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How much electricity can we generate with 1 stroke? 1 kW or ?

Thank you.

- Thread starter johnjz
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How much electricity can we generate with 1 stroke? 1 kW or ?

Thank you.

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russ_watters

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Thank you very much for answering, but I am a bit disappointed, not much energy or power.

I Googled 9.8 m/s/s, and got "A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth)." which would apply to falling water and hydro-electricity but does that also apply to hydraulic-electricity?

1000 kg is pushing on a piston, which is "let go" so in a sense it is free-falling, but it's not moving very fast... and the speed is controllable via a pressure cap, which might control a force of liquid to drive a generator.

PS. If I divide 9800 by 3600 by 1000, I get = 0.0027 which is even a more miserable result :)

I Googled 9.8 m/s/s, and got "A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth)." which would apply to falling water and hydro-electricity but does that also apply to hydraulic-electricity?

1000 kg is pushing on a piston, which is "let go" so in a sense it is free-falling, but it's not moving very fast... and the speed is controllable via a pressure cap, which might control a force of liquid to drive a generator.

PS. If I divide 9800 by 3600 by 1000, I get = 0.0027 which is even a more miserable result :)

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russ_watters

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Yep - it was a first-thing-in-the-morning typo. Your result is correct.Thank you very much for answering, but I am a bit disappointed, not much energy or power.

PS. If I divide 9800 by 3600 by 1000, I get = 0.0027 which is even a more miserable result :)

Really gives you an appreciation for how much energy/power is required to run a light bulb (or a car!), doesn't it!

9.8m/s/s, aka, "g", is the acceleration of an object due to gravity at the surface of the earth. If you plug that value into f=ma (substitute g for a), you get the force required to hold an object still against gravity (f=mg), aka, the weight of the object. Since the work equation is w=fd, inserting the weight equation into it and using height as the distance gives w=mgh. This is the equation for gravitational potential energy. There are lots of different ways to convert it to other forms of energy, but the important thing is that energy is conserved, so w=fd tells you the total energy available for conversion.I Googled 9.8 m/s/s, and got "A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth)." which would apply to falling water and hydro-electricity but does that also apply to hydraulic-electricity?

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