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

Question #161843

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

1
2021-02-09T07:08:03-0500

"Xd= 1.2 pu \\newline\nXq= 0.8 pu\\newline\nRa= 0.025 pu\\newline"

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

"Terminal\\space voltage \\space V=1pu\\newline\nArmature \\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."

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 \nE= cos22. 62+0.51\\times 0.025+0.86\\times1.2\\newline\nE= 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."

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 \nE= cos23.24+0.97\\times 0.025-0.24\\times1.2\\newline\nE= 0.66pu"

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

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