Let’s consider three phase, Y connected, 220 volt (line to line), 7.5 kW, 55 Hz, four pole IM has the following parameter values in Ω/phase referred to the stator: R1=0.294Ω , R2=0.144 Ω, X1=0.503 Ω, X2=0.209 Ω, Xm=13.25 Ω Determine the following for 0.02 slip. a)Electromechanical Torque(Tmech) Electromechanical power(Pmech) b)Maximum electromechanical torque(Tmax),Corresponding speed(nr,max) c)Electromechanical starting torque(Tstart), Corresponding stator load current(I2,start)

Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter3: Power Transformers
Section: Chapter Questions
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Let’s consider three phase, Y connected, 220 volt (line to line), 7.5 kW, 55 Hz, four pole IM has the following parameter values in Ω/phase referred to the stator:

R1=0.294Ω , R2=0.144 Ω, X1=0.503 Ω, X2=0.209 Ω, Xm=13.25 Ω

Determine the following for 0.02 slip.

a)Electromechanical Torque(Tmech) Electromechanical power(Pmech)

b)Maximum electromechanical torque(Tmax),Corresponding speed(nr,max)

c)Electromechanical starting torque(Tstart), Corresponding stator load current(I2,start)

 

X,
X,
R,
ww
i(1)
i,(1)
R,
X_
Fig. Equivalent Circuit of IM
Lest's consider three phase, Y connected, 220 volt(line to line), 7.5 kW, 55 Hz. four pole IM has the following parameter values in /phase
referred to the stator:
R = 0.294 N, R = 0.144 N, X1 = 0.503 N, X2 = 0.209 N, Xm = 13.25 N
Determine the following for 0.02 slip
a) Electromechanical Torque (Tmech), Electromechanical power (Pmech)
b) Maximum electromechanical torque (Tmaz), Corresponding speed (n, maz )
C) Electromechanical statrting torque (Tstart), Corresponding stator load current (I, start)
a. Tmech = 33.2165 N. m, Pmech
5624.6 W, n,maz
1333.51 rpm, Tmaz = 126.956 N. m, I2 start = 150.265 A, Tstart
56.4528 N
b. Tmech = 33.3705 N. m, Pmech
7293.87 W, n7,maz
950.539 rpm, Tmaz = 153.768 N. m, I2,start = 175.257 A, Tztart
70.2798
c. Tmech
35.4105 N. m, Рпеch
3933.77 W, nmaz = 919.728 rpm, Tmaz = 154.791 N. m, I2start = 125.567 A, Ttart = 72.5229 .
d. Tmech = 35.4105 N. m, Pmech =
7872.83 W, n,maz = 825.827 rpm, Tmaz = 154.758 N.m, Istart = 145.657 A, Ttart = 73.7912
e. Tmech = 35.2537 N. m, Pmech
2353.62 W, ny, maz
925.793 rpm, Tmaz = 220.437 N. m, I2 start = 115.562 A, Ttart = 57.7327
Transcribed Image Text:X, X, R, ww i(1) i,(1) R, X_ Fig. Equivalent Circuit of IM Lest's consider three phase, Y connected, 220 volt(line to line), 7.5 kW, 55 Hz. four pole IM has the following parameter values in /phase referred to the stator: R = 0.294 N, R = 0.144 N, X1 = 0.503 N, X2 = 0.209 N, Xm = 13.25 N Determine the following for 0.02 slip a) Electromechanical Torque (Tmech), Electromechanical power (Pmech) b) Maximum electromechanical torque (Tmaz), Corresponding speed (n, maz ) C) Electromechanical statrting torque (Tstart), Corresponding stator load current (I, start) a. Tmech = 33.2165 N. m, Pmech 5624.6 W, n,maz 1333.51 rpm, Tmaz = 126.956 N. m, I2 start = 150.265 A, Tstart 56.4528 N b. Tmech = 33.3705 N. m, Pmech 7293.87 W, n7,maz 950.539 rpm, Tmaz = 153.768 N. m, I2,start = 175.257 A, Tztart 70.2798 c. Tmech 35.4105 N. m, Рпеch 3933.77 W, nmaz = 919.728 rpm, Tmaz = 154.791 N. m, I2start = 125.567 A, Ttart = 72.5229 . d. Tmech = 35.4105 N. m, Pmech = 7872.83 W, n,maz = 825.827 rpm, Tmaz = 154.758 N.m, Istart = 145.657 A, Ttart = 73.7912 e. Tmech = 35.2537 N. m, Pmech 2353.62 W, ny, maz 925.793 rpm, Tmaz = 220.437 N. m, I2 start = 115.562 A, Ttart = 57.7327
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