Can you help with a question please....
Hi there,
This is my first post on the forums (hello all!) and I have a theoretical question that's been bugging me - I'm really hoping somebody here might be able to answer it....
If a one metric tonne weight was raised to a height of 5m and then allowed to descend in a controlled manor whilst connected to a system of cogs & gears which in turn are connected to a turbine/generator - by the time it's descended how much electricity could be generated?
I'm thinking it could be around 0.01 Kwh but could really do with somebody else's input.
Much appreciated,
Simon
This is my first post on the forums (hello all!) and I have a theoretical question that's been bugging me - I'm really hoping somebody here might be able to answer it....
If a one metric tonne weight was raised to a height of 5m and then allowed to descend in a controlled manor whilst connected to a system of cogs & gears which in turn are connected to a turbine/generator - by the time it's descended how much electricity could be generated?
I'm thinking it could be around 0.01 Kwh but could really do with somebody else's input.
Much appreciated,
Simon
Replies
-
xheavenlyxOk, here is a breakdown. good questions though.
Factors determining the total energy:
1. Weight of the ..well, weight. (5 tonne).
2. The hight weight raised to, relative to the "generator machinery". (Example 10 m)
3. The radius of the gear (pulley) and the rotating gear the generator is connected to (Ex: 2:1). PLUS, the transfered torque/force ( here its F = mg => 5Tonnex9.80)
4. The friction of the rotating generator axle and other power losses.
So, in the end you have to connect all the physics formulas to get the "whole equation" containing all the unknowns!
and
5. The most important part: The energy required to raise the weight again to the same hight!! That actually, drops the efficiency down 😀 -
shocky_mcsparkssimon20_08
You have energy stored in the form of a weight held over some datum. The units are ft-lb's (or kg-m excuse me...I'm an old school American engineer). You intend to convert this into mechanical energy by allowing it to fall back to the datum within some time frame.
Get everything into ft-lb/sec and you have horsepower (I'm too busy to do it for you). Multiplied times .746 and you have how many kilowatts that are available in the system.
The amount of energy your system actually creates divided by this number is the efficiency of your system. Your job as the engineer would be to devise a system with the highest efficiency possible. -
simon20_08Hey guys,
Thanks for your feedback, certainly something to get the grey matter going! 👍
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