Distribution Factor MCQ Quiz in मल्याळम - Objective Question with Answer for Distribution Factor - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Winding is chorded by 1 slot, so it is short pitched by $\frac{1}{\frac{66}{6}}=\frac{1}{11}$ or $\frac{{180}^{\circ }}{11}={16.3636}^{\circ }$

Pitch factor, $k_C=\cos \left(\frac{{16.3636}^{\circ }}{2}\right)$

= 0.9898

Slots per pole, $n=\frac{66}{6}=11$

$\beta =\frac{{180}^{\circ }}{11}={16.3636}^{\circ }$

Slots per pole per phase

$=\frac{66}{6\times 3}=3.667$

Distribution factor, $k_d=\frac{\sin \left(\frac{m\beta }{2}\right)}{m\sin \left(\frac{\beta }{2}\right)}$

$=\frac{\sin \left(\frac{11}{3}\times \frac{{180}^{\circ }}{11\times 2}\right)}{\frac{11}{3}\sin \left(\frac{{180}^{\circ }}{11\times 2}\right)}$

$=\frac{\sin {30}^{\circ }}{\frac{11}{3}\sin \left(\frac{{90}^{\circ }}{11}\right)}$

= 0.9582

No. of conductors per phase $=\frac{66\times 6}{3}=132$

No. of turns per phase, $T=\frac{132}{2}=66$

Phase emf generated, $E_{Ph}=4.44k_C{k}_{d}f\varphi T$

Flux per pole, $\varphi =\frac{0.72}{6}=0.12wb$

$\begin{array}{l}f=\frac{6\times 1000}{120}=50H_z\\ E_{Ph}=4.44\times 0.9898\times 0.9582\times 50\times 0.12\times 66\end{array}$

= 1667.56 V

Line Voltage, $E_L=\sqrt{3}{E}_{Ph}=\sqrt{3}\times 1667.56$

= 2888.3 V