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Old   May 7, 2012, 11:37
Question Granular pressure
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Qingang Xiong
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Dear members,
I am incorporating KTGF for sovling the alphaEqn.H in twoPhaseEulerFoam. In this sovler, is the "ppMagf" to be differentiated alpha to particle granular temperature? If so, I think it is so complex. Looking forward to your reply. Thank you!
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Old   May 8, 2012, 02:24
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Alberto Passalacqua
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Quote:
Originally Posted by Qingang Xiong View Post
Dear members,
I am incorporating KTGF for sovling the alphaEqn.H in twoPhaseEulerFoam. In this sovler, is the "ppMagf" to be differentiated alpha to particle granular temperature? If so, I think it is so complex. Looking forward to your reply. Thank you!
The ppMagf term is related to the normal stress modulus of the phase G(\alpha_s). You should check the definition of this term in the relevant literature, to have some background on the model. If you consider a granular phase, with equation of state:

p_s = p_s(\alpha_s, \Theta_s),

the normal stress modulus is

G(\alpha_s, \Theta_s) = \frac{\partial p_s}{\partial \alpha_s}

In twoPhaseEulerFoam ppMagf = G(\alpha_s)/(\alpha_s \rho_s), due to the treatment used in discretizing the phase momentum equation.

Best,
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Last edited by alberto; May 8, 2012 at 12:50. Reason: Re-wrote equations in LaTeX
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Old   May 8, 2012, 11:49
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Quote:
Originally Posted by alberto View Post
The ppMagf term is related to the normal stress modulus of the phase G(alpha). You should check the definition of this term in the relevant literature, to have some background on the model. If you consider a granular phase, with equation of state:

p_s = p_s(alpha_s, Theta_s),

the normal stress modulus is (d indicates partial derivative)

G(alpha_s, Theta_s) = d(p_s)/d(alpha_s)

In twoPhaseEulerFoam ppMagf = G(alpha_s)/(alpha_s*rho_s), due to the treatment used in discretizing the phase momentum equation.

Best,
Dear Alberto,
Thank you very much for your reply. So from your description, Theta_s can be regarded as an independent variable to alpha_s though Theta_s is implicitly obtained from alpha_s. Am I right?
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Old   May 8, 2012, 12:29
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Alberto Passalacqua
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Quote:
Originally Posted by Qingang Xiong View Post
Dear Alberto,
Thank you very much for your reply. So from your description, \Theta_s can be regarded as an independent variable to \alpha_s though \Theta_s is implicitly obtained from \alpha_s. Am I right?
You are, in short, treating the particle phase as a compressible fluid. All you need to compute it 1/\psi, being \psi the compressibility of the particle phase, which is defined as

\psi = \frac{\partial \alpha_s}{\partial p_s}

This is analogous to the case of an ideal gas, where you have

\rho = \frac{p}{RT}

so

\psi = \frac{\partial \rho}{\partial p} = \frac{1}{RT}

The difference is that, in a granular flow, generally \rho_s = const, and the compressibility is due to changes in the phase fraction \alpha_s.

Best,
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Last edited by alberto; May 8, 2012 at 12:46. Reason: Corrected and re-wrote equations with LaTeX
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Old   May 8, 2012, 12:49
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Quote:
Originally Posted by alberto View Post
You are, in short, treating the particle phase as a compressible fluid. All you need to compute it 1/\psi, being \psi the compressibility of the particle phase, which is defined as

\psi = \frac{\partial \alpha_s}{\partial p_s}

This is analogous to the case of an ideal gas, where you have

\rho = \frac{p}{RT}

so

\psi = \frac{\partial \rho}{\partial p} = \frac{1}{RT}

The difference is that, in a granular flow, generally \rho_s = const, and the compressibility is due to changes in the phase fraction \alpha_s.

Best,
Dear Alberto,
I understand what you mean. This point is clear to me. Thank you veru much!

Best
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