I have a house on a small mountain. It is about 60-100 meters higher than the stream. I have seen videos on youtube for building an electric generator from an old washing machine. The guy said his generator produces about 900 watts. Does it make sense to build one of these generators and use it to power an electric waterpump to pump the water up to the top of the mountain? Do these generators produce enough electricity? Would I need a battery?
To raise water 100 meters you'd need 142.2 psi or ~1 MPa of water pressure. A 200 psi pump would deliver 58.8psi at the top of the mountain. My first google result for a 200 psi pump was a 0.8 GPM 200 psi pump on eBay that consumes ~100W. That may deliver about 0.234 GPM (0.87L/min) or 14 GPH if my calculations are correct, so depending on how much volume you'd need, 900W may be plenty depending on pump efficiency. If by washing machine generator you mean the hydroelectric ones, then so long as you use full wave rectifiers for each of the three phases I doubt you'd need a battery or smoothing capacitors to run a pump, just a regulator or inverter for whatever voltage the appropriate pump requires. The Fisher and Paykel washing machine conversion I saw on YouTube generated 21A at 29V so I'd chose the pump appropriately if going that route, and always over size pumps and consider running two in parallel and live with half the flow rate when one fails, same goes for the generator.
By generator, do you mean a gasoline powered generator?
If yes, then what you say is possible, but there may be more efficient ways of accomplishing the same task. Every time you change the form of energy, you lose some efficiency. So let's just list the conversions in your proposed example, and let's assume you get to keep 80% of the input energy of each step, due to conversion inefficiency:
Gasoline to rotary mechanical energy 100% x 0.80 (that's 80% efficiency) = 80% left
Rotary mechanical energy to electricity via generator 80% x 0.80 = 64% left
Electricity to mechanical rotary energy (an electric pump of some kind) 64% x 0.80 = 51.2%
Rotary pump energy to moving water to a higher elevation 51.2% x 0.8 = 40.9%
So, the best case scenario with 4 conversions is the loss of 60% of the original input energy.
In real life, the losses will be much worse. Most small gasoline engines are lucky if they turn 25% of the chemical energy of gasoline into useful rotary torque/energy. That's a 75% loss in the first step. Plugging that in drops it to an 88% loss of the original energy in the gasoline.
One improvement would be to drive the pump directly from the gas engine. This cuts out two of the four conversions.
If you could link to one of the videos you refer to, I could make a better estimate. There are also many many different kinds of pumps you could use, which will dramatically affect how well the system as a whole works.
If you could give me an idea of how many liters or gallons of water per day you would need to move, that would help also.
William Bronson wrote: If you have a storage tank above your point of use,the GPM of the pump becomes moot.
William, I have a storage tank above the point of use. I want to use the water power of the stream to generate electricity. See attached pictures. I am unclear on whether I can just direct use a Pelton Wheel in this situation. And if so, where would the wheel be placed? Does it have to be placed perpendicular to the water flow? How deep would the water need to be? This water is about 8 inches deep.
posted 4 years ago
If the primary goal is to pump water, then turning the kinetic energy of your stream into electricity would not be the -most- efficient way to pump water. After making the electricity, you would have to turn the energy back into mechanical energy to run a pump. You lose efficiency at both ends of the conversion.
But if you have a lot of hydro energy available to you (which I am kind of doubtful about, just looking at the pictures) then you may have other uses for the electricity as well.
You can do a google search for hydro power calculator and that will give you a good first approximation of how much power is actually available to harness. Here is one example:
A more efficient option, if the primary goal is to move water uphill, is a ram pump. Nicely designed, they can be up to 75% efficient. Your proposed solution of going from flowing water, to electricity, to a pump, to flowing water, might be 25% efficient as a system.
Troy Rhodes wrote:If the primary goal is to pump water, then turning the kinetic energy of your stream into electricity would not be the -most- efficient way to pump water.
Troy, why do I need to be concerned with efficiency? The water power is free. I would just need to buy an electric pump to pump the water up.
I looked into ram pumps but it doesn't seem that the drop in elevation is enough to get the water 100 meters up. I think I would have to buy a very long pipe to put in the stream and then it still might not work.
One thing I want to find it is can I combine a ram pump with an electric, use the electric pump to increase the flow rate going into the ram pump?
posted 4 years ago
Let us suppose that your water power can supply 50 watts of electricity, 24/7. That's 1200 watt hours per day.
But there are not many pumps that will run on a measly 50 watts.
But the water supply in our example produces 200 watts of mechanical power to produce those 50 watts.
Thus, a mechanical pump driven directly by the water could be able to move 4 times as much water as the tiny pump that would run on a 50 watts.
Because it is more efficient.
It all depends on how much energy your water could deliver...
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