Calculating the force required to generate 100 kilowatt per hour

[Jaxodius]

Hello to everyone,

I am trying to calculate how much physical force is needed at point B (imagine a person pushing on a bar attached to point A) to generate 100 kw per hour where the radius is 50 meter. If you imagine that at point A there is a system of gears attached to an alternator/generator which transforms 1 rpm from point B into 3600 rpm at point A(4 gears with a gear ration of 15:1 between them), how much force would be needed to generate 100 kilowatts per hour? I understand there will be friction, but at the moment i am just trying conceptualize how much physical work is needed to generate that amount of power.

This is the first time i am trying something like this so if i have made any wrong assumptions, please correct me.

 

[Andrew Mason]

I am trying to calculate how much physical force is needed at point B (imagine a person pushing on a bar attached to point A) to generate 100 kw per hour where the radius is 50 meter. If you imagine that at point A there is a system of gears attached to an alternator/generator which transforms 1 rpm from point B into 3600 rpm at point A(4 gears with a gear ration of 15:1 between them), how much force would be needed to generate 100 kilowatts per hour? I understand there will be friction, but at the moment i am just trying conceptualize how much physical work is needed to generate that amount of power.

You need to give us more information.

Power is work done per unit time. So 100 KW is 100 KJ per second = 100,000 Nm/sec.

In order to determine how much force is required to generate such power, you have to tell us how fast you want to make the bar rotate. For example, if you want to generate 100 KW at a rotation rate of 1 revolution/second (ie. B moves 2∏R = 314 m. in one second) then you would need a force of 100,000/314 = 318 N.

AM

 

[Jaxodius]

Thank you Andrew. you raise a very good question:

[1] i am not sure weather to generate 60 kilowatts per hour ( i am simplifying the numbers so i can understand easier but my actual project is for 100 kilowatts per hour ), if one needs to generate 60 kw per second or 1 kw per minute?

[2] Also, the way i visualize it in my mind, is that a person will have to be applying 318 Newtons per meter for 314 meters to make 1 revolution. is this correct? if so, do i use the 60kw/sec or 1kw/min to calculate the force i would need.

 

[Nugatory]

You are confusing yourself by talking about "kilowatts/sec" and "kilowatts/min"; those terms make no more sense than something like "miles per hour per day". Power, measured in watts, is not an amount of work but the rate at which that work is delivered: one Joule of work per second is one watt.

Your problem is asking for 100 kw, meaning that it will deliver 100,000 Joules/sec for as long as it is operating, and it will take some force in Newtons to do this. Obviously the longer it operates the more total joules it will deliver, but the force in Newtons will be the same; it'll just be operating for a longer time.

 

[Jaxodius]

Thank you Nugatory. I think i really am confusing myself more. let me explain what it is i want to do. i have an oven that is rated at 100 kWh and it need to run for 3 hours. I am trying to determine how much force would be needed to provide 100 kWh to that oven using the set-up i mentioned in my first post ( 4 gears with 15:1 gear ratio attached to a bar with the length of 50 m ).

Clearly i cannot do 314 m in 1 second so i was thinking what if i added more gears so that instead of providing the 318 N force per 1 revolution of point B, i provide it over 1/60 of that one revolution. Therefore :

3600 rpm * 60 = 216000 rpm . If i used 6 gears with ratio of 15:1 between 5 of them and 5:1 between 1 of them making it a total of 253125:1 and then proceeded to make 1 revolution from point , will i be providing the same power every second of that revolution? I know i cannot do 314 meters in 1 minute either lol but am i thinking about this the right way? or is there another way to spread that much force and still provide 100 kw/s?

 

[Nugatory]

Thank you Nugatory. I think i really am confusing myself more. let me explain what it is i want to do. i have an oven that is rated at 100 kWh and it need to run for 3 hours. I am trying to determine how much force would be needed to provide 100 kWh to that oven using the set-up i mentioned in my first post ( 4 gears with 15:1 gear ratio attached to a bar with the length of 50 m ).

You're still confusing power (watts) and energy (joules) here. 100 kWh is a measure of energy. It's the amount of energy a 100 kw power source delivers in one hour, or a 10 kW source delivers in 10 hours, or a 1 kW source delivers in 100 hours. If you remember that there are 3600 seconds in an hour, and one watt is one joule per second, you'll see that all of the above combinations come out to the same amount of energy, 3.6x10⁸ joules, whether it's a little power delivered over a long time or a lot of power delivered over a short time.

(100 kW is also a LOT of power... It's about 130 horsepower, and I suspect that you're misreading the specs of your oven).

3600 rpm * 60 = 216000 rpm . If i used 6 gears with ratio of 15:1 between 5 of them and 5:1 between 1 of them making it a total of 253125:1 and then proceeded to make 1 revolution from point , will i be providing the same power every second of that revolution? I know i cannot do 314 meters in 1 minute either lol but am i thinking about this the right way? or is there another way to spread that much force and still provide 100 kw/s?

Yes, you're thinking about it the right way now: as long as you're maintaining a constant speed and force you're delivering a constant power. But notice that this doesn't change if you play with the gearing or the length of the lever arm; all that does is make it so you're applying the same force over a different distance and (because your speed is constant) a different time, so applying the same amount of power for a longer time and doing more total work.

You're also quite right that you can't do 314 meters in one second - as I said above, 100 kW is a lot of power.

 

[Jaxodius]

Thank you for your prompt and very helpful reply Nugatory. After doing some excel calculations i think i understand it enough to start with some practical experimenting.

Thank you both for your help. I consider this problem solved.

 

[sophiecentaur]

Thank you for your prompt and very helpful reply Nugatory. After doing some excel calculations i think i understand it enough to start with some practical experimenting.

Thank you both for your help. I consider this problem solved.

I'm afraid the problem will not be solved until you understand what the real units of power and energy are. Under even the mildest of scrutiny, what you are writing makes no sense. It would not take too much of your time to learn how to use the appropriate units and it would help so much more for your credibility.
It really does matter.

 

[russ_watters]

Just for emphasis:
There is no such unit as "kw per hour" or "kw per second". I don't believe you understand that yet either, so I'm relatively certain you have not calculated anything of value yet.

 

[Jaxodius]

Sophie please let me know if my understanding of power and energy is still wrong:

Example: when i use a 1kw AA cell in my phone that has the ability to use either 1 watt or 1000 watts every second it is turned on, the 1kw AA cell will either last 1000 seconds or 1 second depending on the amount of energy it draws. So the power in the 1kw AA cell can last depending if the energy drawn by my phone is 1 watt / sec or 1kw/sec.

Power= 1 Kw and Energy= 1kw/s or 1w/s

I spoke to nugatory in pm after his last message because i was getting embarrassed and wanted a clarification. it was after his reply in pm, which clarified things a bit more, that i made my last post.

Russ_watters you wrote:

There is no such unit as "kw per hour" or "kw per second"

I want to quote you something from http://physics.nist.gov/cuu/pdf/sp811.pdf page 24 which reads:

Reference [4: ISO 31-0] suggests that if a space is used to indicate units formed by multiplication,
the space may be omitted if it does not cause confusion. This possibility is reflected in the
common practice of using the symbol kWh rather than kW · h or kW h for the kilowatt hour.

So are you wrong or are you talking about something other than what is mentioned on that page? because when i use the term kwh, i mean x kw per hour. please clarify.

As for not calculating anything of value, i was able to calculate how many people i would require to use x amount of force over x amount of distance to make the whole thing realistic. so for me, it was very valuable.

 

[QuantumPion]

Sophie please let me know if my understanding of power and energy is still wrong:

Example: when i use a 1kw AA cell in my phone that has the ability to use either 1 watt or 1000 watts every second it is turned on, the 1kw AA cell will either last 1000 seconds or 1 second depending on the amount of energy it draws. So the power in the 1kw AA cell can last depending if the energy drawn by my phone is 1 watt / sec or 1kw/sec.

Replace every instance of watt with joule and it is mostly correct.

I want to quote you something from http://physics.nist.gov/cuu/pdf/sp811.pdf page 24 which reads:

So are you wrong or are you talking about something other than what is mentioned on that page? because when i use the term kwh, i mean x kw per hour. please clarify.

Understand that kW*hr (kilowatt-hours) is not the same as kW/hr (kilowatt PER hour). One kilowatt-hour is a unit of energy, and is equal to 3600 kilojoules (1 kW * 1 hr = 1 kW*hr). Kilowatt/hour represents something entirely different. Physically, a kilowatt per hour would be a rate of change of power, for example the ramp rate of a nuclear reactor (which typically have limits on how fast they are allowed to change the power level due to material constraints).

 

[willem2]

Sophie please let me know if my understanding of power and energy is still wrong:

Example: when i use a 1kw AA cell in my phone that has the ability to use either 1 watt or 1000 watts every second it is turned on, the 1kw AA cell will either last 1000 seconds or 1 second depending on the amount of energy it draws. So the power in the 1kw AA cell can last depending if the energy drawn by my phone is 1 watt / sec or 1kw/sec.

Power= 1 Kw and Energy= 1kw/s or 1w/s

The unit of power is indeed Watt (W), or kW = 1000 Watt.
The units of energy is Joule or Watt-second and NOT Watt per second.

A You gett one Watt-second by producing one watt for a second. (or 2 watts for a half second etc.)
You have energy = power * time, E = P T here, and because the units of power is watt, the unit of time is seconds, the unit of energy must be watt second.

You use "per second" in a unit for a rate of change, for example speed,
wich has unit meters per second. you have speed = distance/time, we divide here instead of multiplying, and that's why you have m/s or meters per second, and NOT meter-second.

The capacity of batteries is usually given in Ampere-hour, If you multiply this by the voltage you get watt-hours. You're talking about a 1 kW AA cell as if this is the capacity of the battery, but kW is a unit of power. 1 kW would drive a vacuum cleaner, and is a few hunded times the maximum power a phone would use, or that an AA battery could deliver.

 

[sophiecentaur]

Sophie please let me know if my understanding of power and energy is still wrong:

Example: when i use a 1kw AA cell in my phone that has the ability to use either 1 watt or 1000 watts every second it is turned on, the 1kw AA cell will either last 1000 seconds or 1 second depending on the amount of energy it draws. So the power in the 1kw AA cell can last depending if the energy drawn by my phone is 1 watt / sec or 1kw/sec.

Power= 1 Kw and Energy= 1kw/s or 1w/s

I spoke to nugatory in pm after his last message because i was getting embarrassed and wanted a clarification. it was after his reply in pm, which clarified things a bit more, that i made my last post.

Russ_watters you wrote:



I want to quote you something from http://physics.nist.gov/cuu/pdf/sp811.pdf page 24 which reads:



So are you wrong or are you talking about something other than what is mentioned on that page? because when i use the term kwh, i mean x kw per hour. please clarify.

As for not calculating anything of value, i was able to calculate how many people i would require to use x amount of force over x amount of distance to make the whole thing realistic. so for me, it was very valuable.

I think it would help if, instead of hanging on to your particular version of this, you started way back at the beginning and stuck with the definition of Energy and Power that everyone uses. Some of the figures you are calculating are correct (by chance, because some errors are sort-of cancelling out) but your thesis is just flawed. If you cannot see what you are doing wrong then you have to believe that you are wrong and find out, for yourself, what the problem is.

Start with the definition that Energy is Power TIMES time and that Power is Energy PER unit time.

 

[russ_watters]

Jax, a kWh is kw TIMES hours, not divided by hours. So you actually have two errors here, not one:
1. You don't know what a kWh is.
2. You shouldn't be using it anyway.

Just about everything you've posted should be about power, not energy. kW - just kW - not kWh, not kw per hour, not kWh per hour. Just kW.

Now perhaps you got lucky and your errors canceled each other out, but it is more likely you got the wrong answer.

 

[CWatters]

Just for info... What will be applying force on the bar?

i believe a fit human can generate about 1/3rd of a horsepower (eg 250W) for sustained periods. If you need 100kW that's quite a team of people you will need (400 people).

I read somewhere that to produce one horsepower (750W) you actually need 1.5 real horses (so 200 needed). I've no idea if they can keep that up for three hours though.

PS: Insulate the oven better. They will thank you for it.

 

[Jaxodius]

Thank you sophie, i will keep that in mind.

CWatters i was planning on hiring people to do it because it will only be for 3 hours. and you are right it will be about 400 people. I was thinking that maybe i can divide them into groups and 1 group takes over another.

Honestly, hiring people is not a problem. i am more concerned about the 6 gears with a ratio of 15:1 between them. is a gear system like this even realistic? what kind of problem can arise?

 

[QuantumPion]

What is the purpose of your giant human-powered generator? Hiring 400 people for 3 hours at $10/hr would cost $12,000. I did some googling and found that you could rent a 100 kW generator for a day for $500.

 

[Jaxodius]

It does not take $10/hr to hire someone in china or India :)

Do you think the gears will hold up?

 

[jeffrey c mc.]

Jaxodus;

From just a mechanical perspective. A 50 meter lever seems somewhat excessive. And your discussion of running an oven, and then running a cell phone with a 1kw AA battery does not add to the clarity at all. Besides it would be more instructive if you first told us about the generator, and the rating of same. Irregardless of the gears and the lever used to drive the generator, the power availability would depend on the capabilities of the generator itself. Whether or not the lever and gears can deliver the necessary torque to the generator does not depend on the weather. Sorry, as I have been taken to task for my language and use of same, I may as well make my own observations. Nothing personal!

 

[Jaxodius]

Thank you jeffery. and please do not hesitate to point out any wrong assumptions i have made (as i said in my first post ). this will be my first attempt at anything like this and i would like to get it right. and to clarify, the project is for 100 kw-second ( thank you willem2 for clarifying this ) but i gave the example of the cell phone just for sophie. i know there is no 1kw AA battery lol

Besides it would be more instructive if you first told us about the generator, and the rating of same.

This is the item i had in mind to attach the gear to at point A ( diagram in the first post ). only instead of using fuel, i wanted to calculate the force required to do it manually:

http://www.engine-trade.com/product/cummins-200kw-diesel-generator-set-for-landuse.htm

It has all the details. i would post them here but it would look like a huge spam. at the moment i am not even sure how i would attach the gears to this but it cannot be absolutely impossible.

 

[CWatters]

A 50 meter lever seems somewhat excessive.

I think it's going to be a challenge connecting 400 people to a shaft in such a way that they can all deliver 250W. You need to do more research because I think 250W is the figure for cycling at near ideal cadence (eg most comfortable pedal rpm) and it might be a lot lower for pushing or pulling on a lever.

Worth watching...



Took 78 cyclists to power an electric shower which is probably 8-10kW.

 

[CWatters]

If you assume 400 people can be connected to a 50m lever and are jogging along at 5mph somehow delivering 250W each then the angular velocity of the shaft works out at about 0.044 radians/second and the torque about 2.2 * 10^6 Newton meters. Nearest thing I can think of to something like that is probably a wind turbine but they rotate slightly faster and lower torque.

 

[russ_watters]

...and i would like to get it right. and to clarify, the project is for 100 kw-second...

I doubt it. That's 100 kW for 1 second or 1 kW for a minute and a half. About as much energy as having a lightbulb on while you eat dinner.

It sounds from other posts (as others have gleaned) like you have a 100 kW load that you want to power for 3 hours. That's 300 kWh.

I think the estimates other people are giving on the amount of power you can extract from humans is too high because you aren't talking about athletes but normal people. I expect you'll need at least a thousand.

 

[cjl]

I doubt it. That's 100 kW for 1 second or 1 kW for a minute and a half. About as much energy as having a lightbulb on while you eat dinner.

It sounds from other posts (as others have gleaned) like you have a 100 kW load that you want to power for 3 hours. That's 300 kWh.

I think the estimates other people are giving on the amount of power you can extract from humans is too high because you aren't talking about athletes but normal people. I expect you'll need at least a thousand.

In addition, the 100kW number sounds extremely high - unless it's some kind of industrial oven or similar, chances are that the real power draw is much smaller.

 

[Jaxodius]

Thank you for the video CWatters. you said:

I think it's going to be a challenge connecting 400 people to a shaft in such a way that they can all deliver 250W

I was thinking of having 4 arms/levers attached to point A in such a manner that they form a plus sign. then i can have 50 people per arm. Infact, as the discussions have progressed, something else comes to mind. what if the 250w per person was directed to raise an object which was 5000 kg / m3, such as hematite ( iron ore which is 5095 kg/m3 ) and had 10 m3 of it raised to 100 meters ( it does not strictly have to be raised above ground, it could just as well be pulled up from a shaft dug 100m into the ground. This is the calculation i made. please tell me if it is wrong:

10 m3 blocks of hematite ( 1 on top of another ) = 50000 kg m3

(so 50000 (weight) * .028 m/s (to allow the blocks to fall in 1 hour as 100 meter / 3600 seconds ) * 100 (height)) / 1000 (to convert it into kw ) = 140 kw-s.

Is this correct? using this, for one hour i would be getting 140 kw-s for every second the weight falls?

russ_watters you said:

It sounds from other posts (as others have gleaned) like you have a 100 kW load that you want to power for 3 hours. That's 300 kWh.

That is correct.

I think the estimates other people are giving on the amount of power you can extract from humans is too high because you aren't talking about athletes but normal people. I expect you'll need at least a thousand.

my understanding is that, if i increase the length of the lever to 100 meters, i would have not have to use the same force but only work longer. honestly, as long as i get 300 kwh to discharge every 24 hours, i do not care if it takes 8 hours to get this done.

cji you are correct, it is an industrial oven.

 

[jeffrey c mc.]

Just for emphasis:
There is no such unit as "kw per hour" or "kw per second". I don't believe you understand that yet either, so I'm relatively certain you have not calculated anything of value yet.

Well, I don't know about you, but I pay a certain amount of cash for KW hours. When I have used a KW in an hour, I pay for the going rate of a 'KW hour. A sixty watt light bulb uses sixty watts every hour. This system for AC use is dependent on the specific cyclic rate of the current. Which, where I use AC is 60 hertz. It's a good thing for AC timepieces also, as in clocks. In fact a watt hour meter is just that, a clock that instead of having hour and minute hands, has a series of dials that are base ten, I believe--at least the old ones were before the move to digital.

In electricity; Power, is a function of the induction and the energy potential. Also, Energy potential is a function of induction and resistance. Don't bother refuting Mr Ohm sir. I did not make the preceding up in a fit of Mania, or Asberger-ity either.

 

[russ_watters]

my understanding is that, if i increase the length of the lever to 100 meters, i would have not have to use the same force but only work longer. honestly, as long as i get 300 kwh to discharge every 24 hours, i do not care if it takes 8 hours to get this done.

24 hours? 8 hours?

1. How are you going to store the energy if the oven is only running for 3 hours?
2. It still means multiple shifts. Still a larger number of people, just not all at the same time.

 

[russ_watters]

Well, I don't know about you, but I pay a certain amount of cash for KW hours. When I have used a KW in an hour, I pay for the going rate of a 'KW hour. [emphasis added]

You don't use "a kW in an hour", you use a kW for an hour to get a kWh.

A sixty watt light bulb uses sixty watts every hour.

No it doesn't. It uses 60 Watts all the time, or 60 watt-hours every hour.

 

[jeffrey c mc.]

Russ; yes I think I see my error. Thanks for the correction. In the case of the light-bulb perhaps I should say enlightenment. Oh yeah, I went there, I did.

 

[Dale]

This is completely silly. If it made any kind of economic sense to generate power this way then power companies would be generating it this way.

A kWh costs about $0.16 and since a human's maximum sustained output is 250 W that is at least 4 man hours. Since nobody will work for anything even remotely approaching $0.04 per hour (not even in China or India) it will cost you far more to make power this way than to simply buy it.

 

[gmax137]

This is completely silly...

... Therefore :

3600 rpm * 60 = 216000 rpm . If i used 6 gears with ratio of 15:1 between 5 of them and 5:1 between 1 of them making it a total of 253125:1 ...

This is completely impractical as well.

 

[CWatters]

Jaxodius - The oven will probably only use 100kW when the element is actually on. Once upto temperature the thermostat will turn the element on and off so that the average power supplied matches the losses through the walls of the oven. So your first priority should be to determin actual power consumption and reduces losses by insulating the oven as much as possible. Once upto temperature the losses might be a lot less than 100kW?

 

[Jaxodius]

Dale you said:

This is completely silly. If it made any kind of economic sense to generate power this way then power companies would be generating it this way.

i understand that you and others here think that the whole thing is silly. i understand that all the power companies in the world think it is silly too. that is ok. i still want to try it out and i need some help with the maths of it all. maybe the first day i try it out it will all just go poof in my face. and that is fine too. but i still want to try it out.

gmax137 you said:

This is completely impractical as well.

Can you please clarify how? i have asked this question in my previous posts as well but i was told by one user that i should not worry about the gears but the force being applied.

CWatters ( as in sea water? ) you said:

So your first priority should be to determin actual power consumption and reduces losses by insulating the oven as much as possible

The oven will already be insulated. so i am not too concerned with that at the moment. i just want to be able to get to a point where i can atleast make an attempt at something like this.

russ you said:

24 hours? 8 hours?

yes for me it is acceptable if 1/3 of the whole day is lost as long as i have that power by the end of that 8 hours to discharge for 3 hours every day.

1. How are you going to store the energy if the oven is only running for 3 hours?

from post # 25 :

Infact, as the discussions have progressed, something else comes to mind. what if the 250w per person was directed to raise an object which was 5000 kg / m3, such as hematite ( iron ore which is 5095 kg/m3 ) and had 10 m3 of it raised to 100 meters ( it does not strictly have to be raised above ground, it could just as well be pulled up from a shaft dug 100m into the ground. This is the calculation i made. please tell me if it is wrong:

10 m3 blocks of hematite ( 1 on top of another ) = 50000 kg m3

(so 50000 (weight) * .028 m/s (to allow the blocks to fall in 1 hour as 100 meter / 3600 seconds ) * 100 (height)) / 1000 (to convert it into kw ) = 140 kw-s.

is the maths correct here?

 

[DEvens]

Before you put an advert on Craigslist for 400 people to push your oven, you really should be sure you've got something vaguely correct.

Read the label on your oven. Does it really say 100 kW? The electric circuit on a house is typically not bigger than 200 amps. At 240 volts and 200 amps that's only 48 kW. So you've got something more than twice the power of a fairly hefty house electric service. Whatever are you putting in there? Are you sure it isn't 10kW?

If you are using 400 people that's about 250 watts per person. That's a respectable amount of power to expect a person to continue to put out for three hours, even if they can do it at their most efficient output level.

If you are turning an electric generator, you need to consider efficiency of the generator. Often they are only about 30% efficient. That means, if you put 100 watts of work in, you get only about 30 watts of power out. The rest goes to heating the generator. So you won't need 400 people you will need 1330.

A 100 kW generator is pretty big. Probably need at least a pickup truck to carry it.

This entire project is sounding pretty silly.

 

[QuantumPion]

Dale you said:



i understand that you and others here think that the whole thing is silly. i understand that all the power companies in the world think it is silly too. that is ok. i still want to try it out and i need some help with the maths of it all. maybe the first day i try it out it will all just go poof in my face. and that is fine too. but i still want to try it out.

The reason why we think it is silly is because the only way you could have a net gain of energy is if you have slave labor that you are literally starving to death. Exerting 250 W for 3 hours is 2,700,000 J of energy. In food terms, this is 650 kcal per person, or about 1 hamburger. So let me get this straight - you are going to hire 1000 people and buy them 1000 hamburgers so you can run an oven for 3 hours? Why not just take the 1000 hamburgers and burn them in a furnace? You'd get more heat energy that way compared to all the losses due to human metabolism losses and generator efficiency losses in your scheme.

 

[jeffrey c mc.]

[One watt is the rate at which work is done when an object's velocity is held constant at one meter per second against constant opposing force of one Newton.
The unit, defined as one joule per second, measures the rate of energy conversion or transfer.] From Wiki-media.

So I was thinking in mechanical perspectives. If I have two objects both massing the same, say 5 kg. One is at rest, the other is in motion on a trajectory that will cause it to strike the stationary object. The surface(s) that impact will be considered to be flat, have the same area, and be parallel, and aligned, when they strike. No application of resistance, or opposition will be considered. If the moving object is moving at one meter per second, when it impacts the stationary object, it will cease moving, and the impacted object will begin moving at one meter per second. Please correct this statement if not correct.

I realize that in the definition from wiki, the opposition of one Newton is integral in the definition. Yet since the opposing force; the stationary object; equates with the moving object, I believe this accounts for the force in my textual description

Jeffrey.

 

[sophiecentaur]

[One watt is the rate at which work is done when an object's velocity is held constant at one meter per second against constant opposing force of one Newton.
The unit, defined as one joule per second, measures the rate of energy conversion or transfer.] From Wiki-media.

So I was thinking in mechanical perspectives. If I have two objects both massing the same, say 5 kg. One is at rest, the other is in motion on a trajectory that will cause it to strike the stationary object. The surface(s) that impact will be considered to be flat, have the same area, and be parallel, and aligned, when they strike. No application of resistance, or opposition will be considered. If the moving object is moving at one meter per second, when it impacts the stationary object, it will cease moving, and the impacted object will begin moving at one meter per second. Please correct this statement if not correct.

I realize that in the definition from wiki, the opposition of one Newton is integral in the definition. Yet since the opposing force; the stationary object; equates with the moving object, I believe this accounts for the force in my textual description

Jeffrey.

If the objects are perfectly elastic, all the momentum of the moving one will transfer to the originally stationary one (they 'exchange' momentum). If they coalesc, then they will move off together at half velocity and half of the original Kinetic Energy is lost. Momentum is always conserved but the kinetic energy is not, except for elastic collisions. In your scenario, and in nearly all collision problems, it is not a good idea to try to consider 'forces' when all you want to know is the before and after situations. This is because there are a whole lot of possible forces which will provide the same end result (depending upon the elastic constants of the objects). It's better to use the Momentum approach, which doesn't need that sort of detail but still gives the answer.

 

[jeffrey c mc.]

I see what you mean, after reviewing the terms of 'elastic' and 'coalesc', I realized it was not right to ignore the properties of the two objects, in question. As, in the case of two rubber masses, as compared to two steel bearings. They both bounce, yet the steal bearing, even with being extremely hard and not as easily seen as elastic, would transfer its' momentum in a manner which would transfer more of the kinetic energy of the moving one, to its' counterpart, while two rubber balls would deform more, so there would be a slower transfer of kinetic energy, which would provide more coalesc-ing; which would transfer less of the moving objects inertia, to the stationary object. At least that is what I understood about your discussion, and review of the terms you used.

 

[Jaxodius]

what if the 250w per person was directed to raise an object which was 5000 kg / m3, such as hematite ( iron ore which is 5095 kg/m3 ) and had 10 m3 of it raised to 100 meters ( it does not strictly have to be raised above ground, it could just as well be pulled up from a shaft dug 100m into the ground. This is the calculation i made. please tell me if it is wrong:

10 m3 blocks of hematite ( 1 on top of another ) = 50000 kg m3

(so 50000 (weight) * .028 m/s (to allow the blocks to fall in 1 hour as 100 meter / 3600 seconds ) * 100 (height)) / 1000 (to convert it into kw ) = 140 kw-s.

is the maths correct here? how would i calculate how long it would take for that load to be raised up to 100 meters?

 

[jbriggs444]

is the maths correct here? how would i calculate how long it would take for that load to be raised up to 100 meters?

No.

You are referring here to:

10 m3 blocks of hematite ( 1 on top of another ) = 50000 kg m3

10 blocks times 1 cubic meter per block times 5095 kg per cubic meter = 50950 kg.

The number was acceptable, but the units were wrong.

(so 50000 (weight) * .028 m/s (to allow the blocks to fall in 1 hour as 100 meter / 3600 seconds ) * 100 (height)) / 1000 (to convert it into kw ) = 140 kw-s

That 50000 is not a weight. It is a mass. It is not in units of Newtons. It is in units of kilograms.

The weight of a mass is the mass times the acceleration of gravity. The acceleration of gravity on the surface of the Earth is approximately 9.8 meters per second per second.

50000 kg times 9.8 meters/second² = a weight of 490,000 kg-meters/sec² = 490,000 Newtons. Call it 500,000 Newtons.

100 meters divided by 3600 seconds = .028 meters per second velocity. That was correct.

Power = force times velocity.

500,000 Newtons times 0.028 meters per second = 14,000 Newton-meters/second = 14,000 Watts.

You double-dipped on height, multiplying by the 100 meters once to get your estimate of 0.28 meters per second fall rate and then again by that same factor of 100 for no reason.

You presented the result as 140 kilowatt-seconds. That is a unit of energy. Both the number and the units were incorrect.

You really need to keep track of units in your calculations.

 

[Jaxodius]

Thank you jbriggs and also for pointing out he errors in my calculation. I realize i keep mixing the units. maybe i need to rethink the whole thing.

Thank you to everyone for helping.

 

[tomfy]

I just wanted to comment that a unit like 'kW per hour' is not meaningless, although it is apparently being used incorrectly by jaxodius. As an example of what this unit (kW per hour) would mean, the world produces electric power at a certain rate - something like 2.2 TW in 2008 (or equivalently 2 billion kW. In the year 2000 this rate was more like 1.6 TW, so it has increased by 0.6 TW ( or 600,000,000 kW) in 8 years. 8 years is about 70,000 hours, so the rate of increase of electric power production is
600,000,000 kW / 70,000 hours, or
8600 kW/hr. This is roughly one new large electric power plant (generating 1 gigawatt, i.e. a million kW) every 4 or 5 days.
The situation is very similar to the unit meters/sec/sec (meters per second per second or meters per second squared) which is a unit of acceleration. If you drop a stone near the surface of the Earth it accelerates downward with an acceleration of 9.8 meters per second per second; when you first let go of it its speed is zero, but a second later it will be moving downward with a speed of 9.8 meters per second.

 

[russ_watters]

I'll grant you that, tomfy, but it is pretty rare (and not applicable here). Good point though. And welcome to PF!