Gauge pressure in Momentum conservation of fluids.

In summary: Yes I did but the problem I am facing is if we consider a CV which includes only the fluid in the tube , we get a pressure of 220 kpa acting on left cross section a pressure of 1atm on the cross section where the fluid leaves and then an Fx for horizontal force by pipe and Fy for vertical force by the pipe. Now when we'll apply force balance on pipe we'll get Fx in opposite direction on it by fluid and Fy in opposite direction by fluid in horizontal and vertical direction respectively. Now the question is how can we right the total pressure force...
  • #1
yash_it_is
7
0
I want to ask why is it that we use gauge pressure instead of absolute pressure in CV analysis for momentum conservation of fluids.

I did read that because P(atm) would be present everywhere so it won't have a net effect on the CV but it's highly non intuitive as I can't apply force balance on something like random and satisfactorily say that the net force due to atmospheric pressure is indeed zero and we only need gauge pressure.

Please explain this to me in as much detail as possible.
thanks!
 
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  • #2
yash_it_is said:
it's highly non intuitive
Intuition is not a good reason for accepting or rejecting scientific ideas.
Ambient pressure can have significant effect on the properties of a fluid but it is only differences in pressure that are needed when discussing linear flow.
yash_it_is said:
I can't apply force balance on something like random
Why not? Elementary kinetic theory uses simple statistics to account for the forces on a piston due to random motion of molecules. Where is your problem with this idea? Does this link make any sense to you?
 
  • #3
sophiecentaur said:
Intuition is not a good reason for accepting or rejecting scientific ideas.
Ambient pressure can have significant effect on the properties of a fluid but it is only differences in pressure that are needed when discussing linear flow.

Why not? Elementary kinetic theory uses simple statistics to account for the forces on a piston due to random motion of molecules. Where is your problem with this idea? Does this link make any sense to you?
the link ain't working.
 
  • #4
yash_it_is said:
the link ain't working.
Also my problem is if we had a bent tube let's say and the pressure at one cross section is P1 and the pressure at the outlet is P2 both being absolute and the atmospheric pressure being Patm( considering area of cross sections to change arbitrarily throughout the pipe). why do we write surface force F = (P1-Patm)A at one end and not F= P1A.
 
  • #5
Have you tried to solve any particular CV momentum balance problem using first gauge pressures and then absolute pressures? They should give the same results, plus you will have a better understanding of why this plays out this way.
 
  • #6
Chestermiller said:
Have you tried to solve any particular CV momentum balance problem using first gauge pressures and then absolute pressures? They should give the same results, plus you will have a better understanding of why this plays out this way.
I did but the problem I am facing is when we have arbitrary variation in cross section one cannot solve for force due to pressure so easily but the case of gauge pressure as effective pressure over anybody with any cross section is generalized.
I'd really like a mathematical proof.
 
  • #7
yash_it_is said:
I did but the problem I am facing is when we have arbitrary variation in cross section one cannot solve for force due to pressure so easily but the case of gauge pressure as effective pressure over anybody with any cross section is generalized.
I'd really like a mathematical proof.
Please show what you have done in the analyses of the specific problem.
 
  • #8
Chestermiller said:
Please show what you have done in the analyses of the specific problem.
this was the question i was working on and instead of the gauge pressure they used I used Pabsolute and the solution was different cause I don't understand how Patm force was able to cancel out Pabsolute in the x direction in cross section 1 thereby leading P1-Patm force.
 

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  • #9
yash_it_is said:
this was the question i was working on and instead of the gauge pressure they used I used Pabsolute and the solution was different cause I don't understand how Patm force was the Pabsolute in the x direction thereby leading P1-Patm force.
I have trouble seeing the text because it is small and blurry. But, I get the gist of the problem you are attacking. When you set up the problem using absolute pressures, did you include the atmospheric pressure force on the outside of the pipe?
 
  • #10
Chestermiller said:
I have trouble seeing the text because it is small and blurry. But, I get the gist of the problem you are attacking. When you set up the problem using absolute pressures, did you include the atmospheric pressure force on the outside of the pipe?
Yes I did but the problem I am facing is if we consider a CV which includes only the fluid in the tube , we get a pressure of 220 kpa acting on left cross section a pressure of 1atm on the cross section where the fluid leaves and then an Fx for horizontal force by pipe and Fy for vertical force by the pipe. Now when we'll apply force balance on pipe we'll get Fx in opposite direction on it by fluid and Fy in opposite direction by fluid in horizontal and vertical direction respectively. Now the question is how can we right the total pressure force due to atm pressure in X direction as -patm.A1 where A1 is the are of inlet cross section .
 
  • #11
yash_it_is said:
Yes I did but the problem I am facing is if we consider a CV which includes only the fluid in the tube , we get a pressure of 220 kpa acting on left cross section a pressure of 1atm on the cross section where the fluid leaves and then an Fx for horizontal force by pipe and Fy for vertical force by the pipe. Now when we'll apply force balance on pipe we'll get Fx in opposite direction on it by fluid and Fy in opposite direction by fluid in horizontal and vertical direction respectively. Now the question is how can we right the total pressure force due to atm pressure in X direction as -patm.A1 where A1 is the are of inlet cross section .
You are trying to get the force on the pipe as a result of the flow. Your method correctly determines the force of the pipe on the fluid. But, when you do the force balance on the pipe then, you need to include the force of the atmosphere on the outside of the pipe.
 
  • #12
Chestermiller said:
You are trying to get the force on the pipe as a result of the flow. Your method correctly determines the force of the pipe on the fluid. But, when you do the force balance on the pipe then, you need to include the force of the atmosphere on the outside of the pipe.
yes that's what I'm trying to understand why in the X-direction due to the pressure the force is (P1 - patm)A1 when patm acts all over the pipe how can we say that in the X-direction due to pressure the net force by atmospheric pressure be -PatmA1 when Patm clearly acts everywhere and net Area on which it should act to give a force in X direction isn't visible.
 
  • #13
yash_it_is said:
yes that's what I'm trying to understand why in the X-direction due to the pressure the force is (P1 - patm)A1 when patm acts all over the pipe how can we say that in the X-direction due to pressure the net force by atmospheric pressure be -PatmA1 when Patm clearly acts everywhere and net Area on which it should act to give a force in X direction isn't visible.
Please write out your equations for the absolute pressure case so I can get a better understanding of your issue.
 
  • #15
If there were no fluid flowing through the pipe, and there were end caps placed over the inlet and outlet of the pipe, then the pipe would be in equilibrium (neglecting its weight) under the action of atmospheric pressure. This means that the net atmospheric pressure force acting on the entire portion of the pipe that does not include the end caps must be equal and opposite to the net atmospheric pressure force acting on the end caps. When fluid is flowing through the pipe, the atmospheric pressure force acting on the portion of the pipe that does not include the inlet and outlet has to be the same.
 
  • #16
Chestermiller said:
If there were no fluid flowing through the pipe, and there were end caps placed over the inlet and outlet of the pipe, then the pipe would be in equilibrium (neglecting its weight) under the action of atmospheric pressure. This means that the net atmospheric pressure force acting on the entire portion of the pipe that does not include the end caps must be equal and opposite to the net atmospheric pressure force acting on the end caps. When fluid is flowing through the pipe, the atmospheric pressure force acting on the portion of the pipe that does not include the inlet and outlet has to be the same.

That's a great answer! I spent a few hours being stuck on the same issue. But this saved me. I have a few more thoughts about it now:

Think of a horizontal pipe with a large area (A1) on the left and a smaller one (A2) on the right. Imagine that there is no flow. If you think about the forces acting only on the two ends, they have to be different since P_atm*A1 > P_atm*A2. However, along the length of the pipe, the c/s area changes gradually, due to which the pipe wall makes a certain angle with the horizontal. Consequently, the external force acting on the pipe will have two components, one along the radius of the pipe and one along the pipe axis. The force projected along the pipe axis is a product of P_atm and the pipe's area projected horizontally, i.e. (A1-A2). This force will be transferred from the pipe to the flow as a reaction force. Thus, the net force acting on the flow due to the atmospheric pressure is (P_atm*A1): at the left end, (-P_atm*A2): at the right end and is negative due to being leftward, and [-P_atm*(A1-A2)]: on the pipe wall and is also directed leftward. The net horizontal force due to the atmospheric pressure will be the addition of these three forces, P_atm*A1 - P_atm*A2 - P_atm*(A1-A2), which is zero.

I hope that this mathematical explanation helps understanding the intuition of having net 0 atmospheric pressure force and using only the gauge pressure (rather than the absolute) in the momentum equation.
 

FAQ: Gauge pressure in Momentum conservation of fluids.

What is gauge pressure?

Gauge pressure is the pressure measured relative to atmospheric pressure. It is the difference between the absolute pressure and the local atmospheric pressure.

How is gauge pressure used in momentum conservation of fluids?

Gauge pressure is used in momentum conservation of fluids to calculate the forces acting on a fluid as it moves through a system. It is an important factor in determining the direction and magnitude of fluid flow.

What is the equation for gauge pressure?

The equation for gauge pressure is P = Pabs - Patm, where P is the gauge pressure, Pabs is the absolute pressure, and Patm is the atmospheric pressure.

How is gauge pressure different from absolute pressure?

Gauge pressure is different from absolute pressure because it takes into account the atmospheric pressure, while absolute pressure does not. Gauge pressure is relative, while absolute pressure is measured from a complete vacuum.

What are some common units for gauge pressure?

Common units for gauge pressure include pounds per square inch (psi), bar, pascal (Pa), and kilopascal (kPa). Other units, such as millimeters of mercury (mmHg) and atmospheres (atm), can also be used.

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