Armature dimensions of 50 kW shunt generator

Assumption: the speed is read as 600 rpm (the standard textbook version of this problem). At 6000 rpm a 4-pole machine would have a flux frequency of 200 Hz and a peripheral speed near 57 m/s, which is not practical; the method is the same either way.

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).

Step 1: Armature power

$$\begin{aligned} I_L &= \frac{50000}{220} = 227.27\ \text{A} \\ I_f &= 0.01\times227.27 = 2.27\ \text{A} \\ I_a &= I_L + I_f = 229.55\ \text{A} \\ E &= 220 + 0.03\times220 = 226.6\ \text{V} \\ P_a &= E I_a = 226.6\times229.55 = 52.02\ \text{kW} \end{aligned}$$

Step 2: Loadings and $D^2L$

$$\begin{aligned} B_{av} &= \psi B_m = 0.67\times0.83 = 0.556\ \text{Wb/m}^2 \\ C_0 &= \pi^2\times0.556\times30000\times10^{-3} = 164.7 \\ n &= 600/60 = 10\ \text{rps} \\ D^2L &= \frac{52.02}{164.7\times10} = 0.03159\ \text{m}^3 \end{aligned}$$

Step 3: Square pole face

$$L = \psi\frac{\pi D}{p} = \frac{0.67\pi D}{4} = 0.5262D$$ $$\begin{aligned} D^3 &= \frac{0.03159}{0.5262} = 0.06003\ \text{m}^3 \\ D &= 0.392\ \text{m} \\ L &= 0.5262\times0.392 = 0.206\ \text{m} \end{aligned}$$

Check

• Pole pitch $\tau = \pi\times0.392/4 = 0.308$ m; pole arc $= 0.67\tau = 0.206$ m $= L$.
• Peripheral speed $= \pi Dn = 12.3$ m/s; frequency $= 4\times10/2 = 20$ Hz. Both acceptable.

Answer (600 rpm): $D \approx 0.39$ m, $L \approx 0.21$ m. (If 6000 rpm is used literally: $D^2L = 0.00316$ m³, giving $D \approx 0.182$ m and $L \approx 0.096$ m.)