--- title: "Design 11 kW motor from 15 kW motor loadings" subject: "Electric Machine Design" chapter: 4 key: "semester_5/electric_machine_design:design-11kw-from-15kw-motor" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-11kw-from-15kw-motor site: BCT Engineering --- # Design 11 kW motor from 15 kW motor loadings Electric Machine Design · Chapter 4: Three phase induction motor design - Asked in: 2082 Baishakh, 12 marks; 2081 Bhadra, 12 marks; 2076 Chaitra, 14 marks (asked 3 times) - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/design-11kw-from-15kw-motor - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4 ## Question A 15 kW, 440 V, 50 Hz, 1480 rpm, 3-phase induction motor is built with an inner diameter of stator 25 cm and length 16 cm. The specific loading is 23000 ampere conductors per meter. Estimate the following parameters for a 11 kW, 460 V, 6 pole, 50 Hz delta connected induction motor, assuming same specific loadings as the previous motor with 84% efficiency and power factor 85% for each machine. Assume current density = 4 A/mm², stator slot pitch = 15 to 25 mm, ratio of slot depth to width = 3, flux density in stator core = 1.2 Wb/m², winding factor = 0.955, slot space factor = 0.4. Calculate: (i) Main dimension (ii) No. of stator slots (iii) No. of stator conductors and area of stator slot (iv) Dimension of each stator slot (v) Minimum width of stator teeth ## Short answer Determine C0=84.03 from the 15 kW motor; then for 11 kW D2L=0.0110 m3. With L=τ, D=0.276 m, L=0.1445 m. Stator: 36 slots, 1800 conductors (50/slot), conductor area 2.79 mm2, slot 10.8×32.4 mm, minimum tooth width 5.47 mm. ## Answer **Method:** find the specific loadings (as the output coefficient $C_0$) from the 15 kW motor, then use the same $C_0$ for the 11 kW motor. Output equation: $Q = C_0 D^2 L n_s$, $C_0 = 11 B_{av}\, ac\, K_w \times 10^{-3}$. #### Specific loadings from the 15 kW motor 1480 rpm at 50 Hz means 4 poles, $n_s = 1500/60 = 25$ rps. $$ \begin{aligned} Q_1 &= \frac{15}{0.84 \times 0.85} = 21.01\ \text{kVA} \\ C_0 &= \frac{Q_1}{D^2 L n_s} = \frac{21.01}{0.25^2 \times 0.16 \times 25} = 84.03 \\ B_{av} &= \frac{C_0}{11\, ac\, K_w \times 10^{-3}} = \frac{84.03}{11 \times 23000 \times 0.955 \times 10^{-3}} = 0.348\ \text{Wb/m}^2 \end{aligned} $$ #### (i) Main dimensions of the 11 kW motor 6 poles: $n_s = 1000/60 = 16.67$ rps. $$ \begin{aligned} Q &= \frac{11}{0.84 \times 0.85} = 15.41\ \text{kVA} \\ D^2L &= \frac{Q}{C_0 n_s} = \frac{15.41}{84.03 \times 16.67} = 0.0110\ \text{m}^3 \end{aligned} $$ Assume $L = \tau$ (good overall design), $\tau = \pi D/6$: $$ D^3 = \frac{0.0110 \times 6}{\pi} = 0.02101 \Rightarrow D = 0.276\ \text{m}, \quad L = \tau = 0.1445\ \text{m} $$ #### (ii) Number of stator slots Slot pitch 15–25 mm. With $q = 2$: $S_s = 3 \times 6 \times 2 = 36$, slot pitch $= \pi \times 275.9/36 = 24.1$ mm (within range). **Take 36 slots.** #### (iii) Stator conductors and conductor area $$ \begin{aligned} \Phi &= B_{av}\tau L = 0.348 \times 0.1445 \times 0.1445 = 7.26 \times 10^{-3}\ \text{Wb} \\ T_s &= \frac{E_{ph}}{4.44 f \Phi K_w} = \frac{460}{4.44 \times 50 \times 7.26\times10^{-3} \times 0.955} = 298.9 \end{aligned} $$ (Delta connection, so $E_{ph} = 460$ V.) Conductors $= 6T_s = 1793$; per slot $= 1793/36 = 49.8$, **take 50**. Total conductors $Z_s = 36 \times 50 = 1800$, $T_s = 300$. $$ \begin{aligned} I_{ph} &= \frac{15406}{3 \times 460} = 11.16\ \text{A} \\ a_s &= \frac{11.16}{4} = 2.79\ \text{mm}^2 \ (\text{dia} \approx 1.89\ \text{mm}) \end{aligned} $$ Slot area: $$ A_{slot} = \frac{50 \times 2.79}{0.4} = 348.9\ \text{mm}^2 $$ #### (iv) Slot dimensions $d = 3w$, so $3w^2 = 348.9$: $w = 10.8$ mm, $d = 32.4$ mm. #### (v) Minimum width of stator teeth Maximum tooth density 1.7 Wb/m², net iron length $L_i = 0.9L = 0.130$ m, teeth per pole $= 36/6 = 6$: $$ W_{ts(min)} = \frac{\Phi}{1.7 \times \frac{S_s}{P} \times L_i} = \frac{7.26\times10^{-3}}{1.7 \times 6 \times 0.130} = 5.47\ \text{mm} $$ (Stator core depth with $B_c = 1.2$ Wb/m²: $d_{cs} = \Phi/(2B_cL_i) = 23.3$ mm.) **Answer:** $D = 27.6$ cm, $L = 14.45$ cm; 36 slots; 1800 conductors (50 per slot, $T_s = 300$), $a_s = 2.79$ mm², slot area 349 mm²; slot $10.8 \times 32.4$ mm; $W_{ts(min)} = 5.47$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. 1480 rpm at 50 Hz is a 4-pole motor (ns = 25 rps). What is C0 from the 15 kW motor, with Q = 15/(0.84*0.85) = 21.01 kVA, D = 0.25 m, L = 0.16 m? A. C0 = 21.01/(0.25^2*0.16*25) = 84.03 B. C0 = 21.01/(0.25*0.16*25) = 21.0 C. C0 = 15/(0.25^2*0.16*25) = 60 D. C0 = 21.01*0.25^2*0.16*25 = 5.25 Answer: A. C0 = Q/(D^2 L ns) = 21.01/0.25 = 84.03, and the same C0 is used for the 11 kW motor. (Key point: Output coefficient from the 15 kW motor) 2. For the 6-pole 11 kW motor (ns = 16.67 rps, Q = 15.41 kVA) with L = tau, which is correct? A. D^2L = Q/(C0 ns) = 0.0110 m^3 and D = 0.276 m, L = 0.1445 m B. D^2L = Q*C0*ns, D = 0.5 m C. D^2L = Q/(C0*25) = 0.0073 m^3 D. D = 0.1445 m, L = 0.276 m Answer: A. D^2L = 15.41/(84.03*16.67) = 0.0110; L = tau = pi D/6 gives D^3 = 0.021 and D = 0.276 m. (Key point: Main dimensions) 3. Why is Eph taken as 460 V when finding Ts? A. The motor is star connected so Eph = 460/sqrt3 B. The motor is delta connected so phase voltage equals line voltage C. Eph is always the rated power in volts D. Eph = 460*sqrt3 for delta Answer: B. In delta the phase voltage equals the line voltage, so Ts = 460/(4.44*50*Phi*0.955), about 299, taken as 300 with 50 conductors per slot. (Key point: Turns per phase for delta connection) 4. With 50 conductors per slot, as = 2.79 mm^2 and space factor 0.4, what is the slot area and size (d = 3w)? A. 50*2.79*0.4 = 55.8 mm^2 B. 50*2.79/0.4 = 349 mm^2, so w = 10.8 mm and d = 32.4 mm C. 2.79/0.4 = 7 mm^2 D. 349 mm^2 with w = 32.4 mm and d = 10.8 mm Answer: B. Slot area = copper area / space factor = 348.9 mm^2; 3w^2 = 348.9 gives w = 10.8 mm and d = 32.4 mm. 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