0
$\begingroup$

Let's say there are 2 identical pipes (same diameter, surface roughness) in which a fluid flows.

The static pressure of both the flows are equal. But Flow A is faster than Flow B. Will there be any difference in the pressure exerted on the pipes by these 2 flows?

(I tried to work this out with Bernoulli but that only works if there is no additional energy in the flow. Here Flow A has additional kinetic energy)

Thanks in advance :)

$\endgroup$

1 Answer 1

0
$\begingroup$

I don’t quite understand what you mean by “static pressure.” There’s just one pressure, which obeys the Bernoulli equation, $$ P+\frac12\rho v^2+mgh=\text{constant} $$ I guess you must mean the value of $P$ for stationary fluid at your reference height. But in a flowing fluid, the pressure will change through the flowing region. If the flow is doing work, the pressure difference across a fluid element explains the net force which is responsible for the work done.

In particular, flow rate in a pipe with energy loss requires a pressure difference between the inlet and the outlet. So if you want the “static pressure” to be same for two pipes with different flow, you have to specify where. If both pipes drain to the atmosphere, the increased flow must come from a more powerful pump, and the pressure will be higher everywhere in the more strongly driven pipe. You could also have the two pipes both filling from the same inlet pressure, in which case the more powerful pump (at the outlet) would produce a lower pressure through the pipe.

Imagine your two pipes are parallel to each other, same length, with a small crosspipe joining them in the middle somewhere. The direction of flow in the crosspipe will tell you the direction of the pressure difference. If the faster-flowing pipe has a pump pushing extra fluid through the inlet of faster pipe, it’ll push fluid through the crosspiece. If the pump is pulling more fluid through the outlet of the faster pipe, that pump will also pull fluid through the crosspiece.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.