# Navier-Stokes equation

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 June 20, 2006, 01:44 Navier-Stokes equation #1 Ishaq Guest   Posts: n/a Hi All. Please state the advantage and disadvantage of the Conservative and Non-conservative forms of the Navier-Stokes equations. tamiru likes this.

 June 20, 2006, 06:48 Re: Navier-Stokes equation #2 ganesh Guest   Posts: n/a Dear Ishaq, One of the basic properties of the governing laws of fluid mechanics is conservation. The numerical methods that you employ are also expected to mimick this property. It is for this reason that the conservative form of the governing equations are employed. It can also be shown that conservation is also inherently linked with obtaining the exact shock location. Non-conservative forms do not predict the shock location correctly, if the grids are not very fine (on fine grids the discretisation errors are low), and would also require derviatives one order more than that desired by conservative form of the governing laws, for the same accuracy. Regards, Ganesh

 June 21, 2006, 01:26 Re: Navier-Stokes equation #3 bajjal Guest   Posts: n/a The equtions by looking at them we could say that this are conservative equtions and this are not, is it the classification is done based on above statements as u mentioned,the only problem in non conservative equations is derivatives of one order more is required for same order of accuracy , impementation also is more complex than other but finally is'nt though more tingent both can forms mimic the conservation

 June 21, 2006, 01:35 Re: Navier-Stokes equation #4 bajjal Guest   Posts: n/a I mean is it based on complexity one is said has conservative form and other as non-conservative though both types can mimic numerically conservation(though one form of equations may be bit tingent)

 June 21, 2006, 03:25 Re: Navier-Stokes equation #5 Tom Guest   Posts: n/a A conservation law is any equation of the general form (f.u)_t = div(F) (*) which follows directly from the classical derivation of the equations as a statement of conservation for a fixed volume V; i.e. the rate of change of f.u inside volume V must equal the flux through the surface - the divergence theorem then gives (*) integrated over the volume V. Since this must hold for all choices of V the pointwise equation (*) must also hold. The nonconservative version is obtained simply by rearranging (*) to obtain an equation for the rate of change of u rather than f.u and differentiating out the div(F) term. Which form you use is usually a matter of choice and depends upon the problem you want to solve and the numerical techniques you wish to use.

 June 21, 2006, 05:40 Re: Navier-Stokes equation #6 bajjal Guest   Posts: n/a ganesh:One of the basic properties of the governing laws of fluid mechanics is conservation. The numerical methods that you employ are also expected to mimick this property. It is for this reason that the conservative form of the governing equations are employed. Tom the question is actually why those equations are NAMED as non-conservative equations as per above statement are they are not able to really numerically mimick conservation , but ganesh told they do but with more complexity . Any corrections welcome

 June 21, 2006, 06:22 Re: Navier-Stokes equation #7 Tom Guest   Posts: n/a They fail to mimic exact conservation mainly because of numerical discretization error; i.e. if you expect the integral over the domain to be zero then with a conservative scheme the discrete equations will give zero when summed (trapezoidal rule). The non-conservative scheme will give a value O(h) where h is the discretization error (so when h->0 the nonconservative system approaches the conservative). This would suggest that the conservative scheme was "more accurate" and so you should always use the conservative discretization. However there are counter-examples to this which show that, using a different error indicator such as the maximum difference between the exact and approximate solutions, the nonconservative discretization can be more accurate.

 June 21, 2006, 09:49 Re: Navier-Stokes equation #8 Mani Guest   Posts: n/a I guess it's fair to say that the conservative set of variables lends itself more naturally to construction of conservative numerical methods, although the same can be done, maybe with a little more effort, by using non-conservative variables. As ganesh has pointed out, conservation is especially important to get the right position and strength of discontinuities such as shocks in compressible flow. Which method gives you a more accurate solution surely depends on the type of flow and the discretization method. Tom points out the 'exact' conservation that can be achieved by a conservative method. However, even with conservative methods for compressible flow, for example, the flux balance includes a non-physical (i.e. artificial) dissipation necessary for shock capturing. So, bear in mind that 'exact' conservation does not mean 'exact' solution in any sense. To force conservation through-out the whole iteration process just helps to get the above issues (shock position and strength) straightened out, and with added artificial dissipation also helps to prevent non-physical (but numerically possible) solutions like 'expansion shocks' in inviscid flow. From a numerical perspective I don't see any reason why not to use the conservative variables. They are the natural set of variables, i.e. the variables that seem most suitable to describe nature's fundamental laws. The reason why people often prefer non-conservative variables like pressure and velocity is simply that those are the variables we know how to measure in experiments. Humans have developed a feeling for those variables and understand 'velocity' better than 'momentum', and feel 'pressure' better than 'energy'. I am not aware of any real reasons to prefer non-conservative variables from a numerical point of view. 'Personal preference' probably wins over numerical considerations when conservativeness is really not a huge issue, such as in incompressible flow. Aside from that, I am sure that the history of various branches of CFD does its part in dictating the use of either set of variables, without much deep reasoning behind it.

 June 24, 2006, 04:59 Re: Navier-Stokes equation #9 bajjal Guest   Posts: n/a thanks mani and tom

 October 17, 2013, 08:32 novier stokes #10 New Member   Shima Join Date: Oct 2013 Posts: 4 Rep Power: 4 Hi evrybody, I want to know why novier stokes equation is named non conservative and momentum conservation equation is named conservative law? I want to prove it. Can anyone help me? The equations along X axis have been attached to this thread. A1.docx

October 17, 2013, 11:11
#11
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Filippo Maria Denaro
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Quote:
 Originally Posted by Shima Fallah Hi evrybody, I want to know why novier stokes equation is named non conservative and momentum conservation equation is named conservative law? I want to prove it. Can anyone help me? The equations along X axis have been attached to this thread. Attachment 26135

Navier-Stokes equations (continuity, momentum and energy) can be written both in conservative and non-conservative form

October 17, 2013, 19:34
#12
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Shima
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Quote:
 Originally Posted by FMDenaro Navier-Stokes equations (continuity, momentum and energy) can be written both in conservative and non-conservative form
Thank you FMDenaro,
Did you see the file that have been attached to this quote? My teacher said the first one is conservative and the second one is non-conservative. He saied I should prove it.

Last edited by Shima Fallah; October 18, 2013 at 15:34.

October 17, 2013, 21:02
#13
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Filippo Maria Denaro
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Quote:
 Originally Posted by Shima Fallah Thank you FMDenaro, Did you see the file that have been attached to this quote? My teacher said the first one is conservative and the second one is non-conservative. He say I should prove it.
see http://en.wikipedia.org/wiki/Euler_e...id_dynamics%29,
but I suggest reading a good textbook such as Kundu

 October 18, 2013, 13:02 #14 Senior Member   Reza Join Date: Mar 2009 Location: Appleton, WI Posts: 116 Rep Power: 9 The two equations that you have, are mathematically the same (assuming continuity holds). The only difference is how they are written. Using your notation, the momentum equation comes directly from the application of Reynolds transport theorem to Newton's second law for a infinitesimal fluid element. So, it essentially states a conservation law: Momentum change for a control volume = momentum coming into the CV - momentum going out of the CV + momentum generation inside the domain On the other hand, the Navier-Stokes equation assumes continuity holds and expands some of the partial derivatives to simplify the notation. So, for example, you can use (uv)' = u v' + u' v and so on to expand the lhs, and then you'll see lhs of continuity which is equal to zero. This form cannot be converted to the conservation law like the momentum equation, because it has lost some of the required terms. By the way, there is a missing addition in your first equation (in between first and second terms in lhs). FMDenaro and Shima Fallah like this.

 October 18, 2013, 13:26 #15 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 2,516 Rep Power: 31 I agree, I also suggest studyng the conservation equation starting from the integral-based Reynolds theorem

October 18, 2013, 15:54
#16
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Shima
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Quote:
 Originally Posted by FMDenaro I agree, I also suggest studyng the conservation equation starting from the integral-based Reynolds theorem
Thank you very much.
Could you please upload the PDF of this book here?

October 18, 2013, 15:56
#17
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Shima
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Quote:
 Originally Posted by triple_r The two equations that you have, are mathematically the same (assuming continuity holds). The only difference is how they are written. Using your notation, the momentum equation comes directly from the application of Reynolds transport theorem to Newton's second law for a infinitesimal fluid element. So, it essentially states a conservation law: Momentum change for a control volume = momentum coming into the CV - momentum going out of the CV + momentum generation inside the domain On the other hand, the Navier-Stokes equation assumes continuity holds and expands some of the partial derivatives to simplify the notation. So, for example, you can use (uv)' = u v' + u' v and so on to expand the lhs, and then you'll see lhs of continuity which is equal to zero. This form cannot be converted to the conservation law like the momentum equation, because it has lost some of the required terms. By the way, there is a missing addition in your first equation (in between first and second terms in lhs).
Thank you very much.
Yes, you are right.

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