Main dimensions and poles of 37 kW shunt motor

Data: 37 kW, 230 V, 1400 rpm ($n = 23.33$ rps) shunt motor, $B_{av} = 0.5$ Wb/m², $ac = 22000$ A/m, $\eta = 0.9$, $\psi = 0.7$, square pole face.

Armature power

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

$$P_a = P\,\frac{2+\eta}{3\eta} = 37 \times \frac{2.9}{2.7} = 39.74\ \text{kW}$$

Output coefficient and $D^2L$

$$\begin{aligned} C_0 &= \pi^2B_{av}\,ac\times10^{-3} = \pi^2 \times 0.5 \times 22 = 108.57 \\ D^2L &= \frac{P_a}{C_0n} = \frac{39.74}{108.57 \times 23.33} = 0.01569\ \text{m}^3 \end{aligned}$$

Square pole face

$L = \text{pole arc} = \psi\tau = 0.7\pi D/P$, so $D^3 = D^2L \cdot P/(0.7\pi)$.

Choice of number of poles

Full-load current $\approx 37000/(0.9 \times 230) = 178.7$ A.

• 6 poles: frequency 70 Hz exceeds the usual 25–50 Hz limit (high iron loss).
• 2 poles: frequency 23.3 Hz is below the usual range; the core is long ($L > D$) and armature mmf per pole is high ($ac\,\pi D/2P \approx 4190$ A).
• 4 poles: frequency 46.7 Hz is within range, current per path 45 A, armature mmf per pole $= 22000 \times \pi \times 0.306/8 = 2640$ A, peripheral speed $= \pi \times 0.306 \times 23.33 = 22.4$ m/s (below 30 m/s).

Choose $P = 4$. Then

$$\begin{aligned} L &= \frac{0.7\pi D}{4} = 0.5498D \\ D^3 &= \frac{0.01569}{0.5498} = 0.02853 \Rightarrow D = 0.306\ \text{m} \\ L &= 0.5498 \times 0.306 = 0.168\ \text{m} \end{aligned}$$

Answer: 4 poles, $D \approx 30.6$ cm, $L \approx 16.8$ cm.