Bernoulli vs real life experiment

[gen x]

Homework Statement

testing Bernoulli in real life experiment

Relevant Equations

P1+1/2 ρ V1^2 = P2 + 1/2 ρ V2^2
A1 V1= A2 V2

If have close pipe system with water inside pressurized at P1= 200 000Pa absolute, density 1000kg/m3, wider pipe diameter=2cm, contraction pipe diameter=1.49cm, that is contraction area ratio A1/A2=1.8

a) If water is stationary(pump OFF) and if I drill a hole anywhere at pipe, water will leak out, because pressure(200kPa) inside is higher than atmospheric pressure (101 325Pa).

b)If I turn on pump and water start flowing with with v1=10m/s in A1 wider section, from Bernoulli equation I will get pressure P2 in contraction(section A2) 88 491Pa, that mean, if I now drill hole in narrow section, water will not leak out, air will be sucked in because P2 is lower then atmospheric pressure.

How realistic is this?

 

[erobz]

a) how is pressure being applied if the pump is off?

b) $v_1$ appears out of thin air, where did you get it?

I’d like to explore what you think the entire system is for a bit. Then maybe the computation you could elaborate on. See latex guide for the mathematics formatting here. https://www.physicsforums.com/help/latexhelp/

 

[Lnewqban]

Is the P1 value relative or absolute?

 

[haruspex]

a) how is pressure being applied if the pump is off?

b) $v_1$ appears out of thin air, where did you get it?

I’d like to explore what you think the entire system is for a bit. Then maybe the computation you could elaborate on. See latex guide for the mathematics formatting here. https://www.physicsforums.com/help/latexhelp/

My guess is that there is some head of water providing pressure and the pump merely circulates the water at the bottom in some horizontal loop, the rate through the wider portion being set to $v_1$.
According to the calculation, with the pump on, air will be drawn into the loop and bubble up to the surface (which I take to be open to the atmosphere).

 

[erobz]

My guess is that there is some head of water providing pressure and the pump merely circulates the water at the bottom in some horizontal loop, the rate through the wider portion being set to $v_1$.
According to the calculation, with the pump on, air will be drawn into the loop and bubble up to the surface (which I take to be open to the atmosphere).

In my opinion “the system” must be precisely tuned to achieve relative vacuum in the contraction.

For instance, If it’s jetting to atmosphere at that point, it won’t achieve relative vacuum. If there is a static head keeping water column in the system before the pump is on than in any real system the pressure climbs from there to circulate via pumping. I want to see that they understand some of these “real system” responses before we say blindly “given a,b” a Venturi effect arises when it’s not necessarily the case in real plumbing.

 

[gen x]

a) how is pressure being applied if the pump is off?

b) $v_1$ appears out of thin air, where did you get it?

a) initial pressure is 200kPa, open valve fill the pressure and close the valve. Pump is here for circulation not pressure

b)from pump that is inside pipe system

Why is it important how pressure and speed are achieved?

Is the P1 value relative or absolute?

write in text, absolute

 

[erobz]

Why is it important how pressure and speed are achieved?

See my latest reply to @haruspex.

Post in thread 'Bernoulli vs real life experiment'
https://www.physicsforums.com/threads/bernoulli-vs-real-life-experiment.1083214/post-7291159

Given the thread title there is much to unpack.

 

[haruspex]

In my opinion “the system” must be precisely tuned to achieve relative vacuum in the contraction.

For instance, If it’s jetting to atmosphere at that point, it won’t achieve relative vacuum. If there is a static head keeping water column in the system before the pump is on than in any real system the pressure climbs from there to circulate via pumping. I want to see that they understand some of these “real system” responses before we say blindly “given a,b” a Venturi effect arises when it’s not necessarily the case in real plumbing.

Agreed - I only said "according to the calculation". The pump can only generate a flow by raising pressure. However, once a circulation is established the generated pressure only need be enough to overcome drag.

 

[gen x]

Agreed - I only said "according to the calculation". The pump can only generate a flow by raising pressure. However, once a circulation is established the generated pressure only need be enough to overcome drag.

When pump start working, does static pressure rise after the pump(outlet) or only rise dynamic pressure?
If yes, than we must add this pressure to P1(200kPa)?

 

[erobz]

When pump start working, does static pressure rise after the pump(outlet) or only rise dynamic pressure?
If yes, than we must add this pressure to P1(200kPa)?

View attachment 367728

This system has a static pressure pre charge of 200 kPa. Turning on the pump will accelerate flow through the pump via pressure differential across it until the viscous losses in the loop grow to match the differential.

It’s more complicated than this considering the actual response of the turbo machinery, but that is the general idea.

 

[gen x]

This system has a static pressure pre charge of 200 kPa. Turning on the pump will accelerate flow through the pump via pressure differential across it until the viscous losses in the loop grow to match the differential.

It’s more complicated than this considering the actual response of the turbo machinery, but that is the general idea.

My configuration can achieve P2 lower than atmospheric pressure?

 

[erobz]

My configuration can achieve P2 lower than atmospheric pressure?

In a real system with viscous losses I wouldn’t be able to tell until I had some real system/fluid properties to fiddle with computationally.

 

[jbriggs444]

When pump start working, does static pressure rise after the pump(outlet) or only rise dynamic pressure?
If yes, than we must add this pressure to P1(200kPa)?

View attachment 367728

As drawn, the pressure inside the system is indeterminate. You have a fixed amount of fluid in a container with a fixed volume. No headroom anywhere. No pressure relief valve.

In the ideal world, that would be an insolvable problem. In the real world it means that your setup will be finicky. The pressure won't stay where you expect it to be.


If you have a system like this and you drill one small hole anywhere, nothing will happen. No fluid will leak out. No air will come in. Just like the experiment with the inverted drinking glass full of water.

 

[gen x]

As drawn, the pressure inside the system is indeterminate. You have a fixed amount of fluid in a container with a fixed volume. No headroom anywhere. No pressure relief valve.

In the ideal world, that would be an insolvable problem. In the real world it means that your setup will be finicky. The pressure won't stay where you expect it to be.


If you have a system like this and you drill one small hole anywhere, nothing will happen. No fluid will leak out. No air will come in. Just like the experiment with the inverted drinking glass full of water.

Yes you are right, but I think you look at my system too strictly and too literally, so my point of original question is lost.

If pressure is 200kPa, I doesnt matter how is achieved.

Let suppose I use expansion tank or in water there is a llitttle bit of air that store pressure energy

 

[jbriggs444]

If pressure [somewhere in the full width tube portion] is [maintained at] 200kPa, I doesn't matter how is achieved.

I've taken the liberty of clarifying your assertion.

Yes, I agree that an expansion tank would work for that.

 

[Lnewqban]

write in text, absolute

In that case, your calculated value of 88 kPa is absolute as well and below atmospheric pressure.

The pump does not know or care what is the internal pressure in your loop.
Its job is always to create a downstream-upstream pressure differential.
The higher the volumetric flow that is desired inside the loop, the higher that delta pressure must be to overcome greater friction and other losses of kinetic energy in the fluid.

Because of that, your calculated value for P2 will dramatically change according to the relative location of your restriction pipe respect to the pump: higher if located downstream the pump and vice-verse.
The relative height of the restriction pipe will also affect your calculated value of P2, unless the whole loop is horizontal.

The answer is, yes, depending on its location, the restriction section may suck air into the loop.
Real life loops have automatic air vent devices installed at higher points (where internal pressure stays above atmospheric one) in order to eliminate any air accidentally sucked in through seals and non-hermetic connections.

Please, see:
https://www.caleffi.com/en-us/blog/where-should-air-separator-be-located

 

[DEvens]

About lowering the internal pressure below atmospheric and "sucked in." There is a Star Trek reference.

Riker: "You were right. Somebody blew the hatch and they were all sucked out into space."
Data: "Correction, sir, blown out."
Riker: "Thank you, Data."
Data: "A common mistake, sir."
Riker: "I know! Thank you, Data."