Gravity Electricity Generation?

[Jeffb47]

We've got a scalable sun tracker that requires no electricity. See << Link removed by Moderator >> Though invented for use as a less expensive alternative to the traditional electric motor solar trackers, we have come to realize that it can actually lift a lot of weight. We haven't tested the limits of its lifting capacity yet, but we know it can easily lift 3,000 lbs per day about 3 feet. We also know that we can scale this up and/or use multiple units to lift much larger weights. So the big question for this forum is: How much weight do we need to lift and how high do we need to lift it in a day to equal the output of one 300w solar PV panel for say 8 hours? To make this simpler, assume we can lift the weight 20 feet. Thank you!

 

[berkeman]

We've got a scalable sun tracker that requires no electricity. See << Link Removed by Moderator >> Though invented for use as a less expensive alternative to the traditional electric motor solar trackers, we have come to realize that it can actually lift a lot of weight. We haven't tested the limits of its lifting capacity yet, but we know it can easily lift 3,000 lbs per day about 3 feet. We also know that we can scale this up and/or use multiple units to lift much larger weights. So the big question for this forum is: How much weight do we need to lift and how high do we need to lift it in a day to equal the output of one 300w solar PV panel for say 8 hours? To make this simpler, assume we can lift the weight 20 feet. Thank you!

Welcome to the PF.

Interesting idea using the phase-change material to generate the hydraulic force needed to move the solar panels. Clever

Are you familiar with the equations for Work and Energy? For gravitational potential energy changes, use W = m * g * delta_height. Where m is the mass, and g is the gravitational acceleration g = 9.8m/s^2. For the Energy of the solar panel, use E = Power * time. Set the work required to move the panels equal to the electrical energy produced by the panel in your time t.

 

[Jeffb47]

Thanks for the response, this is very helpful in figuring out the weight needed to move the panels, we'll use that. Big question there is always the amount of work needed to move panels given variable wind conditions. That's why these systems are always way overbuilt.

My original question was actually slightly different. We're thinking of alternatives to moving the solar panels, i.e. can we generate electricity without using solar panels? Two idea are 1) use the torque we create (slow but strong) to run a generator (fast but with little resistance) via a set of gears, OR 2) use our rotational power to raise a large weight that can be dropped like a cuckoo clock attached to a generator. So for this latter idea, how much weight do we need lower (i.e. raise) to generate 300w for 8 hours? I'm guessing it's a lot of weight. So building off your suggestion, I guess my question is really a matter of converting Work into Electricity given delta-height. Just solve for m. I don't know the Work to Electricity equation(s) though.

 

[Bandersnatch]

If you followed berkeman's suggestion, you'd get something like this:

300W for 8 hours equals about 8.6MJ.
Using simplified potential energy equation U=mgh, you can see that you need 60J to lift 1kg up 6m (about 20 feet). So 300W applied to a lifting engine with 0 losses could lift 144 metric tons of material.
Of course, there will always be losses, but even with 10% efficiency you still get ~14 tons per day.

This same efficiency argument goes the other way around - only a fraction of the potential energy from dropping a weight can be converted to electricity, and you'd need to drop more than 144 tons to get 8.6 MJ.
Look up the efficiencies of dams if you want to put a number on it.

In general, though, you're not looking at anything very viable here. Lifting things is very cheap energy-wise, and vice versa. A good visualisation of this is the following factoid: when you heat a cup of water from room temp to boil for tea, you're expending enough energy to lift the same volume of water some 10 km up.

 

[Jeffb47]

Ok, this is very insightful and helpful, and the answer is what I feared. Could we lift 144 tons 20 feet in one day? Probably. But to put an economic spin on this, that may require us to design and build a really really massive drive unit or utilize something like 864 of our existing "HD" drive units (assuming we can lift 1 ton 1m, so (144/1)*6)). Assuming we can build these for $500 a pop, that's $432k to do what a $250 PV panel can do. That's not even in the same galaxy. Looks like we'll stick to our core application unless someone else has any suggestions...