Calculating the PSI needed to generate a certain fluid speed

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Discussion Overview

The discussion revolves around calculating the pressure in PSI required to achieve a fluid speed of 150 mph through a 2 mm round orifice. Participants explore various parameters affecting this calculation, including fluid type, pipe configuration, and flow rates. The context includes practical applications related to a constant pressure spray boom system.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asks for the pressure needed to generate a 150 mph stream from a 2 mm orifice and the corresponding flow rate in GPM.
  • Another participant emphasizes the need for additional information, such as the type of fluid and the configuration of the piping system.
  • Some participants argue that simplifying the problem too much may lead to an unanswerable situation due to the complexities of pressure losses in the system.
  • A participant suggests using water as the fluid and discusses the implications of pressure drop through the orifice.
  • There are discussions about the specific setup involving a constant pressure spray boom and the parameters of the tubing and orifice.
  • One participant provides a calculation suggesting that 79.1295 mph can be achieved at 50 PSI with a flow rate of 2.24 GPM, but questions the accuracy of this in relation to the original query.
  • Another participant mentions the use of the modified Bernoulli equation to calculate the required pressure, considering factors like head loss and flow rate.
  • Some participants express uncertainty about the calculations and seek further clarification or data, including the possibility of using manufacturer datasheets for standard nozzles.
  • There are repeated mentions of the need for specific parameters to arrive at a meaningful answer, including the flow rate required to achieve the desired velocity.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the exact pressure required, with multiple competing views on how to approach the problem and the necessary parameters. The discussion remains unresolved regarding the specific calculations and assumptions needed to determine the required PSI.

Contextual Notes

Limitations include missing assumptions about fluid properties, the configuration of the piping system, and the specific characteristics of the orifice. There are unresolved mathematical steps related to the calculations proposed by participants.

Stetson
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How much pressure in PSI would it take to generate 150mph stream out of a 2 mm perfectly round orifice? And how many GPM would that equate to? Thanks for any help
 
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Your going to need a lot more information than that. What fluid? What is the configuration of the pipe and fitting network before the orifice?
 
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Well let’s not make this any more complicated than we need to... let’s just use water, and the pressure needed at the point it exits the orifice.
 
The problem is that you are making it so simplified that it has no answer. If you want to know the pressure required for a give flow rate, you need to know something about the losses that have to be overcome by that pressure.
 
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Give me all the parameters you need to solve the problem.
 
Or let’s just say, the pipe is 1/4” under 50 psi. reduces to the 2mm orifice that is spraying a straight stream, how fast is that stream of water moving?
 
Stetson said:
150mph stream out of a 2 mm perfectly round orifice? And how many GPM would that equate to?
At least that part of your question looks pretty straightforward. Have you tried doing the calculation? A 2mm diameter cylinder of water traveling at 150mph is what flux of water? There are a couple of unit conversions involved, but it seems pretty straightforward other than that, no?
 
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I've given you all we need to do a problem like this. If you have no pipes or fittings upstream then the pressure drop from the orifice to the orifice is zero. You need zero psi to achieve literally any flow rate. This is why you need to include information about your system.

Now, in your recent post, you did. You gave us a set of tubing/piping (though no material or length) and then a reduction (though no idea about tapered versus sudden) so you are getting to a point where you can have an answer, but not necessarily the answer that will actually pertain to your system.
 
this is on a constant pressure spray boom. Were my nozzles mount to the boom. in place of one of the nozzles i have a pressure gauge, so i know at that point in the system i am getting 50 psi.
It does go through a nozzle body with a diaphragm and would be nearly impossible for me to explain the changes that Happen through that. My idea is to figure out what it would be if it went straight from the boom to the orifice with this scenario. 6.35mm polypropylene tubing 50mm long and reduces down to the 2mm orifice in a funnel shape over the distance of 12.7mm.

At that point i will know what pressure i will need on the exit side of the nozzle body, i can place a pressure gauge there and see what my pressure loss is through the nozzle body and adjust pressure accordingly.

So, original question was, how much PSI does it take to generate a 150 MPH stream out of a perfectly round 2mm orifice?
 
  • #10
berkeman said:
At least that part of your question looks pretty straightforward. Have you tried doing the calculation? A 2mm diameter cylinder of water traveling at 150mph is what flux of water? There are a couple of unit conversions involved, but it seems pretty straightforward other than that, no?

If you can come up with some numbers i would love to see them, i came up with 79.1295 MPH at 50PSI with a flow rate of 2.24GPM.

This is a little bit advanced for me, and really don’t know if i am even in the ballpark. Was hoping to get some answers from the science community on here. This is not just theoretical it’s for a real life application.
 
  • #11
Your supply pressure would be helpful. Replacing your nozzle with a gauge will not give you anything useful as it will stop the flow. Does this mean your supply pressure is 50 psi?
 
  • #12
Stetson said:
If you can come up with some numbers i would love to see them, i came up with 79.1295 MPH at 50PSI with a flow rate of 2.24GPM.
Sorry, this makes no sense for your question that I focused on. You asked about the flow rate for a cylinder of water moving at 150m[h. The answer to that question is pretty straightforward, IMO.
Stetson said:
this is on a constant pressure spray boom.
This is not my area of expertise, but if you are using a standard nozzle that you ;purchase from a manufacturer, isn't there a datasheet proving this info?
 
  • #13
Stetson said:
If you can come up with some numbers i would love to see them, i came up with 79.1295 MPH at 50PSI with a flow rate of 2.24GPM.
Sorry, this makes no sense for your question that I focused on. You asked about the flow rate for a cylinder of water moving at 150m[h. The answer to that question is pretty straightforward, IMO.
Stetson said:
this is on a constant pressure spray boom.
This is not my area of expertise, but if you are using a standard nozzle that you purchase from a manufacturer, isn't there a datasheet proving this info?
 
  • #14
berkeman said:
Sorry, this makes no sense for your question that I focused on. You asked about the flow rate for a cylinder of water moving at 150m[h. The answer to that question is pretty straightforward, IMO.

This is not my area of expertise, but if you are using a standard nozzle that you purchase from a manufacturer, isn't there a datasheet proving this info?

I guess I am having a hard time relaying what I want to find out.

I have since found this much out,

It will take 3.33 GPM moving through a 2mm diameter orifice to reach a velocity of 150mph. What PSI would achieve this?
 
  • #15
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@Stetson are you trying to develop your own nozzle?
 
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  • #17
Stetson said:
I guess I am having a hard time relaying what I want to find out.

I have since found this much out,

It will take 3.33 GPM moving through a 2mm diameter orifice to reach a velocity of 150mph. What PSI would achieve this?
Are you maybe trying to make a Water Jet Cutter? https://en.wikipedia.org/wiki/Water_jet_cutter

Some (pretty high) pressures are listed, but I'm not seeing any jet velocities listed. Why is 150mph the magic number for you?
 
  • #18
So here is the thing. The best way to calculate the pressure required to achieve a certain ends is by using the modified Bernoulli equation below:
\dfrac{p_1}{\rho g} + \dfrac{V_1^2}{2g} + z_1 = \dfrac{p_2}{\rho g} + \dfrac{V_2^2}{2g} + z_2 + h_{\ell}.
In this case, ##h_{\ell}## is the head loss and represents all the losses due to friction and valves and pipe bends and the like. ##p_2## is the exit pressure and is going to be equal to atmospheric pressure, and ##p_1## is going to be your "supply pressure" at some given reference point (either just after your pump or at some other known location in the system. The ##z## terms are heights of the two points to account for gravity.

So you are basically looking for a pressure required to meet your requirements. If you solve for that pressure drop (aka the supply gauge pressure, you get something like this:
p_1 - p_2 = \Delta p = \dfrac{1}{2}\rho\left(V_2^2-V_1^2\right) + \rho g \left(z_2 - z_1\right) + \rho g h_{\ell}.

The other thing is that you rarely know velocity, but flow rate, ##Q=VA## is pretty common to know, so you can replace those velocity terms with areas and ##Q##,
\Delta p = \dfrac{1}{2}\rho Q^2 \left[\left(\dfrac{1}{A_2}\right)^2-\left(\dfrac{1}{A_1}\right)^2\right] + \rho g \left(z_2 - z_1\right) + \rho g h_{\ell}.

And then finally, you actually do have a velocity requirement in that you want a specified ##V_2##, so go ahead and use ##Q=V_2 A_2## to get a final, useful equation,
\boxed{\Delta p = \dfrac{1}{2}\rho V_2^2 \left[1-\left(\dfrac{A_2}{A_1}\right)^2\right] + \rho g \left(z_2 - z_1\right) + \rho g h_{\ell}}.

So, that is the final equation you should be working with here (barring any algebra errors I may have made). You will notice that the pressure you need has several dependencies, all of which require information about changes from one point to another. In your first question, you only gave us one, so these terms are all at best zero and at worst (and in reality) have no defined value. What you need is some indication of the height change and area change between your source pressure (##p_1##) and your outlet. If you had those two things, you could get some sense of the pressure drop required if you assume ##h_{\ell}## is zero or negligible, and your later post seemed to indicate that you do have those values available (though I don't think you provided height). You have to be careful here, though, because ##h_{\ell}## is non-negligible in most cases and will probably be a fairly substantial portion of the pressure requirement here.

The final takeaway here is that you need ##A_2/A_1## and ##z_2-z_1## to get any semblance of an answer, and you really need to consider the ##h_{\ell}## term if you want a good answer. You need to have some way of measuring ##p_1## or providing it at a point, and the remainder of the terms need to take into account everything between that point and the nozzle. Usually that point is located downstream of a pump or at a spigot or other supply line. The ##h_{\ell}## term is actually the sum of many terms for bends and expansions and valves and the like, and you can look up the various forms in books and manuals and handbooks and such. They may all have their own dependencies on velocity, though, so the problem can get messy.

I will also mention that plugging the nozzle end to measure the pressure is not necessarily a valid way to do it. At best, that will give you a good estimate of your supply pressure provided there are no other outlets in the system and that you take height change into account. If there are other outlets, it isn't going to tell you much as far as I can tell.
 
  • #19
Averagesupernova said:
@Stetson are you trying to develop your own nozzle?

Yes, we are working on a few prototypes. running my nozzle next to what I currently have installed on my plane right now, there is a great deal of difference. Its a tip that I believe will reduce physical drift from spray planes. running some ground tests, nothing super scientific about it really, just putting my nozzle and the other nozzle side by side and pressuring up the system you can see a dramatic difference. Thats placing them in a 90 degree crosswind, not a fan or anything like that, just outside in real world conditions that we would fly in.

the reason I was looking for the psi and speed was to be able to run some actual string tests out of the plane, plane moving 150MPH ground speed with droplet moving 150 mph opposite direction, to have the droplet fall with less forces acting on it vs. being subjected to so much wind shear. As the process is now, its like throwing a tomato against a wall. your spectrum of droplets are large to fine and everything in-between. And these nozzles we are using now have been through rigorous testing by the manufacturer and they do have a data sheet on them, but I am simply trying to build something better.
 
  • #20
_1409408506.jpg
Nothing new about this picture that you don't already know, just a glimpse of what we are dealing with. This is only at 50cm boom height. We are generally closer to 8-10feet above the crop canopy. Moving on average of 150MPH ground speed. For herbicide applications I want to be able to 'draw a line' with my airplane. A larger, more consistent droplet spectrum is the way to do this in my opinion, but we have to find a sweet spot between the physics of all of it and the biology of the plant and the mode of action of the chemical. In reality, it would be much easier just to keep doing what I am doing and go spray, but I think I am onto something with my design. Thanks for any help or input you may have
 

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