Design 11 kW motor from 15 kW motor loadings

Method: find the specific loadings (as the output coefficient $C_0$) from the 15 kW motor, then use the same $C_0$ for the 11 kW motor.

Output equation: $Q = C_0 D^2 L n_s$, $C_0 = 11 B_{av}\, ac\, K_w \times 10^{-3}$.

Specific loadings from the 15 kW motor

1480 rpm at 50 Hz means 4 poles, $n_s = 1500/60 = 25$ rps.

$$\begin{aligned} Q_1 &= \frac{15}{0.84 \times 0.85} = 21.01\ \text{kVA} \\ C_0 &= \frac{Q_1}{D^2 L n_s} = \frac{21.01}{0.25^2 \times 0.16 \times 25} = 84.03 \\ B_{av} &= \frac{C_0}{11\, ac\, K_w \times 10^{-3}} = \frac{84.03}{11 \times 23000 \times 0.955 \times 10^{-3}} = 0.348\ \text{Wb/m}^2 \end{aligned}$$

(i) Main dimensions of the 11 kW motor

6 poles: $n_s = 1000/60 = 16.67$ rps.

$$\begin{aligned} Q &= \frac{11}{0.84 \times 0.85} = 15.41\ \text{kVA} \\ D^2L &= \frac{Q}{C_0 n_s} = \frac{15.41}{84.03 \times 16.67} = 0.0110\ \text{m}^3 \end{aligned}$$

Assume $L = \tau$ (good overall design), $\tau = \pi D/6$:

$$D^3 = \frac{0.0110 \times 6}{\pi} = 0.02101 \Rightarrow D = 0.276\ \text{m}, \quad L = \tau = 0.1445\ \text{m}$$

(ii) Number of stator slots

Slot pitch 15–25 mm. With $q = 2$: $S_s = 3 \times 6 \times 2 = 36$, slot pitch $= \pi \times 275.9/36 = 24.1$ mm (within range). Take 36 slots.

(iii) Stator conductors and conductor area

$$\begin{aligned} \Phi &= B_{av}\tau L = 0.348 \times 0.1445 \times 0.1445 = 7.26 \times 10^{-3}\ \text{Wb} \\ T_s &= \frac{E_{ph}}{4.44 f \Phi K_w} = \frac{460}{4.44 \times 50 \times 7.26\times10^{-3} \times 0.955} = 298.9 \end{aligned}$$

(Delta connection, so $E_{ph} = 460$ V.) Conductors $= 6T_s = 1793$; per slot $= 1793/36 = 49.8$, take 50. Total conductors $Z_s = 36 \times 50 = 1800$, $T_s = 300$.

$$\begin{aligned} I_{ph} &= \frac{15406}{3 \times 460} = 11.16\ \text{A} \\ a_s &= \frac{11.16}{4} = 2.79\ \text{mm}^2 \ (\text{dia} \approx 1.89\ \text{mm}) \end{aligned}$$

Slot area:

$$A_{slot} = \frac{50 \times 2.79}{0.4} = 348.9\ \text{mm}^2$$

(iv) Slot dimensions

$d = 3w$, so $3w^2 = 348.9$: $w = 10.8$ mm, $d = 32.4$ mm.

(v) Minimum width of stator teeth

Maximum tooth density 1.7 Wb/m², net iron length $L_i = 0.9L = 0.130$ m, teeth per pole $= 36/6 = 6$:

$$W_{ts(min)} = \frac{\Phi}{1.7 \times \frac{S_s}{P} \times L_i} = \frac{7.26\times10^{-3}}{1.7 \times 6 \times 0.130} = 5.47\ \text{mm}$$

(Stator core depth with $B_c = 1.2$ Wb/m²: $d_{cs} = \Phi/(2B_cL_i) = 23.3$ mm.)

Answer: $D = 27.6$ cm, $L = 14.45$ cm; 36 slots; 1800 conductors (50 per slot, $T_s = 300$), $a_s = 2.79$ mm², slot area 349 mm²; slot $10.8 \times 32.4$ mm; $W_{ts(min)} = 5.47$ mm.