# A motor running at its rated power provides 2.6 lb-ft torque at its rated speed of 2020 rpm. The motor draws 5.6-A from a 230-V supply at a phase angle of 37 degrees. Would the effciency of the motor be 82.7% or 1.6%

The relation between the torque `T` , the motor angular speed `omega` and its power `P` in circular motion is (similar to `P = F*v` in linear motion)

`P_m =T*omega`

we know that `1 lb*ft = 1.3558 N*m` and `omega=2*pi*F`

therefore

`P_m = (1.3558*2.6)*(2*pi*2200/60) =745.66 W`

This is the mechanical power that the motor generates.

The apparent electric power that the motor draws from the AC source is

`P_a =U*I =230*5.6 =1288 VA`

The real electrical power (the part of apparent power that can be transformed into mechanical power) is

`P = P_a*cos(theta) = 1288*cos(37) =1028.6 W`

The efficiency of the motor is just the rapport of the mechanical power produced to the real electric power.

`eta = P/P_a = 745.66/1028.6 =0.725 =72.5%`

The efficiency of the motor is 72.5%.

Approved by eNotes Editorial Team