0
$\begingroup$

enter image description here

I just wanted to enquire, can this system ever be possible for ideal fluid.

I strongly believe it can't be possible since, there has been no increase in the cross section, but still the pressure difference has increased between the points, indicating a change in velocity which would ultimately flout the equation of continuity.

One more thing, can viscous fluid ever achieve this configuration?

P.S. the fluid is same everywhere.

Improve this question
$\endgroup$
0
$\begingroup$

By definition an ideal liquid is inviscid. An inviscid fluid flows through a pipe without any pressure losses because there are no viscous (friction) energy losses.

In your figure the pipe $L,L,L$ is perfectly horizontal, so there are no changes in potential energy either.

In that situation, for an ideal liquid, the levels in the manometers would all be identical.

One more thing, can viscous fluid ever achieve this configuration?

For a viscous, Newtonian liquid in steady flow pressure loss is directly proportional to pipe length (for a pipe of constant cross section). That would yield the pattern shown in your figure. See Darcy-Weisbach equation.

Improve this answer
$\endgroup$
9
  • $\begingroup$ It would also be true for steady turbulent flow. $\endgroup$ – Chet Miller Oct 15 '16 at 2:32
  • $\begingroup$ You're right: it's only the friction coefficient that differs. It's been edited, thanks. $\endgroup$ – Gert Oct 15 '16 at 4:05
  • $\begingroup$ But if I measure $h_i \rho g$ ${i=1,2,3}$ that would yield different pressures beneath $h_i$. Won't that make my system go wrong? $\endgroup$ – user118752 Oct 15 '16 at 9:21
  • $\begingroup$ I know that whenever a fluid (non-viscuous) passes through tube of same crossection, there is no change in kinetic energy that's why no change in pressure (assuming the tube to be horizontal) but when we apply the$ h_i \rho g$ at subsequent points L, there is actually a decrease in pressure. How is that possible? $\endgroup$ – user118752 Oct 15 '16 at 9:55
  • $\begingroup$ @Harsh Sharma It won't make your system go wrong. There is no flow in those vertical tubes, so it merely represents a hydrostatic measurement of the pressure at the three locations. It's the same as if you put a pressure transducer in the wall of the tube at those locations. $\endgroup$ – Chet Miller Oct 15 '16 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy