Main dimensions of 5 kW generator (Bav 0.42)

Data: 5 kW, 250 V, 4 poles, 1500 rpm ($n = 25$ rps) shunt generator, $\eta = 0.87$, $B_{av} = 0.42$ Wb/m², $ac = 15000$ A/m, $\psi = 0.66$, square pole face.

Armature power

Taking armature copper loss as one-third of total losses (generator):

$$P_a = P\,\frac{1+2\eta}{3\eta} = 5 \times \frac{1 + 2 \times 0.87}{3 \times 0.87} = 5 \times \frac{2.74}{2.61} = 5.249\ \text{kW}$$

Output coefficient

$$C_0 = \pi^2B_{av}\,ac\times10^{-3} = \pi^2 \times 0.42 \times 15000 \times 10^{-3} = 62.18$$

$D^2L$

$$D^2L = \frac{P_a}{C_0n} = \frac{5.249}{62.18 \times 25} = 3.377\times10^{-3}\ \text{m}^3$$

Square pole face

Pole arc $= L$ and pole arc $= \psi\tau = 0.66 \times \pi D/4 = 0.5184D$, so $L = 0.5184D$:

$$\begin{aligned} D^3 &= \frac{3.377\times10^{-3}}{0.5184} = 6.514\times10^{-3}\ \text{m}^3 \\ D &= 0.1868\ \text{m} \\ L &= 0.5184 \times 0.1868 = 0.0968\ \text{m} \end{aligned}$$

Checks

• Pole pitch $\tau = \pi D/4 = 0.1467$ m; pole arc $= 0.66\tau = 0.0968$ m $= L$ (square face).
• Peripheral speed $= \pi Dn = 14.7$ m/s (below 30 m/s).
• Frequency $= Pn/2 = 50$ Hz (upper limit of the normal range).

Answer: $D \approx 18.7$ cm, $L \approx 9.68$ cm.