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Type and size pond pump to use? Answered

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)??

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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 ft2

Wetted perimeter P

P = 0.166 + 1 + 0.166 = 1.333 ft

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 R2/3 x S1/2

Q = (1.5/0.017) x 0.166 ft2 x 0.12452/3 ft x 0.01661/2

Q = 88.235 x 0.166 ft2 x 0.2497 ft x 0.1288 = 0.471 cfs (ft3 /s)

Q = (1.5/0.03) x 0.166 ft2 x 0.12452/3 ft x 0.01661/2

Q = 50 x 0.166 ft2 x 0.2497 ft x 0.1288 = 0.267 cfs (ft3 /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.

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.

Pumps are rated in terms of pressure and flow 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 on how 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 be paying for that energy flow somehow. E.g. if the pump is electric, the electricity provider charges by the kilowatt*hour.

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.

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 and maximum flow at 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 higher than 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 than the 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.