Performing an Transient Conjugate Heat Transfer Simulation to check Grid dependency and Effects of Super-Cycling interval

Objective:

Performing Conjugate heat transfer Transient simulation for a Pipe with Outer diameter of 0.02 and inner diameter of 0.015 leaves with thickness of 0.005.Here we are performing grid dependency test and effect of supercycle state interval

Steps involved in running Transient simulation

1)Geomentry Creation

2)Boundary flagging

3)Diagnosing the geomentry

4)Case Setup

5)Solve

5)Post processing the results

CHT Analysis:

The analysis type Conjugate heat transfer (CHT) allows the simulation of the heat transfer between Solid and Fluid domains by exchanging thermal energy at the interfaces between them. It requires a multiregion mesh to have a clear definition of the interfaces in the computational domain.

Typical examples for conjugate heat transfer simulation are heat exchangers.electronic equipment and general-purpose cooling and heating systems.

Conjugate heat transfer corresponds with the combination of heat transfer in solids and heat transfer in fluids. In solids, conduction often dominates whereas in fluids, convection usually dominates. Efficiently combining heat transfer in fluids and solids is the key to designing effective coolers, heaters, or heat exchangers. Forced convection is the most common way to achieve high heat transfer rate. In some applications, the performances are further improved by combining convection with phase change (for example liquid water to vapor phase change).

Heat transfer in solids and heat transfer in fluids are combined in the majority of applications. This is because fluids flow around solids or between solid walls, and because solids are usually immersed in a fluid.

Y+

A non-dimensional wall distance for a wall-bounded flow can be defined in the following way:

Where is the friction velocity at the nearest wall, is the distance to the nearest wall and is the local Kinamatic viscosity of the fluid.

is often refered to simply as y plus and is commonly used in boundary layer theory and in defining the law of wall.it is used to understand the how coarse or fine grid is generated

so Y+ is used to know whenther it is required to use wall fuctions or not to understand more about it let us look into the below figure

If the flow is laminar then we can directly go with no slip condition or if the cell center lies in viscous sub region we caould easily resolve the boundary layer without the use of wall function but this is not the case in turbulent case

U+ is the non dimentional velocity it determines value of velocity at first cell

if you observe the picture more clearly we can see that U+=Y+ curve and logarithmic curve are perfectly matching in viscous sub region but it is getting seperated from buffer layer

***Dont get confused inertial sublayers and defect layers are nothing but turbulent region***

So the curve does nor hols ,we can overcome theis model by using wall function. the standard wall function is the one where the relation ship between u+ and Y+ is linear

u+=(1/k) * ln(Y+) + E

where K is Von Karman\'s constant and is generally 0.41

E is the law of wall constant

if Y+ is less than threshold of 10 then first node (In finite volume method cell center is node) lies in Viscous sub layer

if Y+ lies in between 10 - 30 then first node (In finite volume method cell center is node) lies in Buffer layer

if Y+ is greater than 30 then first node(In finite volume method cell center is node) lies in turbulent region

Step1

Geomentry

By using create-->Shape-->Cylinder we are creating a cylinder of Extenal diameter of 0.02m and internal diameter of 0.015 with 0.2m in lengt along z axis.After that boundary flagging is done by creating 4 boundaries namely Inlet ,Outlet,Solid Outer wall and solid side.

then Diagnosis is performed to check any error are present.If present necessary corrections are made.Then moved to case setup

Step2

Boundary Flagging

Solid Outer wall:To provide heat input

Solid Side:Thickness walls

Inlet:Fluid Enters

Outlet:Fluid Leaves

Interface: Where convection takes place such that heat transfer from solid wall to the fluid

Step3

Diagnosing the geomentry

We will just run the diagnosis for a set of important problems which are going to effect the solution

Step4

Case Setup

Application Type

=>In case setup we choose time based model

Materials

=>In materials tab enable gas and solid simulation with air and Alluminium as materials.Along with them in material tab species mechanism is also enabled.

=>Classification of species such as Aluminium as solid material and Air is as gas is done in species section.

Simulation Parameters

=>Going to simulation parameter we are going to choose Transient and hydrosimulation(The main difference between Hydro and non-hydro is that no navier stokes equtions are solved in non-hydro it is only used to check our mesh changes while running simulation).Then rest of then are set to default.We are going to unselect the Shared memory flag

Start time	0
End time	0.5 sec
Initial time	1.00E-07
Minimum time step	1.00E-07
Maximum time step	1

=>Coming to boundary there are 5 boundaries are flagged with different Boundary ID, they are

Boundary ID	Boundary Name	Boundary Type	Region
1	Solid Outer Wall	wall	Solid Region
2	Solid Sides	wall	Solid Region
3	Inlet	Inflow	Fluid Region
4	Outlet	outflow	Fluid Region
5	Interface	Interface	Fwd:Fluid Region,Rev: Solid Region

>solid outer wall:Velocity boundary condition slip,Temperature boundary condition-->Heat Flux--> -10000W/m2

>Solid sides:Velocity boundary condition slip,Temperature boundary condition-->Zero normal gradient

>inflow:Pressure boundary condition-->Zero normal gradient,velocity boundary condition-->As inlet is along z direction 7.32m/sec is entered calculation shown below.Temperature boundary condition-->300k.Species boundary condition-->Air

calculating velocity with 7000 renolyds number

Kinematic viscosity:1.56*10^-5 m2/sec

Diameter:0.015m

>Outlet:Pressure boundary condition-->101325,velocity boundary condition-->Zero normal gradient.Temperature boundary condition-->300k.Species boundary condition-->Air

Initial conditions

Regions ans initializations

=>Fluid and solid regions are created with strean ID\'s 0 and 1.As per the requirements the boundaries are set to to aluminium and air at 101325 pa and 300KHere were are performing turbulance simulation.

Physical Models

In turbulance model RANS RNG K-epsilon is choosen

Begin super-cycle data	0.05
Solid side heat trasfer solver	Steady-state solver
time length for each cycle stages	0.05
Totala number of stages	1
conduction CFL number	100
Sensible internal energy (SIE) tolerance	1.00E-07
SIE relaxation factor for super cycle	1.4

Output point is taken at outlet to measure the temperature

To perform Supercycle stage we enable super cycle in physical model and set the interval by taking 0.004 mesh as base size with varying intervals such as 0.01,0.02 and 0.03

Grid Control

=>Base grid dx=dy=dz=0.002,0.003,0.004.Then postprocessing output files are set to defaults because the data which we are looking is already in default selections

Output/Post processing files

Post variable selaet is set to default because all the imformation we needed is already their by default.

Time interval for writing 3D output data file:- 0.01s.

Time interval for writing text output:- 1e-6s.

Time interval for writing restarting output:- 0.01s

Step5

Solve

Then we need to export the input files to a folder to start running a simulation.then by using CYgwin we are directing to the input file and simulation is stared by using an executable file and simulation ran by using two processor. After successfully running the simulation we can post process the results and check the general outcomes whether simulation ran properly and are they making any sense

Step6

Post Processing Results

PartI

Case I

0.002m Mesh Size

This figure helps us to understand mesh is generated perfectly for the grid size of 0.002m

Temperature Contour

First looking at temperature scale the maximum temperature is 640K and the lowest temoerature recorded is 300.When the fuel passes along the cyclinder the fluid at the walls are getting heated up

Y+

So here this contour helps to understant that from the nearest wall is nothind but inner wall Y+ is changing along the length because of boundary layer formation we can say that in the first quater length of the pipe we can see the cell center is lying in the buffer region and after that cell center is almost lying in the viscous boundary layer.So we need to use law of wall function because max Y+ value crosses 11 and it is almost all the cells are lying in sub viscous layer and it is recommended to use K-Epsilon Model unless Y+ is less that 10 we are not going to use K-w model(K-w model give good results when Y+ is less than 10)

Case II

0.003m Mesh Size

This figure helps us to understand mesh is generated perfectly for the grid size of 0.003m

Temperature Contour

First looking at temperature scale the maximum temperature is 670K and the lowest temoerature recorded is 300.When the fuel passes along the cyclinder the fluid at the walls are getting heated up

Y+

In order to capture the information near the wall we need to either fine the gird or use the wall function that can be choosen based on Y value. Here the mesh is some what coarse compare to the first case and from the contour we can observa that almost all the region is having Y+ value greater than 25 and the maximum Y+ value is 32. So inorder to capture the results at the boundary more accurately it is better to use the wall function in this case and it is recommended to use K-Epsilon Model unless Y+ is less that 10 we are not going to use K-w model(K-w model give good results when Y+ is less than 10)

Case III

0.004m Mesh Size(Base Grid)

This figure helps us to understand mesh is generated perfectly for the grid size of 0.004m

Temperature Contour

First looking at temperature scale the maximum temperature is 690K and the lowest temoerature recorded is 300.When the fuel passes along the cyclinder the fluid at the walls are getting heated up

Y+

Like i explained in the case 2 we can see the same senario here the mesh is more coarse than the case 2.Here the max of Y+ is 31. So few cell centers lies in the turbulent region and most of the cells lies in buffer region so it is better to use wall function and it is recommended to use K-Epsilon Model unless Y+ is less that 10 we are not going to use K-w model(K-w model give good results when Y+ is less than 10)

Grid Dependent Test

Here are we are looking at mean temperture pattern for a span of 0.5 sec accross different Grid sizes.Mesh is very important factor for attaining maximun accuracy in solution here looking into graph we can say while thhe mesh is geeting finer the more acucrate temperature are obtained and finer mesh is taking more time to converge this is due to more cells to calculate greater the cells greater the time is taken to calculate different attributes

Simply by looking into above graph we can say that finer the grid more the time taken to converge (Due to fining the mesh more elements are generated it takes more time to calculate the elements)but the value accuracy improve as the mesh is getting more finer. At mesh 0.002m the temperature value of solid converged to 640,at mesh 0.003m size the temperature value of solid converged to 660,at mesh 0.004 the temperature value of solid converged to 690.As So theie is a huge deflection in the temperature value. So it is better to have a fine mesh.we can observe the graphs are straight. that is happed because solid takes more time to get heatup than fluid.In order to decrease computational time we are enabling supercycle

In this Above figure Mean temperature for three different mesh sizes are plotted as we can the temperature is increase in all the three plots after 0.1 seconds.In this case we can see that mesh size with 0.002m got converged quickly than 0.003m.their could be multiple reasons.At mesh 0.002m the temperature value of solid converged to 323.2,at mesh 0.003m size the temperature value of solid converged to 323K,at mesh 0.004 the temperature value of solid converged to 325.5K

Part II

Super Cycle Modeling

There are several challenges when solving the fluid and solid domains together. The largest challenge to overcome is the difference in time-scales associated with the fluid and solid. In general, the heat transfer in the solid occurs more slowly than the convective and diffusive time-scales dictated by the fluid flow.In super-cycling, the fluid solver is frozen periodically while the heat transfer in the solid is allowed to progress to steady-state. CONVERGE solves the fluid and solid phases in the same simulation without requiring you to stop and restart the simulation. Super-cycling significantly reduces the computational cost of the CHT simulation

Effect of Super-cycle modeling

Here we are inducted Supercyclic as a new physical model because when we run a conjugate heat transfer problem it helps us to reduce time taken to run the simulation by converging the solution much quicker than the traditional Method.

Here we are checking out how Supercyle stage interval effecting the simulation. For that to be done we are taking different intervals such as 0.01,0.02 and 0.03

At 0.03 time interval

At 0.02 time interval

At 0.01 time interval

Time interval	Total Simulation Time
0.01	385.46
0.02	435.99
0.03	558.59

As we reduce the time intereval we can see that the solution is getting much quicker for the lowest time interval.So we can conclude that as the interval time increases the solution takes more time to converge.

Link for the simulation video

https://youtu.be/zG4GYffMyTw

https://youtu.be/WqzwolKkF58

https://youtu.be/HQVycbdZoVI

Conclusion:

• As the mesh is getting refined accuracy of the solution is getting improved.
• If Mesh grid size is increases simulataneously Y+ also increases
• If Super-cycling time increases the time taken to run the simulation also increases