Question

Consider a permanent magnet DC motor with the following parameters: Ra = 0.2 Ω, La = 0.9 mH, kE = 0.6 V/(rad/s), kT = 0.4 Nm/A, and Jm = 0.045 kg m^2. The rated torque of this motor is 11 Nm. The motor is driving a vehicle at a speed of 2,200 rpm. The load is purely inertial with an inertia of JL = 0.14 kg m^2. Calculate the energy recovered by slowing it down to 750 rpm within a time of 3 s, after the electrical losses in the windings have been accounted for.

Name: consider a permanent magnet dc motor with the following parameters ra 02 la 09 mh ke 06 vrads kt 04 nma and jm 0045 kg m2 the rated torque of this motor is 11 nm the motor is driving a vehic 05444
Uploaded: 2023-09-06T09:50:16-08:00
Duration: 7 min 2 s
Channel: Madhur L
Description: consider a permanent magnet dc motor with the following parameters ra 02 la 09 mh ke 06 vrads kt 04 nma and jm 0045 kg m2 the rated torque of this motor is 11 nm the motor is driving a vehic 05444

          Consider a permanent magnet DC motor with the following parameters: Ra = 0.2 Ω, La = 0.9 mH, kE = 0.6 V/(rad/s), kT = 0.4 Nm/A, and Jm = 0.045 kg m^2. The rated torque of this motor is 11 Nm. The motor is driving a vehicle at a speed of 2,200 rpm. The load is purely inertial with an inertia of JL = 0.14 kg m^2. Calculate the energy recovered by slowing it down to 750 rpm within a time of 3 s, after the electrical losses in the windings have been accounted for.

consider a permanent magnet dc motor with the following parameters ra 02 la 09 mh ke 06 vrads kt 04 nma and jm 0045 kg m2 the rated torque of this motor is 11 nm the motor is driving a vehic 05444

Added by Juana B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

09/06/2023

Step 1

Given: Speed = 2,200 rpm To convert from rpm to rad/s, we use the formula: Speed (rad/s) = Speed (rpm) * 2π / 60 Speed (rad/s) = 2,200 * 2π / 60 Speed (rad/s) = 230.97 rad/s Show more…

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Consider a permanent magnet DC motor with the following parameters: Ra = 0.2 Ω, La = 0.9 mH, kE = 0.6 V/(rad/s), kT = 0.4 Nm/A, and Jm = 0.045 kg m^2. The rated torque of this motor is 11 Nm. The motor is driving a vehicle at a speed of 2,200 rpm. The load is purely inertial with an inertia of JL = 0.14 kg m^2. Calculate the energy recovered by slowing it down to 750 rpm within a time of 3 s, after the electrical losses in the windings have been accounted for.