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Boundary conditions for incompressible (total) energy equation

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Old   December 25, 2015, 04:37
Default Boundary conditions for incompressible (total) energy equation
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Tom-Robin Teschner
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For an incompressible medium, we have the energy equation as follows:

\frac{\partial E}{\partial t} + (u\cdot\nabla)E = -\frac{1}{\rho}u\cdot\nablap + \frac{1}{\rho}u\cdot(\nabla\cdot\tau)

My question is now, how do I define boundary conditions for certain boundary types, i.e. inlet, outlet and walls (lets assume a channel flow). More specifically (I presume I can use Neumann-type boundary conditions for the outlet and walls), do I have to prescribe the energy level at the inlet as a Dirichlet type? Since the total energy is the sum of potential, kinetic and internal energy, I can't see how that should be zero but I couldn't find a way to impose non-zero Dirichlet boundary values for the inlet.
Unfortunately I couldn't find anything related to that here in the forum so I hope someone will know the answer here.
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Old   December 25, 2015, 07:20
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Originally Posted by t.teschner View Post
For an incompressible medium, we have the energy equation as follows:

\frac{\partial E}{\partial t} + (u\cdot\nabla)E = -\frac{1}{\rho}u\cdot\nablap + \frac{1}{\rho}u\cdot(\nabla\cdot\tau)

My question is now, how do I define boundary conditions for certain boundary types, i.e. inlet, outlet and walls (lets assume a channel flow). More specifically (I presume I can use Neumann-type boundary conditions for the outlet and walls), do I have to prescribe the energy level at the inlet as a Dirichlet type? Since the total energy is the sum of potential, kinetic and internal energy, I can't see how that should be zero but I couldn't find a way to impose non-zero Dirichlet boundary values for the inlet.
Unfortunately I couldn't find anything related to that here in the forum so I hope someone will know the answer here.

first, for incompressible flows you have often a decoupled energy equation expressed in terms of the temperature balance....

however, E is sum of kinetic and internal energy, the BC.s for E are consequent to the setting of BC.s for velocity and temperature
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Old   December 25, 2015, 07:37
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ok, what I want to know (did not mention that), can I compute the total energy equation independent from the temperature equation? And if so, can I (at the inlet), prescribe the total energy as the sum of kinetic and internal energy (based on inlet velocity and temperature) as a Dirichlet boundary condition, dropping the potential energy contribution?
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Old   December 25, 2015, 07:49
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Originally Posted by t.teschner View Post
ok, what I want to know (did not mention that), can I compute the total energy equation independent from the temperature equation? And if so, can I (at the inlet), prescribe the total energy as the sum of kinetic and internal energy (based on inlet velocity and temperature) as a Dirichlet boundary condition, dropping the potential energy contribution?

as you see from the equation, if you want to treat as a decoupled equation you need to know a-priori the velocity field and pressure gradient (steady or not). That means you already know the kinetic energy which is a part of the total energy. Therefore, the ony actual unknown field is the internal energy (the temperature)
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Old   December 25, 2015, 18:59
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and what is about the potential energy which also contributes to the total energy (negleted because contribution small?)?

just out of interest then, can I solve the temperature and total energy equation if I already have obtained velocity and pressure beforehand? I mean, the energy equation is the general conservation law and the temperature equation is derived from it. So can I use both or just one equation?
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Old   December 26, 2015, 04:16
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the potential energy is very often disregarded in many problems ...

My opinion is that for incompressible flows you need just to compute the temperature equation, then you sum the resulting internal energy to the (known) kinetic energy and get the total energy.

I think it could be interesting computing the totale energy equation only for some special coupled cases (velocity field depending also on internal energy)
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Old   December 26, 2015, 08:28
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ok, i think that has clarified the topic for me, grazie mille
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Old   December 26, 2015, 08:33
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ok, i think that has clarified the topic for me, grazie mille
di nulla
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