Question 2014.S3.37: A non-salient pole synchronous generator having synchronous ......
A non-salient pole synchronous generator having synchronous reactance of 0.8 pu is supplying 1 pu power to a unity power factor load at a terminal voltage of 1.1 pu. Neglecting the armature resistance, the angle of the voltage behind the synchronous reactance with respect to the angle of the terminal voltage in degrees is —————–
Given that P = 1 pu, V_{ T }=1.1 pu, X_{ s }=0.8 pu , R_{ a }=0, pf =1, So, we have
\begin{gathered}P=\frac{E V}{X_{ s }} \sin \delta \\E=\left[\left(V \cos \phi+I_{ a } R_{ a }\right)^2+\left(V \sin \phi+I_{ a } X_{ s }\right)^2\right]^{1 / 2}\end{gathered}
We know that P=V I \cos \phi \Rightarrow I_{ a }=\frac{1}{1.1 \times 1}=0.91
As \cos\phi=1\Rightarrow\phi=0\Rightarrow\sin\phi=0
\begin{aligned}E & =\left[\{(1.1 \times 1)+(0.91 \times 0)\}^2+\left\{(1.1 \times 0)+(0.91 \times 0.8)^2\right\}^2\right] \\& =\left[(1.1)^2+0.728^2\right]^{1 / 2}=1.319 V \\P & =\frac{1.319 \times 1.1}{0.8} \sin \delta \Rightarrow \delta=33.46^{\circ}\end{aligned}