Question

A 3-phase, 440 V, 50 Hz, six pole salient pole synchronous motor is drawing a 10 A current at 0.8 power factor leading. It has $X_d=10\Omega and{X}_q=7\Omega$. The torque angle $\left(\delta \right)$ is

• A. 9.4${}^o$
• B. 12.1${}^o$
• C. 10.7${}^o$
• D. 13.5${}^o$

Answer 1

The correct option is C 10.7${}^o$

$\overrightarrow{E^{\prime }}=\overrightarrow{V_t}-j\overrightarrow{I_a}{X}_q$

$X_q=7\Omega$
$\text{p.u. value of}{X}_q=\frac{\text{ohmic value}}{\text{base value}}$

$\text{Base ohms}=\frac{\left(\frac{440}{\sqrt{3}}\right)}{10}=25.4\Omega$

$\text{p.u value of}{X}_q=\frac{7}{25.4}=0.2756\text{p.u}$

$\overrightarrow{E}=1-\left(\right(j0.2756)\times 1\angle co{s}^{-1}(0.8\left)\right)$

$=1.18\angle -{10.71}^{\circ }$p.u.

Torque angle $\delta ={10.71}^{\circ }$

$\text{Alternative method:}$
Torque angle, $\delta =ta{n}^{-1}\left[\frac{I{X}_q\cos \theta }{V+I{X}_q\sin \theta }\right]$

$=ta{n}^{-1}\left[\frac{10\times 7\times cos\left({36.87}^{\circ }\right)}{254+(10\times 7sin({36.87}^{\circ }\left)\right)}\right]$

$\delta ={10.71}^{\circ }$