water pumping windmill question 4

[inovermyhead]

Does anyone know the formula for calculating the distance a windmill will pump horizontally?
An 8ft windmill will lift water 180ft with a 1 7/8 cylinder pump. If my static water level is 60ft and the cylinder is @75FT I'm pumping up to the surface 60 ft. I would like to know how far I can pump after that point horizontally. Surely water will move easier horizontally than vertically.
other factors that may influence distance......
1 1/2" pvc pipe at the surface only
drop pipe is 1 1/2" galvanized
I'm at 220 ft above sea level
flow of water is not constant in the pipe
thank you

 

[BigInch]

You say the total pump head is 180 ft. That should be given with a corresponding flowrate. If its given at flow = 0, then that's called the pump's shutoff head and the pump's output head will decrease from that value as flow increases. Anyway, whatever value you have, for now we call it Hp.

I'll assume your pump is a few feet above ground elevation and we'll call the pump centerline elevation Hcl.

The underground lift (water level to the ground level)
60 ft = Hw

Available head, Ha, is what's left after you get the water to the centerline elevation of the pump, so that's

Ha = Hp - Hw - Hcl - Hl

This head is "available" to force water through the discharge piping. What's Hl? There are some friction head losses incurred by running the flowrate (remember that corresponding flowrate I mentioned above?) through the vertical well pipe, so we have to subtract that head loss too. There are pipe head loss tables or online calculators you can use to calculate that. So, with the remaining head Ha of 114 ft, and that corresponding flowrate, you can determine what irrigation supply pipe diameter you need.
Say it's 6" diameter.
Now we check it.

Let's say Ha = 180-60-3.6-2.4 = 114 ft. Looking at the pipe pressure-head loss tables, we find for a 6" you lose 33 ft along the total length, so that's 114 - 34 = 80 ft. head remaining at the end of the supply pipe. If that is greater than what you need to pressure the end-of-the-line sprinkler, you're OK. If not, you need to increase the pipe diameter. and do this check calculation again.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[HEC]

Need a bit more information about the pump and windmill. From your description I am assuming that it is a well mounted single piston style of pump with 1 7/8" piston diameter. What is the stroke of the piston which will provide a volumetric flowrate vs windmill speed.
The windmill characteristic will determin power output and hence how far you can lift and pump the water.

Mark Hutton

 

[inovermyhead]

HEC,
The cylinder has an eight inch stroke and will pump .0960 gallons per stroke. With a 15-20 mph wind the windmill cylinder will pump around 180 gallons an hour and the total lift this cylinder is capable of lifting is around 180 ft. Vertically I will be pumping 60 ft, once at the surface I want to run 1 1/2 inch pvc pipe over to a ground tank. From windmill to where I want to dig the tank there is maybe a 2-3 ft elevation increase. I want to dig my tank a far away as possible so I need to find out how far this thing will pump horizonally. thanks

 

[chicopee]

I've never worked with windmills and this forum is great at introducing our forum audience to new material. I guess the key word is the windmill characteristic curve as pointed out by HEC. This is the technical information the enquirer needs to have.
Also because of the pumping requirements, would the pump be sucking water vapor with a 60'lift afterall atmospheric pressure(appx equivalent to 33'water) is the driver in this lift.

 

[Artisi]

Can we also assume that the mill is capable of lifting water 180 feet from the pump to ground level with little or no residul pressure at the discharge point.

If this is the case it is reasonable to assume that there is still approx 120 ft of head (pressure) left to deliver water thru your pipe line. friction loss thru 1 1/2 PVC pipe is something like 0.2 ft / 100 ft which means you could pump thru something like 550ft of horizontal pipe with an increase of elevation of 2-3 ft.

 

[HEC]

Artsi,
You are correct in estimating the amount of pipe that the windmill will transfer based on the operating data supplied by the OP. Key here is the wind speed this is based on. From my estimates the windmill is generating around 145Watts to move the water (Vertical or horizontal will require the same amount of energy). This is at 13 - 17 knots of wind.
A more relavent question here is what is the expected average wind for the windmill? This will determine the averge performance for the windmill and therefore the pump. The information can be obtained from the Bureau of Meterorology for your area. For example Roxby Downs is not a high wind area and experiences winds of less than 7m/s (13knots) for 97% of time and is an average of 4.7m/s (9 knots). The design data given above by the OP would not be correct for the example location. Because the average wind speed is significantly less than the design wind speed the output would be lower. Assuming a directly proportional relationship for torque and wind turbine speed will give effective turbine power output of 50Watt and from this a flow of 400l/hr (105USG/hr) at 32m (104ft) head (lift). A very diferent system!
Sory about the mixed units. It comes from a mixed past!

Mark Hutton

 

[Artisi]

Mark

Agreed, there are many variables to consider, I have just taken the case as presented 180ft / 180 GPH and assumed a steady flow which in real life it won't be - but at least this gives a starting point for consideration.

 

[RossABQ]

Isn't this essentially a positive displacement pump? The only windmill pump I've ever seen had the piston downhole, and the column had a series of flappers (check valves) above the pump. The flow is not steady state at all.

 

[Apakrat]

Believe the flappers (check valves) mentioned by RossABQ are designed to be forced into a closed position by the static head pressure.

Where will this pressure come from when the pump is in a horizontal position?​

At 74th year working on IR-One2 PhD from UHK - - -

 

[Artisi]

RossABQ - as stated I have assumed steady flow and have acknowledging that it won't be - but the math is far too complicated (for me) to be bothered with tryig to calculate a pulsing flow from a windmill pump subject to an infinite number of variables wind speeds.

Apakrat - the pump is vertically mounted situated 75ft below GL so plenty of static head to close NRV.

 

[inovermyhead]

Ok, I better explain a few things........The windmill pump is located close to the bottom of the well and has three check valves.The windmill pump is the device that actually pumps the water. Water is pumped to the surface when the pump rod raises the piston. The piston check valve closes and holds the water above the piston. As the piston rises, water is moved up the pipe towards the surface. Water is also drawn into the lower section of the pump cylinder through a screen and the two lower check valves. When the pump rod reverses and begins to descend, the lower check valve closes and the piston check valve opens allowing the water in the cylinder to pass through the piston check valve and become trapped above the piston when the check valve closes. This cycle is constantly repeated as the windmill wheel turns, operating the reciprocating mechanism in the gearbox, which operates the pump rod and pump. Now all that being said......I'm not concerned with pressure only distance. A windmill can only be operated as an open ended system, the water must go somewhere or the windmill won't operate. If my distance is too far to the tank my windmill won't turn. My only other option would be to drop down another size on the pump which would give me more lift but reduce my volume of water pumped. Thanks again for all your help.

 

[BigInch]

You nailed it. Seems like as long as the pipe is less than 500 ft or so, Artisi says you'll get some flow.... well, as long as the wind cooperates, and the pipe doesn't freeze up, etc., etc., etc.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[cvg]

not sure I agree with the 500 feet calculation. according to the information given, flow rate is 180 gph or 3 gallons per minute through 1 1/2" pipe. head loss in this size pipe at this flow rate is approximately 1.5 feet per 1,000 feet of pipeline. If you have 120 feet of head being delivered by the pump, than you can certainly pump further than 500 feet.

 

[Apakrat]

Thank You Artisi
Sometimes my stupidity over exposes my true school’n.

Would still like to offer a thought about the windmill delivery flow from the perspective of a proven idiot.

To verify what BigInch & CVG are saying, Ya might wanna check out---------

http://www.engineeringtoolbox.com/pvc-schedule-40-pipe-friction-loss-diagram-d_1147.html

The cylinder has an eight inch stroke and will pump .0960 gallons per stroke. With a 15-20 mph wind the windmill cylinder will pump around 180 gallons an hour

My personal thought is based on----
Five ft. of 1-1/2" PVC has a volume of 0.459 gallons. To reach a flow rate of 5 fps, causing a friction loss of 3 psi per 100 ft of pipe, the windmill must deliver 0.459 gal. per sec.

At 0.459 g/s using 0.096 gal. per stroke, needs 4.78 strokes per second to reach a point of 5 fps flow rate.

The 15-20 mph wind causes a stroke averaging every 0.52 seconds, (180/0.096), or for calc purposes, 17.5 mph wind causes a stroke every 0.52 seconds

Usually, a windmill will be shut down for structural safety at wind speeds of 40 mph or less.

Using 40 mph/17.5 mph = 2.786 strokes per 0.52 seconds equaling 4.395 strokes per second.

The 4.395 strokes X 0.096 gal. per stroke = 0.422 gal. at a flow rate slightly less than 5 fps.

At 74th year working on IR-One2 PhD from UHK - - -

 

[inovermyhead]

Apakrat,
Did you mean that the windmill would be shut down for structual safety at wind speeds above 40 mph or below 40mph? Actually, the only time I shut it down is if a hurricane is approaching. This windmill is almost 90 years old and it runs in above 40 mph winds without failure. Windmills are designed to furl out of the wind so as not to spin at such a high rate of speed. When the wind picks up they will furl out and slow down but continue to pump. Even during a hurricane I will not apply the brakes but put the tail in about a 45 degree angle just to slow it down. So what does everyone think?....... dig the tank at 500ft?

 

[Artisi]

One option worth looking into is to discharge the water at a higher level than ground level and gravity flow the pumped water to your storage tank.

 

[LSThill]

Question why windmill water pumping

Better to go with Solar Water Pumps; Solar Water Pumps livestock producers and many other applications

Replace the (2) windmill water pumping with Solar Water Pumps: no problem, no maintenance problem.

 

[BigInch]

Why wind turbines? Cheaper initial cost.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[BigInch]

Did I mention sometimes they run in the dark? Handy if you're above the Arctic Circle.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/