# Pressure in inclined pipe

## Homework Statement

in the solution , the author gave that volume flow rate = delta P - (rho)(g)(L sin theta) ,
IMO , it's wrong ...

## The Attempt at a Solution

it should be P1 - (rho)(g)(L sin theta) , where (rho)(g)(L) = P2 , correct me if i am wrong . i think so because i consider P1 - (rho)(g)(L sin theta) as delta P

## Answers and Replies

Chestermiller
Mentor
P2 is zero gauge. the pipe discharges to the atmosphere.

foo9008
P2 is zero gauge. the pipe discharges to the atmosphere.
why the author wrote it as delta P - (rho)(g)(L sin theta) ? why not only delta P , which is Pressure at region 1 only ( the P2 = 0 ) ??? why there's an extra (rho)(g)(L sin theta) ?

Chestermiller
Mentor
why the author wrote it as delta P - (rho)(g)(L sin theta) ? why not only delta P , which is Pressure at region 1 only ( the P2 = 0 ) ??? why there's an extra (rho)(g)(L sin theta) ?
Are you familiar with the Navier Stokes equations?

foo9008
Are you familiar with the Navier Stokes equations?
no , i have never heard of that

Chestermiller
Mentor
If the fluid were in static equilibrium, then we would have $$P_1+\rho g z_1=P_2+\rho g z_2$$In this problem, they are taking the datum for elevation z as location 1, the entrance to the pipe section. So ##z_1=0##. So for static equilibrium, our equation becomes:$$P_1=P_2+\rho g z_2$$ The elevation ##z_2## is related to the length of the pipe and the pipe angle by:$$z_2=L\sin \theta$$. Therefore, for static equilibrium, we have:
$$P_1=P_2+\rho g L\sin \theta$$Now, if ##P_1>P_2+\rho g L\sin \theta##, the fluid will flow from point 1 (the inlet) to point 2 (the outlet). Therefore, the driving force for fluid flow from entrance to exit is ##P_1-P_2-\rho g L\sin \theta=\Delta P-\rho g L\sin \theta##.

foo9008