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# Type and size pond pump to use?

I recently installed a 25 foot long river that runs from a pond at the top to my tiny preexisting pond at the bottom. both ponds are very small and I only need a small amount of water running through the river. What size and type of pump do I need to pull water from my bottom pond to my top pond (about thirty feet)??

## Discussions

Best Answer 3 years ago

What is your rise and width?

Answer 3 years ago

Rise is probably about 5 inches, where the water runs the width is about 1 foot.

Answer 3 years ago

Ok so you have about a 0.0166 or just under a 2% slope over 25 feet 1 foot wide, let’s say 2 inches deep of water rubble or stone lining for looks giving a constant of 0.017 to 0.03 or n.

Remember you can use rocks in your river to slow down the flow so you can use a smaller pump.

Area A

A = 1 x 0.166 = 0.166 ft

^{2}Wetted perimeter P

P = 0.166 + 1 + 0.166 = 1.333 ft

Hydraulic radius R

R = A/P = 0.166 / 1.333 = 0.1245 ft

Slope S

S = rise (drop) / run = 20

in over 100 feet or 1.66 feet / 100 feet = 0.0166

Using Manning’s equation Q for discharge,

Q = (1.5/n) x A x R

^{2/3 }x S^{1/2}Q = (1.5/0.017) x 0.166 ft

^{2}x 0.1245^{2/3}ft x 0.0166^{1/2}Q = 88.235 x 0.166 ft

^{2}x 0.2497 ft x 0.1288 = 0.471 cfs (ft^{3}/s)Q = (1.5/0.03) x 0.166 ft

^{2}x 0.1245^{2/3}ft x 0.0166^{1/2}Q = 50 x 0.166 ft

^{2}x 0.2497 ft x 0.1288 = 0.267 cfs (ft^{3}/s)If you use rocks to slow down the flow of water you could probably get away with a 1000 GPH pond pump for about $100. To keep water in your river at all times you will need rocks to slow the flow of water, otherwise you are looking at 3 to 4 gallons a second to keep water in your river.

There is 7.48052 US gallons in a cubic foot.

Just Google pond pumps.

Answer 3 years ago

Awesome, I just found the biggest pump my local hardware store sold was a 1000 gph. I ordered it in and it should be here soon I cant thank you guys enough!!

Answer 3 years ago

We like to help.

We also like to be marked best answer.

Just one last thing, the bottom pool should be large enough so that it will hold the river water and its own water just in case of a blackout. Your river should hold about 15 to 30 US gallons depending on the rocks you use to control the waters flow in your river.

Answer 3 years ago

There is plenty of room for the water from the river , Thanks again for the help!!

Answer 3 years ago

Hey, wow! You're using Manning's equation,

https://en.wikipedia.org/wiki/Manning_formula

and I was just naively assuming there'd be an equilibrium condition where the flow in the channel was exactly equal to the output flow from the pump.

I mean there would have to be, assuming mass is conserved, similar to the assumptions for Kirchhoff's laws.

https://en.wikipedia.org/wiki/Kirchhoff%27s_circui...

Although, for something like this, water is maybe not perfectly conserved. It could going somewhere else, outside the circuit, like maybe evaporating, or seeping into the ground.

However, I suspect the flow rate in the channel will be

close to,approximately equal to, the flow through the pump, in the case where this number, flow rate from the pump, is much larger than the losses due to water loss to evaporation and seepage into the ground.You know: water goes up through a pipe, and comes down through the channel. If those flow rates are unequal, it means water is piling up somewhere.

Answer 3 years ago

If he built it right it should have a plastic liner and on such a small project the evaporation will en negligible. Remember there are two ponds one at each end of the river for pooling so there is leeway.

Answer 3 years ago

There is a good plastic liner, I never even realized the flow rates would be close to equal. That will help alot!! Thank you!!

Answer 3 years ago

The pond at the top is very small so water pooling is minimal, it was jjust put in for more sound as the water fell

3 years ago

Pumps are rated in terms of

pressureandflow rate. Conveniently, pressure is related to the height of a column of liquid.For water, a column 32 feet high (9.8 meters high), corresponds to a pressure difference of about 1 atmosphere,or 100 KPa, or 14.7 PSI (pounds per square inch).

Sometimes when pressure is expressed in units of water height it is called "head", and I don't know why that is.

Anyway, if you have to pump water to a height of 25 feet, you essentially already know the pressure needed for that.

Flow rate can be volume flow rate, or mass flow rate. For a material like water, those numbers are proportional, related by constant factor, which is the density of water.

So the answer depends onhow much flow rate(e.g. in gallons per minute) you want to see.When you go shopping for pumps, they'll probably be rated for such-and-such a flow rate, at such-and-such a pressure, or "head". The units for volume flow rate are unit volume per unit time, e.g gallons per minute, liters per second, cubic feet per hour, etc.

Conveniently, when you multiply those numbers for pressure and volume flow rate, you get

power, the amount of physical work, per unit time, required for to the pump to push that volume through that pressure, or equivalently to lift that volume of water across that height. That number for power is good to know, since you'll probably bepayingfor that energy flow somehow. E.g. if the pump is electric, the electricity provider charges by the kilowatt*hour.Answer 3 years ago

Wait. I think I read that wrong. What is the actual height difference

between the top pond and the bottom pond? You mention a "25 foot long

river", but what is the vertical difference in height, i.e.how far does

the water drop?

Answer 3 years ago

Yah the actual height difference is about 4 or five inches vertical rise from the bottom pond to the top pond.

Answer 3 years ago

Okay. So the pressure needed to lift water 5 inches, is like nothing. Nothing to write home about. ;-) 0.5 feet of water is about 0.23 PSI. I think that's near the (zero pressure, max-flow-rate) side of the pump characteristic.

So the numbers they quote for flow, might be believable for your application, and it should be very, very easy, to find a pump that can lift water such a small height.

Answer 3 years ago

Wow you really know your stuff!! Thank you for the help!!

Answer 3 years ago

I should probably elaborate on this a little bit, in particular the way I am imagining the pump to work.

I am imagining that for a typical water pump, there is a graph relating pressure (e.g. feet of head) and flow rate (e.g. gallons per minute)

An honest pumpmonger will actually have graphs for their pumps. An example of one of these graphs is attached to this post.

The graph looks roughly like a downward sloping line. Maximum flow occurs at zero pumping height. Conversely, at maximum pumping height, the flow rate drops to zero.

A less honest pumpmonger, will tell you about two points on the graph: the maximum flow and maximum pumping height, because those are the biggest, most impressive sounding numbers. But of course, you won't get this pump to produce maximum height

andmaximum flowat the same time, and you know this from looking a graph for a typical pump.So essentially you want to choose a pump capable of pumping water to a height

higherthan the distance you will actually be pumping. If you're just going from these two numbers (if those are the only ones quoted) I would *guess* pumping water to half the maximum height, would give flow at about half the maximum flow rate.If you choose a pump whose maximum pumping height is

less thanthe height difference needed for your application, then that just won't work at all.Another trick salespeople everywhere use for making numbers sound bigger, is to get creative with the units. For example a pump with maximum flow of 10 gallons per minute, sounds more impressive when the quote is 600 gallons per hour, which is, of course, the same number. You may have seen that trick before in other contexts.