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Why does the Ns3 drop off. Fs is equall to W_s*I_s in each phase. Why does the Total Fs not include N_s3

Why does the Ns3 drop off. Fs is equall to W_s*I_s in each phase. Why does the Total Fs not include N_s3

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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Why does the Ns3 drop off. 

Fs is equall to W_s*I_s in each phase.

Why does the Total Fs not include N_s3

 

Let us consider a three-phase stator winding. We will assume that the conductor distribution for the stator windings may be expressed as nas (Øsm) = N31 sin(Pøsm / 2) – N3 sin(3Posm /2) (2.5-1) n (Øsm) = N51 sin(Pøsm / 2 – 2n / 3) – N3 sin(3PO,m / 2) (2.5-2) nes (Øsm) = Ng1 Ssin(Pøsm / 2+ 27t / 3)– N33 sin(3POsm / 2) (2.5-3) us. In (2.5-1)–(2.5-3), the second term to the right of the equal sign (the third harmonic term) is useful in achieving a more uniform slot fill. In order to calculate the stator MMF for this system, we must first find the winding function. Using the methods of Section 2.3, the winding functions may be expressed as 2N s1 2N3 -cos(3РФ,m / 2) ЗР Was (Osm) = -cos(Pom / 2)- P (2.5-4) 2N 53 s cos(POsm /2– 2n / 3)–- P Whs (Osm)= -cos(3РФ,m / 2) ЗР (2.5-5) 2N31 -cos(Posm / 2 + 2n / 3)– P 2N3 -cos(3POsm / 2) ЗР Wes (Øsm)= (2.5-6) From (2.4-17), the total stator MMF may be expressed as Fs (Øsm) = Was (Øsm )i as + W bs (Øsm )ibs + W cs (Øsm )ies (2.5-7) CS
Transcribed Image Text:Let us consider a three-phase stator winding. We will assume that the conductor distribution for the stator windings may be expressed as nas (Øsm) = N31 sin(Pøsm / 2) – N3 sin(3Posm /2) (2.5-1) n (Øsm) = N51 sin(Pøsm / 2 – 2n / 3) – N3 sin(3PO,m / 2) (2.5-2) nes (Øsm) = Ng1 Ssin(Pøsm / 2+ 27t / 3)– N33 sin(3POsm / 2) (2.5-3) us. In (2.5-1)–(2.5-3), the second term to the right of the equal sign (the third harmonic term) is useful in achieving a more uniform slot fill. In order to calculate the stator MMF for this system, we must first find the winding function. Using the methods of Section 2.3, the winding functions may be expressed as 2N s1 2N3 -cos(3РФ,m / 2) ЗР Was (Osm) = -cos(Pom / 2)- P (2.5-4) 2N 53 s cos(POsm /2– 2n / 3)–- P Whs (Osm)= -cos(3РФ,m / 2) ЗР (2.5-5) 2N31 -cos(Posm / 2 + 2n / 3)– P 2N3 -cos(3POsm / 2) ЗР Wes (Øsm)= (2.5-6) From (2.4-17), the total stator MMF may be expressed as Fs (Øsm) = Was (Øsm )i as + W bs (Øsm )ibs + W cs (Øsm )ies (2.5-7) CS
To proceed further, we need to inject currents into the system. Let us consider a balanced three-phase set of currents of the form ins = V21, cos(@,t + ¢; ) (2.5-8) i = V21, cos(@,t + ¢; – 2 / 3) (2.5-9) is = /21, cos(@.t + ¢; +2t / 3) (2.5-10) %3D where I, is the rms magnitude of each phase current, @, is the ac electrical frequency, and ø, is the phase of the a-phase current. Substitution of (2.5-4)–(2.5-6) and (2.5-8)– (2.5-10) into (2.5-7) and simplifying yields -cos(POm /2– @.t – 0,) P Fs = (2.5-11) us.
Transcribed Image Text:To proceed further, we need to inject currents into the system. Let us consider a balanced three-phase set of currents of the form ins = V21, cos(@,t + ¢; ) (2.5-8) i = V21, cos(@,t + ¢; – 2 / 3) (2.5-9) is = /21, cos(@.t + ¢; +2t / 3) (2.5-10) %3D where I, is the rms magnitude of each phase current, @, is the ac electrical frequency, and ø, is the phase of the a-phase current. Substitution of (2.5-4)–(2.5-6) and (2.5-8)– (2.5-10) into (2.5-7) and simplifying yields -cos(POm /2– @.t – 0,) P Fs = (2.5-11) us.
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