# How deep would you have to put a steel pole to generate enough electricity to power one LED using geothermal energy? [closed]

I was thinking about how you might grow plants inside without sunlight powered only by LEDs. But to get the energy for the LEDs, one might drill a hole down into the Earth's magma and put a steel pole. Then the heat from the magma would heat up the pole and you might be able to generate electricity at the top from the heat.

I was wondering, if the pole was for example 1cm in diameter, how deep would you have to drill it to generate electricity for one LED?

This assumes that the heat would make it's way to the top of the steel pole without dissipating. Hopefully it would as steel conducts heat better than rock.

## closed as off-topic by Bill N, John Rennie, Jon Custer, Kyle Kanos, ZeroTheHeroMar 3 at 1:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question appears to be about engineering, which is the application of scientific knowledge to construct a solution to solve a specific problem. As such, it is off topic for this site, which deals with the science, whether theoretical or experimental, of how the natural world works. For more information, see this meta post." – Bill N, John Rennie, Jon Custer, Kyle Kanos, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.

• Mind you, steel is a pretty lousy thermal conductor relative to copper or aluminum (like 10x worse or more). – Jon Custer Feb 28 at 4:15
• yes, but i'm thinking steel might be less likely to melt. Then the copper might just dissolve into the magma. Maybe you could generate the light near the magma and funnel it up with fibre optics. Or maybe the magma itself emits light. – zooby Feb 28 at 4:30
• You need a thermo-electric generator for this application. – David White Feb 28 at 6:06

For thermal conductivity (the rate that you can pull heat up through your rod), we can use $$\frac{Q}{t} = \frac{\kappa A (\Delta T)}{d}$$. The longer the rod, the slower the heat comes up.
If you have to go four times as far to double the temperature difference, you're not getting anything good from the power. Imagine that you could hit something really hot (like 2000K), and it was only a mile down. Then using an aluminum rod we could expect: $$(205 W/m K)\pi (.005m)^2(1700K)/1610m = 0.017 W$$.