30 kW 240 V generator, 2.5% drop

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).

The full-load armature voltage drop $I_aR_a$ is 2.5% of terminal voltage. Field form factor $K_f = B_{av}/B_m = 0.7$ (equal to $\psi$). Since no length ratio is given, the standard assumption of a square pole face (L = pole arc) is used.

Step 1: Armature power from currents and emf

$$\begin{aligned} I_L &= \frac{30000}{240} = 125.0\ \text{A} \\ I_f &= \frac{240}{120} = 2.0\ \text{A} \\ I_a &= I_L + I_f = 127.0\ \text{A} \\ E &= V + I_aR_a = 240 + 0.025\times240 = 246.0\ \text{V} \\ P_a &= E I_a = 246.0\times127.0 = 31.24\ \text{kW} \end{aligned}$$

Step 2: Loadings and $D^2L$

$$\begin{aligned} B_{av} &= \psi B_m = 0.7\times0.85 = 0.595\ \text{Wb/m}^2 \\ C_0 &= \pi^2\times0.595\times20000\times10^{-3} = 117.4 \\ n &= 900/60 = 15\ \text{rps} \\ D^2L &= \frac{31.24}{117.4\times15} = 0.01773\ \text{m}^3 \end{aligned}$$

Step 3: Square pole face

$$L = \psi\frac{\pi D}{p} = \frac{0.7\pi D}{4} = 0.5498D$$ $$\begin{aligned} D^3 &= \frac{0.01773}{0.5498} = 0.03226\ \text{m}^3 \\ D &= 0.3183\ \text{m} \\ L &= 0.5498\times0.3183 = 0.1750\ \text{m} \end{aligned}$$

Check

• Pole pitch $\tau = \pi D/4 = 0.250$ m; pole arc $= 0.7\tau = 0.1750$ m $= L$.
• Peripheral speed $= \pi Dn = 15.0$ m/s; frequency $= pn/2 = 30$ Hz. Both acceptable.

Answer: Armature diameter $D \approx 0.318$ m, armature core length $L \approx 0.175$ m.