--- title: "600 kW generator: dimensions, poles, air gap" subject: "Electric Machine Design" chapter: 5 key: "semester_5/electric_machine_design:dc-generator-600kw-constraints" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-generator-600kw-constraints site: BCT Engineering --- # 600 kW generator: dimensions, poles, air gap Electric Machine Design · Chapter 5: DC Machine Design - Asked in: 2075 Asoj, 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-generator-600kw-constraints - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-5 ## Question 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. ## 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. ## Answer The main dimensions come from the **output equation of a dc machine** (A.K. Sawhney): $$ P_a = C_0 D^2 L n, \qquad C_0 = \pi^2 B_{av}\, ac \times 10^{-3}\ \text{kW/(m}^3\text{ rps)} $$ where $P_a$ = armature power (kW), $D$ = armature diameter, $L$ = core length (m), $n$ = speed (rps). For a generator (Sawhney), $P_a = \dfrac{1+2\eta}{3\eta}P$. #### Step 1: $D^2L$ $$ \begin{aligned} P_a &= \frac{1+2\times0.91}{3\times0.91}\times600 = 619.8\ \text{kW} \\ C_0 &= \pi^2\times0.6\times35000\times10^{-3} = 207.3 \\ n &= 900/60 = 15\ \text{rps} \\ D^2L &= \frac{619.8}{207.3\times15} = 0.1994\ \text{m}^3 \end{aligned} $$ #### Step 2: Number of poles - Armature current $I_a \approx 600000/500 = 1200$ A (field current neglected). - Frequency limit: $f = pn/2 \le 50 \Rightarrow p \le 50\times2/15 = 6.67$. - Current per brush arm (lap winding, $a = p$): $2I_a/p \le 400 \Rightarrow p \ge 2\times1200/400 = 6$. Only $p = 6$ meets both: $f = 6\times15/2 = 45$ Hz and current per brush arm $= 2400/6 = 400$ A. **Poles = 6.** #### Step 3: Diameter limits - Peripheral speed: $\pi D n \le 40 \Rightarrow D \le \dfrac{40}{\pi\times15} = 0.849$ m. - Armature mmf per pole: $AT_a = \dfrac{ac\,\pi D}{2p} \le 7500 \Rightarrow D \le \dfrac{7500\times12}{35000\pi} = 0.819$ m. Take **$D = 0.80$ m** (just inside both limits). $$ L = \frac{D^2L}{D^2} = \frac{0.1994}{0.8^2} = 0.312\ \text{m} $$ Check: $\pi Dn = 37.7$ m/s $< 40$; pole pitch $\tau = \pi\times0.8/6 = 0.419$ m; pole arc $= 0.75\tau = 0.314$ m (nearly square pole face with $L = 0.312$ m). #### Step 4: Length of air gap $$ \begin{aligned} AT_a &= \frac{35000\times\pi\times0.8}{2\times6} = 7330\ \text{A} \\ AT_g &= 0.5\times7330 = 3665\ \text{A} \\ B_g &= \frac{B_{av}}{\psi} = \frac{0.6}{0.75} = 0.8\ \text{Wb/m}^2 \end{aligned} $$ Using $AT_g = 800\,000\,B_g\,k_g\,l_g$ (since $1/\mu_0 \approx 8\times10^5$): $$ l_g = \frac{3665}{800000\times0.8\times1.15} = 4.98\times10^{-3}\ \text{m} \approx 5\ \text{mm} $$ **Answer: Poles = 6, $D = 0.80$ m, $L \approx 0.31$ m, air gap $l_g \approx 5$ mm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which constraints fix p = 6? A. Frequency gives p >= 6 and brush current gives p <= 6.67 B. Frequency gives p <= 6.67 and brush current (2Ia/p <= 400 A) gives p >= 6, so only p = 6 fits C. Peripheral speed fixes p = 6 directly D. Armature mmf gives p = 6 directly Answer: B. f = pn/2 <= 50 gives p <= 6.67; 2 x 1200/p <= 400 gives p >= 6. Hence 6 poles. (Key point: Choosing number of poles) 2. Which constraint is the tighter limit on D, and what D is chosen? A. Peripheral speed (0.849 m); D = 0.85 m B. Armature mmf per pole (0.819 m); D = 0.80 m, inside both limits C. Frequency; D = 0.5 m D. Brush current; D = 1.0 m Answer: B. Peripheral speed gives D <= 0.849 m and armature mmf gives D <= 0.819 m; D = 0.80 m satisfies both. (Key point: Limit on diameter) 3. How are the gap mmf and gap density found for the air-gap length? A. ATg = 0.5 ATa = 3665 A and Bg = Bav/ψ = 0.8 Wb/m^2 B. ATg = 2 ATa and Bg = Bav = 0.6 C. ATg = ATa and Bg = ψ Bav D. ATg = 0.5 ATa and Bg = ψ Bav = 0.45 Answer: A. ATa = ac πD/2p = 7330 A, ATg = 50% of it, and the peak gap density Bg = Bav/ψ = 0.8. (Key point: Air-gap mmf and gap density) 4. Which expression gives the air-gap length? A. lg = ATg x 800000 x Bg B. lg = ATg Bg/(800000 kg) C. lg = 800000 Bg/ATg D. lg = ATg/(800000 Bg kg) = 3665/(800000 x 0.8 x 1.15), about 5 mm Answer: D. From ATg = 800000 Bg kg lg, lg = 4.98 mm, about 5 mm. 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