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Can pressure do work in an incompressible flow? |
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January 15, 2018, 12:48 |
Can pressure do work in an incompressible flow?
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#1 |
New Member
Join Date: Dec 2015
Posts: 27
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I am trying to analyze the energy equation for incompressible flows. I start with N-S and dot it with the velocity vector:
Then I get something like I am interested in the work due to pressure term. Can this be nonzero if the flow is divergence free? Using identities it can be rewritten as Obviously, the first term on the RHS is 0 if the flow is divergence free, but what is the second term? I am having trouble visualizing this one physically. When is it or isn't it equal to 0? I am having trouble understanding how pressure could do work on an incompressible flow. Any elucidation or thoughts would be very appreciated. |
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January 15, 2018, 13:01 |
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#2 |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,768
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I would start directly from the total energy equation . E= (K+U):
d/dt Int[V] (rho*E) dV + Int[S] n.v*(rho*E) dS = Int[S] n.(v.T) dS- Int[S]n.q dS T=-pI + Tau q=-k*Grad T The RHS has two terms, the first one is the total mechanical work (irreversible + the reversible part) the second one is the heat. This equation is nothing else that the equation d Etot/dt = W -Q generally expressed in thermodynamics. If you develop the term for the mechanical work: W= Int[S] n.(v.T) dS = -Int[S] n.(vp) dS + Int[S] n.(v.Tau) dS the first term in the RHS, due to the isotropic part of the stress tensor, is expanded in two terms -Int[V] Div(vp) dV = -Int[V] p Div(v) dV - Int[V] v.(Grad p) dV Then consider that Div(v) is related to the production of volume in a gas system that is zero in incompressible flows. Therefore, you still get a contribution in the mechanical work. However, be careful that what is called "pressure" has no thermodinamic meaning. |
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