10.3 Multimode Heat Transfer with Radiation

Size: px

Start display at page:

Download "10.3 Multimode Heat Transfer with Radiation"

Sharon Randall
5 years ago
Views:

1 0.3 Multimode Heat Transfer with Radiation When radiation is combined with conduction and/or convection, the problems are called conjugate or multimode heat transfer. As an example of a combined-mode radiationconduction problem, consider the geometry and conditions shown in Fig Because the plates are not isothermal, conduction occurs within the plates, and will modify the temperature distributions, which are the unknowns in the problem. Chapter 0: Heat Transfer by Radiation

2 W δ x Surface, k, ε, q (x ) = 0 H T e = 0 K x Surface, k, ε q e (x ) = 00 W/m Figure 0.9 Infinitely long parallel conducting plates with thickness δ and thermal conductivity k. Chapter 0: Heat Transfer by Radiation

3 For this case, the usual net radiation equations are as derived in Example 0.0: q ( x ) ε = σ T σ T F + q x df x x ( ) (, ), rad d, rad A d d ε ε ( x ) q ε = σ T ( x ) σ T ( x ) F + q x df x, x ( ) ( ), rad d, rad A d d ε ε (0.70) (0.7) If conduction is assumed to occur only in the x-direction in the thin plates, then two additional energy equations are written for the conducting plates q = q + kδ,rad q =,rad e kδ d T dx d T dx (0.7) (0.73) Chapter 0: Heat Transfer by Radiation

4 Equations (0.7) and (0.73) contain second-order derivatives, so two boundary conditions are required for each. The appropriate ones are dt dt = = 0 at x, x = W / dx dx dt dt k = ε σ T ( x = 0); k = ε σ T ( x = 0) dx dx The resulting Equations (0.76) are highly nonlinear, involving temperatures to the fourth power as well as second derivatives in temperature. d T W ε d T q + kδ σ T ( x ) σ T ( x ) kδ df e ε = dx 0 ε dx d d d T W ε d T kδ σ T ( x ) σ T ( x ) q + kδ df e ε dx = 0 ε dx d d (0.7) (0.75) (0.76) Chapter 0: Heat Transfer by Radiation

5 To simplify Eqs. (0.76), define kδ T q / ;, T / T ; X x / W ; X x / W ( ) / ref = e σ = θ = 3 ref = = σ Tref W and Eqs. (0.76) become d θ + ( X ) ( X ) ( ) = ε θ ε θ ε dx 0 dx d θ d d d θ ε θ ( X ) ε θ ( X ) ( ε ) + dx = 0 dx d d d θ df df (0.77) (0.78) with boundary conditions [from Eqs. (0.7) and (0.75)] dθ dx X = / X = / dθ dx dθ dx X = 0 dθ = = dx X = 0 ( X 0 ) = ε θ = ( X 0) = ε θ = 0 δ W δ W (0.79) Chapter 0: Heat Transfer by Radiation

6 In terms of the nondimensional variables, the required configuration factor is (using h = H/W), df dx = dx ( ) (0.80) The or modifications of it often appear in radiationconduction problems. Its meaning can be appreciated by multiplying the numerator and denominator by T ref to give kδ T kδ ( Tref / W ref ) h = = σ T W dx X X + h ref ref σ T W 3/ (0.8) Chapter 0: Heat Transfer by Radiation

7 0.3. umerical Methods To solve the radiation-conduction problem of Eqs. (0.78)-(0.80), the integrals in the equations are replaced by summations over elements of width Δx, i.e., x 0 n (, ), f x x dx f x m n n m n n n= Substituting into Eq. (0.78) and using the finite difference form for the second derivatives results in θ, m θ, m + θ, m+ + = ε θ, m Φ X ( X ) n=, m, n θ, M n θ, n + θ, n+ = ε θ, n Φ X ( X ) m=, m, n (0.8) (0.83) Chapter 0: Heat Transfer by Radiation

8 Including the configuration factor relations, the Φs are Φ = Φ ( ε ), n, n, n +, m, n, n ( ) ( ) The boundary conditions become = ε θ ( ε ) +, m, m, m +, m, n, m ε θ h n X m X + h h θ θ + θ m X n X + h M M,( ),( + ),( ),( + ) 3 / 3/ θ θ + θ θ = θ ; θ = θ ; M and even θ θ δ / W,, = ε θ, + Φ X,, n X n = θ θ δ / W (0.8) (0.85) M,, = ε θ, + Φ X, m, X m = Chapter 0: Heat Transfer by Radiation

9 Let ΔX = ΔX (equal sized increments on each plate, so that Eqs. (0.83) - (0.85) become θ θ + θ, m, m, m + + = ε θ, m Φ, m, n n = θ θ + θ Φ = Φ, n, n, n + ( ) ( ) ( ε ) = ε θ, n M m = Φ, m, n, n, n, n +, m, n, n = ε θ ( ε ) + 3/ X, m, m, m +, m, n, m ε θ h n m X + h h m n X + h θ θ + θ 3/ θ θ + θ X (0.86) (0.87) Chapter 0: Heat Transfer by Radiation

10 θ = θ ; θ = θ ; M and even M M,( ),( + ),( ),( + ) θ θ,, δ / W = ε θ X, + Φ,, n X n = M θ θ,, δ / W = ε θ X, + Φ, m, X m = (0.88) The parameter /(ΔX) now appears in Eqs. (0.86) - (0.88), and is a guideline for how to approach a numerical solution. Chapter 0: Heat Transfer by Radiation

11 0.3. Conduction Dominated Problems If conduction dominates (i.e., /(ΔX) >> ), then eq. (0.86) can be rearranged to reflect a conduction problem that is perturbed by the presence of radiation: ( X ) θ θ + θ = ε θ, m Φ X, m, m, m +, m, n n = ( X ) M, n, n +, n + = ε θ, n, m, n X Φ m = θ θ θ (0.89) Chapter 0: Heat Transfer by Radiation

12 Equation (0.89) (including the insertion of the boundary conditions) is a set of equations of the matrix form Aθ(x) = C(x) (0.90) where A is a tridiagonal matrix of coefficients that need only be inverted once. The solution is then - θ(x) = A C(x) (0.9) Equation (0.9) is solved by assuming the distributions of θ,n and θ,m, using these to evaluate Φ,m,n and Φ,m,n, which in turn are used to evaluate C(x). The matrix multiplication indicated in eq. (0.9) gives a new set of θ values; the process is repeated until convergence. Chapter 0: Heat Transfer by Radiation

13 0.3.3 Radiation Dominated Problems If radiation dominates so that /(ΔX) <<, then Eq. (0.86) can be rearranged to θ θ, m, n = = ε ε n= Φ X + (0.9) Again, the set of equations can be arranged as a matrix equation of the form Aθ = C (θ) (0.93) Matrix inversion of A gives θ = A θ θ + θ, m, m, m+, m, n θ θ + θ M, n, n, n+ Φ, m, n X m= ( ) - C (θ) (0.9) Chapter 0: Heat Transfer by Radiation

14 0.3. Problems with Both Modes Significant When is not near either the large or small limit, then the problem is truly nonlinear, and the solution methods described for small or large /(ΔX) values often fail. In that case, Eq. (0.86) can be arranged as ε θ ε θ, m, n θ, m θ, m + θ, m + = + Φ, m, n X ( X ) n = θ, M n θ, n + θ, n+ = Φ, m, n X ( X ) m = (0.95) This equation in matrix form becomes Aθ + Bθ = C θ ( ) We could define so that Eq. (0.96) becomes A (θ) θ = C (θ) 3 (0.96) (0.97) Chapter 0: Heat Transfer by Radiation

15 To avoid this, a modified ewton-raphson iteration method can be used. In this case, Eq. (0.96) is rewritten as M + ( 0 ) (0) j = [ jkθ k + jkθ k ] j residual k = g A B C The residual is a measure of convergence of the solution, and will approach g j = 0 when solution is complete. ext, the function g jk is found (0) 3 ( LHS of Eq.(0.96) jk = jk θ k + jk θ k g A B (0.98) (0.99) and g jk is seen to be the gradient in the residual. ow solve the auxiliary equation g jk [ λ k ] + g j = 0; [ λ k ] = g jk g j (0.00) Chapter 0: Heat Transfer by Radiation

16 The next iterative value for θ k is found from ( p ) ( p ) θ = θ + λ k k k (0.0) Chapter 0: Heat Transfer by Radiation

CHAPTER 4 BOUNDARY LAYER FLOW APPLICATION TO EXTERNAL FLOW

CHAPTER 4 BOUNDARY LAYER FLOW APPLICATION TO EXTERNAL FLOW 4.1 Introduction Boundary layer concept (Prandtl 1904): Eliminate selected terms in the governing equations Two key questions (1) What are the