Question

A $3-\varphi ,50Hz$, star connected alternator with 2 layer winding is running at 1000 rpm. It has 12 turns/coil, 4 slots/poles/phase and a coil pitch of 8 slots. The winding factor at stator will be_________ (Answer upto 3 decimal places)

• A. 0.826

Answer 1

The correct option is A 0.826
$\text{Number of poles,}P=\frac{120\times f}{N}=\frac{120\times 50}{1000}=6$

$\text{Total number of slots}=4\times 6\times 3=72$

$\text{Conductor per phase,}{N}_{ph}=\frac{72\times 12}{3}=283$

$\text{Phase spread =No.of slots per pole per phase}\times \text{Angular slot pitch ..(i)}$

$\text{Also for 3}-\varphi \text{system,}\text{phase spread}=\frac{180}{3}={60}^{\circ }\text{..(ii)}$

$\text{Using equation (i) and (ii)}$

${60}^{\circ }=4\times \gamma$

$\gamma ={15}^{\circ }$

$k_d=\frac{sin\frac{4\times {15}^{\circ }}{2}}{4sin\frac{{15}^{\circ }}{2}}=\frac{sin{30}^{\circ }}{4sin7.5}=0.9576$

$\text{Pole pitch = 12 slots}$

$\text{Coil pitch}=8slots=8\times {15}^{\circ }={120}^{\circ }=({180}^{\circ }-{60}^{\circ })$

$\alpha ={60}^{\circ }$

$\text{Coil factor,}{k}_c=cos\frac{\alpha }{2}=0.866$

$\text{Winding factor}=k_c\times k_d=0.866\times 0.9576$

$=0.829$