--- title: "2.2 kW 440 V cage motor (star-delta) design" subject: "Electric Machine Design" chapter: 4 key: "semester_5/electric_machine_design:design-2-2kw-440v-star-delta" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-2-2kw-440v-star-delta site: BCT Engineering --- # 2.2 kW 440 V cage motor (star-delta) design Electric Machine Design · Chapter 4: Three phase induction motor design - Asked in: 2071 Chaitra, 16 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-2-2kw-440v-star-delta - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4 ## Question For a 2.2 kW, 440 V, 3 phase, 50 Hz, 1430 rpm squirrel cage induction motor using star delta starter and having efficiency of 0.8 and power factor 0.85 at full load the following data are given: Specific magnetic loading = 0.44 Wb/m²; Specific electric loading = 21000 A/m; Slot space factor = 0.4; Ratio of core length to pole pitch = 1.5; Ratio of slot depth to width = 4; Stator slot pitch = 12 to 15 mm; Current density in conductor = 4 A/mm². Assuming all other required data, calculate: i) Main dimensions of the machine ii) No. of stator turns per phase iii) Size of stator conductor iv) No. of stator slots v) Area and dimensions of each stator slot vi) Minimum width of stator teeth ## Short answer Using Q=3.235 kVA and C0=97.07 yields D=0.1042 m, L=0.1228 m. Stator turns 472 (24 slots). Conductor area 0.613 mm² (0.88 mm dia); slot 6.72×26.9 mm; minimum tooth width 3.92 mm. ## Answer **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 | Quantity | Value | |---|---| | $D$, $L$ | 10.42 cm, 12.28 cm | | Turns per phase | 472 | | Conductor area | 0.613 mm² (0.88 mm dia) | | Stator slots | 24 | | Slot area | 180.8 mm² | | Slot size | 6.72 mm × 26.9 mm | | $W_{ts(min)}$ | 3.92 mm | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. 1430 rpm at 50 Hz gives what poles and ns? A. 2 poles, 50 rps B. 6 poles, 16.7 rps C. 4 poles, 23.8 rps D. 4 poles, ns = 25 rps Answer: D. Nearest synchronous speed above 1430 rpm is 1500 rpm. (Key point: Poles from speed) 2. Which sequence gives D^2L? A. Q = kW/(eta cos phi) = 3.235 kVA, then D^2L = Q/(C0 ns) B. D^2L = kW/(C0 ns) C. D^2L = Q x C0 x ns D. D^2L = Q/C0 only Answer: A. The output equation Q = C0 D^2 L ns is rearranged, with C0 = 97.07. (Key point: D^2L calculation) 3. Why is 24 slots chosen rather than 36? A. 36 slots give a pitch above 15 mm B. The slot pitch must be 12-15 mm; 24 gives 13.64 mm while 36 gives 9.09 mm C. 24 is the only multiple of 3 D. Slots must equal poles x 6 Answer: B. With q = 2, slot pitch = pi D/24 is within limits. (Key point: Choosing slot number) 4. Why is Ts changed from 469.4 to 472? A. Because the winding factor changed B. To lower the current density C. Conductors per slot must be a whole number: 118 x 24 = 2832, so Ts = 472 D. To reduce D Answer: C. Conductors per slot must be an integer (117.4 becomes 118). 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