Fluid dynamics in horizontal pipe

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
10 replies · 2K views
Bengo
Messages
46
Reaction score
0
There's a figure that comes with the question but I'm having trouble attaching it so I will describe it the best I can.

There is a large cylinder labeled the reservoir. A horizontal pipe is connected near the base of the reservoir and it is open at the other end so fluid flows out (point B). Then a small vertical horizontal cylinder labeled column 1 that is connected at about halfway of the horizontal pipe.

What will be observed when a more viscous liquid, of the same mass density, is substituted for the less viscous liquid in the system?

Answer: a lower fluid velocity at point B, but an unchanged fluid height in column 1.

I've found 2 threads on this question on another site but I still don't understand how the height of column 1 remains unchanged if the fluid velocity is slower. Wouldn't that mean increased pressure meaning the fluid in column 1 will rise?

Thank you!
 
on Phys.org
What causes the slower velocity?
 
Yes, but what specifically slows it down?
 
Flow rate? According to poiseuilles principle?
 
I don't think anything is changing other than the viscosity and the flow rate. The pressure difference is the same, so is the length and radius.
 
Oooh I think see now. So if there was another column closer to the reservoir the height would be higher compared to the original column? But I still don't understand why Bernoullis equation doesn't apply. That's what I originally used, thinking decreased speed --> higher pressure
 
Well first, Bernoulli's equation doesn't apply to viscous flows, which this clearly is. There are certain corrections that you can make to empirically apply it to things like pipe flow, but you can't do it straight up and you can't do it at all analytically.

Second, Bernoulli's equation doesn't generally apply to comparing points in two different flows originating in different reservoirs because it is really a statement of conservation of energy. So with two different reservoirs, the total energy isn't guaranteed to be the same in the two situations and therefore Bernoulli's equation is not necessarily meaningful. This is related to why it doesn't work for viscous flows since viscosity is dissipative and is going to break this sort of energy balance equation. Now, it just so happens that in this situation, since you held all the other parameters constant, if there was no viscosity, using Bernoulli's equation would have worked, but that result is not general and you shouldn't get into that habit.
 
Wow it's so much clearer now. Thank you so much!