
#19
Nov2712, 04:50 PM

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Same quantity but one is time averaged (e.g. including a thermostat operating) and the other is instantaneous. That first one may look hideous but it could have relevance somewhere.




#20
Nov2712, 09:52 PM

P: 5

ok, thanks for all the replies, I think I got a glimpse at something, the fog seems to be thinning out.
So, is my understandng right if say : I attach a mass of 102 grams (0,102 kg) to a 1 meter long string that is wound around a shaft that turns a small electric generator. (assume a perfect no loss system) When I let the weight go down, it would ''produce'' a power of 1 Watt, or 1 Joule per second. because [itex]0,102kg\times[/itex][itex]\frac{9,81m}{1s^{2}}[/itex][itex]\times1m[/itex] / 1s = 1 Watt. BUT, this works only if the duration is exactly 1 second. How do I make this happen? If I get around the problem and say I lift the thing in 1 second, now it's easy, I control that, but if I have to let gravity do the work, how can you control the duration? Then for watthour: Say I have a very very long string and the generator is up the highest cliff. I let the 102g mass go down for 1 hour. Will it produce 1 watthour of energy? thanks 



#21
Nov2712, 10:17 PM

P: 580

nasu,
Sorry, I misdirected my last post at you. Sorry, I have updated it. Kuin, I think you have it, your calculations look correct. You ask how would you control your gravityweightgenerator mechanism in order to have it produce at a steady rate of 1Watt? This would require some kind of speed regulator, the generator would have to be spinning at a constant rate. The real question is, why would you want/need to do this? In reality, the weight would eventually reach a terminal velocity where the reaction force from the spinning generator was equal to mg, but prior to this, the output rate in watts would not be constant. In other words, power is, or can be, a function of time in which case finding total energy would require integrating over time vs. multiplying by time. 



#22
Nov2712, 10:42 PM

P: 1,909

1 watthour is 3600 J. The weight of your mass is about 1 N. It has to go down 3600 m for the work of gravity to be 3600J or 1Whour. No matter how long it takes. Work= F*d See? no time in here! How fast it goes down determines the power (in watt) and not the energy in Watthour. 



#23
Nov2712, 11:48 PM

P: 100

1 joule is equal to 1 watt second. (Ws)
Does that help? I didn't bother to read all responses yet. 



#24
Nov2812, 01:03 AM

P: 5

So,
in the end, what I'm trying to do is to find a way to get a clear image of the energy used by a typical household over the course of 1 year. A north american typical household of 4 people uses around 10 000kwh every year. but it's hard to figure out what that means so I'm trying to convert that to something we can understand instinctively. I think weights moving up or down give the clearest 'picture' for that purpose. So I'm trying to figure out what 'size' of a weight would have to travel how high to 'work' 10 000kwh. As for the weight itself, water is a good pick, people can easily imagine what a liter of water 'feels' like, and it is also handy since 1 liter of water has a weight of 1kg. so, is this right: 10 000kwh = 36 000 000 000 Nm = 3 670 978 356 kgm so, 3 670 978 356 kg going down 1 meter would produce 10 000 kwh or 36 709 783 kg going down 100 meters so we need to picture a cube of water containing 36 709 783 liters of water going down 100 meter. cubic root of 36 709 783 is 332 liters(10cmx10cm,x10cm) of edge for the cube, that is 332 x 10cm = 3323cm = 33meters So a cube 33 meters of side, filled with water, going down 100 meters represents the energy usage of 1 north american household... 33 meters is about the height of a 10 story building. 100 meters would be a 30 story building. I must be wrong, this would be crazy! this seems a lot. is this right? 



#25
Nov2812, 06:21 AM

P: 821





#26
Nov2812, 07:09 AM

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P: 11,383

Mechanical Work involves very little energy compared with heating things up, in general. We seem not to weight them the same in our appreciation. A 1kW heater would not make a lot of difference to a house if run for an hour. A 1kW power tool could make a very significant difference in the same time.




#27
Nov2812, 07:36 AM

P: 1,909

We use an average 500kwh per month so it will be 6000 kwh per year. But it's a small house. However very little of this is used to move things around. I would say that at least 1/2 if not more is used to heat up things around, mostly water. Why not take sophiecentaur's hint and estimate how much water can be heated up from room temperature to around 7080 degrees, by using this energy?. It would give you a more meaningful image. With the energy equivalent of the heat required to bring the water in your coffee cup to boiling point you can lift the same water some 30 km. Did you ever worried in the morning about how "huge" is this? 



#28
Nov2812, 09:46 AM

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When I was at School, they taught us about the work of Joule. He came up with the concept of 'The Mechanical Equivalent of Heat', based on the heating effect of boring through the barrel of a gun and other such activities. I don't think Energy was seen in quite the same light at the time (Joule's day  no my school days) and the Joule (Work) was measured as equivlent to 4.2 Calories (Heat). This was when they finally realised that the old 'Caloric' theory of heat was nonsense and that heat could not just be treated as a fluid that flowed from place to place.
The calorie is such a small amount of heat that they went and confused everyone by introducing the kiloCalorie  which was then referred to as the Calorie, in the context of Food energy content. That factor of a thousand can often get lost in arguments about eating and exercise. 



#29
Nov2812, 02:59 PM

P: 5

You have the example of rubbing your hands together, but it's hard to get any precise idea of the amount of energy input and of heat output. Thanks again to everyone. 



#30
Nov2812, 04:24 PM

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If you are interested in experiments on converting mechanical energy to heat then there is much info about Joule's early work. It is worth doing the calculations around his experiment on temperature changes as water fell down a waterfall and it is very easy to work out the theoretical rise in temperature you would expect if all the Gravitational Potential Energy went into heating the water at the bottom. It certainly brings home the 'difference' between the two aspects of Energy. 



#31
Nov2812, 05:01 PM

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P: 22,007

I'm pretty sure the remark was just a comment on the fact that people lack intuition for visualizing energy quantities in reallife situations. We've seen time and time again the exact problem discussed above: that people are very surprised at just how little energy there is in lifting a weight vs doing other kinds of electrical work such as producing light or heat.




#32
Nov2812, 05:23 PM

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Yes I agree. I made the same point as you, a few posts back, that it's hard to reconcile the two energy forms. The only way to deal with these counterintuitive things is to do some actual sums and get an answer that can reassure you. There is loads of "accurate data" available.
One useful comparison is between doing light work outside in warm weather and doing the same work out in sub zero conditions. You need an awful lot more food to keep going in the cold. This link could be a way into this subject. It is very complex. 



#33
Nov2812, 07:19 PM

P: 5

hi,
For heat, it's already a little more difficult to grasp. I think, I might be wrong, that for most people, it is difficult to have a clear image of a 1 degree rise in temperature, more so for the electrical energy needed to produce it. The idea is to get super easy understandable examples that will trigger interest where there was none. my problem now, I realize, is that even with kgs and meters, when you deal with such large numbers, it becomes intangible again, it's impossible to represent those amounts. I will have to change scales first I guess, start with the 100W light bulb. Thanks again. 


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