What's the pressure long before the fluid exits a pipe?

  • #1
lost captain
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Thread moved from the technical forums to the schoolwork forums
TL;DR Summary: We know that the fluid pressure at the exit of the pipe is Patm but what about before the exit? How can we calculate that pressure using Bernoulli's equation.

Hello everyone☺️. Please help, i'm seriously stuck 🫠🙇‍♀️
1000013332.jpg


When learning about bernoulli's equation there's a classic problem with a tank draining that asks us to calculate the velocity of the fluid at the exit. Like in the picture. We said that st the exit the pressure is 1atm, so what about before that?
What's the pressure at points 1 2 3 4 and 5 as shown in the picture?
(The fluid is ideal, no friction, no viscosity, constant density)

Thanks in advance for spending time in my silly confusion.
 
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  • #2
Bernoulli is an energy conservation statement. Since you have no friction there is *in this model* no reason for a pressure difference to exist between point 3 and 5, so all these pressures are equal (assuming that point 3 is exactly at the pipe inlet, where the velocity is already equal to the velocity in the pipe). Since it is exactly the friction in a pipe that causes the pressure to become higher. So you've shown that P4 = Patm, which is indeed true *in this model*. In reality it will be a bit higher due to friction.

So in reality there will be a small region in the tank around 3 where the flow accelerates to the velocity in the pipe and here the pressure thus drops from its rho*g*h value to Patm. But to compute this requires detailed flow analysis, which I assume is not part of the task.

You could also assume point 3 to be before the region where the flow starts to accelerate towards the pipe entrance (i.e. it's velocity becomes significant), in which case the velocity would be nearly zero and the pressure thus rho*g*h. From the drawing it is not really clear that point 3 would be exactly *at* the inlet.

For point 1 and 2 you can just compute the pressure with rho*g*h assuming the velocity is zero. You assume zero velocity because the size of the tank is large compared to the flow through the pipe. (So the answer to question 1 is yes, and the velocity at 2 is assumed to be zero).
 
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  • #3
Thank you so much for taking the time to reply🙇‍♂️😊. Did you perhaps mean "lower" here: "Since it is exactly the friction in a pipe that causes the pressure to become higher"
 
  • #4
Depends on which way you reason, I meant 'higher at P4 compared to P5"
 
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  • #5
Ok i understand now. But what if the pipe got narrower at point 5? Then according to the continuity equation the pressure should drop and the velocity increase, but pressure at point 5 is Patm.
 
  • #6
The pressure at point 5 is given, it is equal to Patm, always. So if the pipe gets narrower from point 3 to point 5 it will mean that the pressure at point 3 and 4 will be higher than Patm as per the Bernoulli equation.
 
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  • #7
And then the velocities at points 3 and 4 will be lower?
 
  • #8
lower than what? They will be higher than Patm
 
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  • #9
I said "lower" because i thought that in general, when the pressure at a point increases then it's velocity should decrease. But now i see that's not the case all the time, for example point 5, its velocity increases but the pressure stays the same.
So, lets make this clear:

Velocity 5 will increase ( because of continuity equation)

Pressure 5 will stay the same, = Patm (because at point 5 the pipe is open)

Velocity 4 will stay the same, as it was, (before making the pipe narrower at point 5) because the pipe at point 4 hasn't change, it's still the same diameter, the same area A4 (again because of continuity equation A₄ * V₄ = A₅ * V₅ )

Pressure 4 will increase, it will become bigger than Patm ( because of Bernoulli's equation)

Is the above correct? But i can't understand why Velocity 4 should be higher than Patm as you said in your last reply.
Once again thank you, very much, our conversation gave me a new insight on Bernoulli's and continuity equation
 
  • #10
lost captain said:
But i can't understand why Velocity 4 should be higher than Patm as you said in your last reply.
You've mixed velocity and pressure there.
If the pipe is narrowing from 3 to 5 then the velocity is increasing along it and the pressure reducing along it. That has to follow because it is the reducing pressure that causes the fluid to accelerate.
 
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  • #11
haruspex said:
You've mixed velocity and pressure there.
If the pipe is narrowing from 3 to 5 then the velocity is increasing along it and the pressure reducing along it. That has to follow because it is the reducing pressure that causes the fluid to accelerate.
The pipe doesn't narrow from 3 to 5, the pipe narrows at point 5(or a little bit before 5). Im sorry if i didn't clear this up.
So if the area at point 4 doesn't change, then the velocity V₄ doesn't change either. So the pressure at 4 must then increase. P₄ will be bigger than it was before making the pipe narrower at point 5.
When i say i make the pipenarrower, i mean for example putting a valve there, at 5, and having it half open.
 
  • #12
lost captain said:
I said "lower" because i thought that in general, when the pressure at a point increases then it's velocity should decrease. But now i see that's not the case all the time, for example point 5, its velocity increases but the pressure stays the same.
So, lets make this clear:

Velocity 5 will increase ( because of continuity equation)

Pressure 5 will stay the same, = Patm (because at point 5 the pipe is open)

Velocity 4 will stay the same, as it was, (before making the pipe narrower at point 5) because the pipe at point 4 hasn't change, it's still the same diameter, the same area A4 (again because of continuity equation A₄ * V₄ = A₅ * V₅ )

Pressure 4 will increase, it will become bigger than Patm ( because of Bernoulli's equation)

Is the above correct?

[edit, I overlooked the statement of velocity at 5]
Depends on what you mean with 'the velocity at 5 will increase'. You will need to learn to specify compared to what it increases. It will increase compared to the velocity at p4 if there is a constriction at p5 because of continuity.

However, the velocity at 5 does not change compared to the previous case without constriction, no matter the outlet diameter. This velocity is purely driven by the difference in pressure at the depth of p5 and Patm. So deltaP = 'pressure at the depth of 5 before accelerating' - 'pressure 5 after accelerating' = (Patm + rho*g*h5) - (Patm) = rho*g*h5. This pressure difference is converted to a velocity via Bernoulli. It is independent of the outlet area. You could say that rho*g*h5 gives an available 'budget' by which you can accelerate the flow towards the outlet at point 5.

The other statements are correct.
[/edit]

The confusion starts because it isn't clear to what the pressure is compared against. If you state that pressure 4 increases, you mean increase with respect to the previous case without outlet constriction at 5, which is correct.

If @haruspex states that the pressure from 3 to 5, along 4, decreases because the flow accelerates (he, as wel as me, assumed a continuously decreasing pipe diameter) he means P5 < P4 < P3 (Px meaning 'pressure at x') which is also correct.

lost captain said:
But i can't understand why Velocity 4 should be higher than Patm as you said in your last reply.

(I assume that you mean 'why Pressure 4 should be higher')

The reason the pressure at 4 is higher than Patm is because the flow needs to accelerate from 4 to 5. And at point 5 the pressure is always Patm, no matter what! This is a boundary condition! Only cataclysmic events like the sun stripping away the earth's atmosphere will change the pressure at 5. Assuming that's not part of the model the pressure at 5 will remain Patm. Always.

Since, due to the constriction, the velocity at 5 is higher than 4, the pressure at 4 must be higher than at 5. So pressure is exchanged with velocity, which is exactly what Bernoulli states.

lost captain said:
Once again thank you, very much, our conversation gave me a new insight on Bernoulli's and continuity equation

You're welcome!
 
Last edited:
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  • #13
Arjan82 said:
[edit, I overlooked the statement of velocity at 5]
Depends on what you mean with 'the velocity at 5 will increase'. You will need to learn to specify compared to what it increases. It will increase compared to the velocity at p4 if there is a constriction at p5 because of continuity.

However, the velocity at 5 does not change compared to the previous case without constriction, no matter the outlet diameter. This velocity is purely driven by the difference in pressure at the depth of p5 and Patm. So deltaP = 'pressure at the depth of 5 before accelerating' - 'pressure 5 after accelerating' = (Patm + rho*g*h5) - (Patm) = rho*g*h5. This pressure difference is converted to a velocity via Bernoulli. It is independent of the outlet area. You could say that rho*g*h5 gives an available 'budget' by which you can accelerate the flow towards the outlet at point 5.

The other statements are correct.
[/edit]

The confusion starts because it isn't clear to what the pressure is compared against. If you state that pressure 4 increases, you mean increase with respect to the previous case without outlet constriction at 5, which is correct.

If @haruspex states that the pressure from 3 to 5, along 4, decreases because the flow accelerates (he, as wel as me, assumed a continuously decreasing pipe diameter) he means P5 < P4 < P3 (Px meaning 'pressure at x') which is also correct.



(I assume that you mean 'why Pressure 4 should be higher')

The reason the pressure at 4 is higher than Patm is because the flow needs to accelerate from 4 to 5. And at point 5 the pressure is always Patm, no matter what! This is a boundary condition! Only cataclysmic events like the sun stripping away the earth's atmosphere will change the pressure at 5. Assuming that's not part of the model the pressure at 5 will remain Patm. Always.

Since, due to the constriction, the velocity at 5 is higher than 4, the pressure at 4 must be higher than at 5. So pressure is exchanged with velocity, which is exactly what Bernoulli states.



You're welcome!
Arjan82 said:
[edit, I overlooked the statement of velocity at 5]
Depends on what you mean with 'the velocity at 5 will increase'. You will need to learn to specify compared to what it increases. It will increase compared to the velocity at p4 if there is a constriction at p5 because of continuity.

However, the velocity at 5 does not change compared to the previous case without constriction, no matter the outlet diameter. This velocity is purely driven by the difference in pressure at the depth of p5 and Patm. So deltaP = 'pressure at the depth of 5 before accelerating' - 'pressure 5 after accelerating' = (Patm + rho*g*h5) - (Patm) = rho*g*h5. This pressure difference is converted to a velocity via Bernoulli. It is independent of the outlet area. You could say that rho*g*h5 gives an available 'budget' by which you can accelerate the flow towards the outlet at point 5.

The other statements are correct.
[/edit]

The confusion starts because it isn't clear to what the pressure is compared against. If you state that pressure 4 increases, you mean increase with respect to the previous case without outlet constriction at 5, which is correct.

If @haruspex states that the pressure from 3 to 5, along 4, decreases because the flow accelerates (he, as wel as me, assumed a continuously decreasing pipe diameter) he means P5 < P4 < P3 (Px meaning 'pressure at x') which is also correct.



(I assume that you mean 'why Pressure 4 should be higher')

The reason the pressure at 4 is higher than Patm is because the flow needs to accelerate from 4 to 5. And at point 5 the pressure is always Patm, no matter what! This is a boundary condition! Only cataclysmic events like the sun stripping away the earth's atmosphere will change the pressure at 5. Assuming that's not part of the model the pressure at 5 will remain Patm. Always.

Since, due to the constriction, the velocity at 5 is higher than 4, the pressure at 4 must be higher than at 5. So pressure is exchanged with velocity, which is exactly what Bernoulli states.



You're welcome!
Arjan82 said:
[edit, I overlooked the statement of velocity at 5]
Depends on what you mean with 'the velocity at 5 will increase'. You will need to learn to specify compared to what it increases. It will increase compared to the velocity at p4 if there is a constriction at p5 because of continuity.

However, the velocity at 5 does not change compared to the previous case without constriction, no matter the outlet diameter. This velocity is purely driven by the difference in pressure at the depth of p5 and Patm. So deltaP = 'pressure at the depth of 5 before accelerating' - 'pressure 5 after accelerating' = (Patm + rho*g*h5) - (Patm) = rho*g*h5. This pressure difference is converted to a velocity via Bernoulli. It is independent of the outlet area. You could say that rho*g*h5 gives an available 'budget' by which you can accelerate the flow towards the outlet at point 5.

The other statements are correct.
[/edit]

The confusion starts because it isn't clear to what the pressure is compared against. If you state that pressure 4 increases, you mean increase with respect to the previous case without outlet constriction at 5, which is correct.

If @haruspex states that the pressure from 3 to 5, along 4, decreases because the flow accelerates (he, as wel as me, assumed a continuously decreasing pipe diameter) he means P5 < P4 < P3 (Px meaning 'pressure at x') which is also correct.



(I assume that you mean 'why Pressure 4 should be higher')

The reason the pressure at 4 is higher than Patm is because the flow needs to accelerate from 4 to 5. And at point 5 the pressure is always Patm, no matter what! This is a boundary condition! Only cataclysmic events like the sun stripping away the earth's atmosphere will change the pressure at 5. Assuming that's not part of the model the pressure at 5 will remain Patm. Always.

Since, due to the constriction, the velocity at 5 is higher than 4, the pressure at 4 must be higher than at 5. So pressure is exchanged with velocity, which is exactly what Bernoulli states.



You're welcome!

🔸️"Depends on what you mean with 'the velocity at 5 will increase'. You will need to learn to specify compared to what it increases. It will increase compared to the velocity at p4 if there is a constriction at p5 because of continuity"🔸

Yes i'm sorry, i should have been more specific.

🔸️However, the velocity at 5 does not change compared to the previous case without constriction, no matter the outlet diameter. 🔸


So when there is no constriction Pressure at 4 is the same as Pressure at 5 and equal to Patm, P₄ = P₅ = Patm
and so their velocities should also be the same V₄ = V₅ (Bernoulli's equation)

Now, when there is constriction, when the diameter at point 5 becomes smaller (because of a valve for example), we know that V₅ > V₄ (continuity equation) .
And now, i am thinking to myself, what is V₄ ? Has V₄ changed compared to what it was before the constriction? Is V₄ the same as it was before the constriction?
 
  • #14
lost captain said:
Now, when there is constriction, when the diameter at point 5 becomes smaller (because of a valve for example), we know that V₅ > V₄ (continuity equation) .

Correct

lost captain said:
And now, at this point of the problem, i am thinking what is V₄ ? Has V₄ changed compared to what it was before the constriction? Is V₄ the same as it was before the constriction?

V5 is equal to before the constriction, That means, since A5 is smaller, that the volume flow (A5*V5) is smaller at 5. Because of continuity the volume flow at 4 is equal to 5, this means that....

You should be able to do this yourself now :)
 
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