# Faster, narrower fluid same pressure as slower, wider fluid?

1. May 30, 2015

### yrjosmiel73

Say, I have a garden hose. You know the thing when you cover a part of the hole to make the fluid reach farther because it's faster, right? Say you're pointing it down and there's no air resistance or gravity. You're hitting a surface with the hose and there's a device that measures how much pressure the water is exerting on the surface. Will that have the same pressure as if you didn't do anything at all?

I mean, I think it's the same because the other is just focused on a single spot but more force but the other covering a larger area but weaker force.

2. May 30, 2015

### Staff: Mentor

The pressure will be lower, probably significantly lower. If you block a part of the exit, the amount of water flowing out reduces. This reduces friction in the hose, so pressure at the exit increases - up to the point where you reach the pressure of the water supply. Long before that you are mainly reducing water flow without a large effect on pressure, so the product of both decreases.

3. May 30, 2015

### Staff: Mentor

I'm not sure if I understand the setup correctly, but this sounds like a classic Bernoulli's principle scenario, where the total pressure is the same everywhere. And for a stream of water in air, all is in the form of dynamic or velocity pressure. When it hits something, all is converted (briefly) to static pressure.

4. May 30, 2015

### Staff: Mentor

That is right, but the water flow will depend on the size of the exit. Also, friction is not negligible.

5. May 30, 2015

### Staff: Mentor

Ok, I think I get what he's after -- two different scenarios, one with the thumb on the end of the hose, one without. I was a little thrown by the "if you didn't do anything part"....not focused I guess.

If there were no losses, the pressure would be higher with your thumb over the end, but the force would be the same. In reality though, there are lots of losses. With an open hose, almost all of the pressure/energy is lost in the piping before the water even gets to the end of the hose. So with your thumb over the end, it is probably a higher force, but only "probably" because there will be a lot of loss through the "nozzle" you create with your thumb. If the loss is low enough, the fact that the flow rate is lower means you get a lot less loss through the rest of the piping system.

6. May 31, 2015

### Staff: Mentor

Losses in the hose have the opposite effect.
If there are no losses in the hose (imagine a large bassin with a hole of variable size in it), pressure will always be the same, and the force on the target is proportional to the opening area: F=c*A.
Losses in the pipe reduce with smaller opening area, so the relationship becomes nonlinear (flatter than linear, but not constant).

7. May 31, 2015

I think the real issue is that he is using pressure and force interchangeably. What I interpret the original question as meaning is which plate has a greater applied force: fast and narrow stream or slow and wide stream. If you assume the stream leaving the hose remains at a constant are and that viscous losses are negligible, then you can get a decent approximation here fairly easily.

Assume that scenario 1 is the one without a thumb and that the water velocity leaving the hose is $v_1$ through an area $A_1$. Scenario 2 is the one with the thumb and the water is leaving the hose with $v_2$ through area $A_2$. The total momentum of water leaving the hose in each scenario is $\rho v_1^2 A_1$ and $\rho v_2^2 A_2$ respectively. from conservation of mass, we know that
$$v_1A_1 = v_2 A_2,$$
or
$$v_2 = v_1\dfrac{A_1}{A_2}.$$
That means that the momentum in scenario 2 can be rewritten as
$$\rho v_2^2 A_2 = \rho v_1^2 \dfrac{A_1^2}{A_2^2}A_2 = \rho v_1^2 A_1 \left(\dfrac{A_1}{A_2}\right).$$
So in this case, the momentum in the second scenario differs from that in the first scenario by a factor of $A_1/A_2$, so if you have cut the area in half with your thumb, you are doubling the momentum and therefore the force (since force is a change in momentum).

This is, of course, assuming my interpretation of the question was correct.

8. May 31, 2015

### Staff: Mentor

There is no mass conserved if you change the setup, the flow rate will change.
To see this, imagine a very small area A: do you expect the flow to get supersonic?

9. May 31, 2015

Let's not get too ridiculous here. I'm clearly referring to reasonable area ratios here. That is more than sufficient to cover the original question and whether the force on some plate is equal in the two situations. It's not.

10. May 31, 2015

### Staff: Mentor

I added the extreme case as an example, but the statement in the first line applies to every area.

Right, but not for the reason you gave in post 7.

11. May 31, 2015

It absolutely is the same reason I gave. Even in a lossless system the two situations are not equivalent, which is a good start for any such discussion. Yes, they system is not truly lossless and there is flex in the hose and the stream spreads out after leaving the hose, but it is much more instructive to start with the simple case. Given that not even the most simplified case gives an equivalence between the two scenarios, you wouldn't expect the more complicated case to do so either.

12. May 31, 2015

### rcgldr

In a real world situation, wouldn't the mass flow rate be lower with a restricted exit, such as a thumb or nozzle blocking around 1/2 or more of the opening at the end of a hose?

13. May 31, 2015

### Staff: Mentor

Exactly.
Why do you argue about the point where we agree on?
The mass flow is not the same if you change the area.

14. May 31, 2015

Because we don't agree on the route to get to that point. Sure the mass flow is likely to decrease when you cover part of the exit, but it will likely be almost negligible in magnitude. The pressure loss introduced by your finger is small compared to the ~75 psi water pressure in a typical home. So yes, the mass flow rate is likely to drop some but it won't be much and making the constant mass flow assumption is still instructive. That's my point.

15. May 31, 2015

### Staff: Mentor

Based on what? I disagree with that estimate.
Actually, if you neglect losses in the pipe, mass flow will be proportional to the area. Without losses, the exit velocity is constant and given by the pressure difference between supply and outside only.

16. May 31, 2015

One common way to describe losses in a system is using head loss. A typical garden hose at a house in the US discharges something like 15 gallons per minute through a 5/8" hose at a service pressure of 75 psi, so converting that into a velocity for the unobstructed system, that is about 4.78 m/s. Now, head loss is typically described as
$$h_L = K\left(\dfrac{v^2}{2g}\right).$$
In the case of the hose, let's assume just for a quick calculation that the flow rate is actually constant and the flow rate coming out the constricted hose is double that of the unconstricted hose. The $K$ in the above equation is a coefficient that you can find in many textbooks for a sudden contraction. It is about 0.2 for this flow velocity and contraction ratio (I am assuming that half of the exit is covered by your thumb). To get a pressure loss, you multiply $h_L$ as if it was just a simple height in hydrostatic pressure, $p_{loss} = \rho g h_L$.

So, $K\approx 0.2$, $v\approx 9.56\;\mathrm{m/s}$, and $\rho$ and $g$ are their usual metric values. The end result is that you get a loss of about 1.33 psi. That is less than 2% of the total pressure in the system. Since mass flow rate is directly proportional to the pressure difference, a <2% loss in total pressure is less than a 2% loss in total mass flow rate. I call that an acceptable amount of error for my above calculations based on the constant mass flow assumption.

Honestly, error less than 2% is probably perfectly fine for most engineering purposes, let alone a back-of-the-envelope calculation for illustrative purposes. That is so small that I bet if you went outside with a hose and a 5 gallon bucket and filled it up with and without your thumb, you would have a hard time measuring a difference accurately with a stopwatch.

17. May 31, 2015

### Staff: Mentor

As far as I can see, you assume constant mass flow, which then leads to an approximately constant mass flow. This is not surprising.

To increase water speed from 4.78m/s to 9.56m/s, you need ~70 kJ/m^3 or ~70kPa, about 10 psi, and there is no way to "get this pressure back" (the energy is used for the higher exit velocity). This pressure drop is not available in your pipe any more, so head loss there has to go down by ~13%, which decreases speed by ~6%, not taking losses on the exit into account. This gets significantly worse for smaller exit areas.

18. May 31, 2015

Think about what installs did for a second. I assumed a constant mass flow just to estimate velocity. You could then iterate if you really wanted to do so and decrease my estimated velocity by 2%, then re-run the numbers and get a new pressure loss. Then repeat until the value doesn't change. Of course, lowering the mass flow slightly gives a slightly lower velocity meaning less pressure loss each time, so my estimate is actually conservative. The real value is likely closer to 1%.

19. Jun 2, 2015

### Staff: Mentor

I don't see how you can avoid this 10 psi pressure loss to speed up the water. I think your result violates energy conservation.

20. Jun 2, 2015

### Staff: Mentor

I think the problem is with both the assumption of a 50% area reduction and therefore with the assumption of a constant flow or velocity.

In the real world situation, if covering half of the exit changes very little, why would you do it? The goal of covering the exit with your thumb is to create a MUCH higher velocity, not just a little higher velocity. If covering it by half only yields a 2% pressure loss, then you haven't covered it enough: cover it 90%+ and the loss in the hose/piping becomes negligible, the loss across the nozzle near 100% (both actual loss and conversion to velocity pressure) and the velocity change is by an order of magnitude.

The head loss calculation ignores velocity pressure completely, which you shouldn't do when the whole point of the exercise is to convert static pressure to velocity pressure.