Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
General Engineering
Mechanical Engineering
Electrical Engineering
Aerospace Engineering
Nuclear Engineering
Materials Engineering
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
General Engineering
Mechanical Engineering
Electrical Engineering
Aerospace Engineering
Nuclear Engineering
Materials Engineering
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Engineering
Electrical Engineering
Power System, synchronous generator
Reply to thread
Message
[QUOTE="jim hardy, post: 5439182, member: 327872"] [USER=588734]@LagCompensator[/USER] I'll take a stab at this. Being not a power system guy it'll be amateurish but i hope logical. As textbooks and teaching methods evolve, some things become so refined that we lose track of the lower parts of the learning curve. Okay , a picture might help us grunt this one out from the basics... I think in pictures not formulas. Here's how i envision that circuit. Circle at bottom is phasor diagram, since we don't know power angle of the machine (angle of Ef) i just drew Ef some arbitrary place on a circle with radius 1.4. [ATTACH=full]98920[/ATTACH] What do we know ? Vnetwork i assumed infinite bus since they gave us no network impedance value. I gave it value 1∠0 so it'll be our reference for phasor diagram. Polar for 1∠0 is 1+j0 . We know that I[SUB]line [/SUB]has real component 0.5 because power is 0.5 . So it'll have imaginary component jVars. I'll assume lagging current, so that'll be -jVars. If I'm wrong we'll just get a negative Var current. I[SUB]line[/SUB] = 0.5 -jVars Ohm's law tells us voltage across Z[SUB]line[/SUB] is I[SUB]line[/SUB] X Z[SUB]line[/SUB]. Kirchoff's voltage law tells us that V[SUB]network[/SUB] + I[SUB]line[/SUB] X Z[SUB]line [/SUB] = V[SUB]f[/SUB] but we have a lot of unknown angles... Let's plug in what we know V[SUB]network[/SUB] + I[SUB]line[/SUB] X Z[SUB]line [/SUB] = V[SUB]f[/SUB] 1+j0 + (0.5 -jVars) X (j1.2) = 1.4∠[SUB]unknown[/SUB] 1 +j0.6 +1.2XVars = 1.4∠[SUB]unknown[/SUB] Aha ! Look back at the phasor diagram we have a right triangle with adjacent side (1.0 + 1.2XVars) volts, opposite side 0.6 volts, hypotenuse 1.4 volts and no angles known. Pythagoras let's us find adjacent side (1.0 +1.2XVars) = √(1.4[SUP]2[/SUP] - 0.6[SUP]2[/SUP]) 1.2 X Vars = 1.265 -1 so Vars = (1.265 - 1) / 1.2 = 0.2207 Repeating for real component of I[SUB]line [/SUB] = 0.51 i get 0.2159 That's the simplistic approach i'd use. If we worked numerous problems a shortcut would probably show its head.. Maybe some kind soul will improve(or correct) my explanation... [/QUOTE]
Insert quotes…
Post reply
Forums
Engineering
Electrical Engineering
Power System, synchronous generator
Back
Top