Pressure in inclined pipe

In summary, the author gives that volume flow rate = delta P - (rho)(g)(L sin theta) . IMO, it's wrong because the equation should be P1 - (rho)(g)(L sin theta) .
  • #1
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Homework Statement


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

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
 
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  • #2
P2 is zero gauge. the pipe discharges to the atmosphere.
 
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  • #3
Chestermiller said:
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) ?
 
  • #4
foo9008 said:
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?
 
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  • #5
Chestermiller said:
Are you familiar with the Navier Stokes equations?
no , i have never heard of that
 
  • #6
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##.
 
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What is pressure in an inclined pipe?

Pressure in an inclined pipe refers to the amount of force per unit area that is exerted on the walls of the pipe. It is caused by the weight of the fluid inside the pipe and the force of gravity acting on the fluid.

How does the angle of inclination affect pressure in a pipe?

The angle of inclination of a pipe affects the pressure inside it because it determines the vertical height of the fluid column. The higher the angle of inclination, the higher the pressure due to the increased weight of the fluid.

What is the formula for calculating pressure in an inclined pipe?

The formula for calculating pressure in an inclined pipe is P = ρghsinθ, where P is pressure, ρ is the density of the fluid, g is the acceleration due to gravity, h is the vertical height of the fluid column, and θ is the angle of inclination.

How does the density of the fluid affect pressure in an inclined pipe?

The density of the fluid has a direct impact on the pressure in an inclined pipe. The higher the density of the fluid, the higher the pressure, as there is more weight exerted on the walls of the pipe.

Can pressure in an inclined pipe be greater than atmospheric pressure?

Yes, pressure in an inclined pipe can be greater than atmospheric pressure. This occurs when the angle of inclination is steep, resulting in a higher weight of the fluid and therefore a higher pressure. However, the pressure will still decrease as the fluid moves along the pipe due to frictional losses.

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