--- title: "2.2 kW cage motor design with stator core data" subject: "Electric Machine Design" chapter: 4 key: "semester_5/electric_machine_design:design-2-2kw-winding-factor" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-2-2kw-winding-factor site: BCT Engineering --- # 2.2 kW cage motor design with stator core data Electric Machine Design · Chapter 4: Three phase induction motor design - Asked in: 2070 Asar, 12 marks - Video: [5.5. Stator design in induction machines - Design of stator winding](https://www.youtube.com/watch?v=JrJwhbPJU4U) — Nisha Prakash - 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-winding-factor - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4 ## Question For a 2.2 kW, 400 V, 3 phase, 50 Hz, 1420 rpm squirrel cage induction motor having efficiency of 0.8 and power factor 0.825 at full load the following data are given: Specific magnetic loading = 0.44 Wb/m²; Specific electric loading = 21,000 A/m; Winding factor = 0.955; Slot space factor = 0.4; Ratio of core length to pole pitch = 1.5; Ratio of slot depth to width = 2; Stator slot pitch = 12 to 15 mm; Current density in conductor = 4 A/mm²; Flux density in stator core = 1.2 Wb/m². Assuming all other required parameters, calculate: i) Main dimension of the machine ii) No. of stator conductors and area of stator slot iii) No. of stator slots iv) Area and dimension of each stator slot v) Minimum width of stator teeth ## Short answer Main dimensions: D=10.53 cm, L=12.4 cm. Stator has 2520 conductors (420 turns/phase), conductor area 0.694 mm², in 24 slots. Slot area 182.3 mm² (9.55×19.1 mm). Minimum stator tooth width is 3.96 mm. ## Answer **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With d/w = 2 and slot area 182.3 mm^2, how are w and d found? A. w^2 = 182.3, giving w = 13.5 mm B. 4w^2 = 182.3, giving w = 6.75 mm C. w = 182.3/2 D. 2w^2 = 182.3, giving w = 9.55 mm and d = 19.1 mm Answer: D. Depth is twice the width, so area = w x 2w. (Key point: Slot shape ratio) 2. Why is the design 'just safe' on tooth density? A. Tooth width at gap 13.78 - 9.55 = 4.2 mm is only slightly above the minimum 3.96 mm B. Tooth width is much less than the minimum C. Tooth width equals slot pitch D. The minimum width is 7 mm Answer: A. Actual tooth width must be at least Wts(min); here the margin is small. (Key point: Tooth width check) 3. How is the number of conductors per slot found? A. Ts divided by slots B. Total conductors 6Ts divided by slots, rounded to a whole number (105) C. 6Ts x slots D. Phase current divided by slots Answer: B. 2510/24 = 104.6, rounded to 105. (Key point: Stator conductors) 4. What is the 1.2 Wb/m^2 core density used for? A. To find minimum tooth width B. To find the number of slots C. To fix the stator core depth: dcs = Phi/(2 x 1.2 x Li) D. To calculate Bav Answer: C. Flux splits into two halves through the core, so area = 2 dcs Li. 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