Question

A 4-pole, 50 Hz alternator has double layer winding with 10 turns per coil and number of slots are 60. If each coil is short pitched by an angle of ${36}^{\circ }$ electrical. The fundamental flux per pole is 0.015 Wb. The line to line induced emf for a $3\varphi$ star connection is approximately

• A. 1818.9 V
• B. 1050.2 V
• C. 606.33 V
• D. 350.14 V

Answer 1

The correct option is B 1050.2 V

For double layer winding,
No.of slots = No.of coils
Total number of turns = $60\times 10=600$

Turns per phase =$\frac{600}{3}=200$

Pitch factor $\left(K_c\right)=cos{18}^{\circ }=0.951$

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

$m=\frac{60}{4}\times \frac{1}{3}=5$

$\beta =\frac{180}{60/4}={12}^{\circ }$

$K_d=\frac{sin\frac{5\times 12}{2}}{5sin\frac{12}{2}}=0.9567$

Induced emf, $E_{ph}=\sqrt{2}\pi K_w\varphi f{T}_{ph}$

$E_{ph}=\sqrt{2}\pi \times 0.9567\times 0.951\times 0.015\times 50\times 200$

$E_{ph}=606.33V$

$E_{L-L}=1.05kV$