A 2-pole permanent magnet DC motor is rated at 6V with the following parameters:

Question

A 2-pole permanent magnet DC motor is rated at 6V with the following parameters: Ra=7? LAA=120mH kT=2 oz·in/A J=150 µ oz·in·s 2

assume any value for losses

A 2-pole permanent magnet DC motor is rated at 6V with the following parameters: La,-120mH ky 2 oz in/A ?150 ? ozins The no-load speed is approximately 3350 r/min and the no-load armature current is approximately 0.15A. The machine is operating with rated applied armature voltage and a load torque TL of 0.5oz in. Determine the efficiency of this machine

Explanation / Answer

No load speed=3350 RPM=(3350×2×?/60) rad/s=350.811rad/s

No load Armature current =0.15 A.

Therefore No load torque=0.15×2 oz-in=0.3oz-in

This torque is used up in friction ,windage,inertia of rotor.

No load losses are because of these.

Pno load=0.3 oz-in×350.811.

Now 1 oz-in =0.007061551833 N-m.

So Pno load=0.3×0.007061551833×350.811=0.74317Watts

Also ,Load torque=0.5oz-in

Therefore Total Torque to be developed=Tno load+T load=0.5+0.3=0.8oz-in

Therefore Current on load =0.8oz-in/(2 oz-in/A)=0.4A

Now input power=6V×0.4=2.4 W

Armature resistance losses=0.4^2×7=1.12W

Pno load=0.74317 W

Therefore output power=2.4-1.12-0.74317=0.53682W

Efficiency=(0.53682/2.4)×100=22.367%

Hope this helps.

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