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Euler´s equation for inviscid flow |
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| May16-08, 06:14 AM | #1 |
| May16-08, 04:45 PM | #2 |
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I think you shouldn't omit the last two summands in the formula for the total derivative of velocity. Also you didn't write how you adjusted the pressure after each time step: you should use continuity equation both for mass and energy (to determine the ammount and temperature of the gass in each cell, which are needed to calculate the pressure).
Movement of the molecules towards higher pressure is in fact physically possible, since the pressure only determines the acceleration of the molecules and not the speed: if they have enough speed, then they can move towards higher pressure for some time. |
| May17-08, 03:46 AM | #3 |
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I was told that by laminar flow the last two summands can be ommited
the general idea is> By computing Vx(new) i get the speed of the molecules Then i take each cube and move it to the right direction The source cube should donate molecules(warmth) to other cubes T=T1*n1+T2*n2...../n1+n2.... And then i compute pressure again with new "T" and "n" and do the same in rounds the heat should spread all over the cube.. The process is described on the picture..I used 2D model cube.bmp |
| May18-08, 06:35 PM | #4 |
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Euler´s equation for inviscid flow
I don't understand your calculation of donated molecules. Their amount should increase lineary with time step dt, while it seems to be proportional to dt^2 in your formula. I understood that you took 1 s for your time step. That is probably too much (at least while your typical velocity*dt gives the same order of magnitude as box size a): numerical solving of diferential equations works well only for very small time steps (that cause small relative changes of parameters in each step).
I'm still not convinced about the two omited summands in the total derivatives of velocity: maybe you were supposed to define coordinate system with x axes parallel to velocity (at each box)? (then those parts are of course zero) |
| May19-08, 04:50 AM | #5 |
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Ok,so i can change the time step to 0.01s for example,can you tell me what your suggestion for computing donation of molecules and heat is?..
i thought that the geometrical soulution is good enough for 3D: amouth of molecules in cube/total volume of the cube= amount of molecules for next cube/volume of the molecules for next cube n/V=n1/V1 n1=n*V1/V ..the thing is that whether i omit the last two summands or not i still think there is something wrong with that..one thing can be the time step,but it is not the only.. |
| May19-08, 06:28 PM | #6 |
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The expression for molecules passing from cube [0,0] to [0,1]
nd=n[0,0]*(x1*y1)/a^2 should be replaced by nd=n[0,0]*(x1*(a-y1))/a^2 If you chose small time step dt then y1<<a and the expression becomes aproximately: nd=n[0,0]*x1/a Of course you should sum contributions from all neighbors of each cube in each time step (except diagonal neighbours: their contribution is negligible if you chose small dt, since it is proportional to dt^2). |
| May20-08, 10:34 AM | #7 |
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thank you so much i see your point..
..i tried to compute something on paper for x-axis only,and that seems to be the biggest problem take a look x_axis.bmp i don t like the flow..i don t think it is correct.. btw> i appreciate your help |
| May22-08, 04:35 PM | #8 |
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Here is a simple, but very important advice: always calculate with physical units!
An equation with different units on different sides (or even at different summands on the same side) is certanly wrong. Search for Navier-Stokes equation on the internet and ommit the parts multiplied by viscosity (since your gas is inviscid), then try to compose the correct equation by replacing derivatives with finite diferences. Hint: local acceleration is not really vnew-vold (do you remember the definition of acceleration?). Other parts of the equation must also be corrected to reasonably aproximate the partial derivatives with finite diferences. |
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