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

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

What is turbulence? Part I

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

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

What is turbulence? Energy Cascade

Reynolds Decomposition

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

Boussinesq’s Assumption How to determine eddy viscosity nt?

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

One equation model

Two equations model

K-e turbulence model

K-e turbulence model

Boundary conditions Inlet Conditions

Boundary conditions for a solid wall 1. Wall function

Boundary conditions for a solid wall 1. Wall function

Boundary conditions for a solid wall 2. Two Layer Method

Boundary conditions for a solid wall 2. Two Layer Method

Example – Sudden Expansion Flow ui 0.1 m 0.13 m 1 m 2.5 m

Example – Sudden Expansion Flow – establish mesh

Example – Sudden Expansion Flow – Laminar Flow Case Working fluids – air Density = 1.205 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 0.015 m/s.

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

Example – Sudden Expansion Flow – Results of laminar Flow

Example – Sudden Expansion Flow – Turbulent Flow Case Working fluids – air Density = 1.205 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 = 0.30375, e = 7.859).

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

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.

Example – Forced convection with isothermal boundary ui 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.

Example – Forced convection with isothermal boundary

Example – Forced convection with isothermal boundary

Example – Forced convection with isothermal boundary

Example – Forced convection with isothermal boundary

Example – Forced convection with isothermal boundary

Example – Forced convection with isothermal boundary

Example – Forced convection with isothermal boundary Isothermal contours

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

Example – Natural convection with isothermal boundary

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

Example – Natural convection with isothermal boundary

Example – Natural convection with isothermal boundary

Example – Natural convection with isothermal boundary

Example – Natural convection with isothermal boundary Isothermal contours

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.

Example – Conjugate Heat Transfer .

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

Example – Conjugate Heat Transfer 1 3 2

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

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

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

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

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

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.

Example – Conjugate Heat Transfer Iosthermal contours + Velocity vectors