Sorry I should have said, the residence block has 1500 people living in it. The site is supplied by mains electricity although lately this has been off more than it has been on. Typically we see cuts regularly through the day but they are short. On a night it is possible to have no mains supply...
But there will be 16 carts full of 69.44kg of ore all moving down at any 1 time. So the amount of mass being lowered will be 1111.04kg per meter per second.
I know, but we are backed by a Chinese company and to be honest they couldn't care less if we have power or not on an evening. We asked for a generator for the residence block and were refused point blank. I would love to hire a consultant but I am one up from a grunt (I used to be a grunt in...
Thanks guys, as always I am really appreciative of the help and suitably embarrassed by the simplicity of the answers. I am sticking at it though, physics is fascinating. Someday I would like to answer a question on the forums. On that day I will be a very happy man.
OK so is it 100kW every single hour. So over a day it produces 2400kW. Is that right? The new Nuclear power station will make a fortune. Why does the government need to subsidise it?
I have been looking at the difference and I know one is power and one is energy. I understand that a 2kW kettle on for an hour uses 2kWh but on for 30 minutes uses 1kWh. But with power generation it seems this is constant. If a plant puts out 100kW it does this all day every day regardless of...
I can't find anything like that but...would a 50kW Generator and 4RPM be the same? and 25kW with 8RMP and so on. Would they generate the same power overall or is the power rating its production total regardless of anything?
OK so if there are every any other numpties following in my footsteps...
Bigger blades = bigger coverage area = more wind to interact with = more kinetic energy transferred = more power
Nothing to do with levers, they are quite separate.
So 250 tonnes of rock an hour is 69.44kg of rock per second.
E = 69.44 x 9.81 x 20 = 13624.128 Watts
Oh bugger, that doesn't even work. It has to be more than the power to make it go round.
EDIT: Ok, does the fact that there are 16 x full carts moving down mean I multiply the above by 16...
Sorry, I will give it a go.
The work being done is 16 carts being raised 20 meters. The weight of 16 carts is approx. 3520kg. I don't know enough to apply friction forces yet but I will learn soon.
So here Work (j) = Weight (N) x Height (m) = 34519.4081008128 x 20 = Work = 690388.162016256...
The train is already moving completing a full revolution in a minute although this is slowed with a hydraulic system for timing reasons. It also is moving free of any electrical energy. The weight of the objects dropped into the carts is sufficient to keep the whole loop moving. It just goes...
Another post explained something about blade length too. I was trying to learn about torque and lbs - ft or Nm but could someone please advise.
Say the force needed to rotate a nut is 3.8m Nm. If I attached a big wrench to the nut then the force required decrease every meter away from the nut...