Main dimensions of 600 kW square-pole generator

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 generator (Sawhney), $P_a = \dfrac{1+2\eta}{3\eta}P$.

Step 1: $D^2L$

$$\begin{aligned} P_a &= \frac{1+2\times0.91}{3\times0.91}\times600 = 619.8\ \text{kW} \\ C_0 &= \pi^2\times0.6\times35000\times10^{-3} = 207.3 \\ n &= 900/60 = 15\ \text{rps} \\ D^2L &= \frac{619.8}{207.3\times15} = 0.1994\ \text{m}^3 \end{aligned}$$

Step 2: Number of poles (assumed from usual limits)

• Frequency $f = pn/2 \le 50$ Hz $\Rightarrow p \le 6.67$.
• Armature current $\approx 600000/500 = 1200$ A. With a lap winding, current per brush arm $2I_a/p \le 400$ A $\Rightarrow p \ge 6$.

So $p = 6$ ($f = 45$ Hz, 400 A per brush arm).

Step 3: Square pole face

$$L = \psi\frac{\pi D}{p} = \frac{0.75\pi D}{6} = 0.3927D$$ $$\begin{aligned} D^3 &= \frac{0.1994}{0.3927} = 0.5077\ \text{m}^3 \\ D &= 0.798\ \text{m} \\ L &= 0.3927\times0.798 = 0.313\ \text{m} \end{aligned}$$

Check

• Pole pitch $\tau = \pi\times0.798/6 = 0.418$ m; pole arc $= 0.75\tau = 0.313$ m $= L$.
• Peripheral speed $= \pi Dn = \pi\times0.798\times15 = 37.6$ m/s (below 40 m/s).
• Armature mmf per pole $= ac\,\pi D/2p = 35000\times\pi\times0.798/12 = 7310$ A (reasonable).

Answer: $p = 6$, armature diameter $D \approx 0.80$ m, core length $L \approx 0.31$ m.