Simple mathematical problem?

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 November 10, 2005, 17:16 Simple mathematical problem? #1 Frank Wedburn Guest   Posts: n/a Hello Folks, I've been reading some source material as a prelude to trying a new cfd method. The maths background derives the unsteady Bernoulli equation as an initial step in formulating the technique; there is one point I'm a bit confused over. The derivation for the full technique begins with the continuity equation, 3 momentum equations (invicid) and the isentropic state condition. The problem I'm having is that in the unsteady bernoulli equation the author writes Del=del operator (Nabla) PD = partial derivative rho=density Int=integration P=pressure *** ( Del P)/rho = Del Int( dP/rho) The author claims that this comes from the fact that the isentropic condition means we can write density explicitly as a function of pressure (independent of x,y,z,t). Leibnitz's equation applied to the rhs of the equation does indicate that the equation is correct if we take PD(1/rho)/dx=0 (partial deriv of 1/rho w.r.t x) this partial derivative occurs when we use Leibnitz's rule on the function 1/rho. My problem, however, is that if density is dependent upon pressure then surely it is NOT independent of x,y,z & t. Pressure is surely dependent upon the co-ordinates otherwise the (Del P) term would drop out of Eulers equations in the first place. If density is explicitly dependent upon pressure and pressure dependent upon x,y,z then surely density is dependent upon x,y,z?! If this is the case then the partial derivative of 1/rho is not equal to zero and equation *** must be incorrect? If someone could point me in the right direction I'd be very grateful. I will of course accept what the author is saying and move on but I'd really appreciate it if someone could help me solve this little problem. It's driving me nuts! Thanks in advance, F

 November 10, 2005, 19:00 Re: Simple mathematical problem? #2 Mani Guest   Posts: n/a It really is a mathematical problem. You need to make a distinction between "partial" and "total" derivatives. If density is a function of pressure only (rho=rho(p)), it does not explicitly depend on x, for example. The "partial" derivative of rho with respect to x would be zero, even though density implicitly depends on x, due to p=p(x,y,z,t). The "total" derivative of density with respect to x will not be zero.

 November 10, 2005, 19:11 Re: Simple mathematical problem? #3 Frank Wedburn Guest   Posts: n/a Thanks a lot ) I had assumed it was something like that. I'm not a maths graduate though and am a wee bit rusty on things like that. Cheers for the help, F

 November 10, 2005, 19:19 But surely...... #4 Frank Wedburn Guest   Posts: n/a Now I come to think on it another question arises. Surely if pressure and density are linked via the isentropy condition then pressure may be written explicitly as a function of density. In this case the pressure is explicitly dependent on rho and implicity on x,y & z. The partial derivative of pressure with respect to the co-ordinate variables would be zero? If that was the case then (Del P) term would be removed from the Euler equations which doesn't happen. I'm obviously missing something here. Could someone fill me in? Cheers, F

 November 11, 2005, 14:52 Re: But surely...... #5 Mani Guest   Posts: n/a It get's really confusing, now My intuition tells me that this has something to do with the way you pose your problem. You have to make the (somewhat arbitrary) choice of dependent versus independent variables. You can describe y as dependent on x, or x depending on y. In each case, your problem would be stated differently and you can't have it both ways at the same time. If you were to consider p=p(rho), then you should describe density as rho=rho(x,y,z,t), and your equations may look a little different. In that sense, the partial derivatives depend on how you pose your functional relations. Let's see if we can get a mathematically clean and precise explanation from a mathematician...

 November 11, 2005, 16:47 Re: Simple mathematical problem? #6 andy Guest   Posts: n/a You have 5 conservation equations: mass, momentum and energy with the latter constrained by the laws of thermodynamics plus various constitutives equations like a Newtonian fluid, a gas law, Fouriers law and such depending on the particular fluid. Your independent variables are x, y, z, t and your 5 solution variables are the 3 velocity components and 2 independent thermodynamic properties of state. It doesn't matter which 2 so long as they are independent since all thermodynamic properties can be expressed in terms of just 2. Now your author has assumed constant entropy so it would be wise to choose this as one the thermodynamic variables and to drop the energy equation which is now redundant. This leaves only 3 velocity components, density and pressure as variables which appear in the mass and momentum equations. If you choose pressure as the second independent property then density is a function of pressure and entropy. Since the latter is constant, density will only vary with pressure (but with entropy as a constant specified parameter in your constitutive equations). At a guess, a quick revision of thermodynamic properties might help.

 November 25, 2005, 20:32 Re: Simple mathematical problem? #7 Ananda Himansu Guest   Posts: n/a One of the previous respondents, Mani, put his finger on the answer. Perhaps I can restate his answer and clarify it for Frank, and hopefully not muddy the waters. The key indeed lies in the fact that because of the isentropy assumption, the density, r, can be written as being explicitly dependent solely on the pressure, p. The density is only indirectly dependent on the spatial coordinates, x, through the dependence of the pressure on x. To apply the Leibniz equation, the integral should more clearly (within the restrictions of ascii text!) be written as f = Int {from p0 to p} [1/r(q)] dq, where p0 is some (constant) reference pressure (with a known numerical value), q is a dummy variable of integration that stands in for the pressure, and the density r is completely known as a function of q from the isentropy relation. A basic way to think of the integral is that it defines a function, f(p), solely of pressure. If you specify a particular numerical value of p, then f has some particular numerical value, with no remaining dependence on x. Furthermore, the dependence of f on p enters solely through the upper limit of integration, p. Therefore, df/dx = df(p)/dx = df/dp * dp/dx = 1/r(p) * dp/dx, which is what you desired to prove. Note that the chain rule of differentiation was applied in the second "=" step and that the third "=" step follows from the basic definition of the derivative: df/dp = lim{dp->0}[f(p+dp)-f(p)]/dp, and the additivity of the integral operator with respect to the interval of integration. Note also the transformation from r(q) within the integral to r(p) in the final expression, effectively caused by the evaluation of the integral at its upper limit. The above rather long-winded explanation amounts to saying that when applying the Leibniz equation to df/dx, the dependence of f on x arises solely from the dependence of the upper limit of integration, p, on x; and that PDr/PDx=0, where PD is the partial derivative, so that the second term that is normally present on the RHS of the Leibniz equation vanishes. This vanishing of PDr/PDx is exactly what puzzled Frank, and perhaps it can be more clearly seen as the statement that although dr(p)/dx is not zero (precisely because dp/dx is not zero), it is true that dr(q)/dx is zero (precisely because q is a dummy variable of integration unaffected by any change in x, and r has no other dependence on x). Think of it as: if you change x by an amount dx, the upper limit of integration p will change by an amount dp, but the integrand r(q), for each value of q in the interval of integration, does not change.

 March 23, 2012, 18:26 #9 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 3,356 Rep Power: 38 Bernouilli equation is for irrotational ideal flows, I don't think you are in this condition

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