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Main dimensions of 15 kW 2810 rpm cage motor

Short answer

C0=131.31, D2L=2.885×10-3 m3. From peripheral speed 20 m/s, D=0.1273 m, then L=0.178 m (L/τ≈0.89).

Determine the main dimensions of a 15 kW, 3 phase, 400 V, 50 Hz, 2810 rpm squirrel cage induction motor having efficiency of 0.88 and a full load power factor of 0.9. Assume specific magnetic loading = 0.5 Wb/m² and specific electric loading = 25000 A/m. Take rotor peripheral speed as approximately 20 m/s at synchronous speed.

Data: 15 kW, 400 V, 50 Hz, 2810 rpm. Nearest synchronous speed is 3000 rpm, so P=2P = 2 and ns=50n_s = 50 rps. Assume Kw=0.955K_w = 0.955.

Output coefficient and D2LD^2L

Q=150.88×0.9=18.94 kVAC0=11Bav ac Kw×10−3=11×0.5×25000×0.955×10−3=131.31D2L=QC0ns=18.94131.31×50=2.885×10−3 m3\begin{aligned} Q &= \frac{15}{0.88 \times 0.9} = 18.94\ \text{kVA} \\ C_0 &= 11 B_{av}\, ac\, K_w \times 10^{-3} = 11 \times 0.5 \times 25000 \times 0.955 \times 10^{-3} = 131.31 \\ D^2L &= \frac{Q}{C_0 n_s} = \frac{18.94}{131.31 \times 50} = 2.885 \times 10^{-3}\ \text{m}^3 \end{aligned}

Diameter from peripheral speed

va=πDns=20 m/s⇒D=20π×50=0.1273 mv_a = \pi D n_s = 20\ \text{m/s} \Rightarrow D = \frac{20}{\pi \times 50} = 0.1273\ \text{m}

Core length

L=2.885×10−30.12732=0.178 mL = \frac{2.885\times10^{-3}}{0.1273^2} = 0.178\ \text{m}

Check

Pole pitch τ=πD/2=0.200\tau = \pi D/2 = 0.200 m, so L/τ=0.89L/\tau = 0.89. This lies in the usual range (about 0.6 to 2), so the proportions are acceptable.

Answer: D≈12.7D \approx 12.7 cm, L≈17.8L \approx 17.8 cm.

✓ Answer checked by Claude Sonnet 5.5

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