Irrigation Well Cost of Operation?

[Birken Vogt]

I'd like to know how to figure costs for pumping water for farm use. Right now there is a well that is about 80 feet deep set up for a regular 4" residential pump. The water level is about 20 feet down and if I recall, it would be 20-40 gpm, or possibly more. Head pressure at the well head would be maybe another 20 feet up. 3 phase electrical is available at maybe 20 cents per kwh. Propane and diesel are also options. How can I go about analyzing this data?

Thanks
Birken

[LMJD]

Seems like your best bet would be to consult with one of the well service company reps for a start. Not well drilling, but the pump service outfits. That's not too deep so it might be feasible money wise. Years ago my old boss had a well about 250' deep and he had to buy hay rather than attempt to irrigate his 7 acres.

[griffy71]

Oh theres a lot of unit conversion and bla blah balh. Not sure whaqt you're irrigating but i think you're gonna want a lot more than 20 to 40 gpm. Are you filling ditches or pumping nozzels. If nozzeles then your head will be a lot higher. You mulitply your flow thru your head and run some conversions. This gives you brake horsepower. Then you divide by the efficiency of your pump (60% if you have to guess) and get your horsepower. Then figure your inches per hour per acre and get your ft^3 per hour and convert it to gpm to get your run time. Then multiply your horsepower by .707 kw/hp. Then multiply by your rate. I'll get back to you with the antiipated brake horsepower

[Birken Vogt]

My figures are just for replacement of our current water which is supplied by an irrigation ditch to a pond at the top of the hill, 2-4 miners inches and it is all gravity fed right now.

We feed rain bird sprinklers with it, if I recall the lowest ones are getting 20-something psi and the highest ones barely get enough to work but they are just spaced closely with smaller nozzles to make it work.

Birken

[griffy71]

20 GPM at 80' head would be about 1 HP. 40 GPM Would be about 2 HP. Daily cost running 24 hours would be about $3 a day conservatively. You will have to put a check or foot valve in to hold prime. Let me know how you want to go and I can crunch you more numbers. E-mail me in the PM I just sent you ad I can do it alot faster.

[SmokeyWren]

Originally Posted by Birken Vogt

I'd like to know how to figure costs for pumping water for farm use. Right now there is a well that is about 80 feet deep set up for a regular 4" residential pump. The water level is about 20 feet down and if I recall, it would be 20-40 gpm, or possibly more. ... 3 phase electrical is available at maybe 20 cents per kwh.

Gould makes a variety of submergible pumps for a 4" well. If your well can support a constant 25 GPM without pumping air, then it sounds like you need pump model 25GS. 230 volts, 3-phase, 5 HP. The pump can operate at 8 to 33 GPM if you need to play with pump head pressures or sprinkler head pressure.
http://www.goulds.com/pdf/7310.pdf

5 HP = 3,730 watts or 3.7 KW. (One HP = 746 watts). At $0.20 per KWH, it will cost you about 75 cents per hour to run that pump at full blast, or $18.75 per 24-hour day.

Since your water table is so shallow, you can probably get what you need with 3 or 4 horses instead of 5. Study the charts on the Gould website and decide.

Also check with your electricity provider. You can probably qualify for farm rate which should be less than $0.10 per KWH unless you live in the land of the envirofreaks.

Propane and diesel are also options.

It's been done. 50 or 60 years ago, many oil wells and a lot of farm irrigation pumps were powered by diesel, propane or natural gas internal combustion engines. But it wasn't cost effective in the long run, when you figured in the overhaul or replacement costs of the engines. And back then we paid $0.07 per gallon for propane! Today almost all the pumpjacks and irrigation wells are powered by electric motors. Much bigger motors than the measley little 3-to-5 HP motor you're considering. In the 1960s, Dad was using 25 horse electric motors, because that's all the water we had. But some of the big boys with more water available up on the high plains were running 100-HP electric motors. Talk about a meter going 'round and 'round in a hurry, a hundred-horse electric motor at full song will make it happen.

[Rick M]

I have a 1.5 hp 1ph hanging at 165' with a static water level of 135' that delivers just under 20 gpm free flow. Can't imagine it taking 5 hp to get 20 gpm from a well that has a static level of 20' and is producing 40 psi.

[SmokeyWren]

Originally Posted by Rick M

I have a 1.5 hp 1ph hanging at 165' with a static water level of 135' that delivers just under 20 gpm free flow.

Free flow is a lot different that pumping up the pressure to squirt the water 50' or so out of a sprinkler head in an irrigation system. But this is not my area of expertise, so use those tables in the Gould website to calculate the exact pump, phase, and HP you need to run that exact irrigation system at the pressure and volume of water you need to get the job done.

After you know the HP of the pump you need, then it's easy to convert HP to watts, then to KW, then to KWH, then to KW per day, and finally the cost per hour/day/week to run the pump full time.

[drcampbell]

To calculate pumping power, multiply the volume flowrate by the pressure. That's all there is to it.
(of course, that only works if it's done in a consistent set of units)

Once again, we have managed to hamstring ourselves be clinging to the British system of measure. Once we finally adopt the metric system, this will be almost trivial because there's only one unit for each thing measured. (pressure, volume, flow rate, et al)


25 US gallons per minute = 3.3 cubic feet per minute

I'm going to take a wild guess and say the pump pressure is 30 lb/sq.inch. (10 lb/sq.inch below ground and 20 above) That's 4320 pounds per square foot.

Now that they're in a consistent set of units, (pounds, feet, minutes) just multiply to calculate power:

    cubic feet            pounds              foot*pounds
3.3 ----------  x  4320 -----------  =  14256 -----------
      minute            square foot              minute 

algebra refresher: cubic feet divided by square feet is feet
(feet to the third power divided by feet to the second power is feet to the first power)

1 Hp is defined as 33,000 foot*pounds per minute, so 14,256 foot*pounds per minute is the equivalent of 0.43 Hp. (of actual pumping power)



And in case you're wondering what a "miner's inch" is ...

Miner's inch - Wikipedia, the free encyclopedia
1/60 ft³/s (0.472 L/s) New Zealand
1/50 ft³/s (0.566 L/s) southern California, Idaho, Kansas, Nebraska, New Mexico, North Dakota, South Dakota, Utah, Washington
1/40 ft³/s (0.708 L/s) Arizona, northern California, Montana, Nevada, Oregon
1/38 ft³/s (0.745 L/s) Colorado
1/36 ft³/s (0.787 L/s) British Columbia

[KML]

Is this what you are doing ?

ECO SHOT : LPG for Diesels - PUMPS