synchronous generator expression explaination 1

[poddieR]

Can someone please enlighten me?

I've come across the expression :

P=(3Ef.V/ Xs).sin d

Since I am looking at a generator the p will i presume be negative, however, I can find no expalination which shows the derivation of how this expression indicates the power delivered to the bus bars, and how it can be used to develope an operating chart for the generator.

I've tried to construct vector diagrams to assist but keep getting confused with Vt verses EMF
Can anyone help??

Appologies if I'm appearing slightly dim, I'm a production engineer, but keen to learn.
I'd apreciate any tips & thanks for your time in anticipation!

 

[electricpete]

P=|V1|*|V2|/(XL) * sin(delta) is fundamental expression for real power transfer through an inductance XL. delta is the angle difference between phasor V1 and V2.

To derive the above equation: start with S = V1* conjugate(I). The plug in I = (V1-V2)/jX. Take the real part to get P. Let me know if you want more details on that part.

As far as how that equation is applied to a syncronous generator, I'm not sure. I would assume that Ef=V1 represents an internal voltage and V=V2 represents terminal voltage. Xs represents the inductive reactance between the two voltages corresponding to syncronous reactance. The factor of 3 maybe has something to do with conversion between L-N quantities and L-L quantities on the two voltages?

 

[jbartos]

Suggestions marked ///\\\:
Can someone please enlighten me? I've come across the expression:
P=(3Ef.V/ Xs).sin d
///This equation is closely related to Synchronous Machines theory and performance. Specifically, when you analyze operating characteristics. You may start with Ideal Synchronous machine equivalent circuit to obtain by inspection:
I1a Xa = - Et sind
I1r Xa = V1 - Et cosd
Multiplying through by V1 gives the input active power and reactive electrical powers per phase:
P1 = -(V1/Xa) Et sind
Q1 = (V1/Xa) [Et cosd - V1]
These relate the powers per phase to the rotor excitation level and the load angle for a machine of given reaction reactance connected to an infinite bus of voltage V1.\\ Since I am looking at a generator the p will i presume be negative,
///Yes, for overexcited motor mode or it generator counterpart mode.\\ however, I can find no expalination which shows the derivation of how this expression indicates the power delivered to the bus bars, and how it can be used to develope an operating chart for the generator.
///There are several good books covering it, e.g. Say M.G., "Alternating Current Machines," John Wiley & Sons, 1978, on page 371 Operating Characteristics.\\I've tried to construct vector diagrams to assist but keep getting confused with Vt verses EMF
///Vt is V1 terminal voltage, and EMF is Et.\\

 

[Nabeel]

useing the equivalent circuit, Vt is the constant bus (or terminal) voltage per phase and assum it the reference. Ef is the excitation voltage and d is the angle between Vt and Ef (the power angle).
Zs is the synchronous impedance and Xs is the synchronous reactance. Ia' is the conjugate of the current Ia. S is the per phase complex power at the terminal.
Note: &quot;'&quot; means conj , and &quot;<&quot; means angle
Vt=vt <0
Ef=|ef| <d
Zs = Ra + jXs = |Zs| < th
S = Vt * Ia' --------equation 1
Ia' = | ( Ef - Vt)/Zs | = Ef' / Zs' - Vt' / Zs'

= |Ef|<-d / |Zs|<-th - |Vt|<0 / |Zs|<-th

= [|Ef|/|Zs|] <(th-d) - [|Vt|/|Zs|] <th ------equation 2

sub equation 1 in 2 you get:

S = [|Vt|*|Ef| / |Zs|] < (th-d) - [|Vt|^2 / |Zs|] < th
then the real power P and the reactive power Q per phase are:
P = [|Vt|*|Ef| / |Zs|] cos(th-d) - [|Vt|^2 / |Zs|] sin(th)
Q = [|Vt|*|Ef| / |Zs|] sin(th-d) - [|Vt|^2 / |Zs|] sin(th)

then if Ra is neglected (since it is smaller than Xs), then Zs = Xs and th = 90. Finally, the real power P for 3 phases is:
P = 3 * [|Vt|*|Ef| / |Xs|] sin(d) which is the equation you have. Also it equals to Pmax * sin(d)

 

[Nabeel]

Once again, this equation plots the power characteristics of the machine. Xs should be given from the manufacturer (1 - 1.8 for large machines I believe) while Ef can be calculated knowing that Ia is the current drawn from the machine and it depends on the load (but you can calculate the Ia Full load) where Ia = MVA(rated) / [(3^0.333)*Vt]. then Ef=Vt<0 + Ia * jXs = |Ef|<d. Now, to vary the curve you can change the value of Xs or vary the angle d directly from 0 to 180 (d=0 when Ia *jXs=0), where d should be positive incase of generation.
The plot should tell how the power is changing due to the change of d or Xs(the power angle).