Understanding Bernoulli's Principle

[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.