Skip to main content

Output equation of three phase induction motor

Short answer

The output equation is Q = C0 D² L ns with C0 = 11 Bav ac Kw ×10⁻³, relating kVA input to stator bore diameter D, core length L, synchronous speed ns, and specific loadings Bav and ac, derived from emf and mmf expressions.

Derive the output equation for three phase induction motor.

The output equation relates the kVA input rating QQ of a 3-phase induction motor to its main dimensions (DD = stator bore diameter, LL = core length), its specific loadings (BavB_{av}, acac) and its synchronous speed nsn_s.

Symbols

  • PP = number of poles, ff = frequency, nsn_s = synchronous speed (rps)
  • Φ\Phi = flux per pole (Wb), TphT_{ph} = turns per phase, KwK_w = winding factor
  • EphE_{ph}, IphI_{ph} = phase voltage and phase current
  • BavB_{av} = specific magnetic loading (Wb/m²), acac = specific electric loading (A/m)

Step 1: kVA input

Neglecting the small drop, Vph≈EphV_{ph} \approx E_{ph}:

Q=3EphIph×10−3 kVAQ = 3E_{ph}I_{ph}\times 10^{-3}\ \text{kVA}

Step 2: Induced emf

Eph=4.44 f Φ TphKw,f=Pns2E_{ph} = 4.44\, f\, \Phi\, T_{ph} K_w, \qquad f = \frac{P n_s}{2}

Step 3: Magnetic loading

Total flux around the gap is PΦP\Phi over area πDL\pi D L:

Bav=PΦπDL  ⇒  Φ=BavπDLPB_{av} = \frac{P\Phi}{\pi D L} \;\Rightarrow\; \Phi = \frac{B_{av}\pi D L}{P}

Step 4: Electric loading

Total stator conductors =3×2Tph=6Tph= 3 \times 2T_{ph} = 6T_{ph}, each carrying IphI_{ph}:

ac=6IphTphπD  ⇒  IphTph=ac πD6ac = \frac{6 I_{ph} T_{ph}}{\pi D} \;\Rightarrow\; I_{ph}T_{ph} = \frac{ac\,\pi D}{6}

Step 5: Substitute

Q=3×4.44×Pns2×BavπDLP×Kw×ac πD6×10−3=1.11 π2Bav ac Kw D2L ns×10−3=11 Bav ac Kw×10−3 D2L ns\begin{aligned} Q &= 3 \times 4.44 \times \frac{P n_s}{2} \times \frac{B_{av}\pi D L}{P} \times K_w \times \frac{ac\,\pi D}{6} \times 10^{-3} \\ &= 1.11\,\pi^2 B_{av}\, ac\, K_w\, D^2 L\, n_s \times 10^{-3} \\ &= 11\, B_{av}\, ac\, K_w \times 10^{-3}\, D^2 L\, n_s \end{aligned}

So

Q=C0D2Lns,C0=11Bav ac Kw×10−3Q = C_0 D^2 L n_s, \qquad C_0 = 11 B_{av}\, ac\, K_w \times 10^{-3}

C0C_0 is called the output coefficient. The kVA input is found from the rated shaft output:

Q=kW outputηcos⁡ϕQ = \frac{\text{kW output}}{\eta \cos\phi}

Meaning of the result

  • D2L=QC0nsD^2L = \dfrac{Q}{C_0 n_s}, so the volume of the active part is directly proportional to rating and inversely proportional to speed and specific loadings.
  • Higher BavB_{av} and acac give a smaller, cheaper machine, but power factor, temperature rise and overload capacity limit them.
  • A high-speed motor is smaller than a low-speed motor of the same rating.
  • Typical ranges: Bav=0.3B_{av} = 0.3–0.60.6 Wb/m², ac=5000ac = 5000–4500045000 A/m, Kw≈0.955K_w \approx 0.955 for a full-pitch, 60° phase-spread winding.

The product D2LD^2L is then split into DD and LL using a chosen ratio L/τL/\tau (for example L/τ=1L/\tau = 1 for good overall design) and a check on peripheral speed (πDns≤30\pi D n_s \le 30 m/s for normal designs).

✓ Answer checked by Claude Sonnet 5.5

All Electric Machine Design chapter 4 questions →