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Example 2: a 3-phase SM its armature resistance is 0.50 and syn. Reactance is 0.86692/phase. The supply voltage is 150V and the shaft load including core an friction losses are 6.5 kW. The phase current is 50A. Find the induced emf. Armature winding is (star connected. Peu = 312 Ra = 3x 50²x 0.5= 3750 w P₁ = 6500 + 3750 = 10250w P 10250 cos √3Vxl -1Xs R √3x150x50 tan ¹ 1.732 = 60° 0.789, p = cos ¹0.789 = 37.9° 0= tan At lagging: y=8-= 60-37.9=22.1%, at leading: y= 8+ = 60+37.997.9° E² = V² + Er² − 2V E, cos 22500+2500-150 x 50 x (0.926 or -0.137) Eo=134.3 OR 161.3 V

Example 2: a 3-phase SM its armature resistance is 0.50 and syn. Reactance is 0.86692/phase. The supply voltage is 150V and the shaft load including core an friction losses are 6.5 kW. The phase current is 50A. Find the induced emf. Armature winding is (star connected. Peu = 312 Ra = 3x 50²x 0.5= 3750 w P₁ = 6500 + 3750 = 10250w P 10250 cos √3Vxl -1Xs R √3x150x50 tan ¹ 1.732 = 60° 0.789, p = cos ¹0.789 = 37.9° 0= tan At lagging: y=8-= 60-37.9=22.1%, at leading: y= 8+ = 60+37.997.9° E² = V² + Er² − 2V E, cos 22500+2500-150 x 50 x (0.926 or -0.137) Eo=134.3 OR 161.3 V

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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solvin the queen using equation E=v-I(R+jxs) and find  same ans

Example 2: a 3-phase SM its armature resistance is 0.50 and syn. Reactance is 0.86692/phase. The supply voltage is 150V and the shaft load including core and friction losses are 6.5 kW. The phase current is 50A. Find the induced emf. Armature winding is (star connected. Pcu = 312 R₂ = 3x 50²x 0.5= 3750 w P₁ = 6500 + 3750 = 10250w P 10250 cos √3Vxl Xs R = √3x150x50 tan-¹ 1.732 = 60° 0.789, p = cos ¹0.789 = 37.9° 0= tan At lagging: y=8-= 60-37.9-22.1%, at leading: y= 8+ = 60+37.997.9° E² = V² + Er² − 2V E₂ cos = 22500+2500 -150 x 50 x (0.926 or -0.137) Eo 134.3 OR 161.3 V
Transcribed Image Text:Example 2: a 3-phase SM its armature resistance is 0.50 and syn. Reactance is 0.86692/phase. The supply voltage is 150V and the shaft load including core and friction losses are 6.5 kW. The phase current is 50A. Find the induced emf. Armature winding is (star connected. Pcu = 312 R₂ = 3x 50²x 0.5= 3750 w P₁ = 6500 + 3750 = 10250w P 10250 cos √3Vxl Xs R = √3x150x50 tan-¹ 1.732 = 60° 0.789, p = cos ¹0.789 = 37.9° 0= tan At lagging: y=8-= 60-37.9-22.1%, at leading: y= 8+ = 60+37.997.9° E² = V² + Er² − 2V E₂ cos = 22500+2500 -150 x 50 x (0.926 or -0.137) Eo 134.3 OR 161.3 V
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