Main dimensions and poles of 40 kW shunt motor

The main dimensions come from the output equation of a dc machine (A.K. Sawhney):

$$P_a = C_0 D^2 L n, \qquad C_0 = \pi^2 B_{av}\, ac \times 10^{-3}\ \text{kW/(m}^3\text{ rps)}$$

where $P_a$ = armature power (kW), $D$ = armature diameter, $L$ = core length (m), $n$ = speed (rps).

For a motor, Sawhney takes the armature copper loss as about one-third of the total losses, which gives

$$P_a = \frac{2+\eta}{3\eta}\,P$$

Step 1: Armature power and $D^2L$

$$\begin{aligned} P_a &= \frac{2+0.92}{3\times0.92}\times40 = 42.32\ \text{kW} \\ C_0 &= \pi^2 \times 0.5 \times 22000\times10^{-3} = 108.6 \\ n &= 1450/60 = 24.17\ \text{rps} \\ D^2L &= \frac{42.32}{108.6\times24.17} = 0.01613\ \text{m}^3 \end{aligned}$$

Step 2: Square pole face

Core length = pole arc: $L = \psi\pi D/p = 0.7\pi D/p$, so

$$D^3 = \frac{D^2L\; p}{0.7\pi}$$

Step 3: Choice of number of poles

Try common pole numbers and check frequency ($f = pn/2 \le 50$ Hz) and peripheral speed ($\le 30$ m/s):

• $p = 6$ gives $f = 72.5$ Hz, too high (large iron loss).
• $p = 2$ gives a long, narrow machine ($L > D$) with heavy armature reaction per pole and large yoke; not used for 40 kW.
• $p = 4$ keeps $f$ below 50 Hz with a reasonable $L/D$. Armature current $\approx 40000/(0.92\times240) = 181$ A, so with a lap winding the current per path ($\approx 45$ A) and current per brush arm ($\approx 90$ A) are small.

Choose 4 poles.

Step 4: Main dimensions for $p = 4$

$$\begin{aligned} D^3 &= \frac{0.01613\times4}{0.7\pi} = 0.02934\ \text{m}^3 \\ D &= 0.308\ \text{m} \\ L &= \frac{0.7\pi\times0.308}{4} = 0.170\ \text{m} \\ \tau &= \frac{\pi D}{p} = 0.242\ \text{m}, \quad \text{pole arc} = 0.7\tau = 0.170\ \text{m} \end{aligned}$$

Answer: Number of poles = 4, armature diameter $D \approx 0.308$ m, core length $L \approx 0.170$ m (pole face 0.170 m x 0.170 m).