Main dimensions of 200 kW shunt 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, taking armature copper loss as about one-third of total losses (Sawhney),

$$P_a = \frac{1+2\eta}{3\eta}\,P$$

Step 1: Armature power

$$P_a = \frac{1+2\times0.91}{3\times0.91}\times200 = \frac{2.82}{2.73}\times200 = 206.6\ \text{kW}$$

Step 2: Loadings and output coefficient

$$\begin{aligned} B_{av} &= \psi B_m = 0.67\times0.87 = 0.583\ \text{Wb/m}^2 \\ C_0 &= \pi^2\times0.583\times31000\times10^{-3} = 178.3 \\ n &= 1000/60 = 16.67\ \text{rps} \\ D^2L &= \frac{206.6}{178.3\times16.67} = 0.0695\ \text{m}^3 \end{aligned}$$

Step 3: Use $L/\tau = 0.75$

$$\tau = \frac{\pi D}{p} = \frac{\pi D}{6}, \qquad L = 0.75\tau = \frac{0.75\pi D}{6} = 0.3927D$$ $$\begin{aligned} D^3 &= \frac{0.0695}{0.3927} = 0.1770\ \text{m}^3 \\ D &= 0.561\ \text{m} \\ L &= 0.3927\times0.561 = 0.220\ \text{m} \end{aligned}$$

Check

• Pole pitch $\tau = \pi\times0.561/6 = 0.294$ m; pole arc $= 0.67\tau = 0.197$ m.
• Peripheral speed $= \pi D n = \pi\times0.561\times16.67 = 29.4$ m/s (within the usual 30 m/s limit).
• Frequency $= pn/2 = 6\times16.67/2 = 50$ Hz (acceptable).

Answer: Armature diameter $D \approx 0.561$ m, core length $L \approx 0.220$ m.