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Analytical problem,finite element method (Q2Q1) |
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August 25, 2008, 13:42 |
Analytical problem,finite element method (Q2Q1)
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#1 |
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Hello everybody,
I would be really gratefully if you could help me.(Sorry for the long email, but I try to do a good explanation of the problem) I have to solve the incompressible Navier-Stokes equations in a square domain [0,1]x[0x1]. I calculate velocity and pressure fields with a decoupled scheme, using Q2Q1 pair of elements as spatial discretization, I am using finite element method. As temporal discretization I use Backward Euler Scheme, so itis a scheme of first order. In order to do an error analysis, I simulate an analytical case where the solution, both velocity and pressure,depends on time, something like "sin(a*pi*x)*cos(b*pi*y)*exp(-c*t)". I use a grid of 2^4*2^4 elements, i.e. a spatial step=h=1/16, another one of 2^5*2^5 elements, i.e. h=1/32, another one of 2^6*2^6, i.e. h=1/64 and another grid of 2^7*2^7, i.e. h=1/128. I use the pair element Q2Q1. I use four different temporal steps: incrT=1/64,1/256,1/512 and 1/1024. I think I have programmed everything in the right way. But the problem is that I don't obtain a good result. Well, the pressure error behaves in a good way since less refined the mesh, higher is the error. But, the velocity error doesn't follow this pattern, for example, I obtain: Always with Q2Q1: a)With incrT=1/64. (error with h=1/32) is the lowest. (error with h=1/128)>(error with h=1/64)>(error with h=1/32) b)With incrT=1/256. (error with h=1/32) is the lowest. (error with h=1/128)>(error with h=1/64)>(error with h=1/32) c)With incrT=1/512. (error with h=1/64) is the lowest (error with h=1/128)>(error with h=1/64) d)With incrT=1/1024. (error with h=1/64) is the lowest (error with h=1/128)>(error with h=1/64) Looking at the final solution and comparing it to the exact one, all the previous situations seem to reach a good agreement. i) I have thought that perhaps I should try to implement a temporal discretization scheme of order two, such as Runge-Kutta 2, instead a Backward Euler, but I don't know if this would solve the problem. ii) Besides, the most surprising thing is that the pressure (Q1 element) error has the properly behaviour. Any advice will be welcome. Thanks in advance. Isa |
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August 26, 2008, 03:45 |
Re: Analytical problem,finite element method (Q2Q1
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#2 |
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Isa,
I can't see how your analytic solution satisfy continuity. Maybe it is just because you have not explicitly written the expressions for u, v and p separately. Rami |
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August 26, 2008, 04:23 |
Re: Analytical problem,finite element method (Q2Q1
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#3 |
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Hello,
yes, you are right, I only wanted to show the general shape. The analytical solutions are: u=-cos(2*pi*x)sin(2*pi*y)exp(-8pi*pi*viscosity*t); v=cos(2*pi*y)sin(2*pi*x)exp(-8pi*pi*viscosity*t); p=-0.25*(cos(4*pi*x)+cos(4*pi*y))exp(-8pi*pi*viscosity*t); Best Isa |
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August 26, 2008, 18:29 |
Re: Analytical problem,finite element method (Q2Q1
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#4 |
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Hi Isi,
When you solve the Navier-Stokes eqns if you do not have a b.c. that sets your pressure, then you need to specify the pressure at a point in your domain. You could impose for example that P(0)=1. I would not make any difference if you use the Runge-Kutta method. Actually, the Backward Euler method is a very stable method. Finally, try to make the eqns dimensionless before you solve them. In this way, the Reynolds number will appear, and then you could find the value of the Reynolds number that causes this behavior. |
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August 27, 2008, 08:06 |
Re: Analytical problem,finite element method (Q2Q1
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#5 |
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Hi!
I put b.c. on the velocity variable in all the walls. Backward Euler method is stable but is only first order accurate, it's for that I had thought in implementing Runge-Kutta, because it is of second order. The Reynolds number is 100. Best Isi |
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August 27, 2008, 14:03 |
Re: Analytical problem,finite element method (Q2Q1
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#6 |
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Hi Isi,
Is not enough to specify the velocity in all the boundaries! When you solve the Navier-Stokes eqns and you do not specify the pressure in one of you b.c you need to specify it in your domain. Consider the simplest case of two parallel plates, where the top plate is moving with constant V and the bottom plate is motionless. The solution for the velocity profile is obvious, but does the Navier-Stokes eqns give you any information about the pressure? The answer is no, and the only thing that you learn from the solution of these eqns is that the pressure is constant, and nothing but that! Because of that you need to specify the pressure at a given point in your domain. It won't affect your solution! Yes you are right, backward Euler is a first order method, but its order will affect the rate at which the error is changing. A rule of thumb when you solve a multidimensional problem is to achieve convergence in the spatial domain, and then to achieve convergence in the time domain. If your solution has not converged in the spatial domain, do not solve the eqns for a different value of dt. |
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