# DC motors - Torque/Speed Relationship

Question:

A DC Motor, when supplied from a 195V power supply is observed to rotate under no load conditions at 1730 rpm. A stall test shows that the winding resistance is 1.66 Ω. The motor is connected to a load whose torque-speed characteristic is Torque = 0.128 × ω.

Find the speed (in rpm) at which the motor will drive the load.

So far, I have tried to use the following method:

$$\ K= \frac{V}{\omega} = \frac{195}{1730*\frac{2\pi}{60}} = 1.0764$$

$$\ Torque= K*I = 1.0764* \frac{195}{1.66} = 126.4446$$

And from the question, $Torque = 0.128 \omega$

$$\ \omega = \frac{126.4446}{0.128} = 987.85RPM$$

Can anyone tell me why this is wrong?

• You are using the stall current in place of the no-load motor current, for one thing. – Brian Drummond Apr 26 '18 at 8:47
• I believe that the torque-speed characteristic of the load is given in N-m vs. radians/sec. – Charles Cowie Apr 26 '18 at 12:48
• You are not considering the back-emf created by the rotating shaft. – MathieuL Jan 31 at 20:50

There are two errors:

• the torque you calculate is the torque at stall condition (with a destructive current, if it is maintained for a sufficient long time). When the motor will start rotating, there will be a back electromotive force that will reduce the current;
• omega is in rd/s and not RPM

What you should do:

1. calculate first the torque-speed characteristic of the MOTOR (torque as a function of speed). As explained above, you'll have to take into account the effect of the back emf on the current;
2. combine the MOTOR characteristic with the LOAD characteristic (Torque = 0.128 × ω): the solution will give you both the operating speed and the operating torque