# How Do You Calculate Water Velocity and Force of Impact?

1. May 21, 2015

### layman2015

Hello Everyone,

I am brand new to this forum and know basically nothing about physics, but I am hoping there is someone in these forums that could help a layperson obtain a comprehensible answer to a question that is simply above his head.

I have a home whose water pressure averages between 50-75 psi in the pipes. From the shower head I have a 5/8" hose that extends about 4 feet. I have a spray connection for the hose that has a 1/8" hole at the end. I'd like to find out the velocity of the water as it comes out of the 1/8" hole, as well as find out, if possible, what the force of impact of this stream would be at say, 6".

I hate to say because I am not a plumber, this is pretty much all the information I have. Is this enough information for one of you physics experts to be able to tell me those two numbers?

Thank you.

2. May 21, 2015

### Staff: Mentor

Welcome to PF!

So this is a question that at first glance appears impossibly hard, but in actuality is almost-impossible-to-believe easy. There is only one minor assumption/complication, and besides that, the answer practically just falls off the screen into your lap.

Bernoulli's principle tells us that along a continuous flow stream, energy is conserved and the sum of all different types of pressures are conserved. In this case, you start off with static pressure, convert it to velocity pressure in the nozzle, then convert it back to static pressure as it hits the object. So the Bernoulli's equation description of this situation simply reduces to P=P. Using your max: P=75psi. That's the pressure applied at the impact point. I'll let you calculate the force using the pressure and area of your assumed coherent stream.

Do you still need velocity? That's only a bit more difficult, but you can just use Bernoulli's equation or the definition of force (momentum change over time) to calculate it. I'll let you take a crack.

Now, for that complication/assumption: In the way I've used it, Bernoulli's principle requires conservation of energy. But energy is not conserved in this situation. You have a nozzle and no nozzle is perfect, so there is a loss in accelerating the flow stream from zero to whatever speed it ends up at. So what's the loss? It depends on the nozzle. But a good estimate would be you lose 50% of your starting pressure to it. So use 37 psi and re-calculate.

3. May 21, 2015

### layman2015

Wow, thanks for the response! It certainly sounds like you know what you are doing.

Unfortunately, my knowing little to nothing about physics implies my complete ignorance of the many formulas and principles involved. So letting me know that I can use Bernoulli's principle to calculate velocity kind of leaves me where I started: ignorant and confused. One thing I am not understanding (besides what the heck is a Bernoulli's principle) is that if indeed the resulting impact were equal to the static pressure within the system, then why is it that there seems to be more power to push the hose the smaller the diameter of it's exit? Water seems to come out at a higher velocity, and the hose is physically moved more violently. This event occurs for each reduction of the diameter of the exit until it reaches a certain point where it appears that the flow of the water becomes restricted and begins to lose power.

It also seems to me that the faster an object travels, the greater it's impact upon a surface. If indeed the velocity of the water increases as the diameter of it's exit decreases (to a certain point), then wouldn't it follow that the force of impact in pounds from this accelerated water stream would be greater?

Last edited: May 21, 2015
4. May 22, 2015

### theodoros.mihos

You can see pressure as energy density so pressure loss is the density of energy you have on the exit. That mean:
$$\Delta{p} = \frac{1}{V}\frac{1}{2}mv_0^2 = \frac{1}{2}\rho v_0^2 \Leftrightarrow v_0 = \sqrt{\frac{2\Delta{p}}{\rho}}$$
if the output have the same area to the input. Because mass conservation A v = const so:
$$v = \frac{A_i}{A_f}v_0 = \frac{\phi_i^2}{\phi_f^2} v_0 \Leftrightarrow v = \left(\frac{\phi_i}{\phi_f}\right)^2\sqrt{\frac{2\Delta{p}}{\rho}}$$
You can add the friction coeficient on Δp.

5. May 22, 2015

### layman2015

Whoahoa! Theodoros, thank you for taking time out to respond to my post, but I need things in layman's terms here. I have no idea what all the variables in your formulas represent (nor in what units for that matter), nor would I have the faintest idea as to how to add a friction coefficient to Δp, nor what that coefficient could possibly be, nor what it would actually represent.

I am also not understanding how, if you are equating pressure to energy density, how pressure loss could be the density of energy on the exit. Wouldn't pressure loss be the change in energy density between the static system and the system upon fluid release?

6. May 22, 2015

Inside the system you have only static pressure, outside you have static pressure and velocity of water. So the change in static pressure is the energy density that has gone into velocity, or the flowing of water. Pressure loss is the density of -movement- energy on exit.

In case you are wondering why pressure is the same as energy density, multiply a pascal by a liter and see what you get ;).

7. May 22, 2015

### layman2015

T
Thanks crador for your reply! I found it to be very informative. Thanks for explaining that the energy density has gone into velocity (a transference of energy, right?)

I didn't seem to have a problem in understanding intuitively how pressure could be the same as energy density, but I appreciate the mathematical proof anyway.

Any idea as to how I would arrive at the two numbers I am looking for?

8. May 22, 2015

So basically the first equation above says "the change in pressure is one half times the density of water times the square of the velocity" since movement energy is one half times mass times velocity squared (mass over volume will be density).

If you rearrange that you get the right hand side: velocity is just the square root of two times the pressure change over the density of water.

The second equation is just conservation of mass: output flow is input flow. In your case you just need the approximate ara of your nozzle to get the velocity if you know the flow of water. You already know the velocity from the first equation though, and it seems the flow isn't that important for your application.

Adding the friction coefficient to the pressure change is just multiplying. You lose half the energy to friction -- multiply pressure drop by 1/2 before calculating to cut your result down to a more reasonable answer.

9. May 22, 2015

And when the water hits whatever object you have in its way, all of its movement energy goes back into pressure. Pressure times area will give you force, so if your stream of water is 1/8" thick and circular, divide your friction corrected pressure by the area of the jet to get the force.

10. May 22, 2015

Don't forget to convert to metric!

11. May 22, 2015

### layman2015

crador, you are a true genius! (For me a true genius can take complex issues and break them down so that even a layperson can understand them.) So ρ is the density of water, huh? Who knew?

Theodoros, thanks for the equations! They now become useful to me now that crador has explained them.

crador, so now that just leaves me with the question I have concerning the force of impact. russ_watters states that the force of impact is the same as the static pressure from the system. My question is, how can that be?

If you take a hose and turn it on such that you get a gentle, steady flow of water, and turn it toward an object, lets say a piece of cardboard hanging conveniently in front of you, the water will move the cardboard somewhat. But, if you place your thumb across the opening of the hose such that a powerful stream of water shoots out, aiming that stream at the cardboard causes the cardboard to move more violently. Intuitively, it seems that there is more force of impact with the flow restricted partially by your thumb than with the open hose.

Are you guys saying this is wrong?

12. May 22, 2015

The reservoir of energy you are using is then the hose filled with water, with a choke point before (the tap) slowly replenishing it, and a choke point (your thumb) releasing the energy in bursts or sustained streams.

If you let out the energy faster than it is replenished by the tap, the total pressure in the hose will fall and you will get your gentle gurgle of water. If you open up the tap so that everything is full blast, and you have a strong pump, you may be able to replenish the hose's energy reservoir to annihilate this effect.

A better pressure source would be a barrel of water with a hole poked in it, I think.

Let me know what other problems you can find with this -- pretty fun thought experiment if you ask me. Russ go ahead and jump in if I said something incorrect.

13. May 22, 2015

### layman2015

Wow crador, you are incredibly good at breaking down physics problems and explaining them. Are you by chance a teacher? If not, you should be; you are really good at it!

I understand now that with the faucet not turned on all the way, it itself is restricting the flow such that the pressure in the hose never gets to the level of the static pressure in the pipes of the house, unless I put my thumb over it. It is the increase in pressure within the hose that allows the stream escaping by my thumb to increase in velocity, and the increase in pressure within the hose is reflected by increase in force of impact upon the object.

Correct me please if I am misinterpreting what you said.

14. May 22, 2015