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#21 |
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If we consider the continuity and momentum equations for a general compressible fluid:
![]() ![]() taking the time derivative of the first one and the divergence of the second one: ![]() ![]() and finally substituting the second in the first one, we get: ![]() with: ![]() Now, considering single component flows, we have: ![]() 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 ![]() ![]() And finally moving all the pressure terms to the left side: ![]() 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 ![]() 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 ![]() ![]() ![]() ![]() 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 ![]() ![]() Moreover, with a very lenghty procedure, it can be shown that the divergence of the velocity field is proportional to terms involving ![]() ![]() 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 ![]() 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 |
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#22 | |
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john mckelliget
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#23 |
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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?
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