Conjugate Heat Transfer Simulation

The main purpose of this project is to stimulate the Conjugate Heat Transfer in a pipe using Converge Studio. For this, an airflow is simulated in an Aluminium pipe which receives a constant Heat supply of 10000W/m^2 at the outside wall of the pipe. The main objective is to run the grid dependency test to analyze the results for each test and to analyze the simulation with different supercycle inputs.

Case Setup:

• The Air is used as a fluid and the Aluminimum is used as a solid material in the simulation.
• Transient simulation is used with an initial time step of 1e-7 sec and the run time is set for 0-0.5sec.
• A Heat flux condition is applied to the outer wall of 10000W/m^2.
• Zero normal gradient condition is used for the inlet and outlet walls.
• The velocity at the inlet is set to 3.64m/s to satisfy the condition of Reynolds number=70000.

The density of air at 300k `rho` = 1.184 kg/m^3

Dynamic Viscosity of air `mu` = 1.849*10^-5 kg/(m*s)

Reynolds number Re = (rho*V*D)/(Dynamic viscosity)

Velocity V = 3.64m/s

• The NE boundary condition is given at the Outlet.
• The Interface boundary condition is given for the inside wall of the pipe.
• Two separate regions are created for solid and fluid regions and every boundary is assigned to the specific region to which they belong.
• RNG K-epsilon model is used in turbulence modeling.
• The supercycle time is given as 0.05sec.

Y+:

y+ is a non-dimensional length scale that is normalized using wall shear stress. It is a result of dimensional analysis to provide self-similar solutions to the turbulent boundary layer profile. The turbulent boundary layer consists of 3 layers(viscous sublayer, log layer region, and the buffer region) wherein different relations are observed between the average velocity and this non-dimensional distance y+. y+ helps somewhat identify where these regions lie in the flow domain. It also helps in modeling the wall layer. In order to fully resolve the turbulent boundary layer, the numerical grid wall y+ must be close to 1.

• The inner layer, or sublayer, where viscous shear dominates
• The outer layer, or defect layer, where large scale turbulent eddy shear dominates
• The overlap layer, or log layer, where velocity profiles exhibit a logarithmic variation

A still clear image of the above is shown below:

Source: Applied Computational Fluid Dynamics by André Bakker

where is the free stream velocity and is the boundary layer thickness, is the vertical distance measured from the wall.

It is observed that if we are in the viscous wall region with y+ < 50, there is a direct effect of the viscosity on the shear stress. Conversely in the outer layer with y+ > 50, the effect of viscosity is negligible.

For the sake of completeness, let’s briefly introduce the friction velocity uτ. It should be evident that the viscosity ν and wall shear stress τw are important parameters. From these quantities and ρ, we define the viscous scales being appropriate viscous scales and length scales in the near-wall region.

with

The dimensionless velocity is given by:

Wall Regions and Layers as well as their Properties

1. The viscous sublayer (y+ < 5)
   In the viscous layer, the fluid is dominated by the viscous effect, so it can be assumed that the Reynolds shear stress is negligible. The “linear velocity law” is given by:

1. The logarithmic area (y+ > 30)
   In the logarithmic layer, turbulence stress dominates the flow and velocity profile varies very
   slowly with a logarithmic function along the distance y. The formula describes this region with the Karman constant κ of 0.41 and the constant B=5.2.

1. The Buffer layer (5 < y+ < 30)
   The buffer layer is the transition region between the viscosity-dominated region and the turbulence-dominated part of the flow. Viscous and turbulent stresses are of similar magnitude and since it is complex, the velocity profile is not well defined and the original wall functions avoid the first cell center located in this region.

The Grid dependency test is done for three grid sizes.

Case:1 grid size = 0.004

The above figure shows the meshing for case 1

The velocity magnitude contour for case 1

The Temperature contour for case 1

The y+ contour for case 1

Case 2: grid size = 0.003

The above figure shows the meshing for case 2

The velocity magnitude contour for case 2

The Temperature contour for case 2

The y+ contour for case 2

Case 3: grid size = 0.002

The above figure shows the meshing for case 3

The velocity magnitude contour for case 3

The Temperature contour for case 3

The y+ contour for case 3

For solving the simulation we are using the RNG k-epsilon turbulent model and this model works fine when the y+ ranges in between 10 to 300. If the y+<10 then the k-omega turbulent model yields the best results and the solution gets more accurate at the low y+ values. On considering the range of k-epsilon the CASE-2(grid size=0.003) yields good results because as per the range the y+ value is more than 10 and it is lower than the y+ value of case 1(grid size = 0.004) and regarding the Y+ value of case-3(grid size = 0.002) it is the minimum that is 8.2 but it doesn\'t fall under the range of k-epsilon model so case-2 is chosen as more accurate grid size for this model.

Effect of Supercycle Stage Interval:

In the grid dependency calculations, the supercycle stage interval was set to 0.05s which means the sold portion is treated as a steady-state simulation for every 0.05s. in this part, the simulation is run by changing this interval to 0.03s,0.02s,0.01s.

The above figures show the mean temperature of the solid and fluid regions at different time intervals.

On observing the above two figures it is clearly seen that the end values remain the same for every supercycle interval. As we are using the combination of both steady-state and Transient solver for different time interval we can\'t take the results at the intermediate time steps because it varies for every time interval we can only consider the end results when supercycle is used the main advantage if this is the processing time.

For 0.05s interval

For 0.03s interval

For 0.02s interval

For 0.01s interval

When we compare processing time for all the intervals it shows that 0.03s will take less time for completing the simulation as the time interval decreases the simulation time increases. If we consider the 0.05s interval its run simulation is more than 0.03s this is because for every simulation there will be an optimized time interval at which it takes less time to complete the simulation without losing accuracy If the time interval is more or less than that time step it takes more time for completing the simulation. This optimized time step can be only known by the trial and error method.