A small energy harvester generator needs a mechanical work function of 9.5 Newtons of force moving through 1.5 millimeters distance.
The electrical output of this generator produces 120 microjoules. Questions: 1. How much mechanical energy, in microjoules, is produced to operate the generator by 9.5 Newtons 1.5 millimeters? 2. What is the efficiency of the system?

 

I assume this is homework.

You are given the mechanical work that is done. Question #1 is asking you to express that mechanical work in the same units that output is measured in. Once you know this quantity, can you not calculate the efficiency?

 

MOD NOTE: Moved to Homework Help.

 

I assume this is homework.

You are given the mechanical work that is done. Question #1 is asking you to express that mechanical work in the same units that output is measured in. Once you know this quantity, can you not calculate the efficiency?

 

I know how to calculate the efficiency but I am trying to get another opinion to convince someone that does not believe me. If you could give your opinion I would most appreciate it.

 

So what did YOU get? We'll confirm whether you are correct or not.

 

So what did YOU get? We'll confirm whether you are correct or not.

0.01425Nm input =14250 microjoules Efficiency = 120/14250 = .0084 X100 = 0.84%

 

That's the answer I get, but I must add that I'm uncomfortable with one aspect of the question. To get to this answer, we have to assume that one 'stroke' of the input mechanical energy produces the packet of output energy that is given. That's a reasonable assumption but it is not explicitly stated. The energy is just given as a total and not quoted per stroke or per unit of time.

 

0.01425Nm input =14250 microjoules Efficiency = 120/14250 = .0084 X100 = 0.84%

I agree.

What is the other person saying and what is there reasoning?

 

That's the answer I get, but I must add that I'm uncomfortable with one aspect of the question. To get to this answer, we have to assume that one 'stroke' of the input mechanical energy produces the packet of output energy that is given. That's a reasonable assumption but it is not explicitly stated. The energy is just given as a total and not quoted per stroke or per unit of time.

I agree that it's a reasonable assumption -- and about the only reasonable assumption that can be made with the information given.

 

It is a German company that wants to licence some of my energy harvesters to replace theirs since my patented generators are far superior in power out and duration time. See our UK website where we sell my products.I'm from the UK but live now in NY where I do my R&D work.
https://www.rfsolutions.co.uk/energy-harvesting-c6
They are saying that my generator is less efficient than theirs. WIth mine, it takes 1.2 Nm to push 5 mm and I get an output of 600 microjoules for ~200 msec. They get 120 uJ for 2 msec.

Thanks for the help,

David

 

It would seem to depend on what's important. You are generating 600 uJ / 200 ms for a power output of 3 mW. They are generating 120 uJ / 2 ms for a power output of 60 mW.

Your "1.2 Nm to push 5mm" can't be converted into work since Nm is a not a force, but energy. So you haven't given enough information about yours to calculate an efficiency number.

 

It would seem to depend on what's important. You are generating 600 uJ / 200 ms for a power output of 3 mW. They are generating 120 uJ / 2 ms for a power output of 60 mW.

Your "1.2 Nm to push 5mm" can't be converted into work since Nm is a not a force, but energy. So you haven't given enough information about yours to calculate an efficiency number.

Sorry I meant 1.2 Newtons and I just measured it again and the distance is 3.5 mm.

 

Sorry I meant 1.2 Newtons and I just measured it again and the distance is 3.5 mm.

It would seem to depend on what's important. You are generating 600 uJ / 200 ms for a power output of 3 mW. They are generating 120 uJ / 2 ms for a power output of 60 mW.

Your "1.2 Nm to push 5mm" can't be converted into work since Nm is a not a force, but energy. So you haven't given enough information about yours to calculate an efficiency number.

You are not saying that if I generate 600 ujoules for 200 msec, the 120 ujoules for 2 milliseconds produces more power?

 

You are not saying that if I generate 600 ujoules for 200 msec, the 120 ujoules for 2 milliseconds produces more power?

By definition, yes, the power is much higher for a 120µJ pulse lasting 2ms. The cumulative energy depends on the frequency of the pulses, which we don't know yet.

 

I'll trade you 600 for that 120.....how much you got?

 

You are not saying that if I generate 600 ujoules for 200 msec, the 120 ujoules for 2 milliseconds produces more power?

By definition, power is work per unit time, or the rate at which work is done. Energy (work) and power are not the same thing.

A typical candy bar liberates about the same amount of energy as a stick of dynamite. The energy is the same, but the power is not even comparable.

 

By definition, yes, the power is much higher for a 120µJ pulse lasting 2ms. The cumulative energy depends on the frequency of the pulses, which we don't know yet.

Their pulse is 15 Vp pulse for 2 milliseconds, mine is 23.4 Vp sine wave 200 milliseconds frequency is ~46 Hertz. With their generator, they can only send 1 radio telegram and with mine, we can send 5 telegrams, we use more info in our digitally encoded telegrams because we have a longer transmission.

 

Their pulse is 15 Vp pulse for 2 milliseconds, mine is 23.4 Vp sine wave 200 milliseconds frequency is ~46 Hertz. With their generator, they can only send 1 radio telegram and with mine, we can send 5 telegrams, we use more info in our digitally encoded telegrams because we have a longer transmission.

Not if their pulses are more frequent. What of they can run at 1kHz?

We are still just guessing at the overall application you are dealing with. Better questions get better answers.

 

By definition, yes, the power is much higher for a 120µJ pulse lasting 2ms. The cumulative energy depends on the frequency of the pulses, which we don't know yet.