Answer:
A) slip(s1) = (1800 - 1750) / 1800 = 0.0277
B) 1800 rpm, 1350 rpm, 900 rpm, 450 rpm
Explanation:
Given data
frequency = 60 Hz,
Line - line rms = 440 - V
3 phase induction-motor drive
number of poles = 4
Full-load rated speed = 1750
rated torque = 40 Nm
A) The plot of torque-speed characteristics for the following values of the frequency f : 60 Hz, 45 Hz, 30 Hz, and 15 Hz is attached below
first we calculate the rated speed:
Ns = [tex]\frac{120f}{p}[/tex]
f = 60 Hz . p (number of poles) = 4
Ns[tex]_{1}[/tex] = [tex]\frac{120(60)}{4}[/tex] = 1800
full loaded rated value = 1750.
slip(s1) = (1800 - 1750) / 1800 = 0.0277
considering a linearly condition the slip is low
[tex]\frac{T1}{T2} = \frac{S1}{S2} * \frac{F2}{F1}[/tex]
S1 = 0.0277
f1 = 60 Hz
hence s2 = 0.018 therefore Ns2 = 1500
B) The speeds of operation at : 60 Hz, 45 Hz, 30 Hz, 15 Hz
for 60 Hz :
Ns = [tex]\frac{120f}{p}[/tex] = (120*60) / 4 = 1800 rpm
for 45 Hz:
Ns = 120(f) / p = (120*45) / 4 = 5400 /4 = 1350 rpm
for 30 Hz:
Ns = 120(f) / p = (120*30) / 4 = 3600 / 4 = 900 rpm
for 15 Hz:
Ns = 120(f) / p = (120*15) / 4 = 1800 / 4 = 450 rpm