L14 g 1 pg 6 Table 1: Parameter Хм Example (same as last lecture): Value 100Ω jXeq R₁ R₁ www mm 790Ω 0.4Ω R₁ (1-s) Vph JXM Rc Xea 1.4Ω S R₁ 0.3Ω Given the equivalent circuit parameters shown on Table 1, determine the following for a 3 phase 6 pole machine rotating at 970rpm and operating off a 230V (phase voltage), 50Hz supply: 1. All the Losses (W) 2. Efficiency (%) Rc Rs

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Where did they get 21.9 from?
L14 g 1 pg 6
Table 1:
Parameter
Хм
Rc
Rs
Example (same as last lecture):
Value
100Ω
k
R₁
jXeq
mm
R₁
790Ω
0.4Ω
R, (1-s)
jXMR.
Xeq
1.4Ω
ph
Rr
0.3Ω
Given the equivalent circuit parameters shown on Table 1, determine the
following for a 3 phase 6 pole machine rotating at 970rpm and operating
off a 230V (phase voltage), 50Hz supply:
1. All the Losses (W)
2. Efficiency (%)
Transcribed Image Text:L14 g 1 pg 6 Table 1: Parameter Хм Rc Rs Example (same as last lecture): Value 100Ω k R₁ jXeq mm R₁ 790Ω 0.4Ω R, (1-s) jXMR. Xeq 1.4Ω ph Rr 0.3Ω Given the equivalent circuit parameters shown on Table 1, determine the following for a 3 phase 6 pole machine rotating at 970rpm and operating off a 230V (phase voltage), 50Hz supply: 1. All the Losses (W) 2. Efficiency (%)
LECTURE 14:
EXAMPLE 1:
jxm & Re
Losses are due to:..
Stator
2
R+ >>
3 Rc Iron 2005.
→
1) First, the stator copper loss:
31²R = 3x 21.9² x 0.4
= 575W
Vou
POWER ENGINEERING
ung Ri
m u
Elut
Rs
-MM-
Copper
Rotor Copper
HA
LUTT
Wurs
R₁ (1-5)
S
Transcribed Image Text:LECTURE 14: EXAMPLE 1: jxm & Re Losses are due to:.. Stator 2 R+ >> 3 Rc Iron 2005. → 1) First, the stator copper loss: 31²R = 3x 21.9² x 0.4 = 575W Vou POWER ENGINEERING ung Ri m u Elut Rs -MM- Copper Rotor Copper HA LUTT Wurs R₁ (1-5) S
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