A 60 Hz induction motor has two poles and runs at 3510 rpm.Calculate (a) the synchronous speed and (b) the percent slip.
Answers
Answer:
(a) f = 60\, Hzf=60Hz and p = (2/2) = 1p=(2/2)=1 Hence \textbf{synchronous speed, } n_{s} = (f/p) = (60/1) = 60synchronous speed, n
s
=(f/p)=(60/1)=60 rev/s or 60 \times 60 = 360060×60=3600 rev/min.
(b) Since slip,
s =(\frac{n_{s} \, -\, n_{r}}{n_{s}}) \times 100\%s=(
n
s
n
s
−n
r
)×100%
2 =(\frac{60 \, -\, n_{r}}{60}) \times 1002=(
60
60−n
r
)×100
Hence
\frac{2 \times 60}{100}=60\, -\, n_{r}
100
2×60
=60−n
r
i.e.
n_{r} = 60 \, -\, \frac{2 \times 60}{100}= 58.8n
r
=60−
100
2×60
=58.8 rev/s
i.e. the rotor runs at 58.8 \times 60 = 352858.8×60=3528 rev/min
(c) Since the synchronous speed is 6060 rev/s and that of the rotor is 58.858.8 rev/s, the rotating magnetic field cuts the rotor bars at (60 \, -\, 58.8) =1.2(60−58.8)=1.2 rev/s.
\textbf{Thus the frequency of the e.m.f.’s induced in the rotor bars is 1.2 Hz}.Thus the frequency of the e.m.f.’s induced in the rotor bars is 1.2 Hz.
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