# Calcuating Open-Circuit emf for Alternators

1. Jan 9, 2010

### patjk

My question is not for a specific problem, but in general. I don't quite understand how to calcuate emf for alternators. I will give an example or two to give you an idea.

1. The problem statement, all variables and given/known data
Example 1) A star-connected alternator on infinite bus vars supplies 500MW at 23kV at a power factor of .95 lagging. The synchronous reactance is 2.8 ohms. Calculate a)line current, b) reactive VA, c) the corresponding open-circuit emf, d) the load angle

Example 2) The alternator described in 1) is operated at the same current a a power factor of .95 leading. Calculate a) the power output, b) the reactive VA, c) the open-circuit phase emf, d) the load angle

Please refer only to the bolded portions of those 2 examples.

3. The attempt at a solution
For 1), I don't quite understand how to get the open-circuit emf. I have the solution to this, and for 1c., it has:
Eph^2 = (Vph + Xs*Iph*sin(angle))^2+(Xs*Iph*cos(angle))^2
where Eph is emf, Vph is phase voltage, Iph is phase current, Xs is reactance, and angle is cos^-1(power factor).

I am not sure how to determine this with the given info in the problem. If a phasor diagram was given, I'd be able to know this, but without the diagram, how can determine this?

For 2c., the solution is:
Eph^2 = (Xs*Iph*cos(angle))^2+(Vph-Xs*Iph*sin(angle))^2

And I don't see how that came about.
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In general (for a leading power factor), I think Eph^2 = (Iph*R)^2 + (Iph*X)^2, hence why the above is confusing me.

Load angle is just the angle between Eph and Vph (Iph*R), so if I can figure out Eph, that won't be a problem.

Any ideas/tips/suggestions on this matter is greatly appreciated. Thanks

Last edited: Jan 9, 2010