Power and kinetic energy relationship in a small DC motor

[DamoPhys]

I am trying to relate two concepts, Power and Kinetic Energy, using a small DC motor.

Given the no-load max angular velocity w (rads/s) of the motor and the current drawn (A) at a given voltage (V) the power being used (P) in Watts is:

P = V x A

Given the no-load max angular velocity w (rads/s) and assuming the armature is a cylinder of uniform density, known radius (r) and known mass (m) the kinetic energy (Joules) of the armature is given by

KE = 1/2 I w^2

Where I is the moment of intertia given by

I = 1/2 m r^2.

So I have the Watts (P) being "consumed" and KE in Joules. Can I use this to calculate the efficiency of conversion from electrical to mechanical (under the stated conditions)?
Can I simply express that in 1 second the efficiency is KE / P x 100?

Basically, I am having trouble reconciling the time dimension of Power (J/s)

Thanks in advance.

 

[LightHero]

Yes, that is correct. You can calculate the efficiency of the conversion from electrical to mechanical energy by expressing it as the ratio of the kinetic energy (KE) in Joules per second to the power (P) in Watts multiplied by 100. That way, you are taking into account the time dimension of power.

 

[astrokreng]

I would like to first clarify the relationship between power and kinetic energy. Power is the rate at which energy is being used or transferred, while kinetic energy is the energy an object possesses due to its motion. In the case of a small DC motor, power is the electrical energy being converted to mechanical energy, while kinetic energy is the energy of the rotating armature.

Based on the equations provided, we can see that both power and kinetic energy are dependent on the angular velocity (w) of the motor. This means that as the motor rotates faster, both the power consumption and kinetic energy will increase. However, the relationship between the two is not a direct one, as power also depends on the current (A) and voltage (V) being supplied to the motor, while kinetic energy depends on the mass and radius of the armature.

To calculate the efficiency of conversion from electrical to mechanical energy, we need to consider the time dimension of power. This can be done by dividing the kinetic energy (Joules) by the power (Watts) and multiplying by 100 to get a percentage. This would give us the efficiency of the motor in converting electrical energy to mechanical energy in a given time period.

It is important to note that in a real-world scenario, the efficiency of a motor is affected by various factors such as friction and heat losses. Therefore, the calculated efficiency may not accurately reflect the actual efficiency of the motor.

In conclusion, the relationship between power and kinetic energy in a small DC motor is complex and cannot be simplified to a single formula. It is important to consider all the factors involved and the limitations of the equations when trying to understand and calculate the efficiency of the motor.