Question

At what speed (in RPM) must the armature of a dc motor run to develop a power of 572.4 kW at a torque of 4605Nm ?

Answer 1

The armature of a DC motor must run at approximately 1185.86 RPM to develop a power of 572.4 kW at a torque of 4605 Nm.

Step-by-step explanation:

To determine the speed in RPM that the armature of a DC motor must run to develop a power of 572.4 kW with a torque of 4605 Nm, we can use the power-torque-rotational speed relationship in the following formula:

Power (W) = Torque (Nm) × Angular Speed (rad/s)

Since we are asked to find the speed in RPM (revolutions per minute), we need to convert the angular speed to RPM using the following conversion factor:

1 revolution = 2π radians

and

1 minute = 60 seconds

Let’s first find the angular speed in rad/s:
Power = Torque × Angular Speed
572,400 W = 4605 Nm × Angular Speed
Angular Speed = 572,400 W / 4605 Nm
Angular Speed = 124.33 rad/s

Then, we convert angular speed in rad/s to RPM:
Angular Speed (rad/s) × (60 s/min) / (2π rad/rev) = RPM
124.33 rad/s × (60 / 2π) RPM = 1185.86 RPM

Thus, the armature must run at approximately 1185.86 RPM to develop a power of 572.4 kW at a torque of 4605 Nm.