Answer to Question #161843 in Electrical Engineering for Amos chinwendu Mary

Question

A salient pole synchronous generator has the following per unit parameters;xd=1.2,xq=0.8 ra=00.025. compute the excitation voltage Ef on a per unit basis,when the generator is delivering rated kva at rated voltage and at power factor of 0.8 lagging and 0.8 leading

Answer 1

$Xd= 1.2 pu \newline Xq= 0.8 pu\newline Ra= 0.025 pu\newline$

As it's given that rated kva and rated voltage. We have

$Terminal\space voltage \space V=1pu\newline Armature \space current \space=\space1pu$

1) Power factor $= 0.8 lag$

This implies Power factor angle $, \phi=cos^{-1}(0.8) = 36.87\degree$

We have,

$Tan\psi= \frac{Vsin\phi+IaXq}{Vcos\phi+IaRa}=\frac{0.6+0.8}{0.8+0.025}=1.697.$

This implies $\psi=Tan^{-1}(1.697) =59.49\degree.$

Load angle $(\delta) =\psi-\phi=59.49-36.87=22.62\degree$

We have,

$Id =Iasin\psi=sin59.49=0.86\newline Iq=Iacos\phi= cos59. 49=0.51.$

Hence Excitation voltage is,

$E= Vcos\delta+IqRa+IdXd\newline E= cos22. 62+0.51\times 0.025+0.86\times1.2\newline E= 1.97 pu$

There fore $E\angle\delta=1.97\angle22.62 pu$

2) Power factor is 0.8 lead.

This implies Power factor angle $, \phi=cos^{-1}(0.8) = 36.87\degree$

We have,

$Tan\psi= \frac{Vsin\phi-IaXq}{Vcos\phi+IaRa}=\frac{0.6-0.8}{0.8+0.025}=-0.242$

This implies $\psi=Tan^{-1}(-0.242) =-13.63\degree.$

Load angle $(\delta) =\psi+\phi=-13.63+36.87=23.24\degree$

We have,

$Id =Iasin\psi=sin13.63=0.24\newline Iq=Iacos\phi= cos13.63=0.97$

Hence Excitation voltage is,

$E= Vcos\delta+IqRa-IdXd\newline E= cos23.24+0.97\times 0.025-0.24\times1.2\newline E= 0.66pu$

There fore $E\angle\delta=0.66\angle23.24pu$