Armature size, 7.5 kW 1800 rpm 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).

Step 1: Armature power (losses split at maximum efficiency)

$$\begin{aligned} \text{Input} &= \frac{7.5}{0.83} = 9.036\ \text{kW} \\ I_L &= \frac{9036}{220} = 41.07\ \text{A} \\ I_f &= 0.025 \times 41.07 = 1.027\ \text{A} \\ I_a &= I_L - I_f = 41.07 - 1.027 = 40.05\ \text{A} \\ \text{Total losses} &= 9036 - 7500 = 1536\ \text{W} \end{aligned}$$

At maximum efficiency, variable loss = constant loss, so armature copper loss $= 1536/2 = 768$ W. (Field copper, iron, friction and brush losses form the constant part, so all losses are counted.)

$$\begin{aligned} R_a &= \frac{768}{40.05^2} = 0.479\ \Omega \\ E_b &= V - I_aR_a = 220 - 40.05\times0.479 = 200.8\ \text{V} \\ P_a &= E_b I_a = 200.8 \times 40.05 = 8042\ \text{W} = 8.04\ \text{kW} \end{aligned}$$

Step 2: Specific magnetic loading and output coefficient

The field form factor $K_f = B_{av}/B_m = 0.7$; Sawhney takes $\psi$ (pole arc/pole pitch) $= K_f = 0.7$.

$$\begin{aligned} B_{av} &= K_f B_m = 0.7\times0.9 = 0.63\ \text{Wb/m}^2 \\ C_0 &= \pi^2 B_{av}\, ac\times10^{-3} = \pi^2\times0.63\times30000\times10^{-3} = 186.5 \\ n &= 1800/60 = 30\ \text{rps} \end{aligned}$$ $$D^2L = \frac{P_a}{C_0 n} = \frac{8.04}{186.5\times30} = 1.437\times10^{-3}\ \text{m}^3$$

Step 3: Square pole face

For a square pole face, core length = pole arc:

$$L = \psi\,\tau = \psi\frac{\pi D}{p} = \frac{0.7\pi D}{4} = 0.5498D$$ $$\begin{aligned} D^3 &= \frac{1.437\times10^{-3}}{0.5498} = 2.614\times10^{-3}\ \text{m}^3 \\ D &= 0.1378\ \text{m} \\ L &= 0.5498 \times 0.1378 = 0.0757\ \text{m} \end{aligned}$$

Check

• Pole pitch $\tau = \pi D/4 = 0.108$ m; pole arc $= 0.7\tau = 0.0757$ m $= L$ (square).
• Peripheral speed $= \pi D n = \pi\times0.1378\times30 = 13.0$ m/s (well below 30 m/s).
• Frequency $= pn/2 = 4\times30/2 = 60$ Hz (acceptable for a small machine).

Answer: Armature diameter $D \approx 0.138\ \text{m}\ (138\ \text{mm})$, armature core length $L \approx 0.076\ \text{m}\ (76\ \text{mm})$.