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1000 kW generator: dimensions, poles, air gap

Short answer

With 300 rpm, armature power 1029 kW, D²L=0.7447 m³, p=10 chosen from brush current limit. Square pole face gives D≈1.50 m, L≈0.33 m. Air-gap mmf 50% of armature mmf yields length ≈5.1 mm.

Find the main dimension, number of poles and length of air gap of a 1000 kW, 500 V, 30[?] rpm dc generator. Assume the specific magnetic loading Bav = 0.7 Wb/m², ampere conductor per meter = 40000, square pole face, ratio of pole arc to pole pitch is 0.7. Assume efficiency as 92% and gap contraction factor as 1.15.

Assumptions: the speed is read as 300 rpm (the digit after "30" is unclear; 300 rpm is the usual value for this problem). The air-gap mmf is taken as 50% of the armature mmf per pole (usual design value, not stated), and the limits f≤50f \le 50 Hz and current per brush arm ≤400\le 400 A are used to choose the poles.

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+2×0.923×0.92×1000=1029 kWC0=π2×0.7×40000×10−3=276.3n=300/60=5 rpsD2L=1029276.3×5=0.7447 m3\begin{aligned} P_a &= \frac{1+2\times0.92}{3\times0.92}\times1000 = 1029\ \text{kW} \\ C_0 &= \pi^2\times0.7\times40000\times10^{-3} = 276.3 \\ n &= 300/60 = 5\ \text{rps} \\ D^2L &= \frac{1029}{276.3\times5} = 0.7447\ \text{m}^3 \end{aligned}

Step 2: Number of poles

Armature current Ia≈106/500=2000I_a \approx 10^6/500 = 2000 A. Lap winding, current per brush arm =2Ia/p≤400⇒p≥10= 2I_a/p \le 400 \Rightarrow p \ge 10. Square pole face gives L=0.7πD/pL = 0.7\pi D/p, so D3=D2L p/(0.7π)D^3 = D^2L\,p/(0.7\pi).

ppDD (m)LL (m)ff (Hz)A/brush armATaAT_a/pole
81.3940.3832050010 949
101.5020.330254009 435
121.5960.292303338 355

p=8p = 8 fails the brush current limit. p=10p = 10 is the smallest number that satisfies it, with low frequency and armature mmf below 10 000 A. Choose p=10p = 10.

Step 3: Main dimensions (p=10p = 10)

D3=0.7447×100.7π=3.386 m3D=1.502 mL=0.7π×1.50210=0.330 m\begin{aligned} D^3 &= \frac{0.7447\times10}{0.7\pi} = 3.386\ \text{m}^3 \\ D &= 1.502\ \text{m} \\ L &= \frac{0.7\pi\times1.502}{10} = 0.330\ \text{m} \end{aligned}

Peripheral speed =πDn=π×1.502×5=23.6= \pi Dn = \pi\times1.502\times5 = 23.6 m/s (safe).

Step 4: Air-gap length

ATa=ac πD2p=40000×π×1.50220=9435 AATg=0.5×9435=4718 ABg=Bavψ=0.70.7=1.0 Wb/m2lg=ATg800000 Bgkg=4718800000×1.0×1.15=5.13 mm\begin{aligned} AT_a &= \frac{ac\,\pi D}{2p} = \frac{40000\times\pi\times1.502}{20} = 9435\ \text{A} \\ AT_g &= 0.5\times9435 = 4718\ \text{A} \\ B_g &= \frac{B_{av}}{\psi} = \frac{0.7}{0.7} = 1.0\ \text{Wb/m}^2 \\ l_g &= \frac{AT_g}{800000\,B_g k_g} = \frac{4718}{800000\times1.0\times1.15} = 5.13\ \text{mm} \end{aligned}

Answer: Poles = 10, D≈1.50D \approx 1.50 m, L≈0.33L \approx 0.33 m, air gap lg≈5.1l_g \approx 5.1 mm.

✓ Answer checked by Claude Sonnet 5.5

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