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September 13, 2009, 14:23 
Natural convection with ke turbulence model

#1 
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Hi,
I am currently simulating natural convection in a square cavity with standard ke 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 twodimensional 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 

September 15, 2009, 00:32 

#2 
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JP
Join Date: Mar 2009
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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 HighRayleighNumber Flows When you are solving a highRayleighnumber flow (Ra > 10^8) you should follow one of the procedures outlined below for best results. The first procedure uses a steadystate 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 firstorder 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 higherRayleighnumber solution using the firstorder scheme. 4. After you obtain a solution with the firstorder scheme you may continue the calculation with a higherorder scheme. The second procedure uses a timedependent approach to obtain a steadystate solution [ 139]: 1. Start the solution from a steadystate 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](13.221) 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](13.222) 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] 0.050.09 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 13.221 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. 

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