Timeline for How does the radius of a pipe affect the rate of flow of fluid?
Current License: CC BY-SA 4.0
10 events
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| S Jun 13, 2019 at 19:41 | history | suggested | CommunityBot | CC BY-SA 4.0 |
The integral of r dr is r^2/2, not r^2
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| Jun 13, 2019 at 15:17 | review | Suggested edits | |||
| S Jun 13, 2019 at 19:41 | |||||
| Oct 10, 2017 at 17:21 | comment | added | Floris | @Ali I used "curvature" when I really meant "second derivative" - sloppy of me. I have added the details as requested. | |
| Oct 10, 2017 at 17:20 | history | edited | Floris | CC BY-SA 3.0 |
Added details of $R^2$ derivation as requested in comments.
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| Oct 10, 2017 at 15:35 | comment | added | Ali | I do indeed see it. Hopefully you don't think I'm nitpicking, I just found it missing from your nice answer. Maybe an equation like $v_0 = \frac{P}{\eta}R^2$ following the differential equation would be useful. Another comment, parabolas don't have a constant curvature along them, unless you mean the curvature at its vertex! | |
| Oct 10, 2017 at 15:13 | comment | added | Floris | @Ali I am not sure what I can add beyond the paragraph "It remains to prove..." which shows the equation that allows you to get velocity as a function of radius. Since the curvature is given by $P$ and $\eta$, the shape of the parabola will be the same for all $r$ - but if you double $r$, you will quadruple the height of the parabola (when the ends are kept at 0). Do you see it now or do I need to actually write the math into the answer? | |
| Oct 10, 2017 at 15:03 | comment | added | Ali | Can you elaborate more on how $v_0 \propto R^2$? | |
| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
replaced http://physics.stackexchange.com/ with https://physics.stackexchange.com/
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| Mar 3, 2015 at 11:56 | vote | accept | Tejas Ramdas | ||
| Mar 3, 2015 at 1:44 | history | answered | Floris | CC BY-SA 3.0 |