Pressure enigma.

sbaffini · March 18, 2009, 05:26

If we consider the continuity and momentum equations for a general compressible fluid:

$\frac{\partial \rho}{\partial t} + \frac{\partial}{\partial x_j}\left[ \rho u_j \right] = 0$

$\frac{\partial}{\partial t}\left( \rho u_i \right) + \frac{\partial}{\partial x_j} \left[ \rho u_i u_j + p \delta_{ij} - \tau_{ji} \right] = 0$

taking the time derivative of the first one and the divergence of the second one:

$\frac{\partial^2 \rho}{\partial t^2} + \frac{\partial}{\partial x_j}\left[\frac{\partial \left( \rho u_j \right)}{\partial t} \right] = 0$

$\frac{\partial}{\partial t}\left[\frac{\partial \left( \rho u_i \right)}{\partial x_i} \right] = - \frac{\partial^2}{\partial x_i x_j} \left[ \rho u_i u_j + p \delta_{ij} - \tau_{ji} \right]$

and finally substituting the second in the first one, we get:

$\frac{\partial^2 \rho}{\partial t^2} = \frac{\partial^2}{\partial x_i x_j} \left[ \rho u_i u_j + p \delta_{ij} - \tau_{ji} \right]$

with:

$\tau_{ij} = \mu \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} - \frac{2}{3} \frac{\partial u_k}{\partial x_k} \delta_{ij} \right) + k \delta_{ij} \left( \frac{\partial u_k}{\partial x_k} \right)$

Now, considering single component flows, we have:

$d\rho = \left( \frac{\partial \rho}{\partial S} \right)_p dS + \left( \frac{\partial \rho}{\partial p} \right)_S dp = - \frac{\rho \beta T }{C_p}dS + \frac{1}{c^2} dp$

Obviously, as also stated by someone else, an incompressible fluid doesn't even exists. For the present task we can assume the FLOW (not the fluid) as isothermal and we'll get (which, however, is the same of getting $\beta = 0$ ):

$\frac{1}{c^2} \frac{\partial^2 p}{\partial t^2} = \frac{\partial^2}{\partial x_i x_j} \left[ \rho u_i u_j + p \delta_{ij} - \tau_{ji} \right]$

And finally moving all the pressure terms to the left side:

$\frac{1}{c^2} \frac{\partial^2 p}{\partial t^2} - \frac{\partial^2 p}{\partial x_i x_i} = \frac{\partial^2}{\partial x_i x_j} \left( \rho u_i u_j - \tau_{ji} \right)$

Considering the last equation (which, i think, should be correct) we can say something about the nature of the pressure.

In fact, mathematically speaking, it is similar (somehow and neglecting some lower order terms) to an externally forced mass-spring system with

playing the role of k, the elastic costant of the spring.

So, what happens when k is very high?
In the mass spring-system, after some very fast oscillations (the more k is high the more they are faster), the mass will assume a displacement in equilibrium with the external forcing. The same is true even if the external forcing is not costant but relatively slow, and at every time the mass can be considered in equilibrium with the external force.
In this case the problem is called stiff because the time scale of the system is very different from that of the external forcing.

Going back to the pressure equation, the role of k is now played by

, the speed of sound in the fluid. When

is very high but the fluid velocity is relatively slow compared to

, the problem becomes stiff, that is the thermodynamic is much more faster than the external forcing; this comes out nondimensionalising the right hand terms with $\rho_0 U_{0}^2$ .

Actually, in this case, the thermodynamic nature of the pressure is not changed (why should it be?) but (first of all for computational purposes) we can consider a new equation in which the time derivative term is omitted because of the $\frac{1}{c^2}$ term (being of lower order):

$\frac{\partial^2 p}{\partial x_i x_i} = - \frac{\partial^2}{\partial x_i x_j} \left( \rho u_i u_j - \tau_{ji} \right)$

Moreover, with a very lenghty procedure, it can be shown that the divergence of the velocity field is proportional to terms involving $M^2, \frac{M^2}{Fr}, \frac{M^2}{Str}, \frac{M^2}{Re}, \frac{\beta}{Re Pr}$ , so, in the same hypothesis already made, it also is of lower order and can be assumed to be zero, so we finally get:

$\frac{\partial^2 p}{\partial x_i x_i} = - \frac{\partial^2}{\partial x_i x_j} \left( \rho u_i u_j \right)$

Which is of very different nature respect to the original one, being an equilibrium equation for the pressure (not a time-evolution one), so at every time the pressure is considered in equilibrium or, mathematically speaking, the pressure is considered acting as a Lagrange multiplier for the velocity field; in fact the equation is now a cinematic condition on the velocity field which has to be fulllfilled at every time.

So, what actually changes between the incompressible view and a general one, it's not the pressure nature, which is still thermodynamic, but the parameters affecting it. Actually, in the incompressible view, we are considering the pressure as affected by normal momentum fluxes only and, because the pressure fastly recovers the equilibrium after a momentum change (because of the $M^2 \ll 1$ hypothesis), we just consider it as in equilibrium at each time omitting the lower order time-derivative term. As i said, this is much more a computational necessity (because of the stiffness); in fact, if you are interested in acoustic you will need to retain the time-derivative term and use one of the method of the computational aeroacoustic to treat the equation.

Sorry for the lenghty post but i hope it helps.

jwm · March 18, 2009, 08:55

Quote:

Originally Posted by Jonas Holdeman
;59303

If the equations are rewritten in the stream function-vorticity form or the stream function form, there is no pressure in the problem. The incompressible velocity field exists and can be computed without regard to the pressure. The incompressibility/mass conservation is being maintained without reference to a pressure. Where is a Lagrange multiplier in this case? If one expands the velocity field in terms of a divergence-free basis, the expansion will always be divergence-free no matter what expansion coefficients are chosen. If the expansion coefficients are chosen to best approximate the solution to the NSE, this solution will be divergence-free without regard to the pressure. The incompressibility condition acts like a conservation law. The "pressure" derived from the velocity field (say by the pressure Poisson equation) is that pressure-like quantity consistent with the flow field. How can we say that the pressure drives incompressible flow if there is no pressure appearing in the governing equation? Incompressibility drives the flow and pressure responds!

Bottom line is: the pressure is a physical force whose gradient acts to change the velocity field to ensure mass conservation. It arises from random thermal motions of molecules. I concede that an incompressible fluid is a mathematical abstraction and, consequently, that the pressure field in an incompressible fluid might also be, strictly speaking, a mathematical abstraction and incapable of a stict physical interpretation, but if you add the correct constant to the pressure you get a pretty good approximation to the thermodynamic pressure.

Ford Prefect · July 30, 2009, 04:46

I would like to continue this discussion by asking the following question:

If we have an incompressible fluid in a near vacuum domain then what happens if the "motion driving pressure" differences are large enough to cause the absolute pressure to be negative in some parts of the domain?

Can we really relate the motion driving pressure to the thermodynamic pressure?

March 18, 2009, 05:26		#21
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,151 Blog Entries: 29 Rep Power: 39	If we consider the continuity and momentum equations for a general compressible fluid: $\frac{\partial \rho}{\partial t} + \frac{\partial}{\partial x_j}\left[ \rho u_j \right] = 0$ $\frac{\partial}{\partial t}\left( \rho u_i \right) + \frac{\partial}{\partial x_j} \left[ \rho u_i u_j + p \delta_{ij} - \tau_{ji} \right] = 0$ taking the time derivative of the first one and the divergence of the second one: $\frac{\partial^2 \rho}{\partial t^2} + \frac{\partial}{\partial x_j}\left[\frac{\partial \left( \rho u_j \right)}{\partial t} \right] = 0$ $\frac{\partial}{\partial t}\left[\frac{\partial \left( \rho u_i \right)}{\partial x_i} \right] = - \frac{\partial^2}{\partial x_i x_j} \left[ \rho u_i u_j + p \delta_{ij} - \tau_{ji} \right]$ and finally substituting the second in the first one, we get: $\frac{\partial^2 \rho}{\partial t^2} = \frac{\partial^2}{\partial x_i x_j} \left[ \rho u_i u_j + p \delta_{ij} - \tau_{ji} \right]$ with: $\tau_{ij} = \mu \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} - \frac{2}{3} \frac{\partial u_k}{\partial x_k} \delta_{ij} \right) + k \delta_{ij} \left( \frac{\partial u_k}{\partial x_k} \right)$ Now, considering single component flows, we have: $d\rho = \left( \frac{\partial \rho}{\partial S} \right)_p dS + \left( \frac{\partial \rho}{\partial p} \right)_S dp = - \frac{\rho \beta T }{C_p}dS + \frac{1}{c^2} dp$ Obviously, as also stated by someone else, an incompressible fluid doesn't even exists. For the present task we can assume the FLOW (not the fluid) as isothermal and we'll get (which, however, is the same of getting $\beta = 0$ ): $\frac{1}{c^2} \frac{\partial^2 p}{\partial t^2} = \frac{\partial^2}{\partial x_i x_j} \left[ \rho u_i u_j + p \delta_{ij} - \tau_{ji} \right]$ And finally moving all the pressure terms to the left side: $\frac{1}{c^2} \frac{\partial^2 p}{\partial t^2} - \frac{\partial^2 p}{\partial x_i x_i} = \frac{\partial^2}{\partial x_i x_j} \left( \rho u_i u_j - \tau_{ji} \right)$ Considering the last equation (which, i think, should be correct) we can say something about the nature of the pressure. In fact, mathematically speaking, it is similar (somehow and neglecting some lower order terms) to an externally forced mass-spring system with playing the role of k, the elastic costant of the spring. So, what happens when k is very high? In the mass spring-system, after some very fast oscillations (the more k is high the more they are faster), the mass will assume a displacement in equilibrium with the external forcing. The same is true even if the external forcing is not costant but relatively slow, and at every time the mass can be considered in equilibrium with the external force. In this case the problem is called stiff because the time scale of the system is very different from that of the external forcing. Going back to the pressure equation, the role of k is now played by , the speed of sound in the fluid. When is very high but the fluid velocity is relatively slow compared to , the problem becomes stiff, that is the thermodynamic is much more faster than the external forcing; this comes out nondimensionalising the right hand terms with $\rho_0 U_{0}^2$ . Actually, in this case, the thermodynamic nature of the pressure is not changed (why should it be?) but (first of all for computational purposes) we can consider a new equation in which the time derivative term is omitted because of the $\frac{1}{c^2}$ term (being of lower order): $\frac{\partial^2 p}{\partial x_i x_i} = - \frac{\partial^2}{\partial x_i x_j} \left( \rho u_i u_j - \tau_{ji} \right)$ Moreover, with a very lenghty procedure, it can be shown that the divergence of the velocity field is proportional to terms involving $M^2, \frac{M^2}{Fr}, \frac{M^2}{Str}, \frac{M^2}{Re}, \frac{\beta}{Re Pr}$ , so, in the same hypothesis already made, it also is of lower order and can be assumed to be zero, so we finally get: $\frac{\partial^2 p}{\partial x_i x_i} = - \frac{\partial^2}{\partial x_i x_j} \left( \rho u_i u_j \right)$ Which is of very different nature respect to the original one, being an equilibrium equation for the pressure (not a time-evolution one), so at every time the pressure is considered in equilibrium or, mathematically speaking, the pressure is considered acting as a Lagrange multiplier for the velocity field; in fact the equation is now a cinematic condition on the velocity field which has to be fulllfilled at every time. So, what actually changes between the incompressible view and a general one, it's not the pressure nature, which is still thermodynamic, but the parameters affecting it. Actually, in the incompressible view, we are considering the pressure as affected by normal momentum fluxes only and, because the pressure fastly recovers the equilibrium after a momentum change (because of the $M^2 \ll 1$ hypothesis), we just consider it as in equilibrium at each time omitting the lower order time-derivative term. As i said, this is much more a computational necessity (because of the stiffness); in fact, if you are interested in acoustic you will need to retain the time-derivative term and use one of the method of the computational aeroacoustic to treat the equation. Sorry for the lenghty post but i hope it helps. Last edited by sbaffini; March 18, 2009 at 13:58. Reason: Changing the equation form

July 30, 2009, 04:46		#23
Ford Prefect Senior Member Join Date: Mar 2009 Posts: 151 Rep Power: 17	I would like to continue this discussion by asking the following question: If we have an incompressible fluid in a near vacuum domain then what happens if the "motion driving pressure" differences are large enough to cause the absolute pressure to be negative in some parts of the domain? Can we really relate the motion driving pressure to the thermodynamic pressure? __________________ "Trying is the first step to failure." - Homer Simpson

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