# DC Motor

#### keviiii

Joined Oct 17, 2008
8
hi to all,

im doing a question on DC motors and i'm having some conceptual doubts.

this is a seperately excited DC motor that operates at 1070rpm. a load will be attached to it such that the the new speed of the motor is 950rpm. i have found the new current that the motor will draw, which is 3.274A and also the new back emf that the motor will generate, which is 10.88V. now, i am going to find the rotational loss by the motor, i have used proportion method the find, using (950/1070)*(6.175W) = 5.482W. 6.175W is the rotational loss when the motor is operating without any load. this value of 5.482W is correct.

however, i wondering why is it that this value of rotational loss is different from when i use (3.274A)*(10.88V) to find the rotational loss. this value would be 35.62W. notice the huge diff from the proportional method. why is it so? ):

thanks.

#### mik3

Joined Feb 4, 2008
4,846
(3.274A)*(10.88V) is the electrical power converted to mechanical power. In a frictionless motor (theoretically) its the output mechanical power. However, a in a real motor the mechanical power is a bit less than this electrical power due to friction losses.

#### keviiii

Joined Oct 17, 2008
8
(3.274A)*(10.88V) is the electrical power converted to mechanical power. In a frictionless motor (theoretically) its the output mechanical power. However, a in a real motor the mechanical power is a bit less than this electrical power due to friction losses.
oh i see. can i also say that the difference is due the frictional losses?

so, theoretically, 35.62W IS the rotational loss, just that friction accounts for the difference?

i can also say that using the proportional method to find the rotational loss will take the frictional losses out of consideration?

thank you so much.

#### mik3

Joined Feb 4, 2008
4,846
Yes, the real mechanical is the the V*I of the armature winding minus the rotational losses.