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April 17, 2000, 07:01 
Boussinesq approximation again

#1 
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Hi everybody, I have still a problem concerning the Boussinesq approximation about the density variation connected with the temperature, in the incompressible flows. I know that it's better to use this approximation with little differences of temperature. My question is: if I use this approximation with high differences of temperature, how much is my error, how can I valutate it ? Do you know other approximations for incompressible flows that point their attention to the buoyancy forces? Thank you, Gabriel
P.S. I hope I've been clear enough, but I'm a little bit confused yet. 

April 18, 2000, 04:28 
Re: Boussinesq approximation again

#2 
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Hi, Gabriel.
OK, I'll explain. An error arises due to neglecting certain terms in equations. Consider, for simplicity, 2D steady continuity equation d(\rho u)/dx + d(\rho v)/dy=0 It can be rewritten as (ud\rho/dx + vd\rho/dy)/\rho + du/dx + dv/dy=0 In Boussinesq approximation it becomes du/dx + dv/dy=0 This is a good approximation if (ud\rho/dx + vd\rho/dy)/\rho << du/dx + dv/dy Suppose the characteristic velocity scale is U and length scale is L Then du/dx ~ dv/dy ~ U/L d\rho/dx ~ d\rho/dy ~ delta \rho / L. Substitute and obtain (do it!) that the approximation is good if delta \rho / \rho << 1 This quantity, delta \rho / \rho, is a measure of the relative error. Take a handbook on fluid properties and find what is delta \rho for given temperature variations, divide by \rho and this is it. For gases it does not matter if you use adiabatic formulae or constant pressure formulae since this is a crude estimate only anyway. If the error turns out to be too large you have to use the full equations I am afraid. Rgds, Sergei 

April 20, 2000, 03:55 
Re: Boussinesq approximation again

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Thank you for your explanation. It's very precious. Just a thing: when I make the calculus (delta rho)/rho to measure the relative error, the rho I have to consider is the rho at the lowest temperature in the ambient I'm studing, is that correct ?
Yours, Gabriel. 

May 11, 2000, 09:24 
Re: Boussinesq approximation again

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Hi, Gabriel,
Sorry for not answering earlier. I was a bit busy Let delta\rho = \rho_high  \rho_low > 0 Then, say, error1 = delta\rho / \rho_high=1 \rho_low/\rho_high, error2 = delta\rho / \rho_low =1 \rho_high/\rho_low, Therefore, as soon as error1 << 1 or error2 << 1, \rho_low is quite close to \rho_high Then error1/error2 = \rho_low / \rho_high is quite close to 1. Hence, the answer to your question is: it does not matter. May be it is easier to understand if you notice that delta\rho/\rho is not the exact value of the relative error. It is a crude estimate only. The real error can be several times greater, for example. Yours Sergei 

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