Timeline for Navier Stokes solution doesnt make sense to me
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 25, 2023 at 11:07 | vote | accept | Volpina | ||
| Apr 22, 2023 at 13:12 | answer | added | Volpina | timeline score: 0 | |
| Apr 22, 2023 at 12:34 | comment | added | Tomáš Létal | If diameter changes, mass flow rate would still be constant, so the velocity would be constrained by this. In other words, you would not be able to just pick any velocity distribution at a time. | |
| Apr 22, 2023 at 12:10 | comment | added | Volpina | @TomášLétal the flow velocity is 2 at x=0 there isn't any reason why it couldn't be something else when x not equal 0 | |
| Apr 22, 2023 at 12:06 | comment | added | Volpina | @TomášLétal cant the diameter of the pipe change across x? | |
| Apr 22, 2023 at 12:02 | history | edited | Volpina | CC BY-SA 4.0 |
added 5 characters in body
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| Apr 22, 2023 at 12:01 | comment | added | Tomáš Létal | You are right. But for transient simulation, you basically need known state at the start. So you should have initial velocity distribution at starting time, not just a value at a point. If the fluid is not compressible, the pipe has constant diameter and you are only interested in mean velocity at each cross section, I don't see a possibility for the velocity to be other than constant in the whole pipe, depending just on time. | |
| Apr 22, 2023 at 10:50 | comment | added | Volpina | No It doesn't work like that bcs we know u is only independent of time when x=0. | |
| Apr 22, 2023 at 10:27 | comment | added | Tomáš Létal | If $u(0,t) = 2$, doesn't that mean it will be constant in time? Then $du/dt = 0$, the rest can be separated and integrated and you should end up with velocity of 2 in the whole pipe. | |
| S Apr 22, 2023 at 7:44 | review | First questions | |||
| Apr 23, 2023 at 2:09 | |||||
| S Apr 22, 2023 at 7:44 | history | asked | Volpina | CC BY-SA 4.0 |