# solar powered pump



## Chrisp (Jan 3, 2008)

I’m considering creating a small pondless waterfall using a solar powered pump. The drop will be around 3-4 feet. Will a 5W solar panel, 95GPH pump be powerful enough to handle that elevation? See http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=330342088750&_trkparms=tab%3DWatching for what I had in mind.
Anyone else do a similar project and hace any recommendations?


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## toddalin (Jan 4, 2008)

Probably too much "head" for that pump. You would probably get no more than an dribble.

See the pump head calculator for a more accurate portrayal.

http://www.alaskapump.com/calculator.htm


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## Semper Vaporo (Jan 2, 2008)

The calculator is interesting, but what does it tell me? 
I entered some vaguely plausible numbers (1GPM, .5-in ID, 6-ft pipe, 3-ft elevation) and got an answer of approx "3.26-ftTotal Dynamic Head"... what does that mean? what do I do with it? 

I can only guess that if the answer is positive the pump is moving water and if it goes negative it is not working. 

Or is the number something that I look for on the pump spec sheet... as its possible output or capability?


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## toddalin (Jan 4, 2008)

Posted By Semper Vaporo on 10 Aug 2009 04:15 PM 
The calculator is interesting, but what does it tell me? 
I entered some vaguely plausible numbers (1GPM, .5-in ID, 6-ft pipe, 3-ft elevation) and got an answer of approx "3.26-ftTotal Dynamic Head"... what does that mean? what do I do with it? 

I can only guess that if the answer is positive the pump is moving water and if it goes negative it is not working. 

Or is the number something that I look for on the pump spec sheet... as its possible output or capability? 



The volume is inversely proportional to the head. The pump specs should also include volume at various "heads." 

For example, my "Little Giant" is rated at 475 gallons per hour. But that rating is at 1 foot of head. At 3 feet the flow decreases to 440 gph and at 5 feet it goes to 395 gph. Flow goes to 0 gph at 13 feet of head where it shuts down.


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## SteveC (Jan 2, 2008)

CT

Along with "flow rate," a pump's "head pressure" (or more accurately "pressure-head," "shut-off head," or "shut-off pressure") is a term often used in pump literature to describe how powerful a pump is. "Flow rate" is simply the volume of water a pump can move during a given amount of time. Pump flow rates are usually expressed in gallons per hour or gallons per minute.
"Head pressure" on the other hand is more complicated. "Head" may be simply defined as any resistance to the flow of a pump. When pump manufacturers list the head pressure, they are referring to the vertical discharge pressure head. Described in very simple terms, a pump's vertical discharge "pressure-head" is the vertical lift in height (usually measured in feet of water) at which a pump can no longer exert enough pressure to move water. At this point, the pump may be said to have reached its "shut-off" head pressure. When you look at a flow curve chart for a pump, the "shut-off head" is the point on the graph where the curved line becomes horizontal as the flow rate at that point is zero. The higher a pump's head pressure, the more powerful the pump.
To use flow rate and head pressure to help you select a water pump, you first need to know: 1) how much water you need to move through your filter system (volume of your aquarium and filters combined), 2) how many times per hour you need to turn over the aquarium volume (flow rate - typically three to five times per hour), and 3) how much resistance (head) the pump will encounter as it moves water from point A to point B. The first two are easy. Calculating the resistance a pump will encounter as it does it's job is a science.


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## Semper Vaporo (Jan 2, 2008)

Lemme state this a bit differently... and excuse me if I sound a bit "bitter" in my jumping on this thread this way... I have experience in the following scenario and am still bitter over it! (No, don't ask!)

I want a pond that is,... oh...." 'bout that big", I says as I hold my arms out to indicate what I am contemplating... more like a stream maybe... a "C" shape of about 6 feet in diameter and varies around a foot or two wide at random points and varies from a few inches deep to 18-inches deep (the city ordinance legal limit for a non-fenced yard) where the pump is. I want to pump water from one point in the gap of the "C" to a water fall at other point of the gap that "looks and sounds pretty".

So in the referenced calculator, I can enter the "Differential Elevation" number of, oh say, 2 ft, and I figure I will be using a 3/4 inch garden hose which I am guessing is the inside diameter so I can enter the "Pipe ID" and the open part of the "C" is about 3 ft across so I am guessing the garden hose will be about 8-ft long (2ft of lift plus 1-ft down to the discharge point of the pump, across the 3-ft gap of the "C", plus some slop for fitting it in). And the garden hose is some sort of either plastic or rubber, so I can select that data, too.

But what number goes in the first position of the calculator? I have absolutely no idea how many "Billions of gallons per millisecond" or "teaspoons per fortnight" of water "look pretty" going over a rock that is a foot or so wide and will not have the unpleasant sound of (pardon the unseemly description) a child urinating in a toilet when the water is flowing.

This is where the manufacturers are failing to describe their product in terms that the duffer can understand. (I think they are losing sales because of that.)

I have a very limited budget and thus must purchase the lowest cost pump I can that will still accomplish the desired effect. I cannot, let me repeat, CANNOT, purchase one pump only to find that it is too small for the desired effect and thus have to purchase yet another one (and possibly have to re-dig the pit it is in, for fit), and I do not want to, again let me repeat, WILL NOT, purchase one that has excessive capability (read that as "excessive cost") for the desired effect.

Is there a calculator that will tell me how many gallons is flowing when an X-ft wide rock has a level 1/Y inch of water flowing continuously over it? And is that garden hose too small for what I want?

Now to tie this to the original poster's quest... how many Watts of electricity is required to move this unknown amount of water the desired height?

Will the 5-W "solar panel" generate enough power to lift the water the height requested (and still be moving water)?

Looking at the picture of the referenced pump in operation (assuming it is anything like a "realistic" representation of what it is capable of doing) I am guessing that it is capable of squirting "some" water about 2-ft in the air, thus it could "lift" that amount (what ever it is) of water that high in a hose. How much less water would it lift if the distance were 3-ft instead of 2-ft? Would it wear out quickly if one attempted to move the water that far?


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## Greg Elmassian (Jan 3, 2008)

I had the same quandary when setting up a fountain.... a "Sheer descent", which is like a flat sheet of water... so I pressurized it with a couple of garden hoses... 

When I had the flow rate right, I captured the water and measured it, measured for two minutes I think, and then multiplied the gallons by 30 to get hour's worth of flow... 

actually they gave me the GPH rating of the fountain, but I wanted to be sure... works perfect... 

Regards, Greg


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## Torby (Jan 2, 2008)

I don't imagine a 5 watt pump could do many C*Gallons/Head...


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## toddalin (Jan 4, 2008)

Posted By Semper Vaporo on 10 Aug 2009 08:14 PM 

But what number goes in the first position of the calculator? I have absolutely no idea how many "Billions of gallons per millisecond" or "teaspoons per fortnight" of water "look pretty" going over a rock that is a foot or so wide and will not have the unpleasant sound of (pardon the unseemly description) a child urinating in a toilet when the water is flowing.







Probably best to approximate this using weir calculations:

*To Calculate the water flow in CFS over a Suppressed **Weir*
*To calculate the discharge flow of water over a suppressed weir, the following formula is a good approximation.
3.33 x W (width in feet) x D 3 (depth in feet), and Depth is expressed as the square root of D to the third power.
2 (depth is cubed first and then the square root is taken)*
*e.g. 3.33 x 8 feet wide x 2 feet discharge over weir = 3.33 x 8 x 2.828 = 75.38 CFS or 2nd e.g.*
*3.33 x 6 feet wide x 8 inches overflow = 3.33 x 6 x .64952 (square root of .75 x.75 x .75) = 12.97 CFS*
*Note: The water level must be measured a good 3-4 feet before the discharge weir, and there should be a clear drop of the water into a lower pool of water to be fairly accurate.* 
Also see the link:

http://serc.carleton.edu/microbelif...mgage.html


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## RimfireJim (Mar 25, 2009)

Semper,

Your frustration is understandable, but unfortunately fluid calculations aren't particularly simple. Todd's suggestion of using the weir calculation is one approach, but it still doesn't give one much of a "feel" for the result, unless you have another waterfall to compare it with and measure the depth and width of the flow. Greg's empirical approach is probably the most practical for the topic at hand. Just doing a simple test of a garden hose flowing into a five-gallon bucket will start to give you an idea of what flow rate you need for a small fountain. All it takes is a bucket or tub of known volume and a watch with a second hand.


Re. the output of a pump "squirting" into the air vs. into a pipe or hose: could be a BIG difference. Into air is against virtually no resistance except gravity. Into a pipe could be lots of resistance if you size the pipe too small. Plus, what you see into air is the height at which it "stalls", i.e., has zero flow rate. Even if you used a humongous pipe to that height, the water would come just to the top of the pipe and not flow out. That's where the dynamic head calculator comes into play. You need to know how much water you want to flow and the factors that resist that flow (pipe friction and elevation). Virtually all water pumps for this application have output curves that show a decrease in volume with an increase in head. They won't be damaged by too much head, they just won't be moving much water. (They're called non-positive displacement pumps, unlike gear pumps used for hydraulics.) At the limit, however, they will overheat (the pump end, not the motor) if the flow is zero, because the motor is pumping in power but the pump is not doing any useful work and all the power is going to heating the water, which isn't flowing to carry away the heat.

Play with the input numbers to the calculator and it will help you get an understanding of what the result means. Make the flow rate small, the pipe dia. big, and the result is dominated by elevation. Then, as you increase flow and decrease pipe dia., you'll see the TDH go up even more, showing the resistance (in feet of head) that the pipe friction adds. If you have fittings in the pipe run, you need to add their equivalent in length to the length of the pipe run. There are tables on line that have those numbers. And don't use corrugated pipe/hose - nasty friction losses!


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## Greg Elmassian (Jan 3, 2008)

Yep, I took what my landscaper suggested in pipe diameter, and increased it, and then watched them to make sure they did not use a lot of elbows in all the piping... more resistance... Bought a 2 speed Spa pump and everything works well enough it runs on low speed, saves $$... the outlet is the horizontal black line.... 










Regards, Greg


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