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Air gap and rotor bars, 10 kW motor

Short answer

No-load flux 4.70 mWb, air gap 0.433 mm, stator turns 402. Rotor bar current 744 A (area 148.8 mm²); end-ring current 1776 A (area 355.2 mm²). Bar loss 334.8 W, end-ring loss 133.9 W.

The following design data are provided for an induction motor: Diameter of stator bore (D) = 16 cm, Length of stator core (L) = 8.5 cm, Average flux density (Bav) = 0.44 Wb/m², Power factor = 0.85, Efficiency = 86%, Frequency = 50 Hz, Current density = 5 A/mm², Stator slots = 36, Rotor slots = 30, Length of rotor bar = 15 cm, Mean diameter of end ring = 12 cm, Resistivity of bar conductor = 0.020 Ohm-metre, Power output of 3-phase, 4-pole, 400 V, delta connected = 10 kW. Calculate no-load maximum flux, length of air gap, no. of turns per phase, rotor bar current and area, end ring current and area, losses in bars and end rings.

Data: 10 kW, 400 V, 3-phase, 4 poles, delta, 50 Hz; D=0.16D = 0.16 m, L=0.085L = 0.085 m, Bav=0.44B_{av} = 0.44 Wb/m², η=0.86\eta = 0.86, pf =0.85= 0.85, δ=5\delta = 5 A/mm², Ss=36S_s = 36, Sr=30S_r = 30, bar length 0.15 m, end-ring mean diameter 0.12 m. Assumptions: Kws=0.955K_{ws} = 0.955; resistivity taken as 0.020.02 Ω per m per mm² (i.e. 0.02×10−60.02 \times 10^{-6} Ω·m; "0.020 Ω-m" is a misprint); same current density 5 A/mm² in bars and rings.

Flux per pole

τ=πDP=π×0.164=0.1257 mΦm=BavτL=0.44×0.1257×0.085=4.70×10−3 Wb\begin{aligned} \tau &= \frac{\pi D}{P} = \frac{\pi \times 0.16}{4} = 0.1257\ \text{m} \\ \Phi_m &= B_{av}\tau L = 0.44 \times 0.1257 \times 0.085 = 4.70\times10^{-3}\ \text{Wb} \end{aligned}

Length of air gap

lg=0.2+2DL=0.2+20.16×0.085=0.433 mml_g = 0.2 + 2\sqrt{DL} = 0.2 + 2\sqrt{0.16 \times 0.085} = 0.433\ \text{mm}

Turns per phase

Ts=Eph4.44fΦKw=4004.44×50×4.70×10−3×0.955=401.4T_s = \frac{E_{ph}}{4.44f\Phi K_w} = \frac{400}{4.44 \times 50 \times 4.70\times10^{-3} \times 0.955} = 401.4

Conductors per slot =6×401.4/36=66.9≈67= 6 \times 401.4/36 = 66.9 \approx 67, so Zs=2412Z_s = 2412 and Ts=402T_s = 402.

Stator phase current

Q=100.86×0.85=13.68 kVA,Is=136803×400=11.40 AQ = \frac{10}{0.86 \times 0.85} = 13.68\ \text{kVA}, \qquad I_s = \frac{13680}{3 \times 400} = 11.40\ \text{A}

Rotor bar current and area

Rotor mmf ≈0.85\approx 0.85 of stator mmf:

Ib=0.85×6IsTsKwsSr=0.85×6×11.40×402×0.95530=744.0 Aab=744.05=148.8 mm2\begin{aligned} I_b &= 0.85 \times \frac{6 I_s T_s K_{ws}}{S_r} = 0.85 \times \frac{6 \times 11.40 \times 402 \times 0.955}{30} = 744.0\ \text{A} \\ a_b &= \frac{744.0}{5} = 148.8\ \text{mm}^2 \end{aligned}

End-ring current and area

Ie=SrIbπP=30×744.0π×4=1776 Aae=17765=355.2 mm2\begin{aligned} I_e &= \frac{S_r I_b}{\pi P} = \frac{30 \times 744.0}{\pi \times 4} = 1776\ \text{A} \\ a_e &= \frac{1776}{5} = 355.2\ \text{mm}^2 \end{aligned}

Losses in bars

rb=ρlbab=0.02×0.15148.8=2.016×10−5 ΩPb=SrIb2rb=30×744.02×2.016×10−5=334.8 W\begin{aligned} r_b &= \frac{\rho l_b}{a_b} = \frac{0.02 \times 0.15}{148.8} = 2.016\times10^{-5}\ \Omega \\ P_b &= S_r I_b^2 r_b = 30 \times 744.0^2 \times 2.016\times10^{-5} = 334.8\ \text{W} \end{aligned}

Losses in end rings

re=ρ πDeae=0.02×π×0.12355.2=2.122×10−5 ΩPe=2Ie2re=2×17762×2.122×10−5=133.9 W\begin{aligned} r_e &= \frac{\rho\,\pi D_e}{a_e} = \frac{0.02 \times \pi \times 0.12}{355.2} = 2.122\times10^{-5}\ \Omega \\ P_e &= 2I_e^2 r_e = 2 \times 1776^2 \times 2.122\times10^{-5} = 133.9\ \text{W} \end{aligned}

Total rotor copper loss =334.8+133.9=468.7= 334.8 + 133.9 = 468.7 W.

QuantityValue
Flux per pole4.70 mWb
Air gap0.433 mm
Turns per phase402
Bar current, area744 A, 148.8 mm²
End-ring current, area1776 A, 355.2 mm²
Bar loss334.8 W
End-ring loss133.9 W

✓ Answer checked by Claude Sonnet 5.5

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