Understanding Bernoulli's Principle

[sophiecentaur]

I didn't save them unfortunately. I just did a search for fire hose recovery or loose hose. Here is one that shows a large curve at the beginning but that ends up straightened out by the end. Note when I say straightened, I don't mean "in a perfectly straight line from inlet to outlet". I mean the large curves further from the free end are straightened and only the oscillating end is left. This makes sense in considering the force balances discussed previously.

Much like the link I found. But I think that friction with the ground is probably the reason for damping out the initial big bends. It didn't look very well controlled near the end. The tension certainly has a part to play though.

 

[Chestermiller]

I think this is down to the relative values of the variables. A very small lateral force on a taught string can cause lateral displacement. Likewise, if the hose becomes bent then the force will increase with the displacement so you have a potential runaway situation. I have to wonder if the reports that firemen pull the hose may not be accurate. The hose, under pressure is quite rigid and there may be more 'twisting' or pushing at the outside of the curve involved than they are aware of.

There is a force against the outside of the curve which can increase the displacement and you will get snaking. The movies show what I think I would have expected. (It's easy to 'know' when you've actually seen something happen.)

I think that this description really gets at the heart of what is happening fundamentally. The video referenced in post #8 shows that the section of hose in the region where the firefighters are positioned is curved in such a way that the internal water force (required to compel the water to move along its curved path within the hose) has a horizontal component on the hose in the backward direction. This means that the firefighters have to pull forward on the hose to hold it in position. The illusion is that it is the jet of liquid coming out of the hose that tries to force them back. The reality is that it is the liquid within the curved section of hose that acts to push them back.

 

[sophiecentaur]

The reality is that it is the liquid within the curved section of hose that acts to push them back.

I was impressed by the use of the word "reaction" by the firefighting instructor when describing this force. You don't find N3 in many non physics conversations. "Reaction" is what old people feel against new music.

 

[rcgldr]

It's not a closed system unless you include the power source and whatever that power source effects.

I still haven’t seem a single equation out of you. And who said anything about a closed system?

Somehow lost a post. Simplify the situation to a straight pipe and nozzle attached to a fire hydrant. The pressure of the water keeps the pipe under tension everywhere, except where the water exits the opening in the nozzle. The decreasing pressure gradient corresponds to a "forward" force on the water ejected through the opening of the nozzle. The other part of the Newton third law pair would be a backwards force exerted onto the fire hydrant. The fire hydrant transmits this backwards force onto the earth. The closed system would consist of the earth, water supply, hydrant, pipe, nozzle, and the ejected water.

As posted earlier by sophiecentaur, the mass flow is constant, but the velocity increases, and the momentum per unit mass of water (which is essentially the velocity of a unit mass of water) is higher in the ejected stream than it is in the flow within the pipe.

 

[Chestermiller]

Somehow lost a post. Simplify the situation to a straight pipe and nozzle attached to a fire hydrant. The pressure of the water keeps the pipe under tension everywhere, except where the water exits the opening in the nozzle. The decreasing pressure gradient corresponds to a "forward" force on the water ejected through the opening of the nozzle. The other part of the Newton third law pair would be a backwards force exerted onto the fire hydrant, essentially the water pressure times the cross sectional area of the pipe to hydrant connector. The fire hydrant transmits this backwards force onto the earth. The closed system would consist of the earth, water supply, hydrant, pipe, nozzle, and the ejected water.

As posted earlier by sophiecentaur, the mass flow is constant, but the velocity increases, and the momentum per unit mass of water (which is essentially the velocity of a unit mass of water) is higher in the ejected stream than it is in the flow within the pipe.

This is not what the analysis I presented shows, and what the rest of us have now reached consensus on.
At the nozzle, the exit force is just atmospheric pressure (zero gauge pressure), just as it is in Toricelli jet flow out of a tank. The water has already been accelerated prior to reaching the exit, and there is no additional force.

Plus, the backwards force on the hydrant is the result of the change in direction of the water flow (acceleration) from vertical to horizontal.

 

[rcgldr]

This is not what the analysis I presented shows, and what the rest of us have now reached consensus on. Plus, the backwards force on the hydrant is the result of the change in direction of the water flow (acceleration) from vertical to horizontal.

Consider another example such as a water rocket, in space and free of any external forces. The water does not curve, it's ejected out the opening. The Newton third law pair of forces are a net "backwards" force on the ejected water, and a net "forwards" force on the rocket. The momentum per unit mass of water in the bottle is different than the momentum per unit mass of water in the ejected stream. Momentum of the closed system of rocket and water (including the ejected stream) is conserved. The center of mass of the closed system does not accelerate.

 

[Chestermiller]

Consider another example such as a water rocket, in space and free of any external forces. The water does not curve, it's ejected out the opening. The Newton third law pair of forces are a net "backwards" force on the ejected water, and a net "forwards" force on the rocket. The momentum per unit mass of water in the bottle is different than the momentum per unit mass of water in the ejected stream. Momentum of the closed system of rocket and water (including the ejected stream) is conserved. The center of mass of the closed system does not accelerate.

Are you asking me to model this to show the fundamental difference?

 

[rcgldr]

Are you asking me to model this to show the fundamental difference?

It was just another example. If you want to stick with a fire hydrant like example, consider a large ground level tank of water as the supply source to a pump that pumps the water horizontally, with no curvature of water involved. The output of the pump would go into the straight horizontal pipe with tapered nozzle. The Newton third law pair would be a net forward force ultimately exerted by the pump onto the ejected water, coexistent with a equal but opposing force exerted by the water onto the pump. The momentum per unit mass of water in the pipe would be less than the momentum per unit mass of water in the ejected stream, due to the increase in velocity (using the Earth as a frame of reference), while the mass flow remains the same at all points within the pipe and as the water is ejected out the nozzle.

 

[Chestermiller]

It was just another example. If you want to stick with a fire hydrant like example, consider a large ground level tank of water as the supply source to a pump that pumps the water horizontally, with no curvature of water involved. The output of the pump would go into the straight horizontal pipe with tapered nozzle. The Newton third law pair would be a net forward force ultimately exerted by the pump onto the ejected water, coexistent with a equal but opposing force exerted by the water onto the pump. The momentum per unit mass of water in the pipe would be less than the momentum per unit mass of water in the ejected stream, due to the increase in velocity (using the Earth as a frame of reference), while the mass flow remains the same at all points within the pipe and as the water is ejected out the nozzle.

As I said in several previous responses, if there is something wrong with the analysis I did (which explicitly includes the momentum term you are referring to), please point it out:

Here is the analysis:

From the Bernoulli equation, we have $$P_1+\rho\frac{v_1^2}{2}=P_2+\rho\frac{v_2^2}{2}\tag{1}$$ where the subscript 1 refers to an upstream location within the hose and the subscript 2 refers to the nozzle exit. At the nozzle exit, the pressure is atmospheric (zero gauge pressure), so we have
$$P_1=\rho\frac{v_2^2}{2}\left[1-\left(\frac{A_2}{A_1}\right)^2\right]\tag{2}$$
To get the force that the nozzle exerts on the fluid (and, thus, the force that the fluid exerts on the nozzle) we must perform a macroscopic momentum balance: $$P_1A_1+F=\rho v_2A_2(v_2-v_1)\tag{3}$$ where F is the forward force that the nozzle exerts on the fluid and the right hand side of the equation represents the rate of change of momentum of the fluid passing through the nozzle.

Combining these equations to solve for the force F gives:
$$F=-\rho\frac{v_2^2}{2}A_1\left(1-\frac{A_2}{A_1}\right)^2=-\rho (v_2-v_1)v_2(A_1-A_2)/2$$
This indicates, in agreement with the Chegg Study analysis in post #25, that the hose and nozzle are under axial tension (although the axial nozzle tension does decrease to zero toward the very exit, beyond the converging portion of the nozzle).

 

[rcgldr]

As I said in several previous responses, if there is something wrong with the analysis I did (which explicitly includes the momentum term you are referring to), please point it out:

There's nothing wrong with your analysis, there is a forward force on the nozzle, but the forward force on a closed off nozzle would be greater than the forward force on a open nozzle because with an open nozzle, some of the force goes into accelerating the water through and out of the nozzle.

At hydrant end of a straight pipe, with a closed off nozzle, the backwards force on the hydrant which would be the pressure times the cross sectional area of the pipe, and would be exactly offset by the forward tension in the pipe pulling at the hydrant connection due to the forwards force on the closed off nozzle, so a zero net horizontal force on the hydrant. With an open nozzle, the forward pull by the pipe on the hydrant is less because the forward force on the opened nozzle is less resulting in a net backwards force on the hydrant. As posted before, the Newton third law pair of forces would be a forward force exerted on the ejected water, and a net backwards force exerted on the hydrant.

 

[Chestermiller]

There's nothing wrong with your analysis, there is a forward force on the nozzle, but the forward force on a closed off nozzle would be greater than the forward force on a open nozzle because with an open nozzle, some of the force goes into accelerating the water through and out of the nozzle.

At hydrant end of a straight pipe, with a closed off nozzle, the backwards force on the hydrant which would be the pressure times the cross sectional area of the pipe, and would be exactly offset by the forward tension in the pipe pulling at the hydrant connection due to the forwards force on the closed off nozzle, so a zero net horizontal force on the hydrant. With an open nozzle, the forward pull by the pipe on the hydrant is less because the forward force on the opened nozzle is less resulting in a net backwards force on the hydrant. As posted before, the Newton third law pair of forces would be a forward force exerted on the ejected water, and a net backwards force exerted on the hydrant.

Most of this had nothing to do with the analysis and the points I was making. However, I will say that your analysis of the forces on the hydrant is missing an important component, to wit: the force required to change the direction of the fluid flow from vertical to horizontal (unless you are just including the portion of the hydrant after the change of direction).

I did an analysis of the horizontal component of the reaction force of the ground acting on the hydrant, assuming that the water pipe within the hydrant comes up vertically, and then has a bend within the hydrant, so the exit from the hydrant is horizontal. Assuming that the hose and nozzle attached to the hydrant are perfectly horizontal, I obtain a horizontal reaction force by the ground on the hydrant of $$\rho v_2^2 A_2=\dot{m}v_2$$ where $\dot{m}$ is the mass flow rate of water through the hose.

 

[rcgldr]

Most of this had nothing to do with the analysis and the points I was making. However, I will say that your analysis of the forces on the hydrant is missing an important component, to wit: the force required to change the direction of the fluid flow from vertical to horizontal (unless you are just including the portion of the hydrant after the change of direction).

I was only considering the flow after the change in direction, or considering an idealized case where there is no curvature of flow. I was also trying to focus on the Newton third law pair of forces. The horizontal reaction force on the ground would be the sum of both forces, the force related to curvature of flow and the force related to acceleration of water out of the nozzle. The increase in horizontal velocity is much greater at the nozzle than it is at the hydrant, but the unknowns are the distances over which the water accelerates at the nozzle and at the hydrant.

 

[Chestermiller]

I was only considering the flow after the change in direction, or considering an idealized case where there is no curvature of flow. I was also trying to focus on the Newton third law pair of forces. The horizontal reaction force on the ground would be the sum of both forces, the force related to curvature of flow and the force related to acceleration of water out of the nozzle. The increase in horizontal velocity is much greater at the nozzle than it is at the hydrant, but the unknowns are the distances over which the water accelerates at the nozzle and at the hydrant.

If we neglect viscous drag, then the distances are irrelevant. So, do you agree with the result that I got for the reaction force exerted by the ground on the hydrant or not? (This result is totally consistent with the previous relationships I developed, and can derived either from these previous relationships or from an overall force balance on the combined hydrant and hose/nozzle). You should be thrilled with the result since you finally get the jet momentum that you've been looking for (but only for the reaction force at the hydrant).

 

[rcgldr]

So, do you agree with the result that I got for the reaction force exerted by the ground on the hydrant?

Yes, it seems OK to me.