Solution Electrical Numerical 16

Numerical 16: A 3phase, 4pole 24 slot alternator has its armature coils short-pitched by one slot. Find the distribution factor and pitch factor.

Solution: $\displaystyle \mathrm{n\ =\ \frac{Slots}{Pole}\ =\ \frac{24}{4}\ =\ 6}$ 

$\displaystyle \mathrm{m\ =\ \frac{Slots/pole}{Phase}\ =\ \frac{6}{3}\ =\ 2}$ 

Slot angle, $\displaystyle \mathrm{\beta =\ \frac{180}{n}\ =\ \frac{180}{6}\ =\ 30^o }$ 

Distribution factor $\displaystyle \mathrm{K_d}$, 

$\displaystyle \mathrm{K_d =\ \frac{sin\frac{m\beta }{2}}{m\times sin\frac{\beta }{2}}\ }$ 

$\displaystyle \mathrm{K_d =\ \frac{sin\frac{2\times30}{2}}{2\times sin\frac{30}{2}}\ }$ 

$\displaystyle \mathrm{K_d =\ 0.9659 }$ 

Pitch factor or Coil span factor $\displaystyle \mathrm{K_c}$ ,

$\displaystyle \mathrm{\alpha =\ \beta \times }$ number of slots by which coils are short pitch.

$\displaystyle \mathrm{\alpha =\ 30 \times 1 = 30^o }$ 

$\displaystyle \mathrm{K_c =\ Cos\frac{\alpha }{2} }$ 

$\displaystyle \mathrm{K_c =\ Cos\frac{30}{2} = Cos15^o\ =\ 0.9659 }$