Ideal Pipe Size?

Hello.

First, thanks for the wonderful forum. I have learned a lot from reading here.

I have been planning to build my first laminar nozzle. Towards that end, I figured out the size of arc I was after (7ft high and 7ft across), used that to calculate the minimum pump volume as well as the linear speed at outlet. I was trying to determine the best size of pipe to use for the nozzle. To do that, I used the continuity equation to determine the linear speed inside pipe expansions of 4", 6", 8", and 10". I then used those speeds to calculate the Reynolds number that would exist inside the nozzle. Since the arc was large enough that I understood it would be difficult, or possibly impossible, to achieve with 4" pipe, I expected that I would find the 4" pipe well within the transition zone, maybe 6" in laminar but a bit close to the border, and then 8" and 10" solidly laminar. To my surprise, the Re values were 398, 177, 100, and 64 respectively. Although the larger pipes were more solidly laminar than the small pipe, even the 4" is very far from the 2300 laminar boundary.

From other folks' writings, I take it that it would actually be very unlikely that the 4" pipe would produce that arc without it breaking up near the end. Do any of you have any idea what causes the failure? I have thought that maybe it is a turbulent region where the upward laminar flow collides against the top plate. Maybe that turbulence grows as the flow rate increases and eventually impacts the laminar flow exiting at the jet. If so, any ideas on how it might be mitigated? Regardless of the cause, have any of you seen a real world benefit to going larger than 6"? Would 8" be better? Would 10" be still better? I would expect that there is a point of diminishing returns where bigger diameter really doesn't provide any benefit. I am trying to get a gauge of where that boundary is. If I need anything larger than 4" I will need to order it on-line - so I would rather start with something that has the best chance for success.

Thanks for your help.

-David

Hi, welcome, and thanks for the compliments. We are all learning here so thanks for sharing your thoughts too.

That's interesting. I would like to know more about how you calculated the values. I've noticed that the laminar stream doesn't become turbulent that till after it reaches the top of the arc. Maybe there is something there. Maybe it's because the water isn't moving parallel, and so each particle actually has a different trajectory. I don't know.

The best arcs have been built using an 8" diameter pipe. Before you order online you might want to check your local sprinkler supply stores not the home depot, lowes kind but ones that only deal with sprinklers.

Thanks for the tip on the sprinkler supply stores - I will look into that. I had been looking at hardware stores like the ones you named. They mostly went to 4". It would certainly be easier if I could find something larger around here rather than doing the on-line thing.

For the calculations, I started with getting the linear velocity at some set point in the system. I have done it two ways. Initially, I just assumed a pump size of 1000gph entering through a 1 inch inlet. That inlet size might differ, but seemed like a good starting point. I converted gallons per hour into cubic feet per second:

(1000 gal/hr)(0.13368 ft^3/gal)/(3600 s/hr) = 0.03713 ft^3/s

I then divided that flow rate by the cross sectional area of the inlet:

A = pi * R^2 = (3.14159)(0.5 inch / (12 inch/ft))^2 = (3.14159)(0.04167 ft)^2 = (3.14159)(0.001736 ft^2) = 0.005454 ft^2

v-inlet = (0.03713 ft^3/s)/(0.005454 ft^2) = 6.808 ft/s

When I do the Reynolds calculations, I will SI units, so I converted the v-inlet into m/s:

v-inlet = (6.808 ft/s)(0.03048 m/ft) = 2.075 m/s

That gives you one of the key values you will need to calculate the Reynolds number. I got something similar by using the formulas from this site on pump sizing. Those calculate the v-outlet based on the arc that you want. You could use that number below in the same way that I use v-inlet here. The difference is that you would use the outlet diameter rather than the inlet. The v-outlet is more precise because it is based on the actual arc you want versus my guesstimate of the inlet flow rate. There won't, however, be a huge difference either way. The numbers I provided in my first post were based on the outlet calculations for a specific arc - so numbers from this example based on inlet will be a little different.

Once I had the inlet velocity (or the outlet velocity), I needed to get the linear velocity inside the body of the nozzle. To do that I used the continuity equation. That equation basically states that flowing mass is neither created nor destroyed - so the volume flow rate in will equal the volume flow rate out will equal the volume flow rate at any point inside. So:

Vol1 = Vol2
(v1)(A1) = (v2)(A2)
(v1)(pi)(R1)^2 = (v2)(pi)(R2)^2
(v1)(R1)^2 = (v2)(R2)^2
v2 = (v1)((R1)^2/(R2)^2

For my example, inlet is point 1, so R1 is 1inch/2 = 0.5 inch = 0.04167 ft = 0.0127 m. R2 will be inside the pipe and will vary depending on the size of pipe. For a 6 inch pipe, R2 is 3 inch= 0.25 ft = 0.0762 m. The value v1 is v-inlet calculated above. (And as mentioned, you can do the same thing using the outlet values - but then R1 is 0.25 inch if you have a half inch outlet. Plugging in the numbers gives us:

v2 = (2.075 m/s)(0.0127 m)^2/(0.0762 m)^2 = 0.05764 m/s

A quicker way to do that calculation if you are starting from a 1 inch inlet is just divide the inlet velocity by 36 for a 6 inch pipe, divide by 64 for 8 inch pipe, divide by 100 for 10 inch pipe, etc.

Reynolds number is calculated as:

Re = (density)(v2)(hydraulic diameter)/(dynamic viscosity)

for water at 20-deg C, density = 998 kg/m^3 - it varies a little with temperature, but not too much. Under the same conditions, dynamic viscosity of water is 0.001002 Ns/m^2. The hydraulic diameter of a circular pipe is just its diameter (hydraulic diameter for other shapes has a different definition). In this case the pipes in question are the straws. Our expansion of the nozzle at inlet slows the water down enough that it can be brought into the laminar range, but it is the straws that actually cause the shift to laminar. For an estimate, use 0.005 m for the straw diameter. That gives us:

Re = (998 kg/m^3)(0.05764 m/s)(0.005 m)/(0.001002 Ns/m^2) = 287

Unitless Re values less than 2300 are laminar, values between 2300 and 4000 are transient, and values above 4000 are turbulent. Given that, we are well in the laminar range for this example. I am sure that the sudden jet outlet kicks that value up quite a bit, and nothing is ever ideal, so we are probably a bit higher - but we have a lot of room to play with.


Hopefully that all makes sense.

-David

John,

I was thinking about your comment on the flow not breaking apart until after the top of the arc, and I think you are probably onto something there. Once the water exits the jet, there is nothing to keep it laminar. Probably every little thing reduces its laminar nature (laminarness?... laminarity?.... laminosity?... anyway...). Every little bit of breeze probably cuts it down some. Then you get to the peak of the arc and there will definitely be a difference in the water at the top of the stream versus the water at the bottom. Maybe all of that is enough to slash the laminosity (yeah - I think I will go with that one). Maybe the extra couple hundred points of the 8 inch versus the 4 inch is just enough to allow the arc to complete without breaking down. Interesting idea. It is going to drive me crazy that there is nothing I can calculate to figure out where that breaking point is - but I guess I will have to get over that. Sounds like, experientially (yeah - that one isn't a real word either), 6 inch diameter might cut it, best results with 8 inch, maybe no noticeable improvement in going to 10 inch.


-David

Yeah, I'm not sure what the proper word is...laminarity, laminar....osity. I've often used the word laminosity. Plus I think it sounds cool!

You know you mentioned that there isn't math that can see to describe the behavior of the water, ...yet. We could work on that! That would really be something! We could probably use the Reynolds number but find out where the breaking point of the stream was. It would be interesting. We could build several different models 4, 6, 8, and 10 inch. Measure the flow rate. Verify that it is completely laminar all the way through, then up the flow rate. Continue on until the stream starts to break up, sort of transitional. Then find out when if becomes completely turbulent. There may be a correlation between the different nozzles sizes. We'd never know until we try it.

I'd really like to try a 10" and a 12" and compare some of the results, but alas I don't have unlimited funds. I can think of so many different areas of interesting research if I just had the money.

Why did you divide the inlet dia by 2 to get R1?

For my example, inlet is point 1, so R1 is 1inch/2 = 0.5 inch = 0.04167 ft = 0.0127 m. R2 will be inside the pipe and will vary depending on the size of pipe. For a 6 inch pipe, R2 is 3 inch= 0.25 ft = 0.0762 m. The value v1 is v-inlet calculated above. (And as mentioned, you can do the same thing using the outlet values - but then R1 is 0.25 inch if you have a half inch outlet. Plugging in the numbers gives us:

Hi pbracer,

The volume flow rate (the value that is conserved anywhere in the pipe) is linear flow rate times cross-sectional area. The cross-sectional area of the pipe is a circular area - which is defined as pi times the radius squared. Radius is half of the diameter. Since my inlet size is 1 inch diameter, I need to use 0.5 inch radius for any formulas that want radius instead of diameter. It gets a bit confusing at times since some of the formulas expect the radius, and others (like the Reynolds calculation) expect a diameter. although I used the inlet, I could just as easily have used the outlet - any point where I could know both the pipe size and flow rate.

-David

wow... there was some heavy math numbers here (btw I have not read everything here) ..

But I just wonder, do the SIZE of the output (the diameter of water beam) matter for how long it could stay laminar?
I just got this crazy idea of how big is it possible to make the laminar beam travel ??

What do wee need to make it go far away, as we know high pressure will break up the laminar flow.
So what is the solution, bigger nozzle diameter, or bigger diameter on the brass ring??

I have 2 water pump that I don't use, and I wonder about making a extremely big nozzle for fun..
With 2 input of pumps, and one double size water output brass ring!

Maybe it could travel 4-5 meters away, and 3-4 meters Up ?

I've read somewhere that it is harder to work with larger diameter laminar streams, but I can't find anything on it now.

It would be cool to try and double or triple the current output. That would be really something to see. If I just had enough money and time!!!

That would be very cool to try. As I understand it, it should be possible. The size of output is more related to ability to throw a jet rather than the laminar nature of that jet. The laminar nature (at least on exit from the nozzle) should be based almost entirely on the speed within the nozzle and the diameter of the straws (or effective diameter of other media inside the nozzle). If you increase the size of the outlet, then you need to either increase the size of the inlet (with the same flow rate) or increase the inflow rate (with same inlet size). Of course, you could do some combination of the two as well. The higher inflow, however, might mean you need a larger nozzle diameter to slow the water down enough to make it laminar. I am also still pretty convinced that there is some ugly turbulence at the top plate away from the hole - and I expect that just gets worse with increased flow. So that would make me believe all the more that you will need a larger nozzle to pull it off.

There are a bunch of gotchas there. There might be some degree to which surface tension helps keep the jet laminar once it is out of the nozzle. If that is the case, then the size of the jet might be constrained - too large or too small might not have the right surface tension. As John mentioned early in the thread, there is also the issue of gravity pulling unevenly on the jet - especially at the peak of the arc. That might become more of an issue as the jet is larger. (Though, for all I know, it might become less of an issue rather than more - I am not sure.)

-David

David,

Agreed! I'm not going to attempt it...not yet anyway, but if I did increase the output you would definitly have to increase the nozzle size. If you wanted to scale it up you'd probably have to get something like a 16" diameter pipe! YIKES! That probably would be pretty pricey.

That would also need a very large pump in order to get it too shoot the distance you wanted.

What's your background? Are you an engineer/mathematician/physicist?

For a long time now I have worked in computer network engineering - routers, switches, firewalls - that sort of thing. There is a little bit of math there, but not too much. Mostly the job just comes down to good troubleshooting skills.

Before that, I graduated with a bachelors of science degree in chemical engineering from the University of Minnesota - Institute of Technology. So I guess that does give me a math/science/engineering background even if I don't use it in my daily life. That's quite a ways in the past though.

I suppose I have never quite gotten over wanting to have a mathematical understanding of my hobbies. Don't get me wrong, I do enjoy trying things and just seeing what happens. But I worry that, if I don't know what formulas apply, maybe I will miss the best experiment - because there is only enough time and money to try so many variations. I'll generally try to understand what is going on the best I can before beginning to experiment.

Before we moved I used to do home wine making as another hobby. My wife made great fun of me for using calculus to derive the optimal angle at which to brace the casks to avoid disturbing the precipitate that settles during and after fermentation. Looking at it now I would have to agree that perhaps that shows a slightly neurotic bent.

-David

Hey, there is nothing wrong with trying to apply math to everyday life. What makes math so attractive to me is that these numbers which mean nothing by themselves mean nothing, but when you start applying units to them you can start predicting the behavior of objects. Last summer I made a compress air paper rocket launcher out of PVC. It was really neat. It was fun to build and make paper rockets and shoot them up into the air. Then my mind started wondering how I could make them go higher. So I started wondering what variables were pertinent to the paper rockets. After sometime I was able to develop the equations for the rockets and it gave me some amazing understanding as to the behavior of them. My equations were very close to the actual application too!!!! That's always a bonus.

My wife thinks that I think things through too much sometimes. She is constantly listening to me talk about the same problem over and over and over and well you get the picture, but without understanding you won't be able to solve it.