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

yrjosmiel73

## Main Question or Discussion Point

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.

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mfb
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.

russ_watters
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.

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

russ_watters
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.

yrjosmiel73
mfb
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).

Gold Member
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.

afcsimoes
mfb
Mentor
from conservation of mass, we know that
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?

Gold Member
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?
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.

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

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

Gold Member
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.
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.

rcgldr
Homework Helper
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?

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

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

mfb
Mentor
Sure the mass flow is likely to decrease when you cover part of the exit, but it will likely be almost negligible in magnitude.
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.

Gold Member
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.
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.

mfb
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.

Gold Member
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%.

mfb
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.

russ_watters
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.

mfb
Gold Member
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.
But it is a conservative conservation of energy. In an incompressible fluid flow like this, energy dissipation really only occurs due to the effects of viscosity. Now, it is true that my original post neglected viscosity, but viscosity really only shows up in two places here: the resistance to the flow generated by the walls of the hose, and your finger over the end (since your finger being there involves the generation of eddiess where viscous dissipation is important. The acceleration of the flow, on the other hand, is a result of the constriction of its flow path and the requirement that the mass flow rate upstream of the constriction match the mass flow rate downstream.

So, that acceleration requires energy, but that energy is taken from the potential energy of the flow stored in the form of static pressure and it is essentially a conservative process. This type of energy balance is precisely what is covered by Bernoulli's equation. What is neglected by Bernoulli's equation is any of the dissipative processes that reduce the total pressure. The total pressure is essentially a measure of the total energy available in the flow, and is a sum of kinetic energy (dynamic pressure), essentially spring potential energy (static pressure), and gravitational potential energy (which is often neglected assuming there are no major elevation changes). Dissipation decreases the total pressure in the system. Changing velocity due to an area change does not (except whatever losses are inherent in the method of the area change). So, in fact, you could "get that pressure back" if the flow simply continued passed the obstruction back into the hose of the original diameter (minus the minor losses inherent in the constriction and subsequent expansion). This is precisely how a Venturi tube works.

So then why did I think it was okay to neglect dissipation and its effect on mass flow for my original analysis? So the first source of dissipation here is the hose itself. Assuming that the Reynolds number and wall characteristics are the same in each of the above scenarios, then the losses due to the length of the hose are identical and the equal mass flow assumption is valid. The question then is how much total pressure does your finger remove at the end of the hose. That is why I went through my previous analysis about head loss. It turns out that the first iteration implies that it takes away less than 2% of the total energy in the system, which translates to less than 2% mass flow difference between the two cases. The losses from the hose would be very negligibly different in that case since Reynolds number is linear in velocity.

The result, then, is that for the same total pressure, the mass flow difference between the two scenarios is in the ballpark of 2% to 3% if you were to iterate on this process a few times to reach an answer (aka use it as a predictor-corrector method). That was not enough to invalidate my original approach for most practical purposes.

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.
No, covering half the exit doubles the velocity if you neglect losses. The 2% loss is to mass flow rate (##\rho V A##), not velocity.

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.
It absolutely does not. Generally, head loss is used with Bernoulli's equation to represent the dissipative effects in the system that otherwise would not consider such effects. Essentially, for an incompressible fluid, you have something like this:
$$\dfrac{1}{2g}\rho v_1^2 + \dfrac{p_1}{\rho g} + z_1 = \dfrac{1}{2g}\rho v_2^2 + \dfrac{p_2}{\rho g} + z_2 + h_L.$$
So the head loss represents the loss of total pressure. Now, mass flow must be conserved from one end of a system to the other. So if the area of the duct and the density are not changing and ##\rho V A## must be conserved, then at any two points along the hose before the constriction, all of that dissipated energy from the hose itself comes out of the static pressure (assuming that ##z_1 = z_2##), not the dynamic pressure. At the constriction, mass flow still must be conserved before and after the constriction, so ##\rho V A## is still constant and the change in ##V## is still governed entirely by the change in ##A## and the losses from the constriction still only get taken out of the static pressure.

So no, I did not ignore velocity pressure; it's just that the conservation laws ensure that the viscous losses get taken out of the pool of potential energy rather than the pool of kinetic energy, otherwise mass would either be created or destroyed. The inefficiencies arise in my final answer where there is simply less total pressure and therefore a lower total mass flow rate. Since the velocity is determined by the mass flow rate here, that means it is a slightly lower velocity than predicted by me originally. The magnitude of that difference is only 2% to 3%, though, as I have been trying to show.

mfb
Mentor
We know the exit pressure - it is atmospheric pressure. The total pressure drop in pipe plus exit is constant, if we increase the first one the second one has to go down.
What is neglected by Bernoulli's equation is any of the dissipative processes that reduce the total pressure.
I'm fine with neglecting this part for the exit, as long as we include the more relevant part.

So no, I did not ignore velocity pressure; it's just that the conservation laws ensure that the viscous losses get taken out of the pool of potential energy rather than the pool of kinetic energy
Sure, but you need the pool of potential energy to drive water through the pipe. The more you need for the exit the lower the amount you have for the pipe.

Let's start from scratch, with energy conservation and the values you used on page 1:
75 psi is 517 kPa, let's round it to p0 = 500 kPa. A garden hose will have 100 kPa pressure at the end (atmosphere), so a difference of 400 kPa leads to a velocity of 4.78 m/s. Again, round to 5 m/s for nicer numbers. As ##v^2 \propto \Delta p##, we get ##cv^2 = \Delta p## with ##c=16kPa~\frac{s^2}{m^2}## for the pressure drop in the pipe at a general water velocity v.

At the end of the pipe, we need energy to accelerate the water. We restrict the area to a fraction x of the original area, which means water will exit at a velocity of ##\frac{v}{x}##. This needs an energy per volume of ##\rho \left( \frac{v^2}{x^2} - v^2 \right) = \rho v^2 \left(\frac{1}{x^2}-1\right)##.

Our initial 500kJ/m^3 (plus energy in water velocity) are used for (a) friction in the pipe, (b) acceleration at the exit, (c) overcome atmospheric pressure. As formula:
$$p_0 = cv^2 + \rho v^2 \left(\frac{1}{x^2}-1\right) + 100 kPa$$
Solving for v2,
$$v^2 = \frac{p_0-100kPa}{c+\rho\left(\frac{1}{x^2}-1\right)}$$
Plugging in numbers:
x=1 => v=5 m/s as expected
x=0.5 => v=4.588 m/s (8% reduction)
x=0.25 => v=3.592 m/s (28% reduction)
x=0.1 => 1.865 m/s (63% reduction)
x=0.05 => 0.982 m/s (at this point flow is roughly proportional to x as garden hose losses become negligible and the exit velocity does not change any more)

russ_watters
Mentor
So, let me start off by saying I put together my own reply (now moot) before seeing yours. Yours is better conceptually, but has a couple of issues, one of which took me an embarrassingly long time to identify, and found in an embarrassing way:
Let's start from scratch, with energy conservation and the values you used on page 1:
75 psi is 517 kPa, let's round it to p0 = 500 kPa. A garden hose will have 100 kPa pressure at the end (atmosphere), so a difference of 400 kPa...
Supply pressure is always gauge, so 500 kPa is 500 kPa. Not a huge issue, since 75 psi is pretty big and pretty arbitrary.
...a difference of 400 kPa leads to a velocity of 4.78 m/s.
Just to make sure we're clear: 4.78 was a given, not a calculation.
At the end of the pipe, we need energy to accelerate the water. We restrict the area to a fraction x of the original area, which means water will exit at a velocity of ##\frac{v}{x}##. This needs an energy per volume of ##\rho \left( \frac{v^2}{x^2} - v^2 \right) = \rho v^2 \left(\frac{1}{x^2}-1\right)##.
The error is in there: you derived that from Bernoulli's equation, which is ##p =\frac{1}{2} \rho v^2## for this situation.

You are missing the 1/2.
$$v^2 = \frac{p_0-100kPa}{c+\rho\left(\frac{1}{x^2}-1\right)}$$
Corrected:
$$v^2 = \frac{p_0-100kPa}{c+.5\rho\left(\frac{1}{x^2}-1\right)}$$
Plugging in numbers:
x=1 => v=5 m/s as expected
x=0.5 => v=4.588 m/s (8% reduction)
x=0.25 => v=3.592 m/s (28% reduction)
x=0.1 => 1.865 m/s (63% reduction)
A couple, corrected:
x=0.5 => v=4.78 m/s (9.56 m/s exit)
x=0.1 => v=2.47 m/s (24.7 m/s exit)

The embarassing part is that I sort-of cheated and used an online calculator for velocity=>velocity pressure:
http://www.engineeringtoolbox.com/dynamic-pressure-d_1037.html

The issue is hidden in your numbers because of your different approach from mine (discussed in next post), as you aren't calculating the exit velocity, but rather the pipe velocity. The exit velocty levels-off at 28.25 m/s (the velocity at a Vp of 400 kPa). To make matters worse, I was using English units and had to convert at the end -- we did a lot of mixing and matching here.
x=0.05 => 0.982 m/s (at this point flow is roughly proportional to x as garden hose losses become negligible and the exit velocity does not change any more)
Clarification: since the exit velocity is constant after that, the pipe velocity (not "flow" rate, which usually means volumetric) is proportional to x. IE, since the maximum exit velocity is 28.25, the pipe velocity is 1.41 at a .05x orifice.

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mfb
russ_watters
Mentor
So, what was interesting in all of that is the 3 different approaches for 3 different people. We had this:

Approach 1: Boneh3ad: Assume large orifice, all losses are in the pipe, ignore exit velocity/pressure.
Approach 2: Me: Assume small orifice, pipe loss is negligible, all initial pressure becomes exit velocity pressure.
Approach 3: mfb: solve the entire system equation.

Boneh3ad's and my approaches are exact opposites of each other. He ignores the velocity pressure and assumes all loss is in the pipe (which you can do for a large orifice) and I ignore loss in the pipe and assume all initial pressure becomes velocity pressure (which you can do for a small orifice).

I think this discussion probably says something about physicists vs engineers...

I will say this: mfb's is the most accurate* (assuming the math is done right). However, I still like mine over either of the others because mine is, in my judgement, closer to the point of the OP than boneh3ad's (very high velocity is desired, therefore very small orifice is used, therefore pipe loss is very low, therefore velocity pressure roughly equals supply pressure)...and as an engineer, I find it easier than mfb's (though his his more accurate [when done correctly ;) ]).

We're also still left with interpreting the OP, since we never did get a clarification on force vs pressure. If force is the desired answer, none of these provide it. If pressure is desired, none of these are needed (it is 75 psi...unless that's absolute), though at least we calculated the associated velocity.

*Well...if we are looking for highest possible pressure, the most accurate method requires no calculation (Vp = Sp = 75 psi), but the second most accurate is mine, which also tells you the velocity at that velocity pressure (it's the limit of mfb's as x approaches 0....over x).

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Gold Member
@mfb and @russ_watters,

In order to try to more clearly illustrate the point I've been trying to make, I have solved the full set of equations (iteratively) that are usually used to design these sorts of systems. Here is what I did. I started with the Bernoulli equation modified to incorporate viscous losses through the system:
$$\dfrac{p_1}{\rho g} + \dfrac{v_1^2}{2g} = \dfrac{p_2}{\rho g} + \dfrac{v_2^2}{2 g} + h_{L,\mathrm{major}} + h_{L,\mathrm{minor}}.$$
Here, ##h_{L,\mathrm{major}}## is the major head loss due to the length of the hose, and ##h_{L,\mathrm{minor}}## is the minor loss associated with the sudden contraction. The definition of the minor loss is
$$h_{L,\mathrm{minor}} = K\dfrac{v_2^2}{2g},$$
where ##K## is an experimentally determined coefficient (see https://www.amazon.com/dp/0070621780/?tag=pfamazon01-20&tag=pfamazon01-20). When ##A_2/A_1## is zero, which is the worst case scenario, then ##K = 0.5##, so I used that here to be conservative with this. The definition of the major loss is obtained from the Darcy-Weisebach equation,
$$h_{L,\mathrm{major}} = f\dfrac{\ell}{D}\dfrac{v_2^2}{2g},$$
where ##f## is the Colebrook friction factor, ##\ell## is the length of the hose (100 ft in this example), and ##D## is the diameter (5/8" for most hoses in the US). ##f## can be determined implicitely from the Colebrook formula,
$$\dfrac{1}{\sqrt{f}} = -2\log_{10}\left( \dfrac{\epsilon}{3.7 D} + \dfrac{2.51}{Re\sqrt{f}} \right),$$
where ##\epsilon## is the absolute surface roughness (determined as 1.35 microns according to my previous source's mass flow plot, which is in line with some Googling to check that value) and ##Re## is the Reynolds number.

So, at the end of the day, if you define the volumetric flow rate ##Q = v_1 A_1##, you can use all of these previous equations and combine them to get
$$Q = A_1\sqrt{\dfrac{2(p_{\mathrm{tot}} - p_2)}{\rho\left[ \left( \dfrac{A_1}{A_2} \right)^2\left( 1 + K \right) + f\dfrac{\ell}{D} \right]}}.$$
Using this plus the Colebrook equation in a predictor-corrector scheme, you can plot the flow rate as a function of the area ratio. This relationship is plotted below:

As you can see, for the most conservative case in the contraction, closing off half of the orifice implies that the resulting mass flow through the hose is about 95% that of the uncapped case with the end of the hose completely open. You have to block about 65% of the hose to bring that down to 90%, and almost 90% of the hose to bring that down to 50% of the original mass flow.

The moral of the story is that my original post where ##A_2/A_1 = 0.5## that assumed constant mass flow was only in error by about 5%. I think that makes it a reasonably valid assumption, particularly for illustrative purposes.

russ_watters said:
Approach 1: Boneh3ad: Assume large orifice, all losses are in the pipe, ignore exit velocity/pressure.
I'd also like to point out that this is never what I did. I did assume an arbitrary orifice size for my example, but I made no assumptions about all losses being in the pipe. In fact, I was assuming the losses in the pipe would be nearly identical in the two situations and using that to only calculate the losses due to the contraction at the end. I also didn't ignore the dynamics pressure (or velocity pressure if you prefer that nomenclature) as I previously addressed.

russ_watters said:
Not a huge issue, since 75 psi is pretty big and pretty arbitrary.
It wasn't arbitrary. That was the nominal water pressure supplied to my house that I lived in when I was in graduate school.

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mfb