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Help with power calculation please.

by billbarlow
Tags: calculation, power
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billbarlow
#1
Jan20-14, 12:10 PM
P: 4
I have designed and built a device that harnesses centrifugal force.The power output is in the form of
2 x 10mm steel rod "pistons" that move up and down a distance of 15mm. I have measured the weight these pistons can lift to be 122lbs when the device is operated @ 4000rpm. Each piston moves up and down once per revolution. I have calculated this to be an output of 1.45 hp. ( 15mm=.049125ft x 66.666 times per second=3.27496 ft x 122lbs = 399.54 lbs/ft/sec = .7264hp x 2 = 1.45hp. Can people please either confirm my calculations are correct or point out my error(s).

If I am correct,given that the device is solely powered by a 500w electric drill, am I able to claim the impossible?
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berkeman
#2
Jan20-14, 12:37 PM
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Quote Quote by billbarlow View Post
I have designed and built a device that harnesses centrifugal force.The power output is in the form of
2 x 10mm steel rod "pistons" that move up and down a distance of 15mm. I have measured the weight these pistons can lift to be 122lbs when the device is operated @ 4000rpm. Each piston moves up and down once per revolution. I have calculated this to be an output of 1.45 hp. ( 15mm=.049125ft x 66.666 times per second=3.27496 ft x 122lbs = 399.54 lbs/ft/sec = .7264hp x 2 = 1.45hp. Can people please either confirm my calculations are correct or point out my error(s).

If I am correct,given that the device is solely powered by a 500w electric drill, am I able to claim the impossible?
If you are putting 500W in, then you are not getting more than that amount of power delivered. Can you re-do your calcs in MKS units to double-check?
billbarlow
#3
Jan20-14, 12:45 PM
P: 4
Sorry dont know what MKS units are. 500w = .67hp though??

berkeman
#4
Jan20-14, 12:47 PM
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Help with power calculation please.

MKS = meters, kilograms, seconds.
sophiecentaur
#5
Jan20-14, 04:01 PM
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Quote Quote by billbarlow View Post
I have designed and built a device that harnesses centrifugal force.The power output is in the form of
2 x 10mm steel rod "pistons" that move up and down a distance of 15mm. I have measured the weight these pistons can lift to be 122lbs when the device is operated @ 4000rpm. Each piston moves up and down once per revolution. I have calculated this to be an output of 1.45 hp. ( 15mm=.049125ft x 66.666 times per second=3.27496 ft x 122lbs = 399.54 lbs/ft/sec = .7264hp x 2 = 1.45hp. Can people please either confirm my calculations are correct or point out my error(s).

If I am correct,given that the device is solely powered by a 500w electric drill, am I able to claim the impossible?
You have not defined the output power - or measured it. Moving weights up and down does not imply doing net work. Consider a spinning wheel. One side goes up and then it goes down the other side. No energy is used up at all (except for a bit of friction). Nothing is wrong with Physics. Perpetual motion does not exist.
billbarlow
#6
Jan20-14, 05:13 PM
P: 4
Thank you Sophiecentaur. Is there a way I can measure the power output? Any help much appreciated.
sophiecentaur
#7
Jan20-14, 05:36 PM
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Quote Quote by billbarlow View Post
Thank you Sophiecentaur. Is there a way I can measure the power output? Any help much appreciated.
You seem to be a bit confused about this machine you have made and about some basics of mechanical work and power.
At the moment, there is no power output; you are just making the weights go up and down and 'wasting' some energy, inside it. The machine is doing no useful or identifiable work (e.g. lifting something permanently / pumping water etc.). Moving weights up and down does not constitute a net amount of work except where there are losses such as friction. The descending weights are returning Potential energy to the machine as they fall. If the motor is actually working flat out then you are dissipating 500W in there somewhere as heat. It is unlikely that the motor is actually producing 500W - that is just its maximum rated load because 0.5kW would pretty soon make the device pretty warm. You could measure the current supplied to the motor and that would give a clue as to the actual Power input (VA, actually, rather than Power).
To measure true output power, it is possible to measure power delivered to a shaft by using a 'brake' to measure torque and a timer to measure revs. This will give the output ('the good old Brake Horsepower', of car engine designers). See this link

BTW, what do you mean by "harness centrifugal force"? Centrifugal force is not energy and it is a term which doesn't really apply here, in any case. It is only present in a rotating reference frame.
billbarlow
#8
Jan23-14, 12:37 PM
P: 4
sophiec. Many thanks for your long and detailed reply. I have spent the last two days racking my brains trying to understand why you say what you say. I note from your profile you have a physics degree so I would be a fool not to listen you, but could not comprehend why you say the device has no power output. I think I now have the answer!

My guess is that you assume the pistons are connected to each other by a crankshaft or some other mechanism and the weights are acting as they would on a see-saw? That is not the case. The pistons move totally independently and are not connected in any way. Does this change your opinion of work being done?

The pistons are pushed up and down by centrifugal force acting on them.

Many thanks again.


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