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July 29, 1998, 12:18 |
Scalar Transport Equations
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#1 |
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The properties of a fluid can be represented in Conservative Matrix-Tensor Notation form for both the instaneous NS equations and the Reynolds Averaged/ Favre averaged counterparts (i.e. Turbulent counterparts Depending on which turbulent model is best suited to situation in hand) This matrix notation says that i.e. The state vector + the divergence of both the diffusive and convective flux = source term. How is this normally written and interpreted as a general scalar transport equation (pressure, density, u velocity etc.). If pressure term is which is basically a second scalar quantity is embedded into the divergence of the convective flux. there's two different scalars in the same equation>coupled equations.
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July 30, 1998, 17:35 |
Re: Scalar Transport Equations
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#2 |
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Basically,you have 1)continuity equation ( for mass conservation),2)momentum equation ( for momentum conservation),3)energy equation ( for energy conservation),4)equation of state ( for gas property )and 5) turbulence modeling ( for turbulent flows). There will be more equations for reacting flows. Vector and tensor notations are short-hand notations. The standard approach is to pick a coordinate system ( either Cartesian ,or cylindrical coordinates system) and write down all terms of these equations first. For general coordinate system, you can use chain-rule transformation to derive a new set of equations from these standad equations. From there on, you can use finite-difference method, finite-volume-method, finite element method, or ....to derive your algebraic equations sutable for computer solutions. You can expand the equation in terms of individual terms or you can group them in terms of time-derivative term, convection term, second-order diffusion term, some source terms. Or you can group them in time-derivative term, first-order divergence term, source terms. At the mathematical formulation level, they are all the same.
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July 30, 1998, 18:00 |
Re: Scalar Transport Equations
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#3 |
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Before I forget, the governing equations form a set of coupled, non-linear, partial differential equations. For incompressible flow, the pressure term can be de-coupled from the momentum equations. A second-order pressure equation can be derived. For steady-state compressible flow equations, you can only use iterative method because it's a set of coupled equations. For time-dependent compressible, you have time-derivative term on the left-hand side of the equal sign, and the rest of the terms are moved over to the right-hand side. You can solve this set of equations for density, rho*u, rho*v, rho*w etc.. The pressure can be related to the rho and temperature through equation of state. If you are using explicit method, then every right-hand side terms will be evaluated at the old know time level and the new values can be easily updated. ( In incompressible flow, the flow field can be solved independent of the pressure field.)
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July 31, 1998, 11:21 |
Re: Scalar Transport Equations
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#4 |
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Thanks
But you did not really clarify the situation. For simplicity considering the laminar flow NS equations is it normal to drop the pressure grad. term into the source terms for the momentum equation. I well aware of the fact that for incompressible flows, the continuity equation act's as a scalar transport equation for density and that the equation of state can be used for the correlation of pressure. Also with gauss law and discretization schemes hybdrid etc. and numerical alogthrims the system of unknowns can be solved. But if the pressure gradient is buried into the source terms, all the flow properties obtained to obtain the pressure term are a bit ambiguous to say the least. One more question. Are all turbulence models (Algebraic and First Order Closure Models) based on Prandtel's Hypothesis that the Eddy viscosity can be modelled similar to that of the Molecular Viscosity. Is this not valid only for 2D Flow. Also aren't most of the closure coefficients required to close the turbulence models based on boundary layer approximation via experimental data. |
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July 31, 1998, 12:06 |
Re: Scalar Transport Equations
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#5 |
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I'm not sure I understand your concerns. The mass, momentum, and energy (if important) equations form a coupled set of equations. Although they have some features in common with a general scalar transport equation, they are a distinct animal. As you point out, one of the things that makes the NS equations different from the scalar equation is the appearance of pressure. How this pressure-velocity coupling is handled is extremely important.
It is interesting that although the continuity equation does not actually contain pressure for incompressible flow, it is generally regarded as an equation for pressure. That is, the continuity equation is responsible for obtaining a pressure field which forces the velocity field to satisfy mass. This follows from some of the interesting mathematical properties of div V=0, and forms the basis of segregated solution strategies such as SIMPLE. By segregated I mean that mass and momentum are solved like scalar transport equations, with coupled terms (such as pressure in momentum) lagged. But the fact that some people use a segregated strategy doesn't mean that the equations are scalar equations; they are only treated that way for convenience. But there's no need to... I think it makes more sense to solve them in a coupled fashion, so that, for instance, pressure in momentum is NOT treated as a source term. I use a coupled strategy with my code, and can solve a 2d lid-driven cavity in less than 20 iterations. (By comparison, segregated methods may take a few hundred, though I haven't tried it myself.) Codes such as TASCflow and CFX-5 use a coupled strategy also. Apparently Fluent is also working on this too, but haven't got it working very well yet. The down side of coupled methods is the additional memory required to store the matrix. Anyways, I hope this helps. |
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July 31, 1998, 15:59 |
Re: Scalar Transport Equations
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#6 |
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For the momentum equations, rho*u, rho*v, and rho*w will be the primitive variables ( in incompressible flow equations, it will be u,v,and w). There will be time-derivative term of rho*u , convective term of rho*u, the pressure gradient term ( dp/dx ), and the viscous terms ( anything related to viscous stresses. If you consider this equation as the equation for (rho*u) then you can keep the time-derivative term ( d(rho*u)/dt ) on the left-hand side of the equation and move the rest of the terms to the right-hand side of the equal sign. This will be your equation for rho*u. Mathically, you are free to write the terms on the right-hand side in any form you like. You can expand these terms out into first-order derivatives, second-order derivatives. Or you can group some of them into first-order derivative form and some left-over source terms. For explicit method, all of the right-hand side terms will be calculated based on the known, current time values. You don't really care whether they are viscous terms, pressure term, or source terms, as long as they can be evaluated at the current time. In the implicit method, the situation will be different, because they can not be evaluated at the current time step only. They are also unknown terms. The best way to understand it, is to write your own code. I must say again here that at the mathematical level , it does not matter how you group these terms together. At the finite-difference or finite volume level, the stability of the derived equation depends strongly on the mathematical form of the equation. This is because they will be approximated in completely different ways depending upon the numerical schemes used.
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August 10, 1998, 10:05 |
Re: Scalar Transport Equations
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#7 |
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Thanks
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