Finite Element Analysis of Heat Transfer Problems

Analogy between stress analysis and heat conduction analysis is discussed. Various thermal problems, including steady-state heat transfer and governing differential equations, are explored. Conservation of energy and boundary conditions are detailed for solving thermal analysis problems.

• Finite Element Analysis
• Heat Transfer
• Thermal Problems
• Governing Equations
• Boundary Conditions

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1. CHAP 5 FINITE ELEMENTS FOR HEAT TRANSFER PROBLEMS FINITE ELEMENT ANALYSIS AND DESIGN Nam-Ho Kim 1

2. HEAT CONDUCTION ANALYSIS Analogy between Stress and Heat Conduction Analysis Structural problem Displacement Stress/strain Displacement B.C. Surface traction force Body force Heat transfer problem Temperature (scalar) Heat flux (vector) Fixed temperature B.C. Heat flux B.C. Internal heat generation Thermal conductivity Young s modulus In finite element viewpoint, two problems are identical if a proper interpretation is given. More Complex Problems Coupled structural-thermal problems (thermal strain). Radiation problem 2

3. THERMAL PROBLEM Goals: = [ ]{ } T { } Q K T Thermal load Nodal temperature Conductivity matrix Solve for temperature distribution for a given thermal load. Boundary Conditions Essential BC: Specified temperature Natural BC: Specified heat flux 3

4. STEADY-STATE HEAT TRANSFER PROBLEM Fourier Heat Conduction Equation: Heat flow from high temperature to low temperature dT = q kAdx x Thermal conductivity (W/m/ C ) Heat flux (Watts) Examples of 1D heat conduction problems Thigh Tlow Thigh qx qx Tlow 4

5. GOVERNING DIFFERENTIAL EQUATION Conservation of Energy Energy In + Energy Generated = Energy Out + Energy Increase + = + E E E U in generated out Two modes of heat transfer through the boundary Prescribed surface heat flow Qs per unit area Convective heat transfer h: convection coefficient (W/m2/ C ) ( ) = Q h T T h T Qs dq dx x Qg + D q x x q x A dx 5

6. GOVERNING DIFFERENTIAL EQUATION cont. Conservation of Energy at Steady State No change in internal energy ( U = 0) dq dx ( ) + + + = + q Q P x h T T P x Q A x q x x x s g x E E gen in E out P: perimeter of the cross-section dq Q A dx ( ) = + + hP T T Q P, 0 x L x g s Apply Fourier Law d dT dx + ( ) + + = kA Q A hP T T Q P 0, 0 x L g s dx Rate of change of heat flux is equal to the sum of heat generated and heat transferred 6

7. GOVERNING DIFFERENTIAL EQUATION cont. Boundary conditions Temperature at the boundary is prescribed (essential BC) Heat flux is prescribed (natural BC) Example: essential BC at x = 0, and natural BC at x = L: T(0) T dT kA q dx = = 0 = L x L 7

8. DIRECT METHOD Follow the same procedure with 1D bar element No need to use differential equation Element conduction equation Heat can enter the system only through the nodes Qi: heat enters at node i (Watts) Divide the solid into a number of elements Each element has two nodes and two DOFs (Ti and Tj) For each element, heat entering the element is positive N 2 1 Q1 Q2 QN Q3 j i e ( ) e ( ) e jq iq L(e) Tj Ti xi xj 8

9. ELEMENT EQUATION Fourier law of heat conduction ( ) T T dT dx j L i = = (e) i q kA kA (e) From the conservation of energy for the element (T T) + = (e) i (e) j j L i = + q q 0 (e) j q kA (e) Combine the two equation = T T (e) i (e) j q q 1 1 kA L i Element conductance matrix (e) 1 1 j Similar to 1D bar element (k = E, T = u, q = f) 9

10. ASSEMBLY Assembly using heat conservation at nodes Remember that heat flow into the element is positive Equilibrium of heat flow: T T Q Q 1 1 N i = (e) i 2 2 Q q = [ ] K i T ( ) e 1 = N N T Q N N Same assembly procedure with 1D bar elements Applying BC Striking-the-rows works, but not striking-the-columns because prescribed temperatures are not usually zero Q2 3 1 (1) (2) 2q 2q Element 2 Element 1 2 10

11. EXAMPLE Calculate nodal temperatures of four elements A = 1m2, L = 1m, k = 10W/m/ C Q4 = 200W T1 T2 T3 T4 T5 x 200 C 1 2 3 4 Q2 = 500W Q3 = 0 Q5 = 0 Q1 Element conduction equation T 1 1 10 T 1 1 (1) 1 (1) 2 T T q q (2) 2 (2) 3 1 1 q q 1 = 2 = 10 1 1 2 3 T T T T (4) 4 (4) 5 (3) 3 (3) 4 1 1 1 1 q q q q 4 3 = = 10 10 1 1 1 1 5 4 11

12. EXAMPLE cont. Assembly Q Q Q Q Q T T T T T (1) 1 + + + 1 1 0 0 0 1 2 1 0 0 0 1 2 1 0 0 0 1 2 1 0 0 0 1 1 q 1 1 (1) 2 (2) 3 (3) 4 (2) 2 (3) 3 (4) 4 q q q q q q 2 2 = = 10 3 3 4 4 (4) 5 q 5 5 Boundary conditions (T1 = 200 oC, Q1 is unknown) 200 T T T T 1 1 0 0 0 0 0 0 Q 500 0 200 0 1 1 2 1 2 = 10 0 0 0 1 2 1 3 0 0 1 2 1 4 0 1 1 5 12

13. EXAMPLE cont. Boundary conditions Strike the first row 200 T T T T 1 2 0 0 0 0 0 1 0 0 0 500 0 200 0 2 1 2 1 = 10 3 1 2 1 4 0 1 1 5 Instead of striking the first column, multiply the first column with T1 = 200 oC and move to RHS T T T T 2 1 0 0 0 500 0 200 0 2000 0 0 0 2 1 2 1 3 = + 10 0 0 1 2 1 4 0 1 1 5 Now, the global matrix is positive-definite and can be solved for nodal temperatures 13

14. EXAMPLE cont. Nodal temperatures T= { } { } T 200 230 210 190 190 C How much heat input is required to maintain T1 = 200oC? Use the deleted first row with known nodal temperatures = + + + 300W = Q 10T 10T 0T 0T 0T 1 1 2 3 4 5 Other example 100W 3 4 2 4 5 1 2 1 5 3 x 50 C Q = 0 200W Q = 0 14

15. GALERKIN METHOD FOR HEAT CONDUCTION Direct method is limited for nodal heat input Need more advanced method for heat generation and convection heat transfer Galerkin method in Chapter 3 can be used for this purpose Consider element (e) Interpolation T(x) TN(x) TN (x) = + j i e ( ) e ( ) e iq jq L(e) Tj Ti xi xj i i j j x x x x = = N(x) 1 , N (x) i i i j (e) (e) L L T T i N(x) N (x) = = T(x) { } N T Temperature varies linearly i j j Heat flux dT dx 1 1 = = { } T { } T B Heat flow is constant (e) (e) L L 15

16. GALERKIN METHOD cont. Differential equation with heat generation d dT dx + = kA Q A 0, 0 x L g dx Substitute approximate solution + d dT dx = kA AQ R(x) Residual g dx Integrate the residual with Ni(x) as a weight + x d dT dx j = kA AQ N(x)dx 0 g i dx x i Integrate by parts x x x j dN dT dx dT dx dx j j = kA N(x) kA dx AQ N(x)dx i i g i x x x i i i 16

17. GALERKIN METHOD cont. Substitute interpolation relation x x dN dx dN dx dN dx j j j + = + kA T T dx AQ N(x)dx q(x )N(x ) q(x )N(x ) i i i j g i j i j i i i x x i i Perform integration ( i (e) T L x kA j ) = = + (e) i (e) i (e) i T Q q Q AQ N(x)dx j g i x i Repeat with Nj(x) as a weight ( j i (e) T T L x j = kA (e) j Q AQ N (x)dx ) = + (e) j (e) j Q q g j x i 17

18. GALERKIN METHOD cont. Combine the two equations T T + + (e) i (e) j (e) i (e) j Q Q q q 1 1 kA L = + (e) T (e) (e) [ ]{ } T { } { } k Q q i = (e) 1 1 j Similar to 1D bar element {Q(e)}: thermal load corresponding to the heat source {q(e)}: vector of nodal heat flows across the cross-section Uniform heat source x (e) N(x) N (x) AQ L 1 1 j i g = = (e) { } AQ dx Q g 2 j x i Equally divided to the two nodes Temperature varies linearly in the element, and the heat flux is constant 18

19. EXAMPLE Insulated No heat flow Heat chamber Wall temperature = 200 C Uniform heat source inside the wall Q = 400 W/m3. Thermal conductivity of the wall is k = 25 W/m C. Use four elements through the thickness (unit area) Boundary Condition: T1 = 200, qx=1 = 0. Wall 200 C x 1 m No heat flow x 200 C T1 T2 T3 T4 T5 1 2 3 4 19

20. EXAMPLE cont. Element Matrix Equation All elements are identical T T (1) 1 (1) 2 1 1 50 50 q q 1 = + 100 1 1 2 Assembly Q Q Q Q Q T T T T T (1) 1 + + + 1 1 0 0 0 0 0 0 50 100 100 100 50 q 1 1 (1) 2 (2) 3 (3) 4 (2) 2 (3) 3 (4) 4 1 2 1 q q q q q q 2 2 = = 100 0 0 0 1 2 1 3 3 0 0 1 2 1 4 4 (4) 5 0 1 1 q 5 5 20

21. EXAMPLE cont. Boundary Conditions At node 1, the temperature is given (T1 = 200). Thus, the heat flux at node 1 (Q1) should be unknown. At node 5, the insulation condition required that the heat flux (Q5) should be zero. Thus, the temperature at node 5 should be unknown. At nodes 2 4, the temperature is unknown (T2, T3, T4). Thus the heat flux should be known. 1 2 3 Q1 4 5 Q5 + 200 T T T T 1 1 0 0 0 0 0 0 50 Q 1 1 2 1 100 100 100 50 2 = 100 0 0 0 1 2 1 3 0 0 1 2 1 4 0 1 1 5 21

22. EXAMPLE cont. Imposing Boundary Conditions Remove first row because it contains unknown Q1. Cannot remove first column because T1 is not zero. 200 T T T T 1 2 1 0 0 0 100 100 100 50 2 0 0 0 1 2 1 = 100 3 0 0 1 2 1 4 0 1 1 5 2 T + = 100( 1 200 100(2 T 1 T ) 100 100 2 3 20000 Instead, move the first column to the right. = + 1 T ) 2 3 T T T T 2 1 0 0 0 100 100 100 50 20000 0 0 0 20100 100 100 50 2 1 2 1 3 = + = 100 0 0 1 2 1 4 0 1 1 5 22

23. EXAMPLE cont. Solution = 200 C,T = = 206 C,T = = T 203.5 C,T 207.5 C,T 208 C 1 2 3 4 5 208 207 FEM Exact 206 205 204 T 203 202 201 200 x 0 0.2 0.4 0.6 0.8 1 Discussion In order to maintain 200 degree at node 1, we need to remove heat + = = Q 50 100T = 100T 350 1 1 2 Q 400 W 1 23

24. CONVECTION BC Convection Boundary Condition Happens when a structure is surrounded by fluid Does not exist in structural problems BC includes unknown temperature (mixed BC) Wall qh T T = h q hS(T T) Fluid Temperature Convection Coefficient Heat flow is not prescribed. Rather, it is a function of temperature on the boundary, which is unknown 1D Finite Element When both Nodes 1 and 2 are convection boundary = = q q hAT hAT hAT hAT 1 1 1 2 1 T T 2 2 2 T1 T2 24

25. EXAMPLE (CONVECTION ON THE BOUNDARY) T1 T2 T3 Element equation T T 1 3 1 2 h3 h1 T T T T (2) 2 (2) 3 (1) 1 (1) 2 1 1 1 1 q q q q kA L kA L 2 1 = = 1 1 1 1 3 2 Balance of heat flow = (1) 1 q h A(T T ) Node 1: 1 1 1 + = (1) 2 (2) 2 h A(T q q 0 Node 2: = (2) 3 q T ) Node 3: Global matrix equation 3 3 3 1 1 0 T T T h A(T T ) 1 1 1 0 1 kA L = 1 2 1 2 0 1 1 h A(T T ) 3 3 3 3 25

26. EXAMPLE cont. Move unknown nodal temperatures to LHS kA kA h A 0 L L kA 2kA kA L L L kA kA 0 h A L L + 1 T T T h AT 1 1 1 = 0 2 h AT 3 3 3 + 3 The above matrix is P.D. because of additional positive terms in diagonal How much heat flow through convection boundary? After solving for nodal temperature, use = (1) 1 q h A(T T ) 1 1 1 This is convection at the end of an element 26

27. EXAMPLE: FURNACE WALL Firebrick k1=1.2W/m/oC hi=12W/m2/oC Insulating brick k2=0.2W/m/oC ho=2.0W/m2/oC 16.8 4.8 0 { } {1,411 1,190 = T Insulating brick Firebrick Ta = 20 C Tf = 1,500 C hi x ho 0.12 m 0.25 m 4.8 6.47 1.67 0 T T T 18,000 0 40 1 = 1.67 3.67 2 3 Convection boundary Convection boundary T 552} C No heat flow 1,500 C 20 C x = 1054 W/m = (2) 3 2 q h (T T ) 0 a 3 T1 T2 T3 Tf Ta 1 2 ho hi 27

28. CONVECTION ALONG A ROD Long rod is submerged into a fluid Convection occurs across the entire surface Governing differential equation d dT kA AQ hP T dx dx + ( ) + = = + T 0, 0 x L P 2(b h) g Convection Fluid T b j h ( ) e jq ( ) e iq i xi Convection xj 28

29. CONVECTION ALONG A ROD cont. DE with approximate temperature d dT kA AQ dx dx + ( ) + = hP T T R(x) g Minimize the residual with interpolation function Ni(x) x + d dT dx j + = kA AQ hP(T T) N(x)dx 0 g i dx x i Integration by parts x x x x x j dN dT dx dT dx dx j j j j = kA N(x) kA dx hPTNdx AQ N(x)dx hPT Ndx i i i g i i x x x x x i i i i i 29

30. CONVECTION ALONG A ROD cont. Substitute interpolation scheme and rearrange x x dN dx dN dx dN dx j j j + + + kA T T dx hP(TN TN )N dx i i i j i i j j i x x i i x j = + + (AQ hPT )N dx q(x )N(x ) q(x )N(x ) g i j i j i i i x i Perform integration and simplify ( i j (e) T T hpL L T 6 T 3 kA ) j + + = + (e) (e) i (e) i Q q i x j = + (e) i Q (AQ hPT )N(x)dx g i x i Repeat the same procedure with interpolation function Nj(x) 30

31. CONVECTION ALONG A ROD cont. Finite element equation with convection along the rod T T + + (e) i (e) j (e) i (e) j Q Q q q 1 1 2 1 2 1 (e) kA L hPL i + = (e) 6 1 1 j T + = + (e) T (e) h (e) (e) [ ] [ ] { } { } k k Q q Equivalent conductance matrix due to convection 2 1 2 1 (e) hPL = (e) h k 6 Thermal load vector + (e) (e) Q Q AQ L hPL T 2 1 1 i g = = (e) { } Q j 31

32. EXAMPLE: HEAT FLOW IN A COOLING FIN k = 0.2 W/mm/ C, h = 2 10 4 W/mm2/ C Element conductance matrix + = k k 1 1 2 1 2 1 2 10 4 0.2 200 40 320 40 6 + (e) T (e) h [ ] [ ] 1 1 Thermal load vector Convection T = 30 C 1 1 2 10 4 320 40 30 2 = (e) { } Q 160 mm 1.25 mm Element 1 330 C Insulated x 120 mm T1 T2 T3 T4 3 2 1 32

33. EXAMPLE: HEAT FLOW IN A COOLING FIN cont. Element conduction equation T T (1) 1 (1) 2 1.8533 0.5733 0.5733 1.8533 38.4 38.4 q q 1 = + Element 1 2 T T (2) 2 (2) 3 1.8533 0.5733 0.5733 1.8533 38.4 38.4 q q 2 = + Element 2 3 T T (3) 3 (3) 4 1.8533 0.5733 0.5733 1.8533 38.4 38.4 q q Element 3 3 = + 4 Balance of heat flow = (1) 1 q Q Node 1 1 + = (1) 2 (2) 2 q q 0 Node 2 + = (2) 3 (2) 3 q q 0 Node 3 = (3) 4 q hA(T T ) Node 4 4 33

34. EXAMPLE: HEAT FLOW IN A COOLING FIN cont. Assembly 1.853 + T T T T 38.4 Q .573 3.706 .573 0 0 0 0 1 1 76.8 76.8 hA(T .573 0 0 .573 3.706 .573 1.853 2 = .573 3 38.4 + T ) 4 4 Move T4 to LHS and apply known T1 = 330 + 330 T T T 1.853 .573 0 0 .573 3.706 .573 0 0 0 0 38.4 Q 1 .573 3.706 .573 1.893 76.8 76.8 39.6 2 = .573 3 4 Move the first column to RHS after multiplying with T1=330 3.706 .573 0 T 265.89 .573 3.706 .573 T 76.8 0 .573 1.893 T 39.6 2 = 3 4 34

35. EXAMPLE: HEAT FLOW IN A COOLING FIN cont. Solve for temperature T 330 C, T = = = = 77.57 C, T 37.72 C, T 32.34 C 1 2 3 4 350 300 250 200 150 100 50 0 0 40 80 120 35