5200_L13 - EE 5200 - Lecture 13 Mon Sep 27, 2010 Topics for...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EE 5200 - Lecture 13 Mon Sep 27, 2010 Topics for Today: • Announcements • Matlab - now we are ready to begin using it. • Office hrs: 2:05-2:55pm M,W,F • Office: EERC 614. Phone: 906.487.2857 • Recommended problems from Ch.3, solutions posted • Next: Transmission Line Parameters, Chapters 4,5,6 Synchronous Machines - Chapter 3. • Basic internal structure of machines, cylindrical vs. salient • Field windings • Calculation with Xd and Xq. • Calculation Example(s) • Concepts behind SYNCH exercise set. • S-S behavior - Xd ; Dynamic behavior - Xd’ • Short-circuit behavior - Xd”; s-s, transient, subtransient To: ee5200-l@mtu.edu From: Bruce Mork <bamork@mtu.edu> Subject: d-q synch machine steady-state loading calcs First of all, notation-wise, the internal induced voltage of the synch machine is called Ea in some references (voltage induced on armature windings) and in other references it's called Ef (since induced voltage on armature is due to magnitude of field current according to open-circuit characteristic of machine). In answer to question posed: Yes, Iq by definition is exactly in phase with Ea. Referring to Fig. B-5 in Appendix B reference, 1) determine Ia according to load specified, usually assuming Vt = 1.0 pu at 0°. 2,3) calculate Ea' to find torque angle delta (this is based observation that since jXdId is parallel to Ea, then Vt + IaRa + jXqIa lands you somewhere along the phasor Ea and this allows you to determine delta. 4) knowing delta, resolve Ia into its 2 components Ia = Id + Iq 5) then finally, Ea = Vt + IaRa + jXdId +jXqIq. As a double-check, Ea must end up with the same angle (delta) that you calculated for Ea'. So, the very good thing about this is that there is a double-check built into the calculations, you can immediately see if your answer seems to be correct, i.e. if Ea' and Ea have different angles, then you messed up somewhere along the line... Dr. Mork ...
View Full Document

Page1 / 9

5200_L13 - EE 5200 - Lecture 13 Mon Sep 27, 2010 Topics for...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online