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Electric Machine-check image, all question there. Consider the above equivalent circuit of IM. A three phase, Y connected , 215 V (line to line), 7.5Kw , 60Hz , three pole IM has the following parameter values in Ω/phase referred to the stator R1=0.294Ω , R2=0.144 Ω, X1=0.203 Ω, X2=0.409 Ω, Xm=10.25 Ω The total friction , windage and core losses may be assumed to be constant at 350W independent of load, For a slip s=0.02, compute the followings a)Stator current b)Power factor(pf) c)Rotor speed

Electric Machine-check image, all question there. Consider the above equivalent circuit of IM. A three phase, Y connected , 215 V (line to line), 7.5Kw , 60Hz , three pole IM has the following parameter values in Ω/phase referred to the stator R1=0.294Ω , R2=0.144 Ω, X1=0.203 Ω, X2=0.409 Ω, Xm=10.25 Ω The total friction , windage and core losses may be assumed to be constant at 350W independent of load, For a slip s=0.02, compute the followings a)Stator current b)Power factor(pf) c)Rotor speed

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Electric Machine-check image, all question there.

Consider the above equivalent circuit of IM.

A three phase, Y connected , 215 V (line to line), 7.5Kw , 60Hz , three pole IM has the following parameter values in Ω/phase referred to the stator

R1=0.294Ω , R2=0.144 Ω, X1=0.203 Ω, X2=0.409 Ω, Xm=10.25 Ω

The total friction , windage and core losses may be assumed to be constant at 350W independent of load, For a slip s=0.02, compute the followings

a)Stator current

b)Power factor(pf)

c)Rotor speed

d)Power Output(shaft)

e)Output(shaft) torque

f)Efficiency

x, X, R, i,(1) R, Fig. Equivalent Circuit of IM Consider the above equivalent circuit of IM. A three phase,Y connected, 215 V (line to line), 7.5 kW, 60 Hz, three pole IM has the following parameter values in 2/phase referred to the stator R = 0.294 N, R = 0.144 N, X1 = 0.203 N, X, = 0.409 N, Xm = 10.25 2. The total friction,windage and core losses may be assumed to be constant at 350 W independent of load. For a slip s = 0.02, compute the followings a) Stator Current b) Power Factor(pf) c) Rotor Speed d) Power Output(Shaft) e) Output(Shaft) Torque f) Efficiency(n) O a. I = 12.2135 – 12.2771i A, pf = 0.744373 , n, = 2312 rpm, Pahaft = 5209.49 w, T,haft = 21.1509 N. m, 7=0.86282 O b. I = 11.2135 – 12.2771i A, pf = 0.844373 , n, = 2352 rpm, Pahaft = 5209.49 W, T,haft = 21.1509 N. m, 7=0.79382 %3D О с. — 11.152 — 12.5986i А, pf —D 0.4828 , п, — 1046 rpm, Pahaft = 3040.68 W, T,haft = 21.0056 N. m, 7=0.793137 O d. I = 16.2135 – 12.2771i A, pf = 0.844373 , n, = 2352 rpm, Pshaft = 5209.49 W, Tshaft = 21.1509 N.m, 7=0.86282 O e. I1 = 16.2135 – 12.2771i A, pf = 0.844373 , ng = 2250 rpm, Pahaft = 5009.79 W, Tahaft 22.1509 N. m, 7=0.86282 %3D
Transcribed Image Text:x, X, R, i,(1) R, Fig. Equivalent Circuit of IM Consider the above equivalent circuit of IM. A three phase,Y connected, 215 V (line to line), 7.5 kW, 60 Hz, three pole IM has the following parameter values in 2/phase referred to the stator R = 0.294 N, R = 0.144 N, X1 = 0.203 N, X, = 0.409 N, Xm = 10.25 2. The total friction,windage and core losses may be assumed to be constant at 350 W independent of load. For a slip s = 0.02, compute the followings a) Stator Current b) Power Factor(pf) c) Rotor Speed d) Power Output(Shaft) e) Output(Shaft) Torque f) Efficiency(n) O a. I = 12.2135 – 12.2771i A, pf = 0.744373 , n, = 2312 rpm, Pahaft = 5209.49 w, T,haft = 21.1509 N. m, 7=0.86282 O b. I = 11.2135 – 12.2771i A, pf = 0.844373 , n, = 2352 rpm, Pahaft = 5209.49 W, T,haft = 21.1509 N. m, 7=0.79382 %3D О с. — 11.152 — 12.5986i А, pf —D 0.4828 , п, — 1046 rpm, Pahaft = 3040.68 W, T,haft = 21.0056 N. m, 7=0.793137 O d. I = 16.2135 – 12.2771i A, pf = 0.844373 , n, = 2352 rpm, Pshaft = 5209.49 W, Tshaft = 21.1509 N.m, 7=0.86282 O e. I1 = 16.2135 – 12.2771i A, pf = 0.844373 , ng = 2250 rpm, Pahaft = 5009.79 W, Tahaft 22.1509 N. m, 7=0.86282 %3D
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