How Can I Increase My Lawn Sprinkler System's Flow Rate?

[AJacq]

I have a DIY project for a lawn sprinkler system but my flow rate is insufficient : if my setup is to work properly, the sprinkler manufacturer suggests a flow rate of 2880 litres per hour at a pressure no less than 2 psi.

I measured the flow rate of my current garden faucet, which is connected to 1/2" copper pipe (OD nominal size), and got a flow rate of about 1565 litres per hour.

Here is an inline description of my building's water supply :
A) 1-1/2" copper pipe from the city main to the building's mechanical room
B) 1-1/2" to 1-1/4" copper coupling (don't know why this was done)
C) 1-1/4" copper pipe
D) Pressure gauge indicating 90 psi - 620 kPa
E) Pressure regulator
F) Pressure gauge indicating 65 psi - 450 kPa
G) 1-1/4" copper pipe
H) 1-1/4" to 1-1/2" copper coupling (don't know why this was done)
I) 1-1/2" copper pipe leaving the mechanical room to supply the whole building

I require assistance to figure out what diameter copper pipe I should connect to the building's main line so as to deliver the required flow rate to my sprinkler system.

 

[russ_watters]

Welcome to PF.

Just based on your own pipe's size and flow rate, you should double it to 1". This of course assumes there isn't additional demand on the building mains that would limit the flow.

 

[AJacq]

Welcome to PF.

Just based on your own pipe's size and flow rate, you should double it to 1". This of course assumes there isn't additional demand on the building mains that would limit the flow.

Thanks... I considered the additional demand and I don't expect it to be problematic since the sprinklers would be used around 5:00AM when building water demand is nil or very low.

Is there a formula behind your 1" pipe recommendation ?

 

[Ketch22]

[QUOTE="
Is there a formula behind your 1" pipe recommendation ?[/QUOTE]

There are several formulas that would be involved. The simplest is that the flow in a pipe is based on the area of the circle. As areas are calculated via a square function it would be found that 3/4" pipe has twice the area of 1/2". Thus 2 x 1565 L/Hr would be 3130 L/Hr. Close enough however as the flow rate increases there is a significant increase in turbulence and thus energy required which manifests as pressure reduction. For a selection of formulas that apply here do a search for Reynolds number calculations. But we are working with pipes that are only available in increments. So 1" pipe (next size larger) is actually 4 times the area of 1/2" this would easily move the required amount of water and would do so at a low velocity and thus minimal pressure drop.
The recommendation is a well calculated guess if you want to call it that.

 

[AJacq]

There are several formulas that would be involved. The simplest is that the flow in a pipe is based on the area of the circle. As areas are calculated via a square function it would be found that 3/4" pipe has twice the area of 1/2". Thus 2 x 1565 L/Hr would be 3130 L/Hr. Close enough however as the flow rate increases there is a significant increase in turbulence and thus energy required which manifests as pressure reduction. For a selection of formulas that apply here do a search for Reynolds number calculations. But we are working with pipes that are only available in increments. So 1" pipe (next size larger) is actually 4 times the area of 1/2" this would easily move the required amount of water and would do so at a low velocity and thus minimal pressure drop.
The recommendation is a well calculated guess if you want to call it that.

Thanks... Here is what I got (hope the math is correct)

Assuming type M copper pipes :

Nominal size 1/2 in. ; ID = 0.569 in. ; Area = 0.254 sq. in. @ 1565 L/h

Nominal size 3/4 in. ; ID = 0.811 in. ; Area = 0.516 sq. in. ; Expected flow rate = 3179 L/h

Nominal size 1 in. ; ID = 1.055 in. ; Area = 0.874 sq. in. ; Expected flow rate = 5385 L/h

So a 3/4 in. pipe would deliver the required flow rate of 2880 L/h...

... as for the pressure drop caused by turbulent flow since the building's main line is pressured at 65 psi and I only require a minimum of 2 psi at the sprinklers, I should be fine right ? If nevertheless I wanted to do the math on this, I should calculate de velocity of the water by dividing flow rate by area (making sure I first convert to compatible units), then I could use Reynolds to determine whether I'm in laminar, transition or turbulent flow... is that correct ?

 

[Ketch22]

Yes, your math is correct. Are you using a drip or special flow sprinkler? A required pressure of 2 psi seems a little low for standard styles.
It looks like you have the right concepts and a good enough margin that it will work as you desire. In design of plumbing systems the other item that figures in is head loss (or pressure drop) caused by fittings and valves. There are standard charts from manufactures for this information. You have enough margin that this would not figure in. You could however do the calcs for personal "enjoyment"

 

[AJacq]

Yes, your math is correct. Are you using a drip or special flow sprinkler? A required pressure of 2 psi seems a little low for standard styles.
It looks like you have the right concepts and a good enough margin that it will work as you desire. In design of plumbing systems the other item that figures in is head loss (or pressure drop) caused by fittings and valves. There are standard charts from manufactures for this information. You have enough margin that this would not figure in. You could however do the calcs for personal "enjoyment"

Thanks for the additional information... and you are correct that I do calcs for personal enjoyment and I always learn something new in the process.

I goofed on the pressure (all these different units... when is the whole world going to switch to SI units and be done with it)... the required minimum pressure is 2 bar or 28 psi... good catch Ketch

BTW... this might be a stupid question : when the diameter of a pipe decreases, I figure the velocity of flow will increase and the pressure will decrease, and inversely with an increase in pipe diameter... but in all cases the the flow rate will remain constant... is this correct ?

 

[Ketch22]

[QUOTE="
BTW... this might be a stupid question : when the diameter of a pipe decreases, I figure the velocity of flow will increase and the pressure will decrease, and inversely with an increase in pipe diameter... but in all cases the the flow rate will remain constant... is this correct ?[/QUOTE] This is where the Reynolds number calculations come in. You are essentially correct. Starting with the nozzles that being supplied. Their achieved flow rate is determined by the pressure, unless you have some complicated ones with a pressure compensating orifice. The supply plumbing will flow as required by the dispersal end. However, as the flow rate increases the turbulence in the pipe increases this in turn increases resistance. Once the resistance is sufficient to overcome supply pressure the pressure starts to drop. Keep in mind that the resistance is a cumulative effect by distance of pipe run. Thus long runs of small pipe see a large head loss and larger or shorter runs less. Short large diameter runs see very little change.
Going back to the nozzles with the pressure drop from the connected pipe run it is possible to recalculate the achieved flow at the lower pressure. The lower flow will reduce the turbulence which leads to a minor increase in pressure. This will achieve a state of equilibrium on it's own.

Should you really choose to Geek a little the formula for calculating flows is a little tricky. The base formula is: Q = AV, where Q = Achieved flow in cubic feet per second, A = Area of orifice in Square feet, and V = velocity of liquid in Feet per second. This only actually works out on a true circle orifice on a flat plate with both shoulders being square and sharp. There are correction factors cataloged for K which is a factor to adjust for several standard orifices shapes and configurations which if you can find your orifice style or obtain from manufacture the formula then becomes Q = AVK. In reality for a sprinkler system very little difference will be present and not detectable without special testing.

 

[AJacq]

This is where the Reynolds number calculations come in. You are essentially correct. Starting with the nozzles that being supplied. Their achieved flow rate is determined by the pressure, unless you have some complicated ones with a pressure compensating orifice. The supply plumbing will flow as required by the dispersal end. However, as the flow rate increases the turbulence in the pipe increases this in turn increases resistance. Once the resistance is sufficient to overcome supply pressure the pressure starts to drop. Keep in mind that the resistance is a cumulative effect by distance of pipe run. Thus long runs of small pipe see a large head loss and larger or shorter runs less. Short large diameter runs see very little change.
Going back to the nozzles with the pressure drop from the connected pipe run it is possible to recalculate the achieved flow at the lower pressure. The lower flow will reduce the turbulence which leads to a minor increase in pressure. This will achieve a state of equilibrium on it's own.

Should you really choose to Geek a little the formula for calculating flows is a little tricky. The base formula is: Q = AV, where Q = Achieved flow in cubic feet per second, A = Area of orifice in Square feet, and V = velocity of liquid in Feet per second. This only actually works out on a true circle orifice on a flat plate with both shoulders being square and sharp. There are correction factors cataloged for K which is a factor to adjust for several standard orifices shapes and configurations which if you can find your orifice style or obtain from manufacture the formula then becomes Q = AVK. In reality for a sprinkler system very little difference will be present and not detectable without special testing.

Thank you for taking the time to explain in depth... it's much appreciated !

 

[JBA]

I just saw this discussion and so I am late to the game; but, apart from all of the piping concerns have you determined the maximum flow rating of your pressure regulator?

 

[AJacq]

I just saw this discussion and so I am late to the game; but, apart from all of the piping concerns have you determined the maximum flow rating of your pressure regulator?

Hi JBA... no I have not... I'll look into it tomorrow and get back to you... I'm guessing it will be indicated on the flow regulator itself or I will get a model number and look it up online.

My guess is that it should be ok, as I have stated the regulator is for a whole building and I have never seen a drop in flow at my faucet... and the sprinklers will be running between 5:30AM and 6:00AM... when most (if not all) are asleep and not using the water supply.

If you believe I am off the mark, please tell me... thanks.

 

[JBA]

I don't see any issues with the pipe sizing discussion. Like you, I would expect that the regulator would have plenty of flow capacity, I am simply trying to cover all bases. What I have not seen in the prior discussions is the length of the 1/2" copper line from the supply takeoff to the water faucet where you measured your flow rate or the amount of back pressure applied at the faucet while you were measuring your flow rate. I have a basic liquid piping program that I can use to determine the pressure drop through that line and any other size line if you provide me with the length of the line and any fittings in the line between the supply line connection and the sprinkler system connection point.. On that point, if your flow tester was connected to a standard globe type hose valve, that valve alone can introduce a significant flow restriction as well. Regardless of the size of pipe you end up with, it should be equipped with a full bore ball valve at your sprinkler system piping connection.

 

[AJacq]

I don't see any issues with the pipe sizing discussion. Like you, I would expect that the regulator would have plenty of flow capacity, I am simply trying to cover all bases. What I have not seen in the prior discussions is the length of the 1/2" copper line from the supply takeoff to the water faucet where you measured your flow rate or the amount of back pressure applied at the faucet while you were measuring your flow rate. I have a basic liquid piping program that I can use to determine the pressure drop through that line and any other size line if you provide me with the length of the line and any fittings in the line between the supply line connection and the sprinkler system connection point.. On that point, if your flow tester was connected to a standard globe type hose valve, that valve alone can introduce a significant flow restriction as well. Regardless of the size of pipe you end up with, it should be equipped with a full bore ball valve at your sprinkler system piping connection.

Thanks JBA... I appreciate your offer of help... I believe that I will take measurements and photos (because 1000 of my non-technically savvy words don't come close to a picture)

FYI... the "flow-tester" I used was a 5 gallon bucket in which I put exactly 10 x 1 litre of water with a 1000 ml kitchen measuring cup... I marked the water line with a sharpie... I used an iPhone stopwatch to time how long it take my garden faucet fully opened to fill the bucket to the mark.

 

[gmax137]

... length of pipe...

Yes, this is very important because there is a huge difference in the cost of copper pipe at 1/2 inch or 3/4 inch or 1 inch sizes. So, if you have a long run of pipe out to the sprinkler, the cost difference between 1 inch and 3/4 inch may be quite high; and if 3/4 inch is good enough you will save $$$ over building it with 1 inch pipe. Where I live, 1 inch is about $2.25 per foot compared to $1.30 for 3/4 inch. So it may be worth your time to see if 3/4 inch really is OK. Of course, the longer the pipe is, the more help you get from the larger diameter. It would be a shame to save a few hundred dollars only to find the 3/4 inch system doesn't do the job. Also, look into using something like PEX instead of copper pipe.