How much energy can be generated from a 1000kg weight using a hydraulic system?

AI Thread Summary
A 1000 kg weight lowered by 1 meter can generate approximately 0.0027 kWh of energy, calculated using the formula for gravitational potential energy (mgh). The discussion clarifies that while 1 kW represents power, energy is measured in kWh, highlighting the distinction between the two. The weight's potential energy is converted through a hydraulic system, which can drive a generator, but the overall energy output is relatively low. The conversation emphasizes the efficiency of energy conversion and the significant power requirements for everyday devices. Ultimately, this illustrates the challenges in harnessing energy from such a system effectively.
johnjz
Messages
2
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
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:
johnjz said:
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 :)
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!
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?
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.
 
Thread 'Motional EMF in Faraday disc, co-rotating magnet axial mean flux'
So here is the motional EMF formula. Now I understand the standard Faraday paradox that an axis symmetric field source (like a speaker motor ring magnet) has a magnetic field that is frame invariant under rotation around axis of symmetry. The field is static whether you rotate the magnet or not. So far so good. What puzzles me is this , there is a term average magnetic flux or "azimuthal mean" , this term describes the average magnetic field through the area swept by the rotating Faraday...
Back
Top