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Calculating water pump head and distance
Old 07-21-2020, 02:42 PM   #1
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Calculating water pump head and distance

Every summer when the grass goes dormant/dead from lack of water I look wistfully at the creek behind the house and ponder the feasibility of pumping water from the creek up a rather steep hill, and a horizontal distance of ~80 yards or so, to water the grass and shrubbery. Then I go online and look at gasoline powered water pumps and sort of realize that what I'd like to do is probably not feasible at a price I'd be willing to pay. I'm guessing (and it's only a wild guess) that the vertical distance up the water would have to travel would be about 100 feet or so.

But how does one go about measuring this exactly? I've searched a bit online but I must not be using the right terms because all the results are completely irrelevant to what I'm looking for.

Also, I have a question about water pump specifications. Suppose a pump's specs say it has a suction head of 25 feet and a vertical head of 100 feet. Is the suction head included in that 100 feet, or will the pump actually pump water a total of 125 feet? (I kinda doubt that, given the way marketers play with numbers, but I had to ask.)

Thanks for all the insight you engineering types can offer. Please write slowly and use one-syllable words...
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Old 07-21-2020, 03:10 PM   #2
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I think you want to read this. It has some illustrations to help. It will depend on whether that 100 ft is total head.

https://www.pumpfundamentals.com/what%20is%20head.htm

Quote:

OK smart guy, what if I don't have a tank and I'm pulling water from a lake and the lake is lower than my pump. Ha!


The pump will still produce the same total head but the discharge head will go down. This means you may not have enough pressure to run your devices and you may need to consider getting a pump with a higher total head
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Old 07-21-2020, 03:17 PM   #3
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Just for the record, I'm not a pump guy.

I believe the head pressure is on the suction side of the pump, so unless your pump is sitting 25 ft below the creek intake, I doubt you get to include the 25 ft.

If you had a hose/pipe that was liquid full of water from the creek level up the garden/lawn level and measured the pressure at the creek end, you should be able to convert pressure into ft of elevation change. For example, 100 ft of water head should yield about 43 psi on a pressure gauge. Diameter of hose won't matter, but it needs to be liquid full.
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Old 07-21-2020, 03:27 PM   #4
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If you’re interested in finding the elevation difference between the stream and your lawn, you would need to do some work with a surveying level. This describes the process:


http://www.fao.org/tempref/FI/CDrom/...e/x6707e08.htm

The difference in elevation would be the head above the pump. I believe in your case that the pump heads are listed separately for below and above pump heads.

But the pump head is only part of the puzzle. You need to know what flow the pump provides at that head.

I have a 2” hp, 2” discharge, 2” intake 230 volt pump. 20’ head, probably 150-200’ from pump to farthest sprinkler head. I probably have the intake set about 8’ below pump elevation.

I didn’t do any fancy engineering calculations. I just bought a robust pump, installed the lines - 8 zones with about 6 sprinkler heads per zone. It turned out that I was getting excessive flow out of the pump, so I jumpered zones on my sprinkler controller so that I now have only 3 zones. Some zones now have about 18 heads and I can still run a separate garden hose from the pump at the same time.

So you need to find what flow you’re going to need to water plants, etc. not much if you’re just throwing out a moveable lawn sprinkler. Then get a pump that can provide that flow at your head, discharge pipe size, and distance from the pump.

I don’t know how to calculate flow at a given head. But if your pump says that it can pump up 100’, I assume that means that it can at least run a garden hose or single sprinkler at that head.
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Old 07-21-2020, 03:46 PM   #5
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Popeye's link explains suction head and total head pretty well. Keep in mind it sounds like you are pumping a long distance (300ft?), so pressure loss in the hoses also comes in to play. There are graphs out there to estimate this, but I am too lazy look for them.

I assume you are looking at centrifugal pumps, in which case there should be a pump curve that will tell you flow versus head.
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Old 07-21-2020, 03:58 PM   #6
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Thank you all for the links and the insights.

I've got some studying to do.
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Old 07-21-2020, 04:09 PM   #7
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Our back yard is the Mississippi river. Our neighbor did exactly what you're looking at doing, extending his pipe maybe 20-30' out from shore. Water depth there was roughly 5' +/- with the seasonal changes in flow.

The river is a few hundred feet wide at this point with a few islands nearby. We are on a side channel, not the main one.

Punch line is that he tore it out because he couldn't keep the inlet from clogging. I'm guessing the problem was weeds and twigs but I don't know for sure.
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Old 07-21-2020, 07:00 PM   #8
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Walt - You need a thump pump! As a kid we had an old one in a small creek and the rhythmic noise it made was wonderful. No electricity required. Look up hydraulic ram pump.



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Old 07-21-2020, 11:06 PM   #9
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Walt - You need a thump pump! As a kid we had an old one in a small creek and the rhythmic noise it made was wonderful. No electricity required. Look up hydraulic ram pump.
...
That is totally AMAZING, I would never have believed this was possible if you hadn't posted this.
That you can lift water many times the height (drop) of the input water without using any powered (electric/gas) pump.

It seems to defy the conservation of energy.

I'm trying to think how I could hook one of these up to my small creek now
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Old 07-21-2020, 11:50 PM   #10
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As far as measuring both the distance and elevation from the water source to destination, it seems like a gps mapping website or app would work for you. Use the satellite view, zoom in as close as you can then plot a “route” from Point A to Point B. You’ll get both distance and elevation data as part of the result:

https://www.plotaroute.com/routeplanner
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Old 07-22-2020, 03:22 AM   #11
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The suction ability of a pump is called lift, and is dependent on elevation and pump efficiency, at sea level you cannot lift(suck) more than 25' of elevation here on planet Earth. At 2000' elevation, no more than 22', and that will be diminished by the diameter of your suction line. Your suction line must have a foot valve on it to prevent the line from draining and the pump losing its prime.

IIRC, the rule of thumb for pumps with no lift for easy computation is thus. 5HP pump will pump 100 gpm with 100' head, not counting line loss (line friction).
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Old 07-22-2020, 07:10 AM   #12
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A big mistake with beginners is using a suction line that is too small or too long. Atmospheric pressure is what pushes the water into the pump when the pump pulls a vacuum on it's inlet. If you have too much pressure loss on the inlet pipe, the pump will cavitate. That's why when you look at the pump, it will have a larger connection on the inlet than the outlet. A rule of thumb is to use a pipe about one size larger than the inlet connection if you have much pipe on the inlet.
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Old 07-22-2020, 07:13 AM   #13
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A Ram Pump will not suction lift. It has to be a flooded suction.
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Old 07-22-2020, 07:32 AM   #14
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Are water rights an issue when pulling water from a creek? In the West it is a real thing.
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Old 07-22-2020, 02:52 PM   #15
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Are water rights an issue when pulling water from a creek? In the West it is a real thing.
To my knowledge, no. Here in WV the problem is more often too much water. Lots of places would just LOVE to give you guys their excess! That creek never runs dry and floods several times a year. We're high enough that flooding is not a concern for us.
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Old 07-22-2020, 02:57 PM   #16
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A Ram Pump will not suction lift. It has to be a flooded suction.
Yes. While that's an intriguing idea, there is no "downhill" before I get to the "uphill" so I don't think it'll work for me. Neat pump though, I'd never heard of that before!
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Old 07-22-2020, 03:27 PM   #17
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Here is a trash pump I bought last spring for under 180 bucks. It is a 2 inch 7 hp pump. I pull from a creek 12 feet down and push 75 feet with elevation going up 8 feet. I have absolutely no problem pulling water from 12 down and pushing up hill 8 feet or 75 feet into a 325 gallon tote. I have a small tree nursery at the ranch and use this pump to water the trees. I don't have a picture of my suction in the crick but this unit does an amazing job. I do know yo wouldn't want to be standing in front of that discharge when the water is coming out. It would blow you right down. Lol

I believe my spec on suction was something like 25 feet.
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Old 07-22-2020, 05:37 PM   #18
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Quote:
Originally Posted by Winemaker View Post
The suction ability of a pump is called lift, and is dependent on elevation and pump efficiency, at sea level you cannot lift(suck) more than 25' of elevation here on planet Earth. At 2000' elevation, no more than 22', and that will be diminished by the diameter of your suction line. Your suction line must have a foot valve on it to prevent the line from draining and the pump losing its prime...
Quote:
Originally Posted by Masquernom View Post
A big mistake with beginners is using a suction line that is too small or too long. Atmospheric pressure is what pushes the water into the pump when the pump pulls a vacuum on it's inlet. If you have too much pressure loss on the inlet pipe, the pump will cavitate...

Yes, it is the atmospheric pressure that forces the water to follow the pump suction.

Quote:
Originally Posted by Winemaker View Post
...
IIRC, the rule of thumb for pumps with no lift for easy computation is thus. 5HP pump will pump 100 gpm with 100' head, not counting line loss (line friction).

As to how much power is required, I have no experience here, but get curious to do some simple calculations to make a guess.

The flow of 100 gpm is 1.67 gal per second. That volume of water weighs 13.8 lbs, and lifting it up 100 ft requires the energy of 1380 ft-lbs/sec, which is 1871 watts.

But how 'bout the kinetic energy of the output stream of the pump? Let's say we use a 2" diameter hose, which has a cross section of 0.022 sq.ft. The above flow of 1.67 gal/s is 0.223 c.f./s, and the velocity of the discharge is 0.223/0.022 = 10 ft/s. The kinetic power is 43 ft-lb/sec, or 58 watts. Hmmm, it's not that much compared to the potential energy.

OK, so let's say the power required is roughly 2 kW ideally, and that's about 3 hp. To allow for friction loss and the viscous flow, a 5-hp motor sounds reasonable.
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Old 07-22-2020, 05:50 PM   #19
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Any centrifugal pump has a pump curve. Head vs flow. Why are we guessing on this?
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Old 07-23-2020, 07:34 AM   #20
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100 foot of discharge head is about 44 psi of discharge pressure. I did a quick look online. Cheap gas powered pumps won't produce that much discharge pressure. You can search for "gas powered high pressure pumps" and they have them for sale around $600 - $800.
What flow rate of water do you want? Tell me how many feet above the water the pump will sit, how many gallon per minute of water you want to pump, and how far, and how high above the pump to the discharge, and I can select a pump for you.
Looks like most packaged systems online don't show a pump curve, but we can pick a pump without one.
Looks like most online packages from places like Home Depot are set up to move water from A to B, if they are about the same level. Probably things like pumping out a swimming pool. Centrifugal pumps are good at moving high volumes, not as good at lower flow and high pressure.
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