Timeline for Bernoullis in Parallel pipes
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 3, 2016 at 11:22 | comment | added | Chet Miller | If you want to solve for the actual detailed fluid velocity profiles and pressure profiles throughout the region of flow, you need to solve the Euler equations (for an inviscid fluid), or the Navier Stokes equations for a viscous fluid (assuming turbulence is not present). For turbulent flow, additional approximations need to be made to describe the behavior. Most computational fluid dynamics (CFD) software packages will handle calculations such as these. | |
| Jun 3, 2016 at 7:41 | comment | added | Tom Chester | hi @chestermiller I have been thinking about the above question and have a further query . In the above example we have two different Bernoulli scenarios in each of our pipes at the confluence we will have a differential in static pressures (a pressure gradient perpendicular to the flow) between the two streamlines how does this gradient resolve itself ? What area of fluid mechanics studies this ? | |
| May 16, 2016 at 3:47 | vote | accept | Tom Chester | ||
| May 16, 2016 at 3:43 | comment | added | Tom Chester | Hi @chestermiller thanks for all your correspondence. | |
| May 12, 2016 at 3:18 | comment | added | Chet Miller | For the inviscid flow, the flow splits evenly between pipes a and b. By symmetry, half the flow goes to a and half to b. To understand what's happening at the entrance to the two pipes, it is easier to focus on pipe b. The velocity in pipe b is higher than in conduit 1. So the streamlines for the flow to pipe b have to be converging toward the entrance to pipe b. It's kind of like a funnel. This convergence occurs mainly in the region close to the entrance. The flow convergence causes the fluid velocity to increase and the pressure to decrease. The opposite thing happens for pipe a. | |
| May 12, 2016 at 1:09 | comment | added | Tom Chester | Hi @chestermiller thanks for your explanation , I am interested in why the fluid slows down before it enters pipe a ( and vice versa pipe b) intuitively I know this to be the case but I would love to put a name to the force . Is it due to the orthogonal area presented to the flow by the convergent pipe causes a force impulse . Why I wonder is again this suggests that the Q would deviate to pipe b in an inviscid flow (as you stated) but the k values and head loss suggest that in a viscid flow the Q would deviate to pipe a ... This causes me much confusement | |
| May 11, 2016 at 22:17 | comment | added | Chet Miller | Of course, the discussion in my previous comment applies specifically to the case without friction. | |
| May 11, 2016 at 22:11 | comment | added | Chet Miller | The flow that enters pipe a from the upstream conduit 1 slows down in the immediate vicinity of the entrance to pipe a, with an accompanying increase in pressure. Once inside pipe a, the flow speeds up as it converges toward the exit, with an accompanying decrease in pressure (to a value lower then in either conduit 1 or conduit 2). Once the flow exits pipe a, it slows down to the velocity in conduit 2, with an accompanying increase in pressure, back up to the original value in conduit 1. So the for the flow through a, the pressure first increases, then decreases, then increases again. | |
| May 11, 2016 at 13:38 | comment | added | Chet Miller | That's correct. The fluid flow converging from the big pipe into the entrance of pipe a (and the corresponding pressure change) is different from the fluid flow and pressure change from the big pipe into the entrance of pipe b. In the case of pipe a, the flow is slowing down as it approaches the entrance, and, in the case of pipe b, the flow is speeding up as it approaches the entrance. | |
| May 11, 2016 at 13:32 | comment | added | Tom Chester | hi @chestermiller yes that all makes sense with the literature , but only if p2a does not equal p2b is that your expectation in fact no matter how the flow is split I I can't see any way it can (but it does normalise downstream) | |
| May 11, 2016 at 11:18 | history | answered | Chet Miller | CC BY-SA 3.0 |