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Being an electrical engineer, I'm hoping someone will help me double-check my math. I have a small circuit that I want to power for years that consumes an average 10uA at 3V.

I'd like to power this circuit from a small water turbine with a magnetic coupling that can sustain at most 1 oz-in of torque before slipping.

If the turbine runs 1hr/day at 100rpms (and 1 oz-in of torque), and assuming the generator and regulation circuit run at 100% efficiency, could it power that circuit. I think it's close.

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You're supplying a power of only 30 microwatt, so in a day 2.6 joules will be generated. That's very little.

1 oz-in is about 0.0071 newton meter, 100rpm is about 10.5 rad/s. 0.0071Nm at 10.5rad/s multiplied is 0.074 Watt. With 2.6 Joule you'll be able to do that for about 35 seconds.

And that's assumed there are no losses, anywhere. But there will be. Depending on how you'll store the energy, i think you you'll be able to run it for 5 seconds, if at all. So nowhere near an hour, which you're targeting. You just need more power for that, and tell us how the application is designed, only then guesses can be made about losses.

One tip: get familiar with the metric system, converting an calculating with units is a lot easier.

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your circuit requires (3 volts x 10 microamps) = 30 microwatts to operate. in a 24 hour day it will consume (30 uW x 24hr) = 720 uW-hours of energy.

your pump produces (1 oz-in x 100RPM) units of power (we'll worry about the unit conversion later). in a day it will produce (1 oz-in x 100RPM x 60 minutes) of mechanical energy.

Now go to "Unit Conversions" on the web and find the conversion factors necessary to get the pump output into watt-hours, and equate.

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