--- title: "30 kW 240 V generator, square pole" subject: "Electric Machine Design" chapter: 5 key: "semester_5/electric_machine_design:dc-generator-30kw-240v-square" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-generator-30kw-240v-square site: BCT Engineering --- # 30 kW 240 V generator, square pole Electric Machine Design · Chapter 5: DC Machine Design - Asked in: 2082 Chaitra (new course), 5 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-30kw-240v-square - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-5 ## Question A design is required for a 30 kW, 4 pole, 900 rpm DC shunt generator, the full load terminal voltage being 240 V. Assume that the full load armature voltage is 3% of the terminal voltage. Calculate the main dimensions of the machine if the maximum gap density is 0.85 Wb/m², specific electrical loading of 20,000 ampere conductor/meter and field resistance is 120 Ω. The ratio of pole arc to pole pitch is 0.7. The pole face is square. ## Short answer Ia=127 A, E=247.2 V, Pa=31.39 kW. Co=117.4, D²L=0.01782 m³. For square pole face, D=0.319 m, L=0.175 m. ## 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). Here "armature voltage 3%" means the full-load armature resistance **drop** $I_aR_a = 3\%$ of terminal voltage. The ratio pole arc/pole pitch $\psi = 0.7$ is used as $B_{av}/B_m$. #### Step 1: Armature power from currents and emf $$ \begin{aligned} I_L &= \frac{30000}{240} = 125.0\ \text{A} \\ I_f &= \frac{240}{120} = 2.0\ \text{A} \\ I_a &= I_L + I_f = 127.0\ \text{A} \\ E &= V + I_aR_a = 240 + 0.03\times240 = 247.2\ \text{V} \\ P_a &= E I_a = 247.2\times127.0 = 31.39\ \text{kW} \end{aligned} $$ #### Step 2: Loadings and $D^2L$ $$ \begin{aligned} B_{av} &= \psi B_m = 0.7\times0.85 = 0.595\ \text{Wb/m}^2 \\ C_0 &= \pi^2\times0.595\times20000\times10^{-3} = 117.4 \\ n &= 900/60 = 15\ \text{rps} \\ D^2L &= \frac{31.39}{117.4\times15} = 0.01782\ \text{m}^3 \end{aligned} $$ #### Step 3: Square pole face $$ L = \psi\frac{\pi D}{p} = \frac{0.7\pi D}{4} = 0.5498D $$ $$ \begin{aligned} D^3 &= \frac{0.01782}{0.5498} = 0.03241\ \text{m}^3 \\ D &= 0.3188\ \text{m} \\ L &= 0.5498\times0.3188 = 0.1753\ \text{m} \end{aligned} $$ #### Check - Pole pitch $\tau = \pi D/4 = 0.250$ m; pole arc $= 0.7\tau = 0.1753$ m $= L$. - Peripheral speed $= \pi Dn = 15.0$ m/s; frequency $= pn/2 = 30$ Hz. Both acceptable. **Answer: Armature diameter $D \approx 0.319$ m, armature core length $L \approx 0.175$ m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For 30 kW at 240 V with Rf = 120 Ω, what is the armature current? A. 125 A B. 123 A C. 127 A D. 129 A Answer: C. IL = 30000/240 = 125 A, If = 240/120 = 2 A, so Ia = 127 A. (Key point: Armature current) 2. How is the average gap density obtained here? A. Bav = Bm = 0.85 B. Bav = ψ Bm = 0.7 x 0.85 = 0.595 Wb/m^2 C. Bav = Bm/ψ = 1.21 D. Bav = ψ = 0.7 Answer: B. ψ is used as Bav/Bm, so Bav = 0.595 Wb/m^2, used in C0 = π^2 Bav ac x 10^-3 = 117.4. (Key point: Use of ψ) 3. For a square pole face, how do D and L follow from D^2L = 0.01782 m^3? A. L = 0.5498D, so D^3 = D^2L/0.5498, giving D = 0.319 m and L = 0.175 m, equal to the pole arc B. L = D, so D = 0.261 m C. L = 0.7D, so D = 0.294 m D. L = 0.25D, so D = 0.414 m Answer: A. Square pole face: L = 0.7πD/4 = 0.5498D, so D^3 = 0.03241 and D = 0.319 m. 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