Why do smaller holes shoot higher jets of water?

[Tibriel]

So I know that F = P*A

When you put a nozzle on a hose the area is smaller so does the pressure increase and the force stay the same, does the pressure stay the same and the force increase, or do they both change?

Also if you poke a hole in the bottom of a bucket full of water will the velocity of the water out of the hole (i.e. how far it goes) be related to the size of the hole? Would water shoot out further from a smaller hole than a larger hole?

Thanks

 

[rcgldr]

When you put a nozzle on a hose the area is smaller so does the pressure increase and the force stay the same, does the pressure stay the same and the force increase, or do they both change?

The pressure never increases. A hole reduces the resistance to flow, which reduces the pressure. A smaller hole reduces the pressure less than a larger hole.

Also if you poke a hole in the bottom of a bucket full of water will the velocity of the water out of the hole (i.e. how far it goes) be related to the size of the hole? Would water shoot out further from a smaller hole than a larger hole?

Pressure would be a function of how high the water is above the hole. The higher the water, the higher the pressure at the hole. Because of viscosity, if the hole is too small the stream will be restricted. As the hole gets larger the stream would approach some terminal velocity based on the pressure difference between water at the hole and the air pressure outside.

 

[russ_watters]

I'm not sure I like that explanation...

This issue is all about eliminating dynamic losses in the pipes. When the garden hose is open, the pressure at the outlet is near zero, while the pressure at the main can be up to around 50 psi. The flow comes into equilibrium when the pressure gradient at a particular flow rate "uses up" all of the available 50 psi. By putting your finger over the end of the hose, you reduce the flow rate and effectively eliminate the dynamic losses, making the entire 50 psi available at the end of the hose.

Then you can take that pressure and use Bernoulli's equations to figure out the maximum velocity and height of the stream.

 

[Ralith]

I'm a complete and total layman, but... I think I know what you're talking about. It has to do with how much resistance there is- the higher resist. of a small hole reduces pressure less, but let's less actual water out (normally), it just goes farther.

 

[rcgldr]

This issue is all about eliminating dynamic losses in the pipes. When the garden hose is open, the pressure at the outlet is near zero

That's also because it's accelerating towards an ambient pressure environment beyond the hole. The length and diameter of the hose and flow rate are factors in the amount of dynamic pressure loss. I probably should have explained that better before.

By putting your finger over the end of the hose, you reduce the flow rate and effectively eliminate the dynamic losses, making the entire 50 psi available at the end of the hose.

Not quite, if there's any flow at all, then there are some dynamic losses in the hose.

 

[tankFan86]

If the flux is constant, then the water will have to move faster through a smaller hole in order to achieve the same flux.

 

[russ_watters]

Not quite, if there's any flow at all, then there are some dynamic losses in the hose.

By "effectively eliminated", I mean that since the losses are a square function of flow rate, if you cut down the flow rate enough, you reduce the losses enough to ignore them. Ie, if you have 50 psi available and you cut the flow by a factor of 4, you reduce the losses by a factor of 16, giving you almost 47 psi. Cut the flow rate by 90% and you get 49.5 psi available.

 

[russ_watters]

If the flux is constant, then the water will have to move faster through a smaller hole in order to achieve the same flux.

The flux isn't constant.

 

[tankFan86]

What type of system are we talking about? A tank draining through a hose?

 

[russ_watters]

Whether it is a tank draining through the hose or a household water system with the water coming from a main in the street, the flow will be a function of pressure, area, and loss. In order to make the flux (mass flow rate) constant with a change in area, you'd need to have a way to vary (and regulate) the pressure of the system providing the water.

 

[tankFan86]

Wouldn't a tank draining through a hose be governed by Torricelli (as a first approximation)? He says that the velocity of the jet is proportional to the square root of the height of the water above the jet. Changing the size of the aperture should not affect the velocity of the jet.

 

[russ_watters]

Wouldn't a tank draining through a hose be governed by Torricelli (as a first approximation)? He says that the velocity of the jet is proportional to the square root of the height of the water above the jet. Changing the size of the aperture should not affect the velocity of the jet.

Well you said "flux", which is mass flow rate, not velocity, but in any case, that approximation is true only when there is no loss in the system. Here, the loss is the key to why the velocity changes with the orifice size.

 

[tankFan86]

By `loss' I think you mean an ideal fluid. What do Euler's equations of motion say about this phenomenon? I believe Bernoulli just gives the same as Torricelli.

 

[russ_watters]

I don't know what you mean by "ideal" fluid, but perhaps you are talking about viscocity. Euler's equations deal with inviscid flow (and yes, they do give the same as Bernoulli's* when applied to the whole system). Navier-Stokes deals with viscous flow. Thinking about it a little more, the friction loss and viscocity loss are probably the same thing: with water in a pipe, the velocity at the pipe wall is zero, so the loss is modeled strictly as a viscous friction issue.

*In fluid dynamics, it is critical to know what assumptions you can use and how those assumptions determine what equations you can use. When the garden hose is just open, you cannot simply apply Bernoulli's equation to the water pressure at the main to calculate how fast the water will be moving at the outlet of the hose. Bernoulli's equation requires inviscid flow and conservation of energy and you don't have them in that case. My point in post #7 was that if you can reduce the velocity through the pipe enough to all but eliminate the losses, then Bernoulli's equation can be applied, treating the piping system as a pressure vessel with a single orifice outlet and nothing between them to eat up your pressure.

 

[tankFan86]

I agree that a transparent list of assumptions is critical to any study of fluid dynamics. What assumptions can we make in this case?

Certainly we have incompressibility. Also, the fluid is baratropic, meaning it is under the influence of a conservative potential (gravity). Can we assume an invisid fluid? If so then we could also assume irrotational motion. If not then I think we are stuck with Navier-Stokes.

I think I am still unclear what type of system we are trying to model.

 

[TVP45]

Russ is correct. You can see the effect by filling a drum (say 55 gal) after putting a small nozzle in the bottom. The water "spurt" will no longer depend on whether you put your finger over the nozzle, almost closing it.

 

[russ_watters]

Can we assume an invisid fluid? If so then we could also assume irrotational motion. If not then I think we are stuck with Navier-Stokes.

We certainly cannot assume inviscid flow.

I think I am still unclear what type of system we are trying to model.

When someone says "garden hose" it almost certainly means a real-world garden hose: one connected to a domestic water piping system.

 

[tankFan86]

Do domestic water piping systems drain under gravity, or is pressure added to the system?

 

[russ_watters]

Do domestic water piping systems drain under gravity, or is pressure added to the system?

Most (but not all) domestic water systems are pumped and regulated to maintain a constant pressure at the street. This actually makes them act pretty similar to gravity systems.

 

[tankFan86]

I think I am beginning to understand the phenomenon at work here. In response to the OP's questions, I think it depends (as always).

If you are draining a large enough tank through small enough holes, then I do not think the velocity will be appreciably affected by the size of the holes, only on the depth of the hole beneath the surface. However, as the tank and hole size become similar sizes, the situation gets much more complicated and difficult to predict.

I am still unclear on how domestic water supplies work. Wikipedia's article on `globe valves' offered some insight. As stated before, is seems that a fairly constant pressure is provided to homes. To turn the water on in your kitchen, you must open a valve that causes the water to follow some crazy path that I think reduces the kinetic energy of the fluid from frictional losses. The valve's geometry is such that when the valve is more open, there is less frictional loss, and the water flows faster. I suppose this is an example of how two holes of different sizes can produce jets of different velocities.

I am still struggling with how placing your thumb over the end of hose increases the flow's velocity. My only guess is that there is no appreciable frictional losses due to your finger, and so the flux (mass flux or velocity flux, they are the same in an incompressible fluid) remains constant, requiring an increase in velocity through a smaller aperature. Perhaps russ you could elaborate on your explanation involving pressure?

 

[russ_watters]

I am still struggling with how placing your thumb over the end of hose increases the flow's velocity. My only guess is that there is no appreciable frictional losses due to your finger, and so the flux (mass flux or velocity flux, they are the same in an incompressible fluid) remains constant, requiring an increase in velocity through a smaller aperature. Perhaps russ you could elaborate on your explanation involving pressure?

It really is as simple as I've said: When water flows through several hundred feet of 1/2" to 3/4" pipes, there is a significant amount of viscous friction, which manifests as a loss in static pressure and coressponding reduction in velocity. If you put your finger over the end, you reduce the flow rate, reducing the friction and increasing the static pressure just before the outlet.

 

[tankFan86]

Hey I think I get it. That's really neat! Now how the heck do you write that mathematically?

 

[russ_watters]

Now how the heck do you write that mathematically?

It is extremely difficult, so generally you don't. Piping systems (and ductwork systems for air conditioning, btw), are designed using tables that tell you the loss per 100' of piping and loss for certain types of fitting valves, elbows, flow meters, etc. at a specific flow rate. For smaller domestic systems, the total length isn't enough to go any deeper than that because your pipe size choices are very limited anyway. A house never uses anything other than 1/2, 3/4, or 1" pipes (and 3/8" tubing for a fridge icemaker).

For larger, higher flow rate systems, you might build a spreadsheet that lists every fitting and section of pipe and adds up the pressure loss for each section to give the total.
Whether for air or water, you start with a desired flow rate (arrived at by code requirements and load calculations) and size the system according to that flow. You have to strike a balance between using larger piping or a larger pump (or fan) to get the desired flow rate.

There are CFD programs out there that will model flow through a piping system or air through a ductwork system, but I'm actually using one for a project for the first time ever right now - on a computer lab under-floor supply-air system. It isn't common though. I'm only using it because since there is no ductwork, it is tough to know if the air is going where you want it to go.

 

[Tibriel]

OK so let me see if i understand everything, when you poke a hole in the bottom of a container of water (as long as the hole is not small enough to disrupt flow due to its size) then the exit velocity of the water is directly proportional to the current pressure.

And nozzles work by reducing flow which reduces the loss in pressure... So the smaller the hole the more of the original pressure is available at the flow end and hence the water shoot out further.

 

[tankFan86]

There are CFD programs out there that will model flow through a piping system or air through a ductwork system...

How do these work if the math is too difficult? Do they just look up things in tables as well?

Have you tried solving Navier-Stokes for simple geometries?

 

[russ_watters]

How do these work if the math is too difficult? Do they just look up things in tables as well?

Something that might take weeks for a person to calculate manually takes seconds for a computer.

Have you tried solving Navier-Stokes for simple geometries?

I think I did in college, but I don't remember for sure.

 

[Topher925]

I disagree russ. While I am sure what you described affects the velocity of the water output from the hose I believe what the OP was describing is simply just caused by an increase in velocity of the water. In order for the stream of water to travel higher then there absolutely must be an increase in momentum. And since we can assume water to be incompressible, the only thing that can change its momentum is the velocity.

When you put your finger over part of the area of a hose you are just creating a very crappy nozzle. This causes the velocity of the water to increase and of course also reducing its mass flow rate which will cause a greater pressure at the exit as well. But the overall affect is an increase of momentum compared to having a uniform exit. The "jet nozzles" that are on the sinks in my lab/office/closet do the same thing but much more efficiently.

I'm doing some stuff with the navier-stokes equations but its mostly for open flows. I believe this phenomenon can mostly be modeled with Bernoulli's equation.

 

[russ_watters]

I disagree russ. While I am sure what you described affects the velocity of the water output from the hose I believe what the OP was describing is simply just caused by an increase in velocity of the water.

You're not making a point different from mine here. I'm explaining why velocity increases. I think everyone understands that the spray goes higher because the velocity is higher.

In order for the stream of water to travel higher then there absolutely must be an increase in momentum. And since we can assume water to be incompressible, the only thing that can change its momentum is the velocity.

Ehh, I don't really like saying momentum because it is difficult to work out mathematically the relationship between momentum and height. If you get this problem on a test, the correct method is via kinetic energy per unit mass:

1/2mV^2=mgh
h=V^2/2g

I'm doing some stuff with the navier-stokes equations but its mostly for open flows. I believe this phenomenon can mostly be modeled with Bernoulli's equation.

I said in a previous post that a real modeling of this system is extremely difficult to do from scratch. However, since the problem wasn't really set up for us, we can decide on our own simplifying assumptions. I used assumptions that turned the "after" scenario into a simple Bernoulli situation. The "before" is much more complex.