please do not answer on paper give typed answers A 3-phase 400V 50Hz 6-pole cage
ID: 2081651 • Letter: P
Question
please do not answer on paper give typed answers
A 3-phase 400V 50Hz 6-pole cage induction motor of rating 20 kW, with slip at 4% of the rated load, has a starting torque equal to 120% of the rated torque. The motor is coupled to a pump through a reduction gearbox of efficiency 85%. The breakaway torque is 24N.m at the motor shaft. Assume that the torque available for acceleration remains constant for the entire duration and the motor accelerates to the rated speed in 6 seconds.
Based on this data, please answer the questions below.
a) What is the rated speed of the motor?
b) What are the values of the rated torque and starting torque of the motor?
c) What quantity of torque is available for acceleration?
What is the total Moment of Inertia of the rotating parts of the system (seen at the motor shaft)?
Explanation / Answer
F=PN /120
The rated speed o the motor = N= 120*F /P
= 120*50 /6
= 1000 RPM
Rotor speed = N_r = (1-S)N = (1-0.04) * 1000
= 960 RPM
Rated Torque = T = 9550 * PkW / N_r
= 9550*20/960 =198.96 N-m
Starting Torque means Torque at zero speed . Typically 1.5 times the full load Torque.
= 1.45*196.96 298.43 N - m
Accelarating Torque = 298.43 - 198.96 = 99.47 N - m
The total moment of inertia J_r / p dw/dt = 99.47
J_r = 99.47*20 * 1 / 6 =331.56
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