Armature size, 7.5 kW 1000 rpm motor

Data: 7.5 kW, 220 V shunt motor, 4 poles, 1000 rpm ($n = 16.67$ rps), $\eta = 0.83$, $B_g = 0.9$ Wb/m², $ac = 30000$ A/m, field form factor $\psi = 0.7$, $I_f = 2.5\%$ of rated current, maximum efficiency at full load, square pole face.

Armature power $P_a$

Input and losses:

$$\begin{aligned} P_{in} &= \frac{7500}{0.83} = 9036.1\ \text{W}, \qquad \text{Losses} = 9036.1 - 7500 = 1536.1\ \text{W} \end{aligned}$$

Maximum efficiency at full load means variable (armature copper) loss = constant loss:

$$I_a^2R_a = \frac{1536.1}{2} = 768.1\ \text{W}$$

Currents:

$$\begin{aligned} I &= \frac{9036.1}{220} = 41.07\ \text{A}, \qquad I_f = 0.025 \times 41.07 = 1.03\ \text{A} \\ I_a &= 41.07 - 1.03 = 40.05\ \text{A} \end{aligned}$$

Back emf:

$$\begin{aligned} I_aR_a &= \frac{768.1}{40.05} = 19.18\ \text{V} \\ E_b &= 220 - 19.18 = 200.82\ \text{V} \\ P_a &= E_bI_a = 200.82 \times 40.05 = 8042\ \text{W} = 8.042\ \text{kW} \end{aligned}$$

Output coefficient

$$\begin{aligned} B_{av} &= \psi B_g = 0.7 \times 0.9 = 0.63\ \text{Wb/m}^2 \\ C_0 &= \pi^2 B_{av}\,ac \times 10^{-3} = \pi^2 \times 0.63 \times 30000 \times 10^{-3} = 186.54 \end{aligned}$$

$D^2L$

$$D^2L = \frac{P_a}{C_0 n} = \frac{8.042}{186.54 \times 16.67} = 2.587 \times 10^{-3}\ \text{m}^3$$

Square pole face

Pole arc $= L$, and pole arc $= \psi\tau = 0.7 \times \pi D/4$:

$$L = 0.5498D$$ $$D^3 = \frac{2.587\times10^{-3}}{0.5498} = 4.705\times10^{-3} \Rightarrow D = 0.1676\ \text{m}$$ $$L = 0.5498 \times 0.1676 = 0.0921\ \text{m}$$

Check: peripheral speed $= \pi \times 0.1676 \times 16.67 = 8.8$ m/s (acceptable).

Answer: $D \approx 16.8$ cm, $L \approx 9.2$ cm.