2.2 kW 440 V cage motor (star-delta) design

Data: 2.2 kW, 440 V, 50 Hz, 1430 rpm, so 4 poles and $n_s = 1500/60 = 25$ rps; $\eta = 0.8$, $\cos\phi = 0.85$; $B_{av} = 0.44$ Wb/m², $ac = 21000$ A/m, $L/\tau = 1.5$. Assumptions: winding factor $K_w = 0.955$; star-delta starter, so the motor runs in delta ($E_{ph} = 440$ V); maximum tooth flux density 1.7 Wb/m²; stacking factor 0.9.

i) Main dimensions

$$\begin{aligned} Q &= \frac{\text{kW}}{\eta\cos\phi} = \frac{2.2}{0.8 \times 0.85} = 3.235\ \text{kVA} \\ C_0 &= 11B_{av}\,ac\,K_w\times10^{-3} = 11 \times 0.44 \times 21000 \times 0.955 \times 10^{-3} = 97.07 \\ D^2L &= \frac{Q}{C_0n_s} = \frac{3.235}{97.07 \times 25} = 1.333\times10^{-3}\ \text{m}^3 \end{aligned}$$

With $L = 1.5\tau = 1.5\pi D/4 = 1.178D$:

$$\begin{aligned} D^3 &= \frac{1.333\times10^{-3}}{1.178} = 1.131\times10^{-3} \\ D &= 0.1042\ \text{m}, \qquad L = 1.178 \times 0.1042 = 0.1228\ \text{m} \end{aligned}$$

Peripheral speed $= \pi \times 0.1042 \times 25 = 8.2$ m/s (well within limit).

ii) Stator turns per phase

$$\begin{aligned} \tau &= \frac{\pi \times 0.1042}{4} = 0.0819\ \text{m} \\ \Phi &= B_{av}\tau L = 0.44 \times 0.0819 \times 0.1228 = 4.421\times10^{-3}\ \text{Wb} \\ T_s &= \frac{E_{ph}}{4.44f\Phi K_w} = \frac{440}{4.44 \times 50 \times 4.421\times10^{-3} \times 0.955} = 469.4 \end{aligned}$$

iv) Number of stator slots (needed before rounding turns)

Slot pitch 12–15 mm:

• $q = 2$: $S_s = 24$, slot pitch $= \pi \times 104.2/24 = 13.64$ mm (OK)
• $q = 3$: $S_s = 36$, slot pitch $= 9.09$ mm (too small)

Take 24 slots. Conductors per slot $= 6 \times 469.4/24 = 117.4$, take 118. Total conductors $Z_s = 24 \times 118 = 2832$, so $T_s = 472$.

iii) Size of stator conductor

$$\begin{aligned} I_{ph} &= \frac{Q\times10^3}{3E_{ph}} = \frac{3235.3}{3 \times 440} = 2.451\ \text{A} \\ a_s &= \frac{I_{ph}}{\delta_s} = \frac{2.451}{4} = 0.613\ \text{mm}^2 \end{aligned}$$

Bare diameter $= \sqrt{4 \times 0.613/\pi} = 0.88$ mm (use the nearest standard SWG wire).

v) Area and dimensions of each slot

$$A_{slot} = \frac{\text{cond/slot} \times a_s}{\text{space factor}} = \frac{118 \times 0.613}{0.4} = 180.8\ \text{mm}^2$$

With $d = 4w$: $4w^2 = 180.8$, so $w = 6.72$ mm and $d = 26.9$ mm.

vi) Minimum width of stator teeth

Net iron length $L_i = 0.9 \times 0.1228 = 0.1105$ m; teeth per pole $= 24/4 = 6$.

$$W_{ts(min)} = \frac{\Phi}{1.7 \times \frac{S_s}{P} \times L_i} = \frac{4.421\times10^{-3}}{1.7 \times 6 \times 0.1105} = 3.92\ \text{mm}$$

Tooth width at the gap $= 13.64 - 6.72 = 6.9$ mm $> 3.92$ mm, so tooth density is within limits.

Results