (3.44) 3

24 The boundary conditions for eq. (3.4) can be obtained by substituting eq. (3.39) into eqs. (3.36) and (3.37), i.e., The derivative of Θ is (3.45) (3.46) (3.47) Since J() = and Y () =, C must be zero. Substituting eqs. (3.43) and (3.47) into eq. (3.46) and considering C = we have where n is an integer. Θ () = Θ () = Bi Θ() Θ ( R) = C λj ( λr) C λy ( λr) λ J ( λ ) + Bi J ( λ ) = n n n (3.48) 4

25 The dimensionless temperature with eigenvalue is where C = C C. n 3 θ = C J ( λ ) n n n n R e λ (3.49) For a linear problem, the sum of different θ n for each value of n also satisfies eqs. (3.35) (3.37). n ( ) Fo θ = CnJ λnr e λ (3.5) n= Substituting eq. (3.5) into eq. (3.38) yields C J ( λ R) n n n= Multiplying the above equation by RJ λmr and integrating the resulting equation in the interval of (, ), one obtains Fo = ( ) RJ ( λ R) dr = C RJ ( λ R) J ( λ R) dr m n m n n= λ n 5

26 According to the orthogonal property of Bessel s function, the integral on the right-hand side equals zero if m n but it is not zero if m = n Therefore, we have C m RJ ( λ R) dr = = ( λ ) m m RJ ( λ ) m J m J m mr dr λ λ + λ J ( ) ( ) Changing notation from m to n, we get J( λn ) Cn = λ J ( λ ) + J ( λ ) n n n thus, the dimensionless temperature becomes ( λ ) ( λ ) θ = n= J J R e λ λ λ n n λn Fo n J ( n ) + J ( n ) (3.5) (3.5) 6

27 One-dimensional heat conduction in a spherical coordinate system can be solved by introducing a new dependent variable. Consider a sphere with radius of r o and a uniform initial temperature of T i. It is exposed to a fluid with a temperature of T and the convective heat transfer coefficient between the fluid and finite slab is h. Assuming that there is no internal heat generation and constant thermophysical properties, the governing equation is ( rt ) T r r α t =, < x < r, t > subject to the following boundary and initial conditions T =, r = (axisymmetric) r o (3.53) (3.54) 7

28 T k = h( T T ), r = ro r T = T, < r < r, t = (3.55) (3.56) By using the same dimensionless variables defined in eq. (3.34), eqs. (3.53) (3.56) can be nondimensionalized as ( Rθ ) R Defining a new dependent variable U = Rθ i θ =, < R <, Fo > R Fo θ =, R = R θ = Bi θ, R = R θ =, < R <, Fo = o (3.57) (3.58) (3.59) (3.6) (3.6) 8

29 eqs. (3.57) (3.6) become U U =, < R <, Fo > R Fo θ =, R = R U = (Bi-) θ, R = R U = R, < R <, Fo = (3.6) (3.63) (3.64) (3.65) This problem can be readily solved by using the method of separation of variables. After the solution is obtained, one can change the dependent variable back to θ and the result is θ = R n= 4[sin( λn ) λn cos( λn )] sin( λnr) e λ [λ sin( λ )] n n n λ Fo n (3.66) 9

30 where the eigenvalue is the positive root of the following equation λ cot λ = Bi n n (3.67) 3

31 Nonhomogeneous Problems The solution of a nonhomogeneous problem can be obtained by superposition of a particular solution of the nonhomogeneous problem and the general solution of the corresponding homogeneous problem. Consider a finite slab with thickness of L and a uniform initial temperature of T i as shown in Fig T=T t t > t = T=T i L x Figure 3.7 Heat conduction under boundary condition of the first kind 3

32 Assuming that there is no internal heat generation in the slab and the thermophysical properties of the slab are constants, the energy equation is T x T =, < x < L, t > α t subject to the following boundary and initial conditions T = T, x = T = T, x = L By defining the following dimensionless variables i T = T, < x < L, t = i T Ti x αt θ =, X =, Fo = T T L L i eqs. (3.68) (3.7) will be nondimensionalized as: (3.68) (3.69) (3.7) (3.7) (3.7) 3

33 θ θ =, < X <, Fo > X Fo (3.73) θ =, X = θ =, X = θ =, < X <, Fo = (3.74) (3.75) (3.76) If the steady state temperature is represented by θ s, it must satisfy the following equations: θs =, < X < X θ =, X = which have the following solution: s θ =, X = s θ = X s (3.77) (3.78) (3.79) (3.8) 33

34 To obtain the generation of the problem described by eqs. (3.73) (3.76), a method of partial solution will be employed. In this methodology, it is assumed that the solution of a nonhomogeneous problem can be expressed as θ h where represent the solution of a homogeneous problem. Substituting eqs. (3.73) (3.76) and considering eqs. (3.77) (3.79), we have θ ( X,Fo) = θ ( X ) + θ ( X,Fo) s θh θh =, < X <, Fo > X Fo θ h =, X = θ =, X = h θ h = X, < X <, Fo = h (3.8) (3.8) (3.83) (3.84) (3.85) 34

35 which represent a new homogeneous problem. This problem can be solved using the method of separation of variables and the result is sin( nπ X ) n θh = e π n The solution of the nonhomogeneous problem thus becomes n= sin( nπ X ) θ = X e π n n= ( π ) Fo ( nπ ) Fo (3.86) (3.87) 35

36 If the steady-state solution does not exist, we can use the method of variation of parameter to solve the problem. Consider a finite slab with thickness of L and a uniform initial temperature of T i. At time t =, the left side is subject to a constant heat flux while the right side of the slab is adiabatic (see Fig. 3.8). adiabatic L x Figure 3.8 Heat conduction under boundary condition of the second kind 36

37 Assuming that there is no internal heat generation in the slab and the thermophysical properties of the slab are constants, the energy equation is subject to the following boundary and initial conditions By defining the following dimensionless variables eqs. (3.68) (3.7) will be nondimensionalized as: T x T =, < x < L, t > α t T k = q, x = x T =, x = L x T = T, < x < L, t = i T Ti x αt θ =, X =, Fo = q L / k L L (3.88) (3.89) (3.9) (3.9) (3.9) 37

38 θ θ =, < X <, Fo > X Fo θ =, X = X θ =, X = X θ =, < X <, Fo = (3.93) (3.94) (3.95) (3.96) We will use the method of variation of parameters to solve this problem. This method requires the following steps:. Set up a homogeneous problem by dropping the nonhomogeneous terms,. Solve the homogeneous problem to get eigenvalue and eigenfunctions Θ ( X ) n 3. Assuming the solution of the original nonhomogeneous n problem has the format of = n Θn n= θ ( X,Fo) A (Fo) ( X ) λn 38

39 Θ n 4. Solve for A n (Fo) using orthogonal property of 5. Obtain an ordinary differential equation (ODE) for A n (Fo) and solve for A n (Fo) from the ODE 6. Put together the final solution. We will solve this nonhomogeneous problem be following the above procedure. The corresponding homogeneous problem is: θh θh =, < X <, Fo > X Fo θh =, X = X θh =, X = X θ =, < X <, Fo = h (3.97) (3.98) (3.99) (3.3) 39

40 Assuming the solution of the above homogeneous problem is eq. (3.97) becomes Solving eqs. (3.33) (3.35) yields the following eigenvalues and eigen functions θ = Θ( X ) Γ(Fo) h Θ ( X ) Γ (Fo) = = λ Θ( X ) Γ(Fo) The eigenvalue problem is Θ + λ Θ = Θ () = λ Θ () = n = nπ Θ ( X ) = n cos( nπ X ), n =,,,... (3.3) (3.3) (3.33) (3.34) (3.35) (3.36) (3.37) 4

41 Now, let us assume that the solution of the original nonhomogeneous problem is θ ( X,Fo) A (Fo)cos( nπ X ) Multiplying eq. (3.38) by cos mπ X and integrating the resulting equation in the interval of (, ), one obtains (3.38) (3.39) The integral on the right-hand side of eq. (3.39) can be evaluated as thus, eq. (3.39) becomes = n n= ( ) θ ( X,Fo)cos( mπ X ) dx = A cos( mπ X )cos( nπ X ) dx n n=, m n cos( mπ X )cos( nπ X ) dx = /, m = n, m = n = A (Fo) = θ ( X,Fo) dx, m = (3.3) (3.3) 4

42 Am (Fo) = θ ( X,Fo)cos( mπ X ) dx, m (3.3) Differentiating eq. (3.3) with respect to Fo, one obtains: (3.33) Substituting eq. (3.93) into eq. (3.33) and integrating with respect to X yield da Integrating eq. (3.34) with respect to Fo, we have When Fo =, eq. (3.35) becomes da dfo = θ dx Fo = θ = θ θ = dfo dx X X X = X X = A (Fo) = Fo + C = θ ( X,Fo) dx A () = C = θ ( X,) dx = (3.34) (3.35) (3.36) 4

43 thus, we have A (3.37) (Fo) = Fo Differentiating eq. (3.3) and considering eq. (3.93) yield Using integration by parts twice, the following ODE is obtained: dam θ θ = cos( m X ) dx cos( m X ) dx dfo π = π Fo X (3.38) (3.39) Multiplying eq. (3.39) by an integrating factor ( m e π ) Fo, we have d dfo which can be integrated to get Am dam = ( mπ ) A dfo A e = m e ( mπ ) Fo ( mπ ) Fo m = + C e ( mπ ) ( mπ ) Fo (3.3) (3.3) 43

44 where C is an integral constant that needs to be determined by an initial condition. For Fo =, eq. (3.3) becomes m Substituting eq. (3.3) into eq. (3.3), one obtains therefore, we have = = = A () θ ( X,)cos( mπ X ) dx cos( mπ X ) dx Am C = ( mπ ) = ( mπ ) ( mπ ) Changing m back to n for notation, An = ( nπ ) ( nπ ) e e ( mπ ) Fo ( nπ ) Fo (3.3) (3.33) 44

45 Substituting eqs. (3.37) and (3.33) into eq. (3.38), the solution becomes cos( nπ X ) cos( nπ X ) ( nπ ) Fo θ ( X,Fo) = Fo+ e (3.34) π π When the time (Fourier number) becomes large, the last term on the right-hand side will become zero and the solution is represented by the first two terms only. To simplify eq. (3.34), let us assume the solution at large Fo can be expressed as θ ( X,Fo) = Fo+ f ( X ) (3.35) which is referred to as asymptotic solution and it must satisfy eqs. (3.93) (3.95). Substituting eq. (3.35) into eqs. (3.93) (3.95), we have f ( X ) = (3.36) n n= n= n 45

46 Integrating eq. (3.36) and considering eq. (3.37), we obtain f () =, f () = X f ( X ) = X + C (3.37) (3.38) where C cannot be determined from eq. (3.37) because both boundary conditions are for the first order derivative. To determine C, we can expand f(x) defined in eq. (3.38) into cosine Fourier series, i.e. X = + = + n= f ( X ) X C a a cos( nπ X ) (3.39) After determining a and a n, and considering the f(x) is identical to the second term on the right-hand side of eq. (3.34), we have X cos( nπ X ) X (3.33) + = 3 π n n n = 46

47 Substituting eq. (3.33) into eq. (3.34), the final solution becomes X cos( nπ X ) ( nπ ) Fo θ ( X,Fo) Fo+ X e 3 (3.33) = + π n= n 47

48 Transient Heat Conduction in a Semi-Infinite Body Consider heat conduction in a semi-infinite body (x > ) with an initial temperature of T i. The temperature near the surface of the semi-infinite body will increase because of the surface temperature change, while the temperature far from the surface of the semi-infinite body is not affected and remains at the initial temperature T i. The physical model of the problem is illustrated in Fig. 3.9 T=T t = T=T i x Figure 3.9 Heat conduction in a semi-infinite body 48

49 The governing equation of the heat conduction problem and the corresponding initial and boundary conditions are: T x, t, ( ) T ( x t) = x >, t > x α t T ( x, t) = T x =, t > (3.33) (3.333) (3.334) (3.335) which can be solved by using the method of separation of variables or integral approximate solution. Defining the following dimensionless variables where L is a characteristic length. θ T ( x, t) = T x, t > i T ( x, t) = T x >, t = =, X =, Fo = i i T T x αt T T L L (3.336) 49

50 Eqs. (3.33) (3.335) will be nondimensionalized as θ θ =, X >, Fo > X Fo θ =, X = θ =, X Assuming that the temperature can be expressed as (3.337) (3.338) (3.339) (3.34) (3.34) and substituting eq. (3.34) into eq. (3.337), one obtains whose general solutions are: θ =, X >, Fo = θ ( X,Fo) = Θ( X ) Γ(Fo) Θ ( X ) Γ (Fo) = = λ Θ( X ) Γ(Fo) (3.34) 5

51 Θ = C cosλ X + C sin λ X Γ = C e λ where C, C, and C 3 are integral constants. Substituting eq. (3.34) into eq. (3.338), the following boundary condition of eq. (3.343) is obtained 3 (3.343) (3.344) (3.345) Substituting eq. (3.343) into eq. (3.345) yields and eq. (3.343) becomes Θ = C sin λ X (3.346) Substituting eqs. (3.346) and (3.344) into eq. (3.34), the solution becomes ( ) Fo θ = C( λ)sin λ X e λ (3.347) λ Fo Θ () = 5

52 where C = CC3. The general solution for the problem can be obtained by using linear combination of eq. (3.347) for all possible λ, i.e., λ ( ) Fo θ C( λ)sin λ X e dλ (3.348) = Substituting eq. (3.348) into eq. (3.34), one obtains C( λ)sin ( λ X ) dλ = λ = If we solve the problem by using Laplace transformation, we have sin ( λ X ) sin ( λ X ) dx dλ = λ= π X = Comparing the above two equations, an expression of C is obtained: C( λ) sin ( λ X ) dx = π X = 5

53 The temperature distribution, eq. (3.348), becomes which can be rewritten as Evaluating the integral with respect to yields thus λ Fo θ = e sin( λ X ) dx sin ( λ X ) dλdx π X = λ= λ Fo θ = e [ cos λ( X X ) cos λ( X + X )] dλdx π X = λ= π ( X X ) cos ( ) λ = exp 4Fo 4Fo λ Fo e λ X X d λ= π ( X + X ) cos ( + ) λ = exp 4Fo 4Fo λ Fo e λ X X d λ= ( X X ) ( X + X ) θ = exp dx exp dx 4π Fo X = 4Fo X = 4Fo (3.349) 53

54 Let us define a new variable The first integral in eq. (3.349), becomes Similarly, the second integral can be evaluated by following a similar procedure: Substituting the above two equations into eq. (3.349), we have η = ( X X ) 4Fo X X 4Fo exp dx = 4Fo exp( η ) X = X / 4Fo ( X + X ) 4Fo exp dx = 4Fo exp( η ) X = X / 4Fo / 4Fo X θ = exp( η ) dη exp( η ) dη exp( η ) dη π = X / 4Fo X / 4Fo π dη dη 54

55 which can be written as X θ = erf 4Fo where erf in eq. (3.35) is the error function defined as: z z erf ( z) = e dz π Equation (3.35) can also be rewritten as dimensional form: T T x = erf T T t i 4α The surface heat flux can be obtained by applying the Fourier s law T k( T T ) q t = k = x t i ( ) x= πα (3.35) (3.35) (3.35) (3.353) 55

56 Periodic boundary conditions can be encountered in various applications ranging from heat conduction in a building during day and night to emerging technologies such as pulsed laser processing of materials. Let us reconsider the problem described by eqs. (3.33) (3.335) but replace eq. (3.333) by T = Ti + f ( t) = Ti + Acos( ωt β ), x =, t > (3.354) where A is the amplitude of oscillation, ω is the angular frequency, and β is the phase delay. Introducing excess temperature ϑ = T T i, the governing equation and corresponding boundary and initial conditions become ϑ( x, t) ϑ( x, t) = x >, t > x α t ϑ( x, t) = f ( t) = Acos( ωt β ) x =, t > (3.355) (3.356) 56

57 ϑ ( x, t) = x, t > ϑ ( x, t) = x >, t = (3.357) (3.358). Instead of solving eqs. (3.355) (3.358) directly, we will start with a simpler auxiliary problem defined below: Φ ( x, t) Φ ( x, t) = x >, t > (3.359) x α t τ Φ ( x, t) = f ( τ ) = Acos( ωτ β ) x =, t > Φ ( x, t) = x, t > Φ ( x, t) = x >, t = where in eq. (3.36) is treated as a parameter, rather than time. (3.36) (3.36) (3.36) 57

58 The Duhamel s theorem stated that the solution the original problem is related to the solution of auxiliary problem by t ϑ( x, t) = Φ( x, t τ, τ ) dτ (3.363) t τ = which can rewritten using Leibniz s rule ϑ t ( x, t) = Φ( x, t, ) d ( x, t, ) τ τ t τ τ τ + Φ τ τ = = The second term on the right hand side is Φ ( x, t τ, τ ) = Φ ( x,, τ ) = τ = t therefore, eq. (3.364) becomes t ϑ( x, t) = Φ( x, t τ, τ ) dτ τ = t t (3.364) (3.365) (3.366) 58

59 The solution of the auxiliary problem can be expressed as x f ( τ ) η Φ ( x, t, τ ) = f ( τ ) - erf = e dη 4α t π x / 4αt The partial derivative appearing in eq. (3.366) can be evaluated as x x Φ ( x, t τ, τ ) = f ( τ ) exp t 4 πα ( t ) 4 α( t τ ) 3/ τ (3.367) (3.368) Substituting eq. (3.368) into eq. (3.366), the solution of the original problem becomes x t f ( τ ) x 3/ τ πα = ( t τ ) 4 α( t τ ) ϑ( x, t) = exp 4 dτ (3.369) 59

60 Introducing a new independent variable x ξ = 4 α( t τ ) eq. (3.369) becomes (3.37) For the periodic boundary condition specified in eq. (3.356), we have which can be rewritten as x (, ) = exp( ) x / 4αt ϑ x t f t ξ dξ π 4αξ A x (, ) = cos exp( ) x / 4αt ϑ x t ω t β ξ dξ π 4αξ A x (, ) = cos exp( ) ϑ x t ω t β ξ dξ π 4αξ / 4 A x αt x cos ω t β exp( ξ ) dξ π 4αξ (3.37) (3.37) 6

61 Evaluating the first integral on the right hand side of eq. (3.37) yields / / ω ω ϑ( x, t) = Aexp x cos ωt x β α α / 4 A x αt x cos ω t β exp( ξ ) dξ π 4αξ (3.373) It can be seen that as t, the second term will become zero and the first term represents the steady oscillation. / / s ( x, t) Aexp x ω ω ϑ = cos ωt x β α α ( ) / (3.374) where Aexp x ω /( α) represents the amplitude of oscillation at point x, and x( ω /( α) ) / in the cosine function represents the phase delay of oscillation at point x relative to the oscillation of the surface temperature. 6

62 It is useful here to introduce a concept similar to the thermal boundary layer for convective heat transfer thermal penetration depth. Assuming the thickness of the thermal penetration depth at time t is δ, the temperature of the semi-infinite body at x > δ will be affected but the temperature at will remain unchanged (see Fig. 3.). According to the definition of the thermal penetration depth, the temperature at the thermal penetration depth should satisfy T ( x, t) x = x = δ ( t) T ( x, t) = T x = δ ( t) i Integrating eq. (3.33) in the interval (, ) one obtains T T δ ( t ) T ( x, t) = x x= δ ( t ) x x= α t δ dx (3.375) (3.376) (3.377) 6

63 T T T i δ(t) x Figure 3. Heat conduction in a semi-infinite body with constant wall temperature. The right-hand side of eq. (3.377) can be rewritten using Leibnitz s rule, i.e., T T d δ dδ = Tdx T x= δ x x= δ ( t ) x x= α dt dt which represents the energy balance within the thermal penetration depth. (3.378) 63

64 Substituting eqs. (3.375) and (3.376) into eq. (3.378) yields where T d α = ( Θ Tiδ ) x dt x= (3.379) (3.38) Assume that the temperature distribution in the thermal penetration depth is a third-order polynomial function of x, i.e., where A, A, A, and A 3 and are four constants to be determined using the boundary conditions. δ ( t) Θ ( t) = T ( x, t) dx T( x, t) = A + A x + A x + A x 3 3 (3.38) The surface temperature of the semi-infinite body, T, is not a function of time t, so T( x, t) t = x = (3.38) 64

65 Substituting eq. (3.33) into eq. (3.38) yields, T ( x t) x = x = (3.383) Substituting eq. (3.38) into eqs. (3.333), (3.375), (3.376) and (3.383) yields four equations for the constants in eq. (3.38). Solving for the four constants and substituting the results into eq. (3.38), the temperature distribution in the thermal penetration depth becomes T( x, t) Ti 3 x x = + T T δ δ i (3.384) where the thermal penetration depth, δ, is still unknown. 3 65

66 Substituting eq. (3.384) into eq. (3.379), an ordinary differential equation for δ is obtained: dδ 4α = δ t > dt (3.385) Since the thermal penetration depth equals zero at the beginning of the heat conduction, eq. (3.385) is subject to the following initial condition: δ = t = The solution of eqs. (3.385) and (3.386) is δ = 8α t which is consistent with the result of scale analysis, (3.386) (3.387) δ αt 66

67 3.3. Multidimensional Transient Heat Conduction Systems Consider transient heat conduction in a rectangular bar with dimensions of L L and an initial temperature of T i (see Fig. 3.). At time t =, the rectangular bar is immersed into a fluid with temperature T. y L h,t h,t h,t L x L -L h,t Figure 3. Two-dimensional transient heat conduction 67

68 The energy equation for this problem is T T T + =, < x < L, < y < L, t > x y t (3.388) with the following boundary and initial conditions: T =, x =, < y < L, t > x T k = h( T T ), x = L, < y < L, t > x T =, y =, < x < L, t > y T k = h( T T ), y = L, < x < L, t > y T = T, < x < L, < y < L, t > i (3.389) (3.39) (3.39) (3.39) (3.393) 68

69 Defining the following dimensionless variables T T x y αt αt θ =, X =, Y =, Fo =, Fo = T T L L L L i (3.394) where both Fo and Fo are Fourier numbers but based on different characteristic lengths. The dimensional time will be related to both Fourier numbers, i.e., t = t(fo,fo ) therefore, the right-hand side of eq. (3.388) becomes T θ θ Fo θ Fo = ( Ti T ) = ( Ti T ) + t t Fo t Fo t θ α θ α = ( Ti T ) + Fo L Fo L (3.395) (3.396) 69

70 Nondimensionalizing the left-hand side of eq. (3.388) and considering eq. (3.396), the dimensionless energy equation of the problem becomes θ L θ θ L θ + = + Fo Fo X L Y L (3.397) The boundary and initial conditions, eqs. (3.389) (3.393) are nondimensionalized as θ =, X =, < Y <, Fo >, Fo > X θ = Bi θ, X =, < Y <, Fo >, Fo > X θ =, Y =, < X <, Fo >, Fo > Y θ = Bi θ, Y =, < X <, Fo >, Fo > Y (3.398) (3.399) (3.4) (3.4) 7

71 where Advanced Heat and Mass Transfer by Amir Faghri, Yuwen Zhang, and John R. Howell θ =, < X <, < Y <, Fo = Fo = are Bio numbers for different surfaces. Bi h L h L =, Bi = k k (3.4) (3.43) The idea of product solution is that the solution of the twodimensional problem can be expressed as the product of two one-dimensional problem, i.e., θ = ϕ( X,Fo ) ψ ( Y,Fo ) where ϕ is the solution of the following problem ϕ = X ϕ Fo ϕ =, X =, Fo > X (3.44) (3.45) (3.46) 7

72 and satisfies ψ ϕ = Bi ϕ, X =, Fo > X ϕ =, < X <, Fo = ψ ψ = Y Fo ψ =, Y =, Fo > Y ψ = Bi ψ, Y =, Fo > Y ψ =, < Y <, Fo = (3.47) (3.48) (3.49) (3.4) (3.4) (3.4) It can be demonstrated that eqs. (3.397) (3.4) can be satisfied by eq. (3.44) if ϕ and ψ are solutions of eqs. (3.45) (3.48) and (3.49) (3.4), respectively. The solution of eqs. (3.45) (3.48) is: 4sin λ n n Fo ϕ = cos( λ ) n X e λ (3.43) n= λ + sin λ n n 7

73 Where and the solution of eqs. (3.49) (3.4) is where ψ = m= λ n Bi 4sinν n ν + sin ν m ν m Bi = cot λ (3.44) (3.45) (3.46) Substituting eqs. (3.43) and (3.45) into eq. (3.44), the solution of the two-dimensional problem is obtained. 6sin λ ( ) ( ) n sinν m cos λn X cos ν my e ( λn Fo + ν mfo ) (3.47) θ = n= m= m cos = cotν n m ( ν ) (λ + sin λ )(ν + sin ν ) n n m m m Y e ν m Fo 73