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

Computer Aided Thermal Computer Aided Thermal Fluid AnalysisFluid Analysis

Lecture 10Lecture 10

Dr. Ming-Jyh ChernDr. Ming-Jyh Chern

ME NTUSTME NTUST

Road Map for TodayRoad Map for Today

• What is turbulence?What is turbulence?• Reynolds Averaged Navier-Stokes (RAReynolds Averaged Navier-Stokes (RA

NS) equationsNS) equations• Turbulence modelsTurbulence models• Boundary conditions for turbulence mBoundary conditions for turbulence m

odelsodels

What is turbulence? Part IWhat is turbulence? Part I

What is turbulence? Part IIWhat is turbulence? Part II

Let us see a movie regarding a turbulent flow in a valve.

C:\Documents and Settings\???\My Documents\????????\90v-30-bd.AVI

What is turbulence? Part III – What is turbulence? Part III – Its natureIts nature

• RandomRandom

• Effective MixingEffective Mixing

• High Reynolds numberHigh Reynolds number

• 3-D3-D

• Energy DissipationEnergy Dissipation

• Eddy MotionsEddy Motions

What is turbulence? Energy What is turbulence? Energy CascadeCascade

Reynolds DecompositionReynolds Decomposition

Reynolds Averaged Navier-StoReynolds Averaged Navier-Stokes (RANS) equationskes (RANS) equations

is the so-called Reynolds stress.

Boussinesq’s Assumption Boussinesq’s Assumption

How to determine eddy viscosity t?

Zero equation model Zero equation model

t is assumed to be a constant and depends on various flow fields.

One equation model One equation model

Two equations model Two equations model

K-K- turbulence model turbulence model

K-K- turbulence model turbulence model

Boundary conditions Boundary conditions

Inlet Conditions

Boundary conditions for a Boundary conditions for a solid wallsolid wall

1. Wall function


1. Wall function


2. Two Layer Method


2. Two Layer Method

Example – Sudden Expansion Example – Sudden Expansion FlowFlow

ui0.1 m

0.13 m

1 m

2.5 m

Example – Sudden Expansion Example – Sudden Expansion Flow – establish meshFlow – establish mesh

Example – Sudden Expansion Example – Sudden Expansion Flow – Laminar Flow CaseFlow – Laminar Flow Case

• Working fluids – airWorking fluids – air• Density = 1.205 mDensity = 1.205 m33/s/s• Dynamics viscosity = 1.81e-5 kg/msDynamics viscosity = 1.81e-5 kg/ms• Characteristic length = 0.1 mCharacteristic length = 0.1 m• If we consider a laminar channel flow at If we consider a laminar channel flow at

Re = 100, then the magnitude of inlet velRe = 100, then the magnitude of inlet velocity must be 0.015 m/s. ocity must be 0.015 m/s.

Example – Sudden Expansion Example – Sudden Expansion Flow – Boundary setupFlow – Boundary setup

Inlet boundary

Symmetry boundary

Symmetry boundary

Outlet or constant pressure boundary

Example – Sudden Expansion Example – Sudden Expansion Flow – Results of laminar FlowFlow – Results of laminar Flow

Example – Sudden Expansion Example – Sudden Expansion Flow – Turbulent Flow CaseFlow – Turbulent Flow Case

• Working fluids – airWorking fluids – air• Density = 1.205 mDensity = 1.205 m33/s/s• Dynamics viscosity = 1.81e-5 kg/msDynamics viscosity = 1.81e-5 kg/ms• Characteristic length = 0.1 mCharacteristic length = 0.1 m• If we consider a turbulent channel flow at Re = If we consider a turbulent channel flow at Re =

30,000, then the magnitude of inlet velocity m30,000, then the magnitude of inlet velocity must be 4.5 m/s. ust be 4.5 m/s.

• kk and and at the inlet boundary ( at the inlet boundary (kk = 0.30375, = 0.30375, = 7. = 7.859).859).

Example – Sudden Expansion Example – Sudden Expansion Flow – Results of Turbulent Flow – Results of Turbulent FlowFlow

Contours of k

Simulation of Heat Transfer Simulation of Heat Transfer

• Forced convection or natural Forced convection or natural convection?convection?

• Boundary conditions, a. isothermal Boundary conditions, a. isothermal boundary, b. constant heat flux. boundary, b. constant heat flux.

• Conjugate heat transfer? Heat Conjugate heat transfer? Heat sources should be imposed inside sources should be imposed inside solids. solids.

Example – Forced convection Example – Forced convection with isothermal boundarywith isothermal boundary

ui0.1 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.








Isothermal contours

Example – Natural convection Example – Natural convection with isothermal boundarywith isothermal boundary

T = 294 K

T = 293 K

Adiabatic boundaryAdiabatic boundary

g

0.01 m

0.01 m


3LTTgRa LH

01.0

then

8.9

10216.2

10569.1

293

11

K 1

10 If

5

5

4

L

g

T

TT

Ra

L

LH


• Boussinesq’s approximation: assume the buoBoussinesq’s approximation: assume the buoyant force yant force ff in N-S equations is in N-S equations is

refTTgf





Isothermal contours

Example – Conjugate Heat Example – Conjugate Heat TransferTransfer

• Heat conduction in a solid and Heat conduction in a solid and convection in a fluid are considered convection in a fluid are considered in conjugate heat transfer. in conjugate heat transfer.

• At least, two materials shall be At least, two materials shall be defined as a fluid and a solid in the defined as a fluid and a solid in the model, respectively. model, respectively.


. .


T = 294 K

T = 293 K

Adiabatic boundaryAdiabatic boundary

g

0.01 m

0.01 m

air

Al

Example – Conjugate Heat TransferExample – Conjugate Heat Transfer

1

2

3


4. Choose a solid material from the table or creat a new one. Do not forget to click apply.


5. Use C> /NEW / Zone to select cells into cset.


6. Click Tools/Cell Tools to set Type 2 Solid to Material 2


7. Use cell list to change cells in cset to the type 2 solid


8. Check if there are two different kinds of cells. Red one is fluid 1. Green one is solid 2.


9. Go back to STAR Guide. Click Thermal Options. Click Heat Transfer ON.

10.The rest procedures for simulation of natural convection are as same as the previous example.


Iosthermal contours + Velocity vectors