Bernoulli's Principle and pressure

[nitrOgenX]

Hi guys,

I have what I hope is an easy question.

When you put your thumb over the end of a water hose, why does the water coming out feel like it has a HIGH pressure (meaning, it will hurt somebody being sprayed by it) EVEN THOUGH Bernoulli's Principle says that the pressure should decrease when you decrease the cross-sectional area?

I don't understand?

I hope I make sense. Thanks for your help.

 

[tiny-tim]

welcome to pf!

hi nitrOgenX! welcome to pf!

yes, the pressure alongside your thumb will decrease …

conservation of mass means that a narrower flow must have a higher speed, and then Bernoulli's equation (which is essentially a conservation of energy equation) says that higher speed (and KE) means lower pressure …

but the static pressure on the top of your thumb, where it faces back along the pipe, should be approximately the same as in the rest of the pipe

EDIT: the total pressure, of course, is greater: total pressure on a stationary object = static (ordinary) pressure plus dynamic pressure, 1/2 ρv²

(if you could push your thumb through the side of a pipe, and if your thumb was squishy, that reduced pressured should tend to elongate your thumb )

 

[Turbodog]

Another way to look at it is that even though the fast moving water jet is at a lower pressure, the kinetic energy has increased. When you spray your friend with this jet, there is a momentum change of the fast moving low pressure water when it stikes him so he will feel that force.

 

[nitrOgenX]

@ tiny-tim
Thanks for the warm welcome =]

So it's basically the term ½ρv^2 (dynamic pressure) that accounts for the kinetic energy right?

Thank you to both of you for your responses!

 

[tiny-tim]

hi nitrOgenX!

(try using the X² icon just above the Reply box )

So it's basically the term ½ρv^2 (dynamic pressure) that accounts for the kinetic energy right?

yes …

Bernoulli's equation is a conservation of energy equation, and the dynamic pressure (½ρv²) is the KE term

 

[Phrak]

@ tiny-tim
Thanks for the warm welcome =]

So it's basically the term ½ρv^2 (dynamic pressure) that accounts for the kinetic energy right?

Thank you to both of you for your responses!

The terminology is weird. Dynamic pressure is not really pressure at all, but potential pressure. It is the pressure that could be obtained if the kinetic energy a unit volume where stopped by a barrier and compressed. It would better be called potential pressure.

Bernoulli's Principle is a restatement of: Potential Energy + Kinetic Energy = Constant.

In fact, Bernoulli's Principle comes in various forms. For example, sometimes gravitational potential energy is included, and sometimes not.

Pressure is to Dynamic Pressure as Potential Energy is to Kinetic Energy.

 

[nitrOgenX]

The terminology is weird. Dynamic pressure is not really pressure at all, but potential pressure. It is the pressure that could be obtained if the kinetic energy a unit volume where stopped by a barrier and compressed. It would better be called potential pressure.

Bernoulli's Principle is a restatement of: Potential Energy + Kinetic Energy = Constant.

In fact, Bernoulli's Principle comes in various forms. For example, sometimes gravitational potential energy is included, and sometimes not.

Pressure is to Dynamic Pressure as Potential Energy is to Kinetic Energy.

I understand your analogy. However, if the dynamic pressure is related to potential energy, which part of Bernoulli's is related to kinetic energy? Isn't it the ½ρv^2 I mentioned earlier?

 

[Phrak]

I understand your analogy. However, if the dynamic pressure is related to potential energy, which part of Bernoulli's is related to kinetic energy? Isn't it the ½ρv^2 I mentioned earlier?

Yes, exactly. This part is kinetic energy per unit volume: $\frac{1}{2}mv^2/V$. V is the volume of the fluid having mass m. The dynamic term is pressure derived potential energy per unit volume.

Bernoulli's equation comes in various forms, sometimes including a gravitational potential energy term. Which are you using?

 

[nitrOgenX]

Yes, exactly. This part is kinetic energy per unit volume: $\frac{1}{2}mv^2/V$. V is the volume of the fluid having mass m. The dynamic term is pressure derived potential energy per unit volume.

Bernoulli's equation comes in various forms, sometimes including a gravitational potential energy term. Which are you using?

I use:

P⁄ϒ+v^2/2g + z

Thanks a lot for all your responses.

 

[russ_watters]

None of these answers actually address the root question in the OP: the phenomena the OP noticed has nothing to do with Bernoulli's principle. The reason you feel a higher pressure when you put your thumb over the end of a hose is there is a higher pressure (behind a restriction) when you restrict the flow through a pipe. Why? Because when you restrict the flow you reduce the dynamic losses through the pipe.

Out at the street, your water may be at 30 psi, but at the end of an open hose, it may only be 1 or 2 psi (total). All the other 28 or 29 psi is lost due to friction inside the pipes. When you restrict the flow, you reduce that loss and feel all of the 30 psi on your thumb.

This is the reason why when you remove your thumb, the water rushes out for a second: you've allowed the pressure to build in the system by making the pipes expand (particularly the garden hose) and that residual pressure is released when you first remove your thumb. And if you have the water flowing at only a trickle and you put your thumb over the end, you can also feel the pressure slowly building as the hose fills with water.

 

[nitrOgenX]

None of these answers actually address the root question in the OP: the phenomena the OP noticed has nothing to do with Bernoulli's principle. The reason you feel a higher pressure when you put your thumb over the end of a hose is there is a higher pressure (behind a restriction) when you restrict the flow through a pipe. Why? Because when you restrict the flow you reduce the dynamic losses through the pipe.

Thank you for your response. I understand the pressure building up on your thumb but what I don't understand is this:

If you spray someone with this water from the hose, why does the person being sprayed "FEEL" like there is a lot pressure acting on them?

Out at the street, your water may be at 30 psi, but at the end of an open hose, it may only be 1 or 2 psi (total). All the other 28 or 29 psi is lost due to friction inside the pipes. When you restrict the flow, you reduce that loss and feel all of the 30 psi on your thumb.

Assuming the water WAS at 30 psi, where does the "pressure" that the person being sprayed feels IF your thumb was feeling all of the 30 psi?

I think what we concluded in the previous posts was that you increased the water's kinetic energy by restricting the flow with your thumb. That's what causes that feeling of pressure. The water has enough kinetic energy to do work on you. Is this correct?

BTW... Cool website, you've got some pretty sweet photos on there.

 

[russ_watters]

If you spray someone with this water from the hose, why does the person being sprayed "FEEL" like there is a lot pressure acting on them?

For the person being sprayed, that is sort of Bernoulli's principle. High velocity = high total pressure = high energy. Just like throwing a baseball harder at someone hurts more because the velocity means higher energy. Just note that it's higher than when your thumb isn't on the nozzle because your thumb being on the nozzle increases the pressure and causes the water to spray with a higher velocity.

Assuming the water WAS at 30 psi, where does the "pressure" that the person being sprayed feels IF your thumb was feeling all of the 30 psi?

If the total pressure behind your thumb is 30psi then the total pressure felt by the person being sprayed is 30psi (minus any loss in the nozzle and while the water is flying through the air).

I think what we concluded in the previous posts was that you increased the water's kinetic energy by restricting the flow with your thumb. That's what causes that feeling of pressure. The water has enough kinetic energy to do work on you. Is this correct?

Basically, yes. Bernoulli's equation/principle is a conservation of energy statement. It relates kinetic and potential energy. Water with high static pressure and low velocity (behind your thumb) has a high potential energy and as it travels past your thumb, that potential energy is converted to kinetic energy. Just remember though, that the total energy available at the nozzle (per unit mass) is higher when your thumb is there than when it isn't.

BTW... Cool website, you've got some pretty sweet photos on there.

Thanks!

 

[nitrOgenX]

For the person being sprayed, that is sort of Bernoulli's principle. High velocity = high total pressure = high energy. Just like throwing a baseball harder at someone hurts more because the velocity means higher energy. Just note that it's higher than when your thumb isn't on the nozzle because your thumb being on the nozzle increases the pressure and causes the water to spray with a higher velocity. If the total pressure behind your thumb is 30psi then the total pressure felt by the person being sprayed is 30psi (minus any loss in the nozzle and while the water is flying through the air).
Basically, yes. Bernoulli's equation/principle is a conservation of energy statement. It relates kinetic and potential energy. Water with high static pressure and low velocity (behind your thumb) has a high potential energy and as it travels past your thumb, that potential energy is converted to kinetic energy. Just remember though, that the total energy available at the nozzle (per unit mass) is higher when your thumb is there than when it isn't. Thanks!

I think this answers my question, thanks!

 

[wyu1229]

Does Bernoulli's equation apply for fluids that are not constricted within a pipe? For example, let's say that a fluid were flowing in a pipe and suddenly shoots out in an opening in the pipe. Would the equation P_1 + .5(rho)(v_1)^2 + (rho)(g)(y_1) = P_2 + .5(rho)(v_2)^2 + (rho)(g)(y_2) apply even for a point in the water's trajectory (outside of the pipe)?

 

[firavia]

It is the kinetic energy what you are feeling as pressure , because all the energy of water will suddenly stop when it touch your thumb so all dynamic pressure is converted to static pressure.
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