Question

A 3-phase, 100 MVA star connected, salient pole synchronous motor has following specification.
Excitation voltage: $E_f=1.5p.u$
Terminal voltage: V = 1 p.u.
Direct axis synchronous reactance,$X_d=1p.u.$
Quadrature axis synchronous reactance, $X_q=0.5p.u.$
Keeping that motor does not loose synchronism, the maximum mechanical power output supplied by motor is

• A. 1.74 p.u.
• B. 3.55 p.u.
• C. 1.55 p.u.
• D. 2.14 p.u.

Answer 1

The correct option is A 1.74 p.u.

Salient pole motor's output power is given by

$P=\frac{E_{f}V}{X_d}\sin \delta +V^2.\frac{(X_d-X_q)}{2X_d{X}_q}\sin 2\delta$

$P=1.5\sin \delta +\frac{(1-0.5)}{2\times 1\times 0.5}\sin 2\delta$

$P=1.5\sin \delta +0.5\sin 2\delta$
For maximum power output,

$\frac{dP}{d\delta }=0$
$\frac{dP}{d\delta }=1.5\cos \delta +\cos 2\delta =0$

$2{\cos }^2\delta -1+1.5\cos \delta =0$

Let $\cos \delta =x$
Then, ${\cos }^2\delta =x^2$
$2x^2+1.5x-1=0$
$x=\frac{-1.5\pm \sqrt{2.25+8}}{4}$
$x=0.4253,-1.1757\text{(Not possible)}$
Then $\cos \delta =0.4253$
$\delta ={\cos }^{-1}\left(0.4253\right)$
$\delta ={64.84}^{\circ }$
$\therefore P_{max}=1.5sin\left({64.84}^{\circ }\right)+0.5sin\left({129.68}^{\circ }\right)$
$=1.74p.u.$