1. Computer Aided Thermal Fluid Analysis Lecture 10
Dr. Ming-Jyh Chern
ME NTUST

2. Road Map for Today What is turbulence?
Reynolds Averaged Navier-Stokes (RANS) equations
Turbulence models
Boundary conditions for turbulence models

3. What is turbulence? Part I

4. What is turbulence? Part II
Let us see a movie regarding a turbulent flow in a valve.

5. What is turbulence? Part III – Its nature
Random
Effective Mixing
High Reynolds number
3-D
Energy Dissipation
Eddy Motions

6. What is turbulence? Energy Cascade

7. Reynolds Decomposition

8. Reynolds Averaged Navier-Stokes (RANS) equations
is the so-called Reynolds stress.

9. Boussinesq’s Assumption
How to determine eddy viscosity nt?

10. Zero equation model
nt is assumed to be a constant and depends on various flow fields.

11. One equation model

12. Two equations model

13. K-e turbulence model

14. K-e turbulence model

15. Boundary conditions
Inlet Conditions

16. Boundary conditions for a solid wall
1. Wall function

17. Boundary conditions for a solid wall
1. Wall function

18. Boundary conditions for a solid wall
2. Two Layer Method

19. Boundary conditions for a solid wall
2. Two Layer Method

20. Example – Sudden Expansion Flowui
0.1 m
0.13 m
1 m
2.5 m

21. Example – Sudden Expansion Flow – establish mesh

22. Example – Sudden Expansion Flow – Laminar Flow Case
Working fluids – air
Density = m3/s
Dynamics viscosity = 1.81e-5 kg/ms
Characteristic length = 0.1 m
If we consider a laminar channel flow at Re = 100, then the magnitude of inlet velocity must be m/s.

23. Example – Sudden Expansion Flow – Boundary setup
Outlet or constant pressure boundary
Symmetry boundary
Symmetry boundary
Inlet boundary

24. Example – Sudden Expansion Flow – Results of laminar Flow

25. Example – Sudden Expansion Flow – Turbulent Flow Case
Working fluids – air
Density = m3/s
Dynamics viscosity = 1.81e-5 kg/ms
Characteristic length = 0.1 m
If we consider a turbulent channel flow at Re = 30,000, then the magnitude of inlet velocity must be 4.5 m/s.
k and e at the inlet boundary (k = , e = 7.859).

26. Example – Sudden Expansion Flow – Results of Turbulent Flow
Contours of k

27. Simulation of Heat Transfer
Forced convection or natural convection?
Boundary conditions, a. isothermal boundary, b. constant heat flux.
Conjugate heat transfer? Heat sources should be imposed inside solids.

28. Example – Forced convection with isothermal boundaryui
0.1 m
0.13 m
1 m
2.5 m
T = 313 K
The constant wall temperature is 293 K, except for the orange region at which the temperature is 313 K.

29. Example – Forced convection with isothermal boundary

30. Example – Forced convection with isothermal boundary

31. Example – Forced convection with isothermal boundary

32. Example – Forced convection with isothermal boundary

33. Example – Forced convection with isothermal boundary

34. Example – Forced convection with isothermal boundary

35. Example – Forced convection with isothermal boundary
Isothermal contours

36. Example – Natural convection with isothermal boundary
T = 293 K g
0.01 m
Adiabatic boundary
Adiabatic boundary
0.01 m
T = 294 K

37. Example – Natural convection with isothermal boundary

38. Example – Natural convection with isothermal boundary
Boussinesq’s approximation: assume the buoyant force f in N-S equations is

39. Example – Natural convection with isothermal boundary

40. Example – Natural convection with isothermal boundary

41. Example – Natural convection with isothermal boundary

42. Example – Natural convection with isothermal boundary
Isothermal contours

43. Example – Conjugate Heat Transfer
Heat conduction in a solid and convection in a fluid are considered in conjugate heat transfer.
At least, two materials shall be defined as a fluid and a solid in the model, respectively.

44. Example – Conjugate Heat Transfer.

45. Example – Conjugate Heat Transfer
T = 293 K
air g
0.01 m
Adiabatic boundary
Adiabatic boundary
0.01 m Al
T = 294 K

46. Example – Conjugate Heat Transfer1 3 2

47. Example – Conjugate Heat Transfer
4. Choose a solid material from the table or creat a new one. Do not forget to click apply.

48. Example – Conjugate Heat Transfer
5. Use C> /NEW / Zone to select cells into cset.

49. Example – Conjugate Heat Transfer
6. Click Tools/Cell Tools to set Type 2 Solid to Material 2

50. Example – Conjugate Heat Transfer
7. Use cell list to change cells in cset to the type 2 solid

51. Example – Conjugate Heat Transfer
8. Check if there are two different kinds of cells. Red one is fluid 1. Green one is solid 2.

52. Example – Conjugate Heat Transfer
Go back to STAR Guide. Click Thermal Options. Click Heat Transfer ON.
The rest procedures for simulation of natural convection are as same as the previous example.

53. Example – Conjugate Heat Transfer
Iosthermal contours + Velocity vectors