Question #237176

A 10 pole, 50 Hz, Y connection 3-phase induction motor having a rating of 60 kW and 415V.

The slip of the motor is 5% at 0.6 power factor lagging. If the full load efficiency is 90%,

calculate:

(i) Input power

(ii) Line current and phase current

(iii) Speed of the rotor (rpm)

(iv) Frequency of the rotor

(v) Torque developed by the motor (if friction and windage losses is 0)


1
Expert's answer
2021-09-17T15:33:17-0400

(i)

η=O/P  powerInput  powerInput  power=O/P  powerη=60×10000.9=66.66  kWη = \frac{O/P \;power}{Input \;power} \\ Input \; power = \frac{O/P \;power}{η} \\ = \frac{60 \times 1000}{0.9} = 66.66 \;kW

(ii)

P=3V2I2cosθI2=P3V2cosθ=66.66×10003×415×0.6=154.75  AI2=3IphIph=I23=88.24  AP = \sqrt{3}V_2I_2cosθ \\ I_2 = \frac{P}{\sqrt{3} V_2 cosθ} \\ = \frac{66.66 \times 1000}{\sqrt{3} \times 415 \times 0.6} \\ = 154.75 \;A \\ I_2 = \sqrt{3}I_{ph} \\ I_{ph} = \frac{I_2}{\sqrt{3}} = 88.24 \;A

(iii)

N=Ns(1s)Ns=120fP=120×5010=600  rpm=600(10.05)=570  rpmN= N_s(1-s) \\ N_s = \frac{120f}{P} = \frac{120 \times 50}{10} = 600 \;rpm \\ = 600(1-0.05) \\ = 570 \;rpm

(iv)

fr=sf=0.05×50=2.5  Hzf_r = sf = 0.05 \times 50 = 2.5 \;Hz

(v)

P=τωτ=Pω=60×1000×602π×570=1005.65  NmP =τω \\ τ = \frac{P}{ω} = \frac{60 \times 1000 \times 60}{2 \pi \times 570} = 1005.65 \;Nm


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS