# Force required to hold a fire hose

1. Sep 6, 2014

### toothpaste666

1. The problem statement, all variables and given/known data

A fire hose exerts a force on the person holding it. This is because the water accelerates as it goes from the hose through the nozzle.

How much force is required to hold a 6.7-cm-diameter hose delivering 490L/min through a 0.80-cm-diameter nozzle?

2. Relevant equations
A1V1 = A2V2

P1 + .5ρV1^2 + ρgY1 = P2 + .5ρV2^2 + ρgY2

3. The attempt at a solution

First I converted the diameters into meters, divided it by 2 to get the radius and then found the areas of the hose and nozzle. A1 (the hose) = .0036 m^2 A2 (the nozzle) = 5.0x10^-5 m^2. Then i converted V1 (the velocity in the hose) to m^3/s. V1 = .0082 m^3/s.
Next I must find V2.

A1V1 = A2V2

V2 = A1V1/A2 = (.0036)(.0082)/(5.0x10^-5) = .59 m^3/s

Then I must use bernoulli's equation to find the pressure in the hose and multiply by the area to find the force. the part of the hose that connects to the nozzle and the nozzle are at the same height so Y1 and Y2 = 0. So I am left with

P1 + .5ρV1^2 = P2 + .5ρV2^2

This is where I am stuck. I know the density of water is around 1000 kg/m^3 and I know both of the velocities , but i still have two unknowns (P1 and P2). since the hose/nozzle is a confined space, can I say that P1=P2 or does that only work for non moving fluid?

anyway after that I would have to find P1 (pressure in hose) and multiply by A1 to find the force the hose puts on the person, which would be equal in magnitude to the force needed to hold it. How does my solution look so far?

2. Sep 7, 2014

### SteamKing

Staff Emeritus
This is a round-about way to do this: For a circle,

$A = \pi r^{2} = \pi D^{2}/4$

What units are used to measure velocity?

Why must you use Bernoulli's equation? It doesn't seem particularly applicable to your problem.

Think about what is happening in the nozzle. You are flowing a certain amount of water thru the hose, which means it is travelling at constant velocity. When the flow reaches the nozzle, what happens to the velocity?

Hint: Try re-writing Newton's Second Law, F = m a, for the nozzle.

3. Sep 7, 2014

### toothpaste666

ok I see my first mistake. V1 = .0082/A1 = .0082/.0036 = 2.3 m/s. V2 = .0082/(5.0x10^-5) = 164m/s. When it gets to the nozzle the area gets smaller so the velocity has to go up which is an acceleration. When there is acceleration there is a force so F = ma = m(Δv/Δt)

Δv = V2-V1 = 164 - 2.3 = 162 m/s

I will choose a 1 sec interval so Δt = 1

a = (162 m/s)/s = 162m/s^2

to find m multiply the volume by the density

m = (.0082)(1000) = 8.2 kg

F = ma = (162m/s^2)(8.2 kg) = 1300 kgm/s^2 = 1300 N

4. Apr 10, 2015

### Andy Barrette

Bernoulli's equation relates pressure (F/A) and flow velocity. You can solve for A and flow velocity, and the goal is to find F. Why do you say that Bernoulli's equation is not applicable here? It is misleading.

Last edited: Apr 10, 2015
5. Dec 20, 2017

### William Fleming

All of this is wrong. If a fire hose is perfectly straight there is no backward thrust at all. Only if the hose is curved is there a backward force. In fact, the movement of the water stretches the hose, moving it forward. This is easily demonstrated.

This same erroneous explanation is given for why rockets work. There is no thrust at the nozzle of a rocket. The thrust is at the forward inside wall of the combustion chamber where there is high pressure from impacting molecules. The nozzle serves only to impede the flow of gases so that the desired pressure is maintained for a given burn rate. There is reverse thrust at the nozzle, which can be minimized by the curved walls of the throat. The curved walls cause the gases to flow to the center of the throat. High speed and pressure along the center means low speed and pressure on the walls, resulting in less drag and greater efficiency.

Newton’s Third Law is in effect for the overall system, but that is just an observation which explains nothing.

6. Dec 20, 2017

### conscience

Exactly .

I am actually puzzled that almost everywhere, in all fire hose problems the thrust force is assumed to be vdm/dt or ρav2 . This is the case even when the fire hose is mentioned to be straight in the problem statement or the diagram depicts a straight pipe near the hose .

There has to be change in momentum so as to have a backward thrust force.

What do you think @haruspex ?

7. Dec 21, 2017

### haruspex

Well, there is a change in momentum, so there is a backward thrust from the jet. The question is, is that backward thrust felt by the hose holder? As William describes, if the hose is straight all the way from its immoveable source, the high pressure in the hose will make it fairly rigid, so the thrust can be referred all the way through the water back to that source.

8. Dec 21, 2017

### William Fleming

I can see that, even for a perfectly straight hose that there would be momentum changes at every turn in the plumbing leading to the hose. Wouldn’t the thrust at those turns just be the effective area times the pressure? I’m just asking. Also, think of a pipe coming straight down from an overhead water tank. What force would there be on the pipe except drag against the pipe wall. There’d be the weight of the water in the pipe pulling it down that would vary depending on impedance from a nozzle at the bottom, if there were one. Hmm...

9. Dec 21, 2017

### Staff: Mentor

The way that the straight-hose problem works is this:

From the Bernoulli equation, we have $$P_1+\rho\frac{v_1^2}{2}=P_2+\rho\frac{v_2^2}{2}$$ 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
$$P_1=\rho\frac{v_2^2}{2}\left[1-\left(\frac{A_2}{A_1}\right)^2\right]$$
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: $$P_1A_1+F=\rho v_2A_2(v_2-v_1)$$ 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:
$$F=-\rho\frac{v_2^2}{2}A_1\left(1-\frac{A_2}{A_1}\right)^2$$
This means that the hose and nozzle are exerting a force in the backward direction on the fluid (i.e., pulling back), and the fluid is exerting a force in the forward direction on the nozzle and hose. So the fluid is not exerting a backward force on the nozzle. The forward force of the fluid on the nozzle and hose causes the hose to become taught. The magnitude of the force F represents the tension in the hose.

Last edited: Dec 21, 2017
10. Dec 21, 2017

### CWatters

The problem statement doesn't actually say which direction the force acts.

11. Dec 21, 2017

### haruspex

Not in so many words, but it is clearly implied here:

12. Dec 21, 2017

### haruspex

The hose becomes taut. It is us students that are taught.

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