- #1
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
Calculating the PSI needed to generate a certain fluid speed
In summary, the conversation discusses the calculation of pressure in PSI needed to generate a 150mph stream of water from a 2mm perfectly round orifice. The conversation also touches on the flow rate and parameters needed for the calculation, such as the type of fluid and the configuration of the pipe and fitting network. The ultimate goal is to determine the pressure needed at the point of exit for the orifice, in order to adjust the pressure for a constant pressure spray boom. The conversation also mentions the use of a standard nozzle and its datasheet for reference. Finally, it is mentioned that 3.33 GPM is required to achieve a velocity of 150mph, and the question is posed about the corresponding PSI needed for this flow rate.
- #2
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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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- #3
Stetson
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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.
- #4
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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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- #5
Stetson
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Give me all the parameters you need to solve the problem.
- #6
Stetson
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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?
- #7
berkeman
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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?Stetson said:
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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.
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.
- #9
Stetson
- 19
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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?
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
Stetson
- 19
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berkeman 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.
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.
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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
berkeman
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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 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?Stetson said:
- #13
berkeman
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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 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?Stetson said:
- #14
Stetson
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berkeman 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?
- #15
berkeman
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As I said, this is not my specialty, but it sure seems like the nozzle datasheets should give you that kind of information...Stetson said:
https://www.google.com/search?q=water+nozzle+flow+datasheet&ie=utf-8&oe=utf-8&client=firefox-b
- #16
Averagesupernova
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@Stetson are you trying to develop your own nozzle?
- #17
berkeman
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Are you maybe trying to make a Water Jet Cutter? https://en.wikipedia.org/wiki/Water_jet_cutterStetson said:
Some (pretty high) pressures are listed, but I'm not seeing any jet velocities listed. Why is 150mph the magic number for you?
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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:
[tex]\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}.[/tex]
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:
[tex]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}.[/tex]
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##,
[tex]\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}.[/tex]
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,
[tex]\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}}.[/tex]
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.
[tex]\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}.[/tex]
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:
[tex]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}.[/tex]
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##,
[tex]\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}.[/tex]
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,
[tex]\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}}.[/tex]
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
Stetson
- 19
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Averagesupernova said:
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
Stetson
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- #21
Tom.G
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It seems that the wind drift is highly dependent on droplet size. This would be due to the lower terminal speed of smaller droplets. If that is the case, uniform droplet size would be the next optimization target. Here is a somewhat dense paper on terminal speed:
http://www.atmo.arizona.edu/students/courselinks/fall10/atmo551a/DropletFallSpeed.doc
Along with half a million other results, this was found with a search of:
https://www.google.com/search?&q=terminal+speed+versus+droplet+size
http://www.atmo.arizona.edu/students/courselinks/fall10/atmo551a/DropletFallSpeed.doc
Along with half a million other results, this was found with a search of:
https://www.google.com/search?&q=terminal+speed+versus+droplet+size
Related to Calculating the PSI needed to generate a certain fluid speed
1. How do I calculate the PSI needed to generate a certain fluid speed?
To calculate the PSI (pounds per square inch) needed to generate a certain fluid speed, you will need to use the Bernoulli's equation. This equation takes into account the fluid's velocity, density, and pressure. You will also need to know the area of the opening or nozzle where the fluid is being forced through. Once you have these values, you can plug them into the equation to determine the required PSI.
2. What is the Bernoulli's equation and how does it relate to calculating the PSI?
The Bernoulli's equation is a fundamental equation in fluid dynamics that relates the energy of a fluid to its pressure, velocity, and elevation. It states that the sum of the pressure, kinetic energy, and potential energy per unit volume of a fluid in a steady flow is constant. This equation is used to determine the required PSI needed to generate a certain fluid speed.
3. Can I use any unit of measurement to calculate the PSI?
No, it is important to use consistent units of measurement when calculating the PSI. The most commonly used unit for pressure is PSI, but other units such as bar, pascal, or kilopascal can also be used. Just be sure to convert all values to the same unit before plugging them into the Bernoulli's equation.
4. How does the shape of the opening or nozzle affect the required PSI?
The shape of the opening or nozzle can affect the required PSI needed to generate a certain fluid speed. A smaller, more narrow opening will require a higher PSI to achieve the same fluid velocity as a larger, wider opening. This is because the fluid must travel through a smaller area, increasing the velocity and therefore the required PSI.
5. Are there any other factors that can affect the required PSI?
Yes, there are other factors that can affect the required PSI in addition to the shape of the opening or nozzle. These include the viscosity of the fluid, the length of the piping system, and any obstructions or bends in the system. These factors can cause friction and turbulence, which will increase the required PSI needed to achieve the desired fluid speed.
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