Will this pump work for my artificial irrigation system?

[Pumpquestions]

Hi all,

I am designing an artificial rain system and was thinking about using http://www.dannermfg.com/Store/Products/Danner/PID-02720.aspx to power it. The goal of the system is to deliver roughly 1 gallon of water per hour to each of the 8 exit points (I will measure this by using a bucket and timer). There will be 1" polytubing (chart for head loss) connected to the pump that runs down to a manifold. The http://store.rainbird.com/drip-low-volume/drippers-emitters-microsprays/uxb360025-xeri-bubbler-full-circle-pattern-360-degree-umbrella-1-4-in-barb.html#product_tabs_additional_tabbedI will be using require 15-30psi. There is a 10-20ft elevation decrease from the water tank to the exit points, so that the tank is higher than the plots being watered. The idea is to have the system automated (hence the electrical timer) to run 4 times a day. I will be manually filling the 55-100 gallon water tank once a day.

My questions are:

1) Will the pump I am using actually work for this design? The specs state that "950 GPH Maximum Flow with a 12'10" Shut-Off."

2)If that pump will not work, can you please point me in the right direction for one that would? (I prefer the easiest/lowest cost as this will be semi-off grid)

3) The tube I am using can take a max capacity of 240gph, how can I use a pump that will deliver the needed pressure and flow and not overpower the tubing? I know there are pressure/flow regulators, but I am not sure how that effects the pressure/flow to my 8 exit points.

I really appreciate any help you can offer. I've been working on this for quite a while and am hoping to figure out these last few pieces of the puzzle.

Thanks!
John

 

[oz93666]

Hi ... I've just been installing something similar this week... The first thing I would say is the sprinkler heads shouldn't need 15psi, my feeling is the elevation of the tank is more than enough to drive them with no pump.

Most importantly, because the tank is higher , nearly all pumps will still deliver water even when the power to the pump is switched off!

I would advise you just set it up to start with , with no pump , just a manual tap, to get a feeling of what is required.

4 times a day watering seems excessive. If you can drive the sprinklers with no pump then it would be better to design a system that automatically triggers when the tank is full , a bit like a flushing toilet, no pumps or electric valves required, (electric valves cost as much as pumps)

If you do need an electric valve or pump , ebay is the best place to look.

 

[Pumpquestions]

Hi ... I've just been installing something similar this week... The first thing I would say is the sprinkler heads shouldn't need 15psi, my feeling is the elevation of the tank is more than enough to drive them with no pump.

Most importantly, because the tank is higher , nearly all pumps will still deliver water even when the power to the pump is switched off!

I would advise you just set it up to start with , with no pump , just a manual tap, to get a feeling of what is required.

4 times a day watering seems excessive. If you can drive the sprinklers with no pump then it would be better to design a system that automatically triggers when the tank is full , a bit like a flushing toilet, no pumps or electric valves required, (electric valves cost as much as pumps)

If you do need an electric valve or pump , ebay is the best place to look.

Thanks for the reply. I actually am doing this as an experiment so it has to be 4 times a day and I am hoping to have similar output at each plot. I was wondering how elevation would play out. I know 10-20ft difference will give about 4-8psi (formula used). Can you explain why you don't think the sprinkler heads will need 15psi? That would be great!

 

[jack action]

Let's look at your needs:

• «1 gallon of water per hour to each of the 8 exit points» -> That's 0.133 gpm (= 8 / 60);
• 15-30 psi at the outlet;
• According to your chart for a 1" tube, with the small flow you have, you can assume the head loss is essentially 0 (unless it is ridiculously long). Otherwise you would need to add this pressure to your outlet pressure needed;
• With a 10-20 ft elevation (although your drawing shows 65 ft), you get 4-8 psi «free» (Note: If your tank was at least at an elevation of 35 ft, you would get your 15 psi and you wouldn't require a pump). For the purpose of calculations, let's assume 6 psi;
• So the pressure needed at the pump is 9-24 psi (= 15 - 6 & = 30 - 6)
• The pump power (in hp) needed is GPM X PSI * 0.0007. This assumes a 83 % pump efficiency. To convert hp to watts, you multiply by 746. So the minimum power of the pump would be 0.6-1.7 W (= 0.133*9*0.0007*746 - 0.133*24*.0007*746). That is really small (think of the power to illuminate a 1 W light bulb).

Your pump:

• It has 93 W (way more than you need);
• It can produce a maximum head height of 12' 10" (which translate into a 5.5 psi pressure) and a maximum flow of 950 gph. Since you need 9-24 psi, we already know the pump won't work. But you should know that it can produce either 950 gph @ 0 psi or 0 gph @ 5.5 psi ... or anything in between. All pumps are like that. See the http://www.dannermfg.com/Store/images/instructions/ZG100.pdf to get the actual flow vs head height specifications (You have the model 9.5).

http://www.mcmaster.com/mv1441053944/#4182k5/=ys9uxv seems to fit more your needs. This is just based on a quick search on the web, not a recommendation (I'm not a pump expert).

A pump's flow rate in gallons per minute (gpm) at a given pressure in feet of head is known as max. flow @ feet of head. As pressure increases, flow decreases. When the liquid reaches a height where the pressure is greater than the pump's power, flow stops entirely. This point is called max. ft. of head. Pumps generally operate most efficiently at around 80% of the max.ft. of head.

http://www.mcmaster.com/#about-submersible-pumps/=ysagyt

 

[oz93666]

... Can you explain why you don't think the sprinkler heads will need 15psi? That would be great!

The sprinklers I installed were on uneven ground, and the pump (2.5kw) was not sufficient for the 100 sprinklers, some were working and others were not delivering even a dribble, the difference in height was about 3 ft so my type of sprinkler is ok at 3ft pressure, doesn't throw very far... at 6 or 8 ft(water hight) they would throw far, but every sprinkler type is different, I think you need to experiment ... The problem with watering often is the the thing your growing will keep it's roots very close to the surface, and so is vulnerable. If watering is seldom but intense the plant/grass will develop deep roots and so is more resilient if it should go a day without water.
But the main point is , virtually all pumps will deliver water if the tank is higher than out let, even when not powered.
For your rate of flow 1/2 inch will be sufficient. Timers and pumps are always a haste, somehow water or ants get in and mess things up, an automatic siphon would be a more elegant solution, as the tank reaches full, a siphon is automatically established , which empties the tank through the sprinklers, and then resets automatically.

 

[Pumpquestions]

Let's look at your needs:

• «1 gallon of water per hour to each of the 8 exit points» -> That's 0.133 gpm (= 8 / 60);
• 15-30 psi at the outlet;
• According to your chart for a 1" tube, with the small flow you have, you can assume the head loss is essentially 0 (unless it is ridiculously long). Otherwise you would need to add this pressure to your outlet pressure needed;
• With a 10-20 ft elevation (although your drawing shows 65 ft), you get 4-8 psi «free» (Note: If your tank was at least at an elevation of 35 ft, you would get your 15 psi and you wouldn't require a pump). For the purpose of calculations, let's assume 6 psi;
• So the pressure needed at the pump is 9-24 psi (= 15 - 6 & = 30 - 6)
• The pump power (in hp) needed is GPM X PSI * 0.0007. This assumes a 83 % pump efficiency. To convert hp to watts, you multiply by 746. So the minimum power of the pump would be 0.6-1.7 W (= 0.133*9*0.0007*746 - 0.133*24*.0007*746). That is really small (think of the power to illuminate a 1 W light bulb).

Your pump:

• It has 93 W (way more than you need);
• It can produce a maximum head height of 12' 10" (which translate into a 5.5 psi pressure) and a maximum flow of 950 gph. Since you need 9-24 psi, we already know the pump won't work. But you should know that it can produce either 950 gph @ 0 psi or 0 gph @ 5.5 psi ... or anything in between. All pumps are like that. See the http://www.dannermfg.com/Store/images/instructions/ZG100.pdf to get the actual flow vs head height specifications (You have the model 9.5).

http://www.mcmaster.com/mv1441053944/#4182k5/=ys9uxv seems to fit more your needs. This is just based on a quick search on the web, not a recommendation (I'm not a pump expert).

Thanks, Jack Action. I have a couple of followups.

1) The elevation difference is 10-20ft, but the pipe will be 65 ft long and then connect to a manifold that is around another 50ft. Sorry for the confusion.

2) Is a 0 gph @ 5.5 psi a scenario like there is too much height to get the water through, but the pump is still creating pressure on the piping?

3) Can you please explain why the pump you suggestion seems more like what I need. Do you think it would be a problem providing water through a 60ft mainline and then through a 50ft secondary(manifold) line?

 

[jack action]

2) Is a 0 gph @ 5.5 psi a scenario like there is too much height to get the water through, but the pump is still creating pressure on the piping?

When the pump reaches its maximum pressure, it is because it becomes inefficient to the point where instead of transferring energy to displace the water, it gets out through heat loss. For example, imagine a pump that has a piston and a rubber seal. When the outlet pressure will be so high that the seal will deform (It takes energy to deform the seal) and create a leak pass the piston, the flow will stop because the fluid will go pass the piston instead. As it leaks, the pressure tends to decrease, the seal takes its original form and the piston pushes the fluid again. The pressure rises again, and the whole scenario repeats itself, leaving you with pressure but no flow. All pump have this limit which depends solely on their design.

3) Can you please explain why the pump you suggestion seems more like what I need.

It is all about the minimal pressure required by your problem. You project need at least 9 psi of pressure at the pump outlet, or a head of 21 ft ( = 9 / 0.433). The pump you suggested cannot go over 12' 10" or 5.5 psi (= 12.833 * 0.433). And even at that pressure, you have a flow of 0 gpm.

The pump I suggested, can create a head of 24.5 ft (or 10.6 psi) which is a little more than your minimum requirement. It has to be higher, because at that pressure, the pump produces no flow (by definition). We also know that this pump produces 10.2 gpm @ 10 ft head. We also know that you need 0.133 gpm. So by doing a simple interpolation (no actually true but close enough) we can estimate that at 0.133 gpm, the will have a head of 24.3 ft (= [10 - 24.5] / [10.2 - 0] * [0.133 - 0] + 24.5).

Like the text I quoted mentions, a pump is usually most efficient when the pressure is at about 80% of its maximum pressure (so 19.6 ft in this case). But then you would have to maintain a flow of 3.4 gpm (= [10 - 24.5] / [10.2 - 0] * [19.6 - 10] + 10.2). You could do that with a pressure regulator for example (although 19.6 ft is a little smaller than your 21 ft requirement). The fact is that this pump is still a little big for your intended use (flow wise), but at least it would be able to do the job (flow and pressure wise).

Do you think it would be a problem providing water through a 60ft mainline and then through a 50ft secondary(manifold) line?

If you refer to your own link about head loss, for a 1" tube with a flow of 1 gpm, you lose 0.03 psi of pressure for every 100 ft of tubing.

With a 110 ft of tubing @ a flow of 0.133 gpm you will get something like 0.004 psi of pressure loss (= 0.133 / 1 * 0.03 * 110 / 100). So instead of 15-30 psi, you need 15.004-30.004 psi; Not a big difference in calculations.

And from your drawing the 60 ft line (which I assumed to be 1" as well) connects in the middle of your manifold: This means you actually have 60 ft with 0.133 gpm (loss: 0.0024 psi [= 0.133 / 1 * 0.03 * 60 / 100]) and then 25 ft with 0.0665 gpm (loss: 0.0005 psi [= 0.0665 / 1 * 0.03 * 25 / 100]), for a total loss of 0.0029 psi. And the flow even drops further in your manifold as you pass each port.

I'm just calculating this for demonstration because your tubing is so large for the flow you have, you can consider that you have no restriction at all.

 

[Pumpquestions]

Thanks, Jack Action. I am hoping you don't mind me asking you some other questions. I swear I have been reading up on all of this, but some things still confuse me.

It is all about the minimal pressure required by your problem. You project need at least 9 psi of pressure at the pump outlet, or a head of 21 ft ( = 9 / 0.433). The pump you suggested cannot go over 12' 10" or 5.5 psi (= 12.833 * 0.433). And even at that pressure, you have a flow of 0 gpm.

The pump I suggested, can create a head of 24.5 ft (or 10.6 psi) which is a little more than your minimum requirement. It has to be higher, because at that pressure, the pump produces no flow (by definition). We also know that this pump produces 10.2 gpm @ 10 ft head. We also know that you need 0.133 gpm. So by doing a simple interpolation (no actually true but close enough) we can estimate that at 0.133 gpm, the will have a head of 24.3 ft (= [10 - 24.5] / [10.2 - 0] * [0.133 - 0] + 24.5).

Like the text I quoted mentions, a pump is usually most efficient when the pressure is at about 80% of its maximum pressure (so 19.6 ft in this case). But then you would have to maintain a flow of 3.4 gpm (= [10 - 24.5] / [10.2 - 0] * [19.6 - 10] + 10.2). You could do that with a pressure regulator for example (although 19.6 ft is a little smaller than your 21 ft requirement). The fact is that this pump is still a little big for your intended use (flow wise), but at least it would be able to do the job (flow and pressure wise).

1) What determines if the head changes? For instance, for the pump you provided, @ 1ft of head you get 13.5 gpm and @10 ft of head you get 10.2 gpm. What would I do to change it.

2) Additionally, in my case, I would like to have much less water exiting per minute. Could I manipulate this flow by restricting flow at the main-manifold valve? How would this affect the pressure, pump, and main line? Basically, it seems impossible to maintain a high enough pressure for the sprinklers to work, while keeping a low enough flow rate. I will probably have a 55-100 gallon source water tank and I don't want it to drain in 5 or 10 minutes.

3) I was told by the manufacturer that the tubing has a maximum of 250gph. Is there a way to accomplish appropriate pressure and flow rates? Since the sprinkler output can be modified, will that regulate the flow?

4) Am I correct that something like this pump provide sufficient pressure, while having a lower gpm?

If you refer to your own link about head loss, for a 1" tube with a flow of 1 gpm, you lose 0.03 psi of pressure for every 100 ft of tubing.

With a 110 ft of tubing @ a flow of 0.133 gpm you will get something like 0.004 psi of pressure loss (= 0.133 / 1 * 0.03 * 110 / 100). So instead of 15-30 psi, you need 15.004-30.004 psi; Not a big difference in calculations.

And from your drawing the 60 ft line (which I assumed to be 1" as well) connects in the middle of your manifold: This means you actually have 60 ft with 0.133 gpm (loss: 0.0024 psi [= 0.133 / 1 * 0.03 * 60 / 100]) and then 25 ft with 0.0665 gpm (loss: 0.0005 psi [= 0.0665 / 1 * 0.03 * 25 / 100]), for a total loss of 0.0029 psi. And the flow even drops further in your manifold as you pass each port.

I'm just calculating this for demonstration because your tubing is so large for the flow you have, you can consider that you have no restriction at all.

Sorry if I missed something, but I got a little lost here. For the pump you gave you linked, the flow would be between 10-13 gpm. In that case, wouldn't my pressure loss be around 2.3-3.3 psi? Thanks, again.

 

[oz93666]

What measures are you taking to stop a continual flow of water to the sprinklers, even when the pump is not powered. The storage tank is 15 ft. higher than outlet!

 

[jack action]

For your questions 1 through 3, the answer is the same: the restriction of your system will dictate the pressure you will have. The thing that makes everything complicated to analyze, is that the restriction of your system depends on the flow inside the pipes (see your link about pressure loss inside pipe).

You can set the restriction of your system by designing it properly (pipe length and diameter) or you can adjust it with a pressure regulator or a valve.

The pressure regulator sets the restriction to have the desired output pressure and whatever excess flow comes from the pump, it is sent back to the tank.

The valve sets the restriction size. The pressure you will get will depend on the pressure/flow characteristic of your pump. If the valve is wide open, the flow will be high, but the pressure will be close to zero. If the valve is close, there is no flow, but the pressure is at its maximum behind in the pipe behind the valve.

Two things to remember:

1. A pressure regulator or a valve can only add restriction to your system, i.e. if you have long pipes and/or small diameter after the valve/regulator that already creates, say, 3 psi of restriction, the valve/regulator can only increase that value.
2. The maximum pressure or flow of the system is defined by the pump. If the pump cannot exceed, say, 10 psi, a valve/regulator will not create a larger pressure than that. The flow will depend on the design of the pump and will vary according to its pressure (high pressure -> low flow; low pressure -> high flow).

Since your sprinklers require a certain pressure to function properly, you need a pump that can produce that pressure. Remember that the sprinklers are a restriction (like a valve) and that at their working pressure, a certain flow is associated with it. In your case it is 0-35 gph @ 15-30 psi which, IMO (but I maybe wrong), means you need at least 15 psi to get 0 gph and you'll get 35 gph at 30 psi. Higher than 30 psi, the sprinkler will probably break. With the low flow you need, you probably want to hang on the 15 psi side.

Since you want a certain flow, you need a pump that can deliver this flow AT the desired pressure. Ideally it would deliver exactly the flow you want at the pressure you want.

If you would use a pump that has a larger flow than you need at the pressure required, the higher flow would increase the restriction in the system, thus the pressure as well, and you would get whatever flow the pump would give at that new pressure. In that case, you can use a pressure regulator that will only allow a certain pressure in your system and any excess flow that would tend to increase the pressure, will be diverted back to the tank.

4) Am I correct that something like this pump provide sufficient pressure, while having a lower gpm?

This is a transfer pump that is used to drain something. In this case, the pump doesn't feed any system, the water is just thrown out in the open air, so essentially no restriction. The only spec needed is then the maximum flow, which is given. In the case you would drain water from a basement that was flooded, you would need to pump the water out at ground level. This is why they also give the flow at 10 ft of head. Unfortunately, they don't give the maximum pressure the pump can give, i.e. when flow is zero (and there is one) because buyers of this kind of pump usually don't care. Unless the spec is provided, your guess would be as good as mine.

Sorry if I missed something, but I got a little lost here. For the pump you gave you linked, the flow would be between 10-13 gpm. In that case, wouldn't my pressure loss be around 2.3-3.3 psi? Thanks, again.

The specs for the pump in my link defines 13.5 gpm @ 1 ft and 10.2 gpm @ 10 ft, but also 0 gpm @ 24.5 ft; That is what «maximum feet of head» means. It's you that decide how much flow that you want. You want 0.133 gpm, you will get 0.133 gpm everywhere in your system. What you are now trying to do is to design your system to achieve that goal.

Look at the following graph to visualize how pressure relates to flow with a pump (any pump):

​

 

[Pumpquestions]

What measures are you taking to stop a continual flow of water to the sprinklers, even when the pump is not powered. The storage tank is 15 ft. higher than outlet!

If I used polyvinyl tubing, I was going to insert a small 1/4" barbed elbow joint that is pointed back into the barrel. However, I am now considering having the portion from the barrel be PVC and create a split-off. One avenue would recirculate excess flow (controlled with a valve) to go back into the barrel, and the other would connect to the poly tubing. This would alleviate the 250 gph maximum of the poly tubing. Do you have a recommendation for an anti-siphon mechanism polyvinyl hose or PVC? I have to keep polytubing as the surface I will be placing it is highly irregular.