Flutterby™! : Face/Twit/Whatever status 2009-04-03 00:16:24.20342+00

Face/Twit/Whatever status

2009-04-03 00:16:24.20342+00 by Dan Lyke 11 comments

Dan: Silly Q: Water mains at 95PSI, house could run at 35. How much energy could we extract from daily water use?

Yeah, this should be fairly easy if I can ballpark water use. Must find water bill.

I believe this might be relevant to your ideas. I can see this being useful in powering your irrigation system, maybe, but where are you going to find a turbine type pressure reducer that will meet code?

More idle musing than anything practical, just sprung from a conversation yesterday about pumped storage facilities.

Nice to see lunch got the wheels turning but I was talking about much larger scale like http://www.rentricity.com/ type projects. I think Meuon nailed why it would not be practical for home use. With 60psi differential you have enough energy to drive a low head micro turbine but your not going to have enough flow to make it worth your effort (let alone sunk cost). Lunch was good for me too, I put TOPGEAR in the tivo cue and decided Jeremy Clarkson http://www.topgear.com/us/the_show/bios/jeremy_clarkson is a good role model.

Yeah, my first reaction is to think of it on the Raccoon Mountain scale, but it seems like there are a bunch of places where just a little bit of power would be useful. For instance, in m's irrigation system, could he skip the photovoltaics and (given some priming) charge the computer/valve battery off a small turbine which could function as the pressure reducer as well? How about for sensors?

And Top Gear-wise, somewhere under my Stig exterior is a James May looking to get out.

Offhand I don't think it would produce very much. One model for this would be the the energy it would take to to keep a bladder inflated in a water tank as water was released. To take a gallon of water out of a 95psi system to a pressure of 35psi, would gain the same energy that it would take to inflate the bladder from 35psi to 95psi. A gallon of air at 35psi is about 9 grams at 0C. The specific heat is 0.23cal/gm, and a kilocalorie/hour is roughly 1.16 watts.

To increase the pressure of a gallon of air from 35psi to 95psi would require a temperature change from 0C (305K) to 554C (828K). So the 100% efficiency energy required would be 1.15Kcal, or 1.33 watt hours. There are about 7.5 gallons in a cubic foot, possibly 10 watt hours from a cubic foot. Assuming 50% efficiency (wild guess), it would take 200cubic feet of water to produce 1 KWH. That is a lot of water. When I lived in a house about the size of yours, with two adults, I rarely used more than about 300 cubic foot/ month.

Its been a very long time since I did a physics word problem, so you might want to check the logic/arithmetic before embarrassing yourself to anyone else with my discussion.

m, yeah, I guess I've got some issues with conflating applications, 'cause you're paying for all the energy. Here in suburbia (semi-urbia?) we've got that 60PSI of wasted energy (stored here in water tanks on the tops of hills).

Dan, The numbers apply to your water supply scenario -- 95lbs to 35lbs, calculated on a 60lb pressure differential. If you use 300 cubic feet a month, you could only hope to save 1.5KWH a month at best. Unless my model assumption or arithmetic is wrong (very possible), payback for you would take forever and longer

My house water pressure is about 32-33lbs, and my garden supply drops it to 10+, so I would do even worse.

#Comment Re: made: 2009-04-03 21:34:10.850137+00 by: andylyke [edit history]

Consider lifting the water 120 feet (pressure in water ~= 0.5psi/ft of head) The acceleration due to gravity in SI units is around 9.8 m/s^2 (not that I know what a square second looks like!)

So if we lift a kg of something 36.1 m, then it has a potential energy of E=mgh e=9.8 x 36.1 Joules/kg ~= 328 Joules/kg or about 0.32 W-hrs per Gallon, if I've followed the numbers correctly.

That's the total potential, but it comes glassfuls at a time, or at best bathtubfuls at a time. Maybe it's better to consider solar panels on the roof?

Andy's solution is certainly the more elegant, and thus most likely to be correct.