--- title: "2.2 kW 400 V cage motor (star-delta) design" subject: "Electric Machine Design" chapter: 4 key: "semester_5/electric_machine_design:design-2-2kw-400v-star-delta" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-2-2kw-400v-star-delta site: BCT Engineering --- # 2.2 kW 400 V cage motor (star-delta) design Electric Machine Design · Chapter 4: Three phase induction motor design - Asked in: 2075 Chaitra, 12 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-400v-star-delta - 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 using star delta starter and 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 = 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) 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 Main dimensions from output equation: D=0.1053 m, L=0.124 m (4-pole, 25 rps). Stator: 420 turns/phase, 24 slots, conductor 0.694 mm² (0.94 mm dia). Slot 6.75×27.0 mm; minimum tooth width 3.96 mm. ## Answer **Data:** 2.2 kW, 400 V, 50 Hz, 1420 rpm, so 4 poles and $n_s = 25$ rps; $\eta = 0.8$, $\cos\phi = 0.825$. **Assumptions:** $K_w = 0.955$; star-delta starting, so the stator runs in **delta** ($E_{ph} = 400$ V); tooth flux density limit 1.7 Wb/m²; iron factor 0.9. #### i) Main dimensions $$ \begin{aligned} Q &= \frac{2.2}{0.8 \times 0.825} = 3.333\ \text{kVA} \\ C_0 &= 11 \times 0.44 \times 21000 \times 0.955 \times 10^{-3} = 97.07 \\ D^2L &= \frac{3.333}{97.07 \times 25} = 1.374 \times 10^{-3}\ \text{m}^3 \end{aligned} $$ $L = 1.5\tau = 1.5 \times \pi D/4 = 1.178D$: $$ D^3 = \frac{1.374\times10^{-3}}{1.178} = 1.166\times10^{-3} \Rightarrow D = 0.1053\ \text{m}, \quad L = 0.124\ \text{m} $$ #### ii) Stator turns per phase $$ \begin{aligned} \tau &= \pi \times 0.1053/4 = 0.0827\ \text{m} \\ \Phi &= 0.44 \times 0.0827 \times 0.124 = 4.51 \times 10^{-3}\ \text{Wb} \\ T_s &= \frac{400}{4.44 \times 50 \times 4.51\times10^{-3} \times 0.955} = 418.3 \end{aligned} $$ #### iv) Number of stator slots Slot pitch 12–15 mm. $q = 2$: $S_s = 3 \times 4 \times 2 = 24$; slot pitch $= \pi \times 105.3/24 = 13.78$ mm (OK). **24 slots.** Conductors per slot $= 6 \times 418.3/24 = 104.6$, take **105**: $Z_s = 2520$, **$T_s = 420$**. #### iii) Size of stator conductor $$ I_{ph} = \frac{3333.3}{3 \times 400} = 2.778\ \text{A}, \qquad a_s = \frac{2.778}{4} = 0.694\ \text{mm}^2 $$ Bare diameter $= \sqrt{4 \times 0.694/\pi} = 0.94$ mm. #### v) Area and dimensions of each slot $$ A_{slot} = \frac{105 \times 0.694}{0.4} = 182.3\ \text{mm}^2 $$ $d = 4w$: $4w^2 = 182.3$, $w = 6.75$ mm, $d = 27.0$ mm. #### vi) Minimum width of stator teeth $L_i = 0.9 \times 0.124 = 0.1116$ m; teeth per pole $= 24/4 = 6$: $$ W_{ts(min)} = \frac{4.51\times10^{-3}}{1.7 \times 6 \times 0.1116} = 3.96\ \text{mm} $$ Actual tooth width at the gap $= 13.78 - 6.75 = 7.0$ mm, so tooth density is safe. **Answer:** $D = 10.53$ cm, $L = 12.4$ cm; $T_s = 420$; $a_s = 0.694$ mm² (0.94 mm dia); 24 slots; slot 182.3 mm² ($6.75 \times 27.0$ 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 a star-delta starter, why is Eph taken as 400 V? A. Star is used continuously, so Eph = 400/sqrt3 B. Eph = 400 x sqrt3 C. Eph is half of line voltage D. In running the stator is delta connected, so phase voltage equals line voltage Answer: D. Star-delta starting means the machine runs in delta; star only reduces voltage at start. (Key point: Delta connection phase voltage) 2. With L/tau = 1.5 and 4 poles, L in terms of D is A. L = 1.5 pi D/4 = 1.178D B. L = 1.5D C. L = 1.5 pi D D. L = pi D/6 Answer: A. tau = pi D/P = pi D/4, so L = 1.5 tau. (Key point: Separation of D and L) 3. How is slot area obtained? A. A = conductors per slot x area x space factor B. A = (conductors per slot x conductor area)/space factor C. A = conductor area/space factor D. A = Is/delta Answer: B. Space factor is copper area divided by slot area, so slot area = copper/0.4. (Key point: Slot area) 4. How is the conductor area found for delta? A. Iph = Q x 10^3/(sqrt3 x 400) B. Iph = Q/400 then divide by 3 C. Iph = Q x 10^3/(3 x 400) = 2.778 A, then divide by delta = 4 A/mm^2 D. Use line current without dividing by current density Answer: C. In delta, Iph = line current/sqrt3 = Q/(3 Eph); as = Iph/delta = 0.694 mm^2. 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