How to explain Flow direction?

[Sacaq]

Hi~

I have questions about the flow direction.
In a straight pipe flow with incompressible viscous fluid, the flow flows from a higher-pressure location to a lower-pressure location.
However, in a manifold pipe system, with a shape like:

flow inlet
↓
├─────→flow outlet
├─────→flow outlet
├─────→flow outlet
│
dead end

Higher pressure can be found in the dead end.

How can it be explained by using fluid dynamics?

Thank you!

 

[hydroman247]

As there is no flow at the dead end, there is no pressure drop. If there was flow, there would be a pressure drop but then by definition, it would not be a dead end. Remember flow is just an arbitrary representation of the rate in which force can be created in hydraulics.

 

[zdas04]

If this is not to be a trivial question, then we should assume that the header is vertical with the dead end on the bottom.

P[sub]total[/sub]=P[sub]static[/sub]+P[sub]dynamic[/sub]

Static pressure in a vertical system is made up of

P[sub]static[/sub]=P[sub]imposed[/sub]+ρ*g*h

Dynamic pressure is a function of the momentum of the fluid. If the hydrostatic pressure is greater than momentum (quite normal when fluid velocity is less than 0.3 Mach) then the pressure at the dead end is properly the highest pressure in the system as drawn (we can't draw any conclusions about where the outlet pipes go).

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. ùGalileo Galilei, Italian Physicist

 

[Sacaq]

Thank you all~
I try to explain this by the thought: pressure drop lead to flow.
However, it doesn't work in a Venturi tube.
In a Venturi tube, the flow goes through the tube with pressure variations of decrement and increment.
Maybe the flow is not in the same streamline, so that Bernoulli equation should not apply at here.
If this is the case, how do I explain this?
Flow separation?
dead air zone?
laminar mixing?
energy dissipation due to viscosity?