This is not homework, I am just curious and don’t know the math. If I had a tower 100 feet tall how much weight would I have to raise to the top of the tower to produce 1 KW of electricity? Just say 50%.efficiency.

A weight at the top of a tower has a certain quantity of potential energy. 1 kilowatt is a measure of power, which is energy delivered per unity of time. So any weight, if allowed (braked) to fall at a high enough constant speed, could deliver 1 kilowatt of mechanical energy; a small weight falling very fast would do it, but so would a heavy weight falling slowly.

I think you’re going to need to restate the scenario a bit. Maybe you meant 1 kilowatt-hour (a unit of energy, e.g. a 1-kilowatt system operated for one hour)?

Yes, I meant 1 KWH!, I am thinking about 20,000# but that doesn’t seem like enough for some reason.

OK, 1 kilowatt-hour = 2,655,000 foot-lbf. Divide by 100 feet, and your weight then needs to be 26,550 pounds. That assumes 100% efficiency in converting gravitational potential energy into electrical energy.

I cannot convert your weird units in my head, but there is a very easy approximation you can use: let *g* = 9.8 m/s^{2}, the “standard” gravitational acceleration near the surface of the earth. Then the difference in potential after lifting it to a height *h* is about *mgh*, where *m* is the mass of your brick.

Ground-level gravitational potential energy E in J is E=mgh with m mass in kg, g as 9.81 m/s^2, and h height in m. Your 1 kWh is 3.6 MJ. Some math yields 12 Mg as the required mass at 30 m (ignoring efficiency, so divide by 50% if you like). That’s about 12 tons at 100 ft.

If you were to use water, that’s 12 m^3 or about 3200 gal.

And if you insist on using American units, note that pounds are a measure of weight (or other force), not mass. So if you want an answer in pounds, then you’re solving for mg, not m.

Hey @HoneyBadgerDC , here is a hydro-electric dam power online calculator.

You may play around with some numbers there to get what you want

For example : A 100ft tall tower (gives a pressure head of 90ft accounting for pressure losses in the pipe), with 2 gallons per second of water flowing out and a turbine efficiency of 50% (i think this is much higher in real life) will give you a kW of power.

If you want 1 kWh - then you need 2 gallons x 60 secs x 60 mins = 7200 gallons of water. This is the same as 60,000 lbs or 30 US tons.

Have you heard of this idea - well business today?

I’m doing this in metric units, and leave the commentary between SI to Imperial to others. I’ve paid that tax a long time ago and have my excemption stamp.

For ease of calculation, I am making some basic assumptions:

- The tower is now 100 meters (~330 feet).
- There is no air resistance or other parasitic losses.
- Because I’m extra-lazy I simplified the gravitational acceleration constant to 10 meters per second squared.
- All “braking” (converting the kinetic energy of the water to electrical energy) is done in one second.
- These simplifications allow us to use the basic potential energy equation: E=10mh, where m is the mass in kilograms, g is now 10 m/s^2 (and already written in), h is the height in meters. To do this in US units, E=70.2mh*Ef, where m is your mass in pounds, h is your height in feet, and Ef is your target efficiency ratio. It’s accurate enough for back of the napkin math.

The apparatus is a tower, from which we’re dropping one kilogram shots of water. There’s a turbine at the bottom that will get spun up by the water which we’ll use to generate electricity.

A one kilogram shot of water will generate 100 joules of energy in a perfectly efficient system. I’m assuming, of course, that the water transfers its energy to the turbine over exactly one second.

Your target of one kilowatt at 50% efficiency requires 20 kilograms dropped from 100 meters.