Week 6: Conjugate Heat Transfer Simulation

Objective:-

To simulate the Conjugate Heat Transfer over pipe using Super- Cycling in CONVERGE and determine Y+

Theory:-

Conjugate Heat Transfer:-

• Conjuate Heat Transfer correspond with the combination of heat transfer in solid and heat transfer in fluid.
• In solid, conduction often domain whereas in fluid convection usually domain.
• Conjugate heat transfer is observed in many situation. For egs, heat sinks are optimized to combined heat transfer by conduction in the heat sink with the convection in the surrounding field.

Super-Cycling Method:-

• In Super-Cycling method, the fluid solver working on convection problem is paused for a small time period allowing the solid state conduction solver catch up, this method significantly reduce the number of cycle and simulation time.

Output Y + :-

• Wall function are useful in telling the solver how to approch the solution near a wall. These used to predict near wall result accurately even at coarse grid.
• The Y + value is a non-dimensional term that can be used to understand how coarse grid are fine, this can then be used to determine if wall function need to be used or not.

For Calculation of Velocity:-

Reynold Number can be given as :-

`Re = frac{rho*v*d}{mu}`

where,

Re - Reynold's Numvber - 7000

`rho` - density of air - 1.225 `frac{kg}{m^3}`

v - velocity - to be calculate

d - diameter - 0.015 m

`mu` - 18.45 * `10^(-6)` Pa.s (dynamic viscosity)

Therefore, Velocity can be calculated as

`v = frac{Re*mu}{rho.d}`

`v = frac{7000*18.45*10^(-6)}{1.225 * 0.015}`

v = 7.028 m/s

Geometry:-

Geometry

Outer Diameter = 0.02 m

Inner Diameter = 0.015 m

Length = 0.2 m

Case Setup:-

Application type:-

1. Time based

Material:-

1. Gas simulation - Air
2. Solid simulation - Aluminium
3. Species - on
4. Reaction Mechanism - off

Global Transport Parameter:-

1. Turbulent Prandtl Number - 0.9
2. Turbulent Schemidt Number - 0.78

Simulation Parameter:-

1. Solver - Transient State Solver

Simulation time parameter:-

1. Start time - 0.0
2. End time - 0.5
3. Initial time step - 1e-7
4. Minimum time step - 1e-7
5. Maximum time step - 1

Region and Initialization:-

1. Fluid Region:-

1. Stream ID - 0
2. Solid - OFF
3. Temperature - 300 K
4. Pressure - 101325 Pa
5. Turbulent Kinetic Energy - 1 `m^2/s^2`
6. Turbulent Dissipation - 100 `m^2/s^3`
7. Species - Air - O2 - 0.23
8. Species - Air - N2 - 0.77

2. Solid Region:-

1. Stream ID - 1
2. Solid - ON
3. Temperature - 300 K
4. Pressure - 101325 Pa
5. Turbulent Kinetic Energy - 1 `m^2/s^2`
6. Turbulent Dissipation - 100 `m^2/s^3`
7. Species - Aluminium - 1

Boundary Condition:-

1. Solid wall - Wall - Stationary - Slip - Heat Flux - (-10000 W/m2) (Solid Region)
2. Solid side - Wall - ZNE (Solid Region)
3. Inlet - Inflow - Pressure (ZNE) (Fluid Region)
4. Inlet - Inflow - Velocity (DI) - z - 7.028 m/s - Species - Air (Fluid Region)
5. Outlet - Outlflow - Pressure (DI) - 101325 Pa (Fluid Region)
6. Interface - Interface - forward - Fluid region - Law of wall
7. Interface - Interface - Reverse - Solid region - Slip - Aluminium

Physical Model:-

1. Turbulance Model - k epsilion
2. Super Cycle Model - Super cycle Interval - 0.05
3. Super Cycle Model - CFL - 100
4. Super Cycle Model - Relaxation - 1.4

CASE 1:-

For Mesh size 0.004 m

MESH:-

TEMPERATURE CONTOUR:-

VELOCITY CONTOUR:-

Y PLUS :-

CASE 2:-

For Mesh Size 0.003 m

MESH:-

TEMPERATURE CONTOUR:-

VELOCITY CONTOUR:-

Y PLUS:-

CASE 3:-

For Mesh size 0.002 m

MESH:-

TEMPERATURE CONTOUR:-

VELOCITY CONTOUR:-

Y PLUS:-

PLOT:-

Average variation of temperature at Outlet:-

Cell Count:-

• From the above results, we can find that as the air flows through the pipe it gets heated up gradually until at the outlet the temperature ranges from 380 and 383 K for different mesh sizes.
• From the Y plus result, we can find as the mesh becomes more refined the value of y plus decrease thus we obtained a transition region in each of the case (For transition region the value of Y + should be in between 10 and 30 ) and whereas value for y + greater than 30 then it is turbulent region

SUPERCYLING EFFECT:-

To test the effect of the supercycling interval in the total simulation the mesh size of 0.004 m is selected and for supercycle interval of the 0.03 s, 0.02s and 0.01s.

The point at which supercycle is maintain is (-0.012763, 0.07896, 0.18)

The time taken for the simulation for various value of supercycling time interval are as follows:-

1. For the 0.03s super cycling time interval the total simulation is 843.196534 seconds.
2. For the 0.02s super cycling time interval the total simulation is 856.350098 seconds.
3. For the 0.01s super cycling time interval the total simulation is 896.367845 seconds.

As the super cycling time interval decrease less time is avaliable for the solid steady solver for calculating the result, thus it will increase the total simulation time.

ANIMATION:-

TEMPERATURE CONTOUR:-

VELOCITY CONTOUR:-