1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hydraulic to electricity

  1. Sep 17, 2009 #1
    I have 1000kg of weight (it can be a heavy brick) suspended on a platform which I can lower (say 1 meter) and drive a hydraulic piston. A hydraulic accumulator will store the hydraulic liquid. The accumulator can then drive on-demand a hydraulic pump into a rotary motion and drive a generator.

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

    Thank you.
  2. jcsd
  3. Sep 17, 2009 #2


    User Avatar

    Staff: Mentor

    A kW is power, not energy. kWh is energy. mgh is also energy. So 1000 kg * 1m * 9.8m/s/s = 9800 joules. A watt is 1 joule per second, so that's 9800/3600/1000=0.027 kWh.
  4. Sep 17, 2009 #3
    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 :)
    Last edited: Sep 17, 2009
  5. Sep 17, 2009 #4


    User Avatar

    Staff: Mentor

    Yep - it was a first-thing-in-the-morning typo. Your result is correct.

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Hydraulic to electricity
  1. Hydraulics question (Replies: 1)

  2. Hydraulic Systems (Replies: 1)

  3. Hydraulic energy (Replies: 9)