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 tonk May 27th 2014 11:42 PM

Gravity Generator

I have created a short video describing my question about a gravity generator

 ChipB May 28th 2014 11:46 AM

A 19,000 Kg weight falling 15.24 meters experiences a change in potential energy according to:

Delta PE = mgh = 19000 Kg x 9.8m/s^2 x 15.24m = 2.84 x10^6 Joules.

Given that it takes 10 minutes = 600 seconds to do this, the power produced is

P = Delta PE/time = 2.84 x 10^6 Joules/600 s = 4730 N-m/s = 4730 watts.

This assumes that the generator is 100% efficient - in practice you would p[robably find that the electrical power actually generated is perhaps 90% of that figure.

As to the size of the generator: it must be big enough to be able to generate 4.73 KW. I haven't attempted to source one of an appropriate size - I'm sure a few google searches will help you find one.

One point missing from your video is any mention of gearing. The generator must be geared so that its core spins at an efficient rate, whihc will depend on the generator selected. You asked about the diameter of the cog - that by itself doesn't matter; what's important is that the gearing between the cog and the generator gives an efficient RPM to the generator. You can't simply attach the cog directly to the inpurt shaft of the generator as it will turn too slowly. The formula for the cog turning speed in radians/second for a given radius of the cog R is:

w (rad/sec) = 15.24m/600s/R = 0.0254m/s /R.

If R = 0.2m then w = 0.127 radians/sec, or 1.21 RPM. This is too slow to function effectively as direct input to the generator shaft; hence the need for gearing.

Hope this helps.

 tonk May 28th 2014 12:17 PM

Gravity Generator

Thank you, it helps immensely. Can I figure that if you were to multiply the 4.7 Kw figure by six, that the output would be 28.2KWH

 ChipB May 28th 2014 01:11 PM

Quote:
 Originally Posted by tonk (Post 25845) Thank you, it helps immensely. Can I figure that if you were to multiply the 4.7 Kw figure by six, that the output would be 28.2KWH
Indeed 6 times 4.7 is 28.2. Are you thinking about having 6 such machines? Or speeding up the rate of derscent of the weights by a factor of 6? Or 6 times the weight?

 tonk May 28th 2014 01:34 PM

Gravity Generator

No, I just figured if it took 10 minutes to produce 4.7 Kw and there are 60 minutes in an hour so the weight would make the trip 6 times over an hour. The premise was you didn't need to factor in the energy required to get the weight back to the top it just instantly returned to its starting point once it dropped the 15.24 meters.

 ChipB May 28th 2014 01:59 PM

You are misunderstanding what power is all about. Power is the rate that energy is produced, measured in energy per unit time (Joules/second, or watts). Your contraption produces a constant 4.7KW of power as long as it runs. Each cycle of dropping a 19000 Kg weight produces a certain amount of energy - I calculated 2/84 x 10^6 Joules. Over the course of an hour the weight drops 6 times, for a total of 6 x 1.84 x 10^6 = 17.0 x 10^6 J. The power output is then calculated by dividing this energy produced by the time taken to produce it:

P= 17.0 x 10^6 J/3600 seconds = 4730 Watts.

So again - power output is constant, though the total amount of energy produced over the course of an hour is 6 times the energy produced in 10 minutes. Hope this helps clear it up.

 tonk May 28th 2014 02:32 PM

Gravity Generator

OK, then let me try and ask the question this way. The output for this device is 4.7 Kw. The average house in America use about 10,000 KWh over the course of a year. How many hours would this device have to operate in order to produce 10,000 KWh? Thanks again for you help and for putting up with my lack of knowledge in this area.

 ChipB May 29th 2014 06:21 AM

10,000 KW-Hr/4.73 KW = 2114 hours, or about 88 days. Stated another way - this device could provide sufficient power on average for 4 homes (ignoring the fact that there are peaks and valleys in energy demand).

 tonk Jun 17th 2014 02:05 PM

Greetings, I had another question on a different subject. I'm not sure if it is in your field of expertise, but if it is:

Premise: Let's say you have a 10,000 gallon pressure tank that was empty, you would have approximately 1338 cubic feet of air at 0 PSI.
What would the pressure be after you added 255 cubic feet of air into that tank? And if after that you wanted to add another 255 cubic feet of air into the tank, what would the pressure be? More importantly, I'm looking for the formula(s) you used to answer both questions.

Cheers,

Kenji

 ChipB Jun 17th 2014 03:08 PM

The formula to use is the Ideal Gas Law:

PV=nRT

where P=pressure, V=volume, n = amount of gas (in moles, or grams, or as in your case cubic feet at 1 atnmosphere), R is a constant, and T is temp in Kelvins. Adding more gas while maintaining the same size container and assuming the temperature in the tank is equalized with outside temp (and hence assumed to be constant) leads to:

P_2 = P_1(n_2/n_1)

In other words the pressure increases by the amount of gas you pump in. Double the amount of gas in the tank and you double the pressure. Going from 1338 ft^3 at 1 atmosphere to 1593 ft^3 will make the pressure:

P_2 = 1 atm(1593/1338) = 1.19 atmosphere.

Adding another 255 ft^3 brings it to:

P_2 = 1 atm(1848/1338) = 1.38 atmosphere. You can see that every time you add 255 ft^3 you increase the pressure by 0.19 atmospheres.

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