--- title: "Stator design of 120 kW 1480 rpm motor" subject: "Electric Machine Design" chapter: 4 key: "semester_5/electric_machine_design:stator-120kw-1480rpm" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/stator-120kw-1480rpm site: BCT Engineering --- # Stator design of 120 kW 1480 rpm motor Electric Machine Design · Chapter 4: Three phase induction motor design - Asked in: 2076 Asoj, 10 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/stator-120kw-1480rpm - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4 ## Question Calculate: i) diameter ii) length iii) Ts iv) Full load current and as v) I²R loss of stator of 3 phase, 120 kW, 2200 V, 50 Hz, 1480 rpm, star connected slip ring induction motor from the following particulars: Bav = 0.8 Tesla, ac = 26000, η = 92%, p.f. = 0.88, L = 1.25τ, kw = 0.955, δ = 5 A/mm², mean length of stator conductors = 75 cm, ρ = 0.021 Ω per meter and mm² section. ## Short answer Using output equation and L=1.25τ: D=30.2 cm, L=29.7 cm. From flux and EMF: turns per phase 106. Full-load current 38.9 A, conductor area 7.78 mm². Stator I²R loss 1.95 kW. ## Answer **Data:** 120 kW, 2200 V, 50 Hz, 1480 rpm, star connected. Nearest synchronous speed 1500 rpm, so $P = 4$ and $n_s = 25$ rps. $B_{av} = 0.8$ T is used as given (it is higher than the usual 0.3–0.6 Wb/m²). #### i) and ii) Diameter and length $$ \begin{aligned} Q &= \frac{120}{0.92 \times 0.88} = 148.22\ \text{kVA} \\ C_0 &= 11B_{av}\,ac\,K_w\times10^{-3} = 11 \times 0.8 \times 26000 \times 0.955 \times 10^{-3} = 218.50 \\ D^2L &= \frac{148.22}{218.50 \times 25} = 0.02713\ \text{m}^3 \end{aligned} $$ $L = 1.25\tau = 1.25 \times \pi D/4 = 0.9817D$: $$ D^3 = \frac{0.02713}{0.9817} = 0.02764 \Rightarrow D = 0.302\ \text{m}, \quad L = 0.297\ \text{m} $$ #### iii) Turns per phase $T_s$ $$ \begin{aligned} \tau &= \pi \times 0.3024/4 = 0.2375\ \text{m} \\ \Phi &= B_{av}\tau L = 0.8 \times 0.2375 \times 0.2968 = 0.05639\ \text{Wb} \\ E_{ph} &= 2200/\sqrt3 = 1270.2\ \text{V} \\ T_s &= \frac{1270.2}{4.44 \times 50 \times 0.05639 \times 0.955} = 106.2 \approx 106 \end{aligned} $$ #### iv) Full-load current and conductor area $a_s$ $$ \begin{aligned} I_s &= \frac{148221}{\sqrt3 \times 2200} = 38.90\ \text{A} \\ a_s &= \frac{38.90}{5} = 7.78\ \text{mm}^2 \end{aligned} $$ #### v) Stator $I^2R$ loss Conductors per phase $= 2T_s = 212$, each 0.75 m long, so length per phase $= 212 \times 0.75 = 159$ m. $$ \begin{aligned} R_s &= \frac{0.021 \times 159}{7.78} = 0.429\ \Omega \\ \text{Loss} &= 3I_s^2R_s = 3 \times 38.90^2 \times 0.429 = 1948\ \text{W} \end{aligned} $$ **Answer:** $D = 30.2$ cm, $L = 29.7$ cm, $T_s = 106$, $I_s = 38.9$ A, $a_s = 7.78$ mm², stator copper loss $\approx 1.95$ kW. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. 1480 rpm at 50 Hz: which poles and ns? A. 2 poles, 50 rps B. 6 poles, 16.7 rps C. 4 poles, 24.7 rps D. 4 poles, 25 rps Answer: D. Nearest synchronous speed is 1500 rpm. (Key point: Poles) 2. With Bav = 0.8, ac = 26000, Kw = 0.955, what is C0? A. 218.5 B. 131.1 C. 157.6 D. 97.1 Answer: A. C0 = 11 x 0.8 x 26000 x 0.955 x 10^-3 = 218.5. (Key point: Output coefficient) 3. How is the stator current found for the star machine? A. Is = Q x 10^3/(3 x 2200) B. Is = Q x 10^3/(sqrt3 x 2200) = 38.9 A C. Is = 120 x 10^3/2200 D. Is = Q/2200 x sqrt3 Answer: B. In star, line current equals phase current. (Key point: Full-load current) 4. How is the stator copper loss computed? A. Is^2 Rs only B. 3 Is Rs C. 3 Is^2 Rs, where Rs = rho x (2Ts x 0.75)/as D. 3 Is^2 Rs using Ts x 0.75 length Answer: C. Conductors per phase = 2Ts, each 0.75 m long; this gives 0.429 ohm and about 1.95 kW. 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