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Old   April 17, 2000, 07:01
Default Boussinesq approximation again
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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.
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Old   April 18, 2000, 04:28
Default Re: Boussinesq approximation again
Sergei Chernyshenko
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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


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
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Old   April 20, 2000, 03:55
Default 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.
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Old   May 11, 2000, 09:24
Default Re: Boussinesq approximation again
Sergei Chernyshenko
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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


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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