Can pressure do work in an incompressible flow?

toodles · January 15, 2018, 12:48

I am trying to analyze the energy equation for incompressible flows. I start with N-S and dot it with the velocity vector:

$\bigg( \rho \frac{D\mathbf{u}}{Dt} - \nabla^2 \mathbf{u} + \nabla p \bigg) \cdot \mathbf{u} = 0$

Then I get something like

$1/2 \rho \frac{\partial \mathbf{u}^2}{\partial t} - \Phi + \mathbf{u}\cdot \nabla p = 0$

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

$\mathbf{u} \cdot \nabla p = -p \nabla \cdot \mathbf{u} + \nabla \cdot (p\mathbf{u})$

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.

FMDenaro · January 15, 2018, 13:01

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.

January 15, 2018, 12:48	Can pressure do work in an incompressible flow?	#1
toodles New Member Join Date: Dec 2015 Posts: 27 Rep Power: 10	I am trying to analyze the energy equation for incompressible flows. I start with N-S and dot it with the velocity vector: $\bigg( \rho \frac{D\mathbf{u}}{Dt} - \nabla^2 \mathbf{u} + \nabla p \bigg) \cdot \mathbf{u} = 0$ Then I get something like $1/2 \rho \frac{\partial \mathbf{u}^2}{\partial t} - \Phi + \mathbf{u}\cdot \nabla p = 0$ 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 $\mathbf{u} \cdot \nabla p = -p \nabla \cdot \mathbf{u} + \nabla \cdot (p\mathbf{u})$ 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		#2
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,768 Rep Power: 71	I would start directly from the total energy equation . E= (K+U): d/dt Int[V] (rhoE) dV + Int[S] n.v*(rhoE) dS = Int[S] n.(v.T) dS- Int[S]n.q dS T=-pI + Tau q=-kGrad 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.