What HP Can Be Generated from a 120 CFS Water Wheel with 20-Foot Drop?

[notinthebox]

I would like to findout what speed and HP I could get from a kind of water wheel.
The unit will have a gear on top and a gear on the bottem
I have over 120 cubic foot of water a second flow to play with.
the containers of water will fill and lower on drive chains and rails
They will loop around at the bottem and drainand go back up
The drop is 20 feet down no angles bot the flip over at the top and bottem let's say a 5 foot gear and no force from the water flow is added
I will have 7440 pounds of water what is the best hp I could pul from a system like that


Think of a water fall and bolting on a water system right on flat to the wall

 

[Mech_Engineer]

Have you considered just looking at the potential energy of the water? Given a mass flow rate mdot (volume flow rate multiplied by density) and the height change, you can calculate a best case power input if all of the water's potential energy were converted.

Power = mdot*g*(h1-h2)

 

[russ_watters]

I'd probably assume 25-50% efficiency along with that.

 

[notinthebox]

sorry error
The power loss looks a bit high (75 to 50%)?
14880 pounds I would like to have it speed down and around at 1 foot per second and each bucket would hold 12 cubic feet of water and there will be 20 of them.
power will be pulled (electric power from gears and or right off the drive chains
so what hp would I be able to get out of this system

 

[Mech_Engineer]

We gave you everything you need to answer the question- take the mass flow rate of the water you're using (kg/s), multiply by gravity (9.81 m/s^2) and the height difference (m). This gives you the raw power in watts, and then take maybe 25% of that value to account for generator efficiency and efficiency losses in the flow.

This is a very rough estimate of the power output you might expect, not taking into account a lot of other factors which might hurt you.

 

[notinthebox]

So am I looking at best a 25% of power output for power input?

 

[Mech_Engineer]

So am I looking at best a 25% of power output for power input?

That's Russ and I's "guesstimate" at an overall efficiency, but isn't based on a calculation. If you assume the generator is about 60% ("typical" value according to Wikipedia, so take it with a grain of salt), and losses in your wheel are around 50% (water leakage, release water before it travels the full vertical distance, friction, etc.) you're looking at 60% of 50% which is (do the math now) 0.6 * 0.5 = 0.3 = 30%.

I think 25% is probably in the ball park for what your real-world efficiency might be, but isn't the maximum efficiency theoretically possible.

 

[notinthebox]

generator efficiency is about 90% new it will go down with age
water loss is zero

One foot drop per second is wanted to set the size of the gear boxes

that or I might run gears on the chains out the sides to a big gear to power the generator

I will use or make 4 30 hp drive chains ( I will need to find out the size of the rods that will hold the buckets and water over est about 1600 pounds each) . That will set the size of the drive chain.

there are going to be 16 bearing

 

[russ_watters]

Water loss is not zero: the best turbines are around 80% efficient and you won't get anywhere near that for a water wheel. The wiki says 60% is the max.

With the generator loss and mechanical transmission loss, 50% overall for the high end of the range was too high.