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In summary, the equation for calculating flywheel inertia is I=1/2(mass*radius²), but it is more complex than that.

- #1

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- #2

NascentOxygen

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Is this a homework question? If so, then I'm sure you were provided with more details than you have shown here.

If not a homework question, then how has it arisen?

- #3

Emna

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I'm trying to work on a project, a system that helps minimize the electricity consumption . for example using a 1kw motor , few belt pulleys, flywheel and a generator you can get a 4 to 5KW of electricity for the daily consumption.NascentOxygen said:

Is this a homework question? If so, then I'm sure you were provided with more details than you have shown here.

If not a homework question, then how has it arisen?

- #4

Nidum

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Emna said:for example using a 1kw motor , few belt pulleys, flywheel and a generator you can get a 4 to 5KW of electricity for the daily consumption.

Not going to work . You can't get more power out of a system than you are putting in . Generally you get less useful power out than you put in .

- #5

malemdk

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You cannot get more power from than the input power, it's fundamental law of nature , law of energy conservation.

Efficiency of any machine always less than 100%,so the output from machine will be less than your input

- #6

Khashishi

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Well now, that depends on exactly what the Emna means to do. If you only need 5kW for a short time, you can use the 1kW to spin up the flywheel, then turn on the generator only when you need the burst of power. You can certainly get more power out than you put in; just not continuously.Nidum said:Not going to work . You can't get more power out of a system than you are putting in . Generally you get less useful power out than you put in .

I=1/2(mass*radius²) is correct for a point mass or ring. For a 3D solid object, you can think of it as being built out of a bunch of point masses. Sum or integrate over each point that makes up the object.

##I = \frac{1}{2} \int \rho r^2 dV## where ##\rho## is the density (as a function of position).

- #7

russ_watters

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That equation only works for a solid, uniform wheel. For a more complicated shape you may be able to take it in simple pieces and calculate the moment of initial of them individually.Emna said:Hi,

I couldn't find how to calculate this flywheel inertia , any help please?! I know that I=1/2(mass*radius²) but it seams more complex then that!

- #8

Emna

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you can increase the angular speed by adding belt pulley and the flywheel will increase the torque as much as you add mass to it!Nidum said:Not going to work . You can't get more power out of a system than you are putting in . Generally you get less useful power out than you put in .

- #9

Emna

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well I'm going to recycle a flywheel of a bicycle, it's made of some kind of plastic an add it some mass like it shown on the picture, i don't know how to find the density of the existed flywheel!Khashishi said:Well now, that depends on exactly what the Emna means to do. If you only need 5kW for a short time, you can use the 1kW to spin up the flywheel, then turn on the generator only when you need the burst of power. You can certainly get more power out than you put in; just not continuously.

I=1/2(mass*radius²) is correct for a point mass or ring. For a 3D solid object, you can think of it as being built out of a bunch of point masses. Sum or integrate over each point that makes up the object.

##I = \frac{1}{2} \int \rho r^2 dV## where ##\rho## is the density (as a function of position).

- #10

Emna

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I need to do calculate it as a proof for the jury members, and also i need it so i see how much the torque became and depending on the torque needed for the generator i ill know how much mass i should add to the flywheel so the system works perfectlymalemdk said:

You cannot get more power from than the input power, it's fundamental law of nature , law of energy conservation.

Efficiency of any machine always less than 100%,so the output from machine will be less than your input

- #11

Emna

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- #12

russ_watters

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Adding mass to a flywheel does not increase the torque on a constant speed system. A flywheel doesn't have any effect on a constant speed system besides adding friction and loss.Emna said:you can increase the angular speed by adding belt pulley and the flywheel will increase the torque as much as you add mass to it!

- #13

russ_watters

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By jury, you mean people grading you for a school project? Please stop and pick another project because if you present this one with a wrong conclusion you will likely get a bad grade.Emna said:I need to do calculate it as a proof for the jury members...

- #14

Emna

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I'm looking for another opportunity too I'm not liking it at all, i just got an answer from an industry today to work with them on a project , I have an interview on Saturday, hope it's an interesting one. Pray for me :) thank you xDruss_watters said:By jury, you mean people grading you for a school project? Please stop and pick another project because if you present this one with a wrong conclusion you will likely get a bad grade.

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- #15

CWatters

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Emna said:I need to do calculate it as a proof for the jury members, and also i need it so i see how much the torque became and depending on the torque needed for the generator i ill know how much mass i should add to the flywheel so the system works perfectly

Can you explain how the system is intended to work?

- #16

Emna

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the motor has an angular speed of 1500rpm and 1kw power, the belt pulley will make the generator turn 4 times the motor and the flywheel is supposed to increase the torque as much as you add it mass. i need to study the system to make the generator provides 4kw of powerCWatters said:Can you explain how the system is intended to work?

- #17

Khashishi

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- #18

Emna

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as long as the system is plugged in it must provide 4kw , if the flywheel helps stocking enough energy for at least 15min so the motor be shut down for that long it would be better.Khashishi said:

- #19

Khashishi

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- #20

Emna

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the belt pulley will do all the work , and it's ok if the flywheel don't stock that energy , it at least ill smooth the energy variation , all i need id how to calculate it's inertia while it has a complicated shape and i got that i should calculate each part of it alone , there's no other way.Khashishi said:

- #21

Khashishi

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- #22

Emna

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exactly , thank you so much for being around n you and all the others ;)Khashishi said:

- #23

jbriggs444

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You have a 1kw motor putting energy into the system. You want to be able to get 4kw out of the system for 15 minutes at a time. And you do not need anything to store energy.Emna said:the belt pulley will do all the work , and it's ok if the flywheel don't stock that energy , it at least ill smooth the energy variation , all i need id how to calculate it's inertia while it has a complicated shape and i got that i should calculate each part of it alone , there's no other way.

That does not makes sense. You are ignoring the part that is impossible and concentrating on the easy parts that you merely do not know how to do.

- #24

NascentOxygen

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This being along the lines of an experimental investigation, perhaps a better approach might be to construct something and thenEmna said:I couldn't find how to calculate this flywheel inertia

Once you know its inertia, you can do some calculations to determine just how much energy it will store at your proposed revs/min. This will show whether your idea is going to be feasible.

- #25

JLT

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What CAD software are you using? that really is the easiest way to find I.

- #26

malemdk

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Presume the 1Kw motor spinning the flywheel at 1700rpm, then Inertia of the wheel will be 0.12kgm2.

As noted by others it will work for short bust of times

- #27

CWatters

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In order for the flywheel and generator to deliver 4kW for 15min, the 1kW motor will have to run for at least 60min.

Just wanted to make sure you understand that basic fact.

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malemdk

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- #29

CWatters

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- #30

Nidum

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I've just read back through all of the postings on this thread .

From #16 : This is the picture that describes the proposed system :

The picture shows a continuous action power multiplying system . Supposedly 1 KW in gives 4 KW out .

From #16 : This is the picture that describes the proposed system :

The picture shows a continuous action power multiplying system . Supposedly 1 KW in gives 4 KW out .

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- #31

malemdk

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Absolutely not possible

- #32

malemdk

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Flywheel inertia is a measure of an object's resistance to changes in its rotational motion. It is determined by the mass and distribution of that mass around the axis of rotation.

The formula for calculating flywheel inertia is I = mr^2, where I is the moment of inertia, m is the mass of the object, and r is the radius of the flywheel.

Calculating flywheel inertia is important for understanding the behavior of rotating systems. It can help engineers design more efficient and stable machines, and it is also important in fields such as physics and mechanics.

The mass and radius of a flywheel both play a role in determining its inertia. The greater the mass, the greater the inertia, and the farther the mass is from the axis of rotation (i.e. larger radius), the greater the inertia as well.

Flywheel inertia is typically measured in units of kilogram-meters squared (kg*m^2). However, other units such as pound-feet squared (lb*ft^2) may also be used depending on the system of measurement being used.

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