2.2 kW cage motor design with stator core data

Data: 2.2 kW, 400 V, 50 Hz, 1420 rpm (4 poles, $n_s = 25$ rps), $\eta = 0.8$, $\cos\phi = 0.825$, $K_w = 0.955$, $L/\tau = 1.5$, $d/w = 2$. Assumptions: delta-connected stator ($E_{ph} = 400$ V); maximum tooth flux density 1.7 Wb/m²; stacking factor 0.9.

i) Main dimensions

$$\begin{aligned} Q &= \frac{2.2}{0.8 \times 0.825} = 3.333\ \text{kVA} \\ C_0 &= 11 \times 0.44 \times 21000 \times 0.955 \times 10^{-3} = 97.07 \\ D^2L &= \frac{3.333}{97.07 \times 25} = 1.374\times10^{-3}\ \text{m}^3 \end{aligned}$$

$L = 1.5 \times \pi D/4 = 1.178D$:

$$D^3 = 1.166\times10^{-3} \Rightarrow D = 0.1053\ \text{m}, \quad L = 0.124\ \text{m}$$

ii) Stator conductors and area of conductor

$$\begin{aligned} \tau &= \pi \times 0.1053/4 = 0.0827\ \text{m} \\ \Phi &= 0.44 \times 0.0827 \times 0.124 = 4.51\times10^{-3}\ \text{Wb} \\ T_s &= \frac{400}{4.44 \times 50 \times 4.51\times10^{-3} \times 0.955} = 418.3 \end{aligned}$$

Stator conductors $= 6T_s \approx 2510$; after choosing slots (below), $Z_s = 24 \times 105 = 2520$ ($T_s = 420$).

$$I_{ph} = \frac{3333.3}{3 \times 400} = 2.778\ \text{A}, \qquad a_s = \frac{2.778}{4} = 0.694\ \text{mm}^2$$

iii) Number of stator slots

Slot pitch 12–15 mm. $q = 2$ gives $S_s = 24$ and slot pitch $= \pi \times 105.3/24 = 13.78$ mm. 24 slots, $2510/24 = 104.6$, so 105 conductors per slot.

iv) Area and dimensions of each slot

$$A_{slot} = \frac{105 \times 0.694}{0.4} = 182.3\ \text{mm}^2$$

$d = 2w$: $2w^2 = 182.3$, $w = 9.55$ mm, $d = 19.1$ mm.

v) Minimum width of stator teeth

$L_i = 0.9 \times 0.124 = 0.1116$ m; teeth per pole $= 6$:

$$W_{ts(min)} = \frac{4.51\times10^{-3}}{1.7 \times 6 \times 0.1116} = 3.96\ \text{mm}$$

Tooth width at the gap $= 13.78 - 9.55 = 4.2$ mm $> 3.96$ mm, so it is just safe.

(The given core density 1.2 Wb/m² fixes the stator core depth: $d_{cs} = \Phi/(2 \times 1.2 \times L_i) = 16.8$ mm.)

Answer: $D = 10.53$ cm, $L = 12.4$ cm; 2520 conductors ($T_s = 420$), $a_s = 0.694$ mm²; 24 slots; slot 182.3 mm² ($9.55 \times 19.1$ mm); $W_{ts(min)} = 3.96$ mm.