Conjugate heat transfer using super-cycling method by Converge

In this project conjugate heat transfer problem in a cylindrical pipe is simulated using super-cycling method of Converge software.

The dimensions of the cylinder used for the simulations are given below:

Inner diameter: 0.015m

Outer diameter: 0.02m

Length: 0.2m

The image of the same cylinder is attached for reference:

Super cycling method of simulation is utilised here. Generally when there is a heat transfer problem that includes both conduction and convection/diffusion, the convection/diffusion takes place faster compared to conduction which results in large simulation time and very high number of cycles which can make these simulations expensive, to avoid these problem Converge provides a method called Super cycling. In super cycling method the fluid solver working on convection/diffusion problem is paused for a small time period allowing the solid state conduction solver to catch up, this method significantly reduces the number of cycles and simulation time. Here, transient solver is used for fluid domain simulation and steady state solver is utilised for the solid domain. The solid material used here is aluminium and the fluid material used in this project is air. The turbulence model used is the k-epsilon model. The simulation is carried out for three mesh sizes for performing grid dependence test for these three meshes simulations were conducted for the super cycling interval of 0.05s. The mesh sizes used are 0.004m, 0.003m and 0.002m. The simulations are run for the time period of 0.5s.

Image showing solid and fluid domain:

Idreg 1 is solid aluminium and idreg 0 is the fluid air.

The boundary conditions are taken as follows:

For inlet : A velocity of 7.3176950535m/s is given as input which corresponds to a Reynold's number of 7000. The velocity is calculated using the formula Re = `(rhovL)/mu` where the density is found to be 1.1766 kg/m3 and dynamic viscosity is found to be 18.45 x 10^-6 Pa.s in air at 300k. The characteristic length for the internal flow in a pipe is its diameter, so here the value of L is 0.015m which is the inner diameter of the pipe and inlet temperature is given as 300k.

For outlet : A pressure of 101325 Pa is given with a backflow temperature of 300k.

For the solid outer wall: Taken as a stationary wall with a heat flux of -10000 W/m2 given to heat up the pipe ('-' indicates the heat flowing into the pipe), therby also heating up the air flowing inside.

For the solid side walls: Designated as a stationary wall and is given with slip condition.

For interface: Again interface is also designated as stationary wall and the forward boundary is taken for fluid region and reverse boundary is taken for solid region. 

Mesh size 0.004m: 

Results obtained from 0.004m size mesh:

Temperature contour:

Velocity contour:

Y+ value:

Mesh size 0.003m:

Results obtained for 0.003m size mesh:

Temperature contour:

Velocity contour:

Y+ value:

Mesh size 0.002m:

Results obtained for 0.002m size mesh:

Temperature contour:

Velocity contour:

Y+ value:

Average temperature plot for various mesh sizes:

Total cells for the various mesh sizes:

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 350k and 359k for different mesh sizes. From the results obtained we find that as the mesh size becomes finer the y+ keeps decreasing because the y+ values obtained here are at transition region (i.e y+ value between 10 and 30), making the mesh coarser will push it closer to turbulent region i.e to y+ greater than 30. 

To test the effect of the super cycling interval in the total simulation time, out of the three mesh sizes the mesh with mesh size 0.003m is selected and in that mesh size simulations were conducted for the super cycle intervals (time length for each cycle stage) of the 0.05s, 0.03s, 0.02s an 0.01s. The results are attached below:

The temperature taken at the monitor point ( -1.631826E-002m, 6.321729E-003m, 2.000000E-001m) for various values of super cycling interval is:

The time taken for the simulation for the various values of super cycling time interval are as follow:

For the 0.05s super cycling time interval the total simulation time is 591.351104 seconds.

For the 0.03s super cycling time interval the total simulation time is 593.367845 seconds.

For the 0.02s super cycling time interval the total simulation time is 597.196534 seconds.

For the 0.01s super cycling time interval the total simulation time is 605.350098 seconds.

As we see here as the super cycling time interval decreases less time is avilable for the solid steady state solver to catch up resulting in the increase of the total simulation time.

Conjugate heat transfer temperature contour animation: