RPM of 100% Efficient Motor?

Question

This is considering an ideal scenraio where we have a 100% efficient motor at all RPMs.

If we start with $$\omega=\frac{P_{Mech}}{\tau}$$

Since the motor is provided with eletrical power, the mechanical power will be some multiple of this power. $$\omega=\frac{VI\eta}{\tau}$$ Now we let \$\eta=1\$ and notice that \$\frac{I}{\tau}=\frac{1}{k_{T}}\$ where \$k_{T} \$ is the torque constant. Therefore; $$\omega=\frac{V}{k_{T}}$$

So lets say we provide 5V and the no-load angular velocity is \$100 rad/sec\$. With this motor, according to the equation above the angular velocity should remain \$100 rad/sec\$ irrespective of the load we apply to the shaft of the motor. So the angular velocity of a 100% efficient motor is constant regardless of load applied?

Looking at the same equation but for a non-ideal motor; $$\omega=\frac{V\eta}{k_{T}}$$ At a constant voltage, when load torque is applied the efficiency decreases (Friction, internal power losses etc) which results in the decrease of angular velocity. But with no losses in electrical power this angular velocity should remain constant, right?

Answer 1

But with no losses in efficiency this angular velocity should remain constant, right?

No. Efficiency and speed relation to voltage are not the same thing.

Let's consider an ideal DC motor wound with superconducting wire, with a wound field, that can be either in series with or parallel to the armature, driving a load torque, or twice that load torque.

The parallel wound motor, which is equivalent to a permanent magnet motor, will spin at a speed controlled by the applied voltage, and draw a current proportional to the torque. As there is no winding resistance, when the load torque and hence the current doubles, the speed will stay the same. The mechanical power delivered and the electrical power absorbed will both double, and the lossless motor stays 100% efficient.

The series wound motor, often used for cranes and car starter motors for its high starting torque, has a very variable speed, due to the magnetic field being proportional to the current drawn. This means the torque is proportional to current squared, or current proportional to square root of torque. As the load torque on this motor doubles, the current increases to 1.4 times, reducing the speed to 0.7 of what it was due to the increase in field. The output mechanical power and the input electrical power both increase by 40%, and the lossless motor stays 100% efficient.

Answer 2

Yes your analysis is correct. In a 100% efficient motor the output speed would not change with load applied. As you varied the torque the current pulled from the electrical supply would change in direct proportion to the torque to maintain the relationship Pmech = Pelec.

You can actually see from this how an real motor works, because the above statement assumes that the motor's power supply is also ideal and will not reduce its voltage output when the current is increased. If, instead, there is a resistance between the output of the supply and the motor, then when current is drawn a voltage drop will appear across this. The ideal motor's angular velocity will then have to slow down because the terminal voltage has reduced.

This is exactly how a real motor is modelled in practice - as an ideal motor in series with a resistor. The ideal motor is just a energy converter with RPM proportional to angular velocity and torque proportional to current.