Question

A 3-phase synchronous generator has a direct axis synchronous reactance of 0.8 p.u and quadrature axis synchronous reactance of 0.5 p.u. The generator is supplying full load at 0.8 lagging p.f. at 1.0 p.u. terminal voltage____p.u will be the no load voltage if the excitation remains unchanged?

• A. 1.6

Answer 1

The correct option is A 1.6
Given,
$V_t=1.0p.u.$
$I_a=1.0p.uat0.8pflagging$
$\varphi =co{s}^{-1}0.8={36.86}^{\circ }$
$X_d=0.8p.u.$
$X_q=0.5p.u.$

As we can use the relation,
$tan\psi =\frac{V_{t}sin\varphi +I_a{X}_q}{V_{t}cos\varphi +I_a{r}_a}$

$\because Here{r}_a=0$

$tan\psi =\frac{1\times 0.6+1\times 0.5}{1\times 0.8+0}$

or $tan\psi =1.375$
$\psi ={53.97}^{\circ }$

$\because$ Power angle, $\delta =\psi -\varphi$
$={53.97}^{\circ }-{36.86}^{\circ }$
$={17.11}^{\circ }$

We can write,
No load voltage, $E_f=V_{t}cos\delta +I_d{X}_d$

$=V_{t}cos\delta +\left(I_{a}sin\psi \right)X_d$

$=1\times cos{17.11}^{\circ }+(1\times sin{53.97}^{\circ })\times 0.8$

$=1.602p.u\approx 1.60p.u.$