--- title: "2.2 kW cage motor designed for best power factor" subject: "Electric Machine Design" chapter: 4 key: "semester_5/electric_machine_design:design-2-2kw-best-power-factor" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-2-2kw-best-power-factor site: BCT Engineering --- # 2.2 kW cage motor designed for best power factor Electric Machine Design · Chapter 4: Three phase induction motor design - Asked in: 2080 Bhadra, 12 marks; 2082 Chaitra (new course), 8 marks (asked 2 times) - Video: [Numerical on main dimension of three phase IM to give best power factor](https://www.youtube.com/watch?v=VBZlKnRPuY0) — Sinhgad Institute Pandharpur - 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-best-power-factor - 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 having efficiency of 0.8 and power factor 0.85 designed for its best power factor. Given Bav = 0.44 Wb/m², ac = 21000 A/m, slot space factor = 0.4, ratio of slot depth to width = 4, stator slot pitch = 12 to 15 mm, δs = 4 A/mm². Calculate (i) Main dimensions (ii) Ts (Turns per phase of stator conductor) (iii) as (Area of stator conductor) (iv) No. of stator slots (v) Area and dimension of each stator slot (vi) Minimum width of stator teeth (vii) Stator slot loading. Note: For best power factor, use τ = √(0.18 L) ## Short answer Compute D2L=1.333×10-3 m3; using τ=√(0.18L) gives D=0.1404 m, L=0.0676 m. Stator: 36 slots, Ts=636 turns/phase, conductor area 0.613 mm2, slot 6.37×25.5 mm (162.4 mm2), minimum tooth width 3.53 mm, slot loading 260 A-conductors. ## Answer **Data:** 2.2 kW, 440 V, 50 Hz, 1430 rpm (4 poles, $n_s = 1500$ rpm $= 25$ rps), $\eta = 0.8$, $\cos\phi = 0.85$, $B_{av} = 0.44$ Wb/m², $ac = 21000$ A/m. **Assumptions:** $K_w = 0.955$; delta-connected stator ($E_{ph} = 440$ V); tooth flux density limit 1.7 Wb/m²; iron stacking factor 0.9. #### (i) Main dimensions $$ \begin{aligned} Q &= \frac{2.2}{0.8 \times 0.85} = 3.235\ \text{kVA} \\ C_0 &= 11 B_{av}\, ac\, K_w \times 10^{-3} = 11 \times 0.44 \times 21000 \times 0.955 \times 10^{-3} = 97.07 \\ D^2L &= \frac{Q}{C_0 n_s} = \frac{3.235}{97.07 \times 25} = 1.333 \times 10^{-3}\ \text{m}^3 \end{aligned} $$ Best power factor: $\tau = \sqrt{0.18L}$, so $L = \tau^2/0.18 = (\pi D/4)^2/0.18$: $$ \begin{aligned} D^2 \cdot \frac{\pi^2 D^2}{16 \times 0.18} &= 1.333\times10^{-3} \\ D^4 &= 3.889 \times 10^{-4} \Rightarrow D = 0.1404\ \text{m} \\ L &= \frac{1.333\times10^{-3}}{0.1404^2} = 0.0676\ \text{m} \end{aligned} $$ Check: $\tau = \pi \times 0.1404/4 = 0.1103$ m $= \sqrt{0.18 \times 0.0676}$. #### (ii) Stator turns per phase $$ \begin{aligned} \Phi &= B_{av}\tau L = 0.44 \times 0.1103 \times 0.0676 = 3.281 \times 10^{-3}\ \text{Wb} \\ T_s &= \frac{440}{4.44 \times 50 \times 3.281\times10^{-3} \times 0.955} = 632.6 \end{aligned} $$ #### (iv) Number of stator slots Slot pitch 12–15 mm. With $q = 3$: $S_s = 3 \times 4 \times 3 = 36$, slot pitch $= \pi \times 140.4/36 = 12.26$ mm (OK). **36 slots.** Conductors per slot $= 6 \times 632.6/36 = 105.4$, take **106**, so $Z_s = 3816$ and **$T_s = 636$**. #### (iii) Area of stator conductor $$ I_{ph} = \frac{3235.3}{3 \times 440} = 2.451\ \text{A}, \qquad a_s = \frac{2.451}{4} = 0.613\ \text{mm}^2 $$ (Bare diameter $\approx 0.88$ mm.) #### (v) Area and dimensions of each slot $$ A_{slot} = \frac{106 \times 0.613}{0.4} = 162.4\ \text{mm}^2 $$ $d/w = 4$: $4w^2 = 162.4$, $w = 6.37$ mm, $d = 25.5$ mm. #### (vi) Minimum width of stator teeth $L_i = 0.9 \times 0.0676 = 0.0608$ m, teeth per pole $= 36/4 = 9$: $$ W_{ts(min)} = \frac{\Phi}{1.7 \times 9 \times L_i} = \frac{3.281\times10^{-3}}{1.7 \times 9 \times 0.0608} = 3.53\ \text{mm} $$ #### (vii) Stator slot loading $$ I_{ph} \times \text{cond/slot} = 2.451 \times 106 = 259.8\ \text{A-conductors} $$ (Well below the usual limit of about 1000–1200.) **Answer:** $D = 14.04$ cm, $L = 6.76$ cm; $T_s = 636$; $a_s = 0.613$ mm²; 36 slots; slot 162.4 mm² ($6.37 \times 25.5$ mm); $W_{ts(min)} = 3.53$ mm; slot loading 260 A-conductors. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is Q for the 2.2 kW motor with efficiency 0.8 and pf 0.85? A. 2.2*0.8*0.85 = 1.50 kVA B. 2.2/0.85 = 2.59 kVA C. 2.2/(0.8*0.85) = 3.235 kVA D. 2.2/0.8 = 2.75 kVA Answer: C. Input kVA = output kW/(efficiency x pf) = 2.2/0.68 = 3.235. (Key point: kVA input) 2. How are D and L found using tau = sqrt(0.18 L)? A. L = tau, D^3 = D^2L*4/pi B. L = tau^2/0.18 with tau = pi D/4 substituted into D^2L = 1.333x10^-3, giving D = 0.1404 m, L = 0.0676 m C. L = 0.18 tau, giving D = 0.30 m D. D = L = 0.11 m Answer: B. From tau^2 = 0.18 L, L = (pi D/4)^2/0.18; substituting gives D^4 = 3.889x10^-4 and D = 0.1404 m. (Key point: Best power factor main dimensions) 3. With 4 poles and q = 3, how many slots and is the slot pitch acceptable (range 12-15 mm)? A. Ss = 3*4*3 = 36, slot pitch 12.26 mm (acceptable) B. Ss = 4*3 = 12, slot pitch 36 mm C. Ss = 36, slot pitch 25 mm D. Ss = 24, slot pitch 18 mm Answer: A. Ss = 3Pq = 36 and the pitch pi*140.4/36 = 12.26 mm lies within 12-15 mm. (Key point: Number of stator slots) 4. With 106 conductors per slot, as = 0.613 mm^2, space factor 0.4 and d/w = 4, what are the slot dimensions? A. Area = 106*0.613*0.4 = 26 mm^2 B. Area = 106*0.613/0.4 = 162.4 mm^2, w = 6.37 mm, d = 25.5 mm C. Area = 162.4 mm^2, w = 25.5 mm, d = 6.37 mm D. Area = 0.613/0.4 = 1.5 mm^2 Answer: B. Slot area = total copper/space factor = 162.4 mm^2; 4w^2 = 162.4 gives w = 6.37 mm and d = 25.5 mm. 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