Energy balance from mechanical to electrical energy

[PhiliosKassin]

A mass is winded via string aroung a flywheel and allowed to fall a specific height before detachement. The flywheel is connected to a small DC motor via gears and starts generating electricity once the mass is allowed to fall..

Now without the DC motor ( generator) the equation of energy balance is:

mgh = (0.5mv^2+ (0.5Iω^2) + ( 0.5Iω^2 x (n1/n2)

where I is the moment of inertia of flywheel and ω is the angular velocity ( which should be equal to the angular max velocity at point of mass detachment)

n1 = number of revs of flywheel before detachement of mass
n2 = number of revs of flywheel after detachement

the third term in the equation is the work done to overcome friction.

Now the question is :

What form is the equation going to have when a DC motor is attached to the flywheel to be allowed to transform rotational energy to electrical? mgh = ?

the electrical energy is in the form of UIt

 

[wywong]

It seems that one has to assume the friction is independent of the load.

Let n3 = number of revs of flywheel after detachment, and E = electrical energy.

I think it is not hard to see

E = (0.5Iω^2)(n2-n3)/n2

Is that what you are after?

 

[Bob S]

Suppose the height of initial release of the mass minus the height of detachment is h. Then if the mass falls freely (without flywheel), the total kinetic energy of the mass is mgh = (1/2)mv². But in the case where it is attached to the flywheel, its final velocity at point of detachment is v₁. So its energy at release is now (1/2)mv₁². So the missing energy is (1/2)m[v² - v₁²]. This was all transferred to the flywheel.

If the flywheel were frictionless and the generator perfect, all the flywheel energy would be converted to electric power. All inertial energy remaining in the flywheel after release of the mass would also be converted to electric power. To maximize the total electric energy, the objective then is to make the final velocity of the mass v₁ as small as possible. If the flywheel were massive and v₁ were small, then the electrical energy would be nearly mgh. If mgh were expressed in joules, then mgh/3600 would be the energy in watt-hours, or in volt amp hours.

 

[osama]

pls what is the energy balance equation in electric machines?

 

[sardar]

, where U is the voltage, I is the current, and t is the time. Therefore, the equation would be:

mgh = (0.5mv^2 + 0.5Iω^2 + 0.5Iω^2 x (n1/n2)) + UIt

This equation takes into account the mechanical energy of the falling mass (mgh), the kinetic energy of the flywheel (0.5mv^2), the rotational energy of the flywheel (0.5Iω^2), the work done to overcome friction (0.5Iω^2 x (n1/n2)), and the electrical energy generated by the DC motor (UIt). The addition of the DC motor allows for the transformation of mechanical energy into electrical energy, which can then be used for various purposes. This highlights the principle of energy conservation, where energy cannot be created or destroyed, but can only be transformed from one form to another.