Solution to Quiz Eight of BEE2-A Electrical Engineering Course: Induction Motor Analysis, Exercises of Electromechanical Systems and Devices

The solutions to quiz eight of the bee2-a electrical engineering course, focusing on the analysis of a 3-phase, 10hp, 208v, 60hz, 6 poles wound rotor induction machine. The solutions include calculations for the operating slip, voltage induced in the rotor per phase, frequency of the induced voltage, and the rotor field rpm with respect to both the rotor and the stator.

Typology: Exercises

2011/2012

Uploaded on 07/31/2012

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Class: BEE2-A
Time Allowed: 10 mins
Maximum Points: 10
Reg. No:
QUIZ EIGHT - SOLUTION
NOTE: Please draw labelled circuit diagrams where appropriate, show complete calculations, and reasoning
to get maximum marks.
1. A 3 phase, 10hp, 208V, 60Hz, 6 poles wound rotor induction machine has a stator to rotor turns ratio of
1:0.5 and both stator and rotor are connected in Y.The motor speed is 1140 rpm. Determine:
a. The operating slip. [2]
b. The voltage induced in the rotor per phase. [3]
c. The frequency of the induced voltage. [1]
d. The rpm of the rotor field with respect to the rotor. [2]
e. The rpm of the rotor field with respect to the stator. [2]
SOLUTION
a. nsync =120fe
P=120×60
6= 1200rpm
s=nsyncnm
nsync
×100% = 12001140
1200 ×100% = 0.05
b. E2=sE2o=sE1
a
E2= 0.05 ×1
2×208
3= 3V
c. fr=sfe= 0.05 ×60 = 3Hz
d. nrotorf ield =nstatorf ield =nsync = 1200rpm
nrotor =nm= 1140rpm
The rpm of the rotor field w.r.t. rotor = 1200 - 1140 = 60 rpm
e. nrotorf ield = 1200rpm, nstator = 0rpm [Stator is stationary].
The rpm of the rotor field w.r.t. stator = 1200 - 0 = 1200 rpm
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Class: BEE2-A Time Allowed: 10 mins Maximum Points: 10 Reg. No:

QUIZ EIGHT - SOLUTION

NOTE: Please draw labelled circuit diagrams where appropriate, show complete calculations, and reasoning to get maximum marks.

  1. A 3 phase, 10hp, 208V, 60Hz, 6 poles wound rotor induction machine has a stator to rotor turns ratio of 1:0.5 and both stator and rotor are connected in Y .The motor speed is 1140 rpm. Determine: a. The operating slip. [2] b. The voltage induced in the rotor per phase. [3] c. The frequency of the induced voltage. [1] d. The rpm of the rotor field with respect to the rotor. [2] e. The rpm of the rotor field with respect to the stator. [2] SOLUTION a. nsync = (^120) Pf e= 1206 × 60 = 1200rpm

s = nsync nsync−n m× 100% = 12001200 −^1140 × 100% = 0. 05

b. E 2 = sE 2 o = s E a^1 E 2 = 0. 05 × 12 × 208 √ 3 = 3V

c. fr = sfe = 0. 05 × 60 = 3Hz

d. nrotorf ield = nstatorf ield = nsync = 1200rpm nrotor = nm = 1140rpm → The rpm of the rotor field w.r.t. rotor = 1200 - 1140 = 60 rpm

e. nrotorf ield = 1200rpm, nstator = 0rpm [Stator is stationary]. → The rpm of the rotor field w.r.t. stator = 1200 - 0 = 1200 rpm

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