# Calculate power

1. May 28, 2006

### optimizer

Hi,
If I have a weight, an example could be the weights in a Pendulum type clock (weight could be pounds, or tons if we're talking Big Ben), is there a way to calculate the maximum amount of electricity the weight could be generating (kilowatts etc) if the clock was replaced with a generator of some sort.

Any links to examples scaled from micro to gigantic (to hammer it into my head) are welcome.
Thanks

2. May 28, 2006

### dav2008

Well take the two extremes: the weight at its highest point, h1 and the weight at its lowest point, h2.

If allowed to fall from h1 to h2 (assuming this is around the surface of the earth) then the maximum work you could obtain would be $$mg\Delta h$$ where $$\Delta h$$ is h1-h2. (The change in gravitational potential energy)

This means that you don't have a pendulum just swinging. You have a pendulum at a height h1, you connect it to some sort of generator, let the weight fall to h2, at which point the change in gravitational potential energy will have gone to either doing work or wasted to friction.

A pendulum swings forever if there is absolutely no friction or anything else to stop it. As soon as you introduce a generator, it will cause the pendulum to slow down so the maximum work you'll get is just the change in gravitational potential energy.

Last edited: May 28, 2006
3. May 28, 2006

### optimizer

Ah, sorry, I forgot to note I'm a Physics Dummy (one of those aged ones that's ready to learn after school days).

Thanks for the answer, it's getting me there.

Yes, so it's "maximum work" that I'm looking for, is it.
I don't know the mg-h formula (willing to research, but probably quicker to ask/learn).

I suppose I would like to see an example calculation that shows a 1 ton weight dropping from h1 (100') to h2 (0') and how much work that would acheive,
and a calculation to show how many kilowatts you could get from that work with a specific type of generator.

No pendulum!

Thanks

4. May 29, 2006

### dav2008

mgh is the gravitational potential energy of an object on the earth, where m is its mass, g is the acceleration due to gravity (9.8 m/s2, and h is the height of the object. The place where you put your height to be 0 is arbitrary so it's the change in height that really matters. Also the value of 9.8 m/s2 is at about the surface of the earth so this formula isn't really accurate if you're dealing with large distances from the Earth. See: http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html#mgh

This problem is basic conservation of energy. If an object is dropped a distance $$\Delta h$$ then the object's gravitational potential energy decreases by $$mg\Delta h$$. Conservation of energy says that energy can't just disappear. Where does it go? If you just drop an object it will go into kinetic energy. If in your case you want the object to somehow turn a generator to generate electricity then that energy will go into generating that electricity. Obviously in the real world some of the energy will go into things like friction so you can't perfectly convert all of the gravitational potential energy to useful work.

In your example, the weight of the object is 2000 lb (weight is equal to mg. pounds are a measure of weight, while kilograms are a measure of mass) The height difference is 100 feet. This means that the maximum work you can obtain would be (2000 lb)(100 ft) = 200,000 lb-ft.

Kilowatts are a measure of energy per time. One watt means one joule per second. lb-ft and Joules are a measure of energy.

5. May 29, 2006

### Staff: Mentor

To expand, you could have the mass fall at any rate permitted by gravitational acceleration and produce any amount of power within those limits for that time.

For longer times, you can consider the acceleration's affect negligible and simply divide the work by the time to get average power. Ie, if you let it fall for 10 seconds, you'd get 200,000/10 = 20,000 lb-ft/sec for 10 seconds and if you let it fall for 20, you'd get 200,000/20= 10,000 lb-ft/sec for 20 seconds.

If you tried to absorb the energy much faster than about 5 seconds, you'd find that in order to get the mass to the ground in 5 seconds or less, you'd have to let it accelerate on its own, and not extract any power from it, then when it got near the ground, extract extra power to decelerate it. You'd end up with much lower power at the beginning and higher power at the end (but still the same total amount of work, of course).