CFD Online Discussion Forums

CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
-   FLUENT (http://www.cfd-online.com/Forums/fluent/)
-   -   Natural convection with k-e turbulence model (http://www.cfd-online.com/Forums/fluent/68283-natural-convection-k-e-turbulence-model.html)

hello September 13, 2009 14:23

Natural convection with k-e turbulence model
 
Hi,
I am currently simulating natural convection in a square cavity with standard k-e turbulence model with enhanced wall treatment for Rayleigh number of 10^11. The setup is a linear temperature gradient applied at the horizontal bottom wall with other walls insulated.
The setup is two-dimensional unsteady flow and working fluid is water. I have also did some mesh refinement along the walls however the solutions I obtained does not converge.
Help please :(

srjp September 15, 2009 00:32

Did you make sure the following?
1. Gravity was on.
2. Temperature dependent density was defined for the material
3. Used laminar model (are you sure the flow is turbulent for your condition?)
4. Given small enough time step

From Fluent Help:
Guidelines for Solving High-Rayleigh-Number Flows

When you are solving a high-Rayleigh-number flow (Ra > 10^8) you should follow one of the procedures outlined below for best results.
The first procedure uses a steady-state approach:
1. Start the solution with a lower value of Rayleigh number (e.g., [IMG]file:///C:/Fluent.Inc/fluent6.3.26/help/html/ug/img5396.gif[/IMG]) and run it to convergence using the first-order scheme.
2. To change the effective Rayleigh number, change the value of gravitational acceleration (e.g., from 9.8 to 0.098 to reduce the Rayleigh number by two orders of magnitude).

3. Use the resulting data file as an initial guess for the higher Rayleigh number and start the higher-Rayleigh-number solution using the first-order scheme.
4. After you obtain a solution with the first-order scheme you may continue the calculation with a higher-order scheme.
The second procedure uses a time-dependent approach to obtain a steady-state solution [ 139]:
1. Start the solution from a steady-state solution obtained for the same or a lower Rayleigh number.
2. Estimate the time constant as [ 32]

[IMG]file:///C:/Fluent.Inc/fluent6.3.26/help/html/ug/img5397.gif[/IMG](13.2-21)


where [IMG]file:///C:/Fluent.Inc/fluent6.3.26/help/html/ug/img5398.gif[/IMG] and [IMG]file:///C:/Fluent.Inc/fluent6.3.26/help/html/ug/img5399.gif[/IMG] are the length and velocity scales, respectively. Use a time step [IMG]file:///C:/Fluent.Inc/fluent6.3.26/help/html/ug/img5400.gif[/IMG] such that

[IMG]file:///C:/Fluent.Inc/fluent6.3.26/help/html/ug/img5401.gif[/IMG](13.2-22)


Using a larger time step [IMG]file:///C:/Fluent.Inc/fluent6.3.26/help/html/ug/img5402.gif[/IMG] may lead to divergence.
3. After oscillations with a typical frequency of [IMG]file:///C:/Fluent.Inc/fluent6.3.26/help/html/ug/img5403.gif[/IMG] 0.05-0.09 have decayed, the solution reaches steady state. Note that [IMG]file:///C:/Fluent.Inc/fluent6.3.26/help/html/ug/img5404.gif[/IMG] is the time constant estimated in Equation 13.2-21 and [IMG]file:///C:/Fluent.Inc/fluent6.3.26/help/html/ug/img5405.gif[/IMG] is the oscillation frequency in Hz. In general this solution process may take as many as 5000 time steps to reach steady state.

enayath May 23, 2014 11:53

srjp,

Could you please upload the figures again....I can not see any pictures...
Also, would you please name the reference 32 for the estimation of time step?

Thank you


All times are GMT -4. The time now is 17:41.