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600 kW generator: dimensions, poles, air gap

Short answer

Using output equation, armature power 619.8 kW gives D²L=0.1994 m³. Constraints limit p=6, D=0.80 m, L≈0.31 m. Air-gap mmf 50% of armature mmf yields length about 5 mm.

Determine the main dimensions, number of poles and length of air gap of a 600 kW, 500 V, 900 rpm generator. Assume average gap density as 0.6 Wb/m² and ampere conductors per meter as 35,000 A/m. The ratio of pole arc to pole pitch is 0.75 and the efficiency is 91%. The following are the design constraints: peripheral speed ≤ 40 m/s, frequency of flux reversal ≤ 50 Hz, current per brush ≤ 400 A, and armature mmf per pole ≤ 7500 A. The mmf required for air gap is 50% of armature mmf and gap contraction factor is 1.15.

The main dimensions come from the output equation of a dc machine (A.K. Sawhney):

Pa=C0D2Ln,C0=π2Bav ac×10−3 kW/(m3 rps)P_a = C_0 D^2 L n, \qquad C_0 = \pi^2 B_{av}\, ac \times 10^{-3}\ \text{kW/(m}^3\text{ rps)}

where PaP_a = armature power (kW), DD = armature diameter, LL = core length (m), nn = speed (rps).

For a generator (Sawhney), Pa=1+2η3ηPP_a = \dfrac{1+2\eta}{3\eta}P.

Step 1: D2LD^2L

Pa=1+2×0.913×0.91×600=619.8 kWC0=π2×0.6×35000×10−3=207.3n=900/60=15 rpsD2L=619.8207.3×15=0.1994 m3\begin{aligned} P_a &= \frac{1+2\times0.91}{3\times0.91}\times600 = 619.8\ \text{kW} \\ C_0 &= \pi^2\times0.6\times35000\times10^{-3} = 207.3 \\ n &= 900/60 = 15\ \text{rps} \\ D^2L &= \frac{619.8}{207.3\times15} = 0.1994\ \text{m}^3 \end{aligned}

Step 2: Number of poles

  • Armature current Ia≈600000/500=1200I_a \approx 600000/500 = 1200 A (field current neglected).
  • Frequency limit: f=pn/2≤50⇒p≤50×2/15=6.67f = pn/2 \le 50 \Rightarrow p \le 50\times2/15 = 6.67.
  • Current per brush arm (lap winding, a=pa = p): 2Ia/p≤400⇒p≥2×1200/400=62I_a/p \le 400 \Rightarrow p \ge 2\times1200/400 = 6.

Only p=6p = 6 meets both: f=6×15/2=45f = 6\times15/2 = 45 Hz and current per brush arm =2400/6=400= 2400/6 = 400 A. Poles = 6.

Step 3: Diameter limits

  • Peripheral speed: πDn≤40⇒D≤40π×15=0.849\pi D n \le 40 \Rightarrow D \le \dfrac{40}{\pi\times15} = 0.849 m.
  • Armature mmf per pole: ATa=ac πD2p≤7500⇒D≤7500×1235000π=0.819AT_a = \dfrac{ac\,\pi D}{2p} \le 7500 \Rightarrow D \le \dfrac{7500\times12}{35000\pi} = 0.819 m.

Take D=0.80D = 0.80 m (just inside both limits).

L=D2LD2=0.19940.82=0.312 mL = \frac{D^2L}{D^2} = \frac{0.1994}{0.8^2} = 0.312\ \text{m}

Check: πDn=37.7\pi Dn = 37.7 m/s <40< 40; pole pitch τ=π×0.8/6=0.419\tau = \pi\times0.8/6 = 0.419 m; pole arc =0.75τ=0.314= 0.75\tau = 0.314 m (nearly square pole face with L=0.312L = 0.312 m).

Step 4: Length of air gap

ATa=35000×π×0.82×6=7330 AATg=0.5×7330=3665 ABg=Bavψ=0.60.75=0.8 Wb/m2\begin{aligned} AT_a &= \frac{35000\times\pi\times0.8}{2\times6} = 7330\ \text{A} \\ AT_g &= 0.5\times7330 = 3665\ \text{A} \\ B_g &= \frac{B_{av}}{\psi} = \frac{0.6}{0.75} = 0.8\ \text{Wb/m}^2 \end{aligned}

Using ATg=800 000 Bg kg lgAT_g = 800\,000\,B_g\,k_g\,l_g (since 1/μ0≈8×1051/\mu_0 \approx 8\times10^5):

lg=3665800000×0.8×1.15=4.98×10−3 m≈5 mml_g = \frac{3665}{800000\times0.8\times1.15} = 4.98\times10^{-3}\ \text{m} \approx 5\ \text{mm}

Answer: Poles = 6, D=0.80D = 0.80 m, L≈0.31L \approx 0.31 m, air gap lg≈5l_g \approx 5 mm.

✓ Answer checked by Claude Sonnet 5.5

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