Calculating the force required to generate 100 kilowatt per hour

[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!