--- title: "30 kW 240 V generator, 3% drop" subject: "Electric Machine Design" chapter: 5 key: "semester_5/electric_machine_design:dc-generator-30kw-240v-3pct" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-generator-30kw-240v-3pct site: BCT Engineering --- # 30 kW 240 V generator, 3% drop Electric Machine Design · Chapter 5: DC Machine Design - Asked in: 2072 Chaitra, 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-30kw-240v-3pct - 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 dimension of the machine. Given that: Maximum gap flux density = 0.85 Wb/m²; Specific electric loading = 20000 ampere conductor per metre; Field resistance = 120 Ω; Ratio of pole arc to pole pitch = 0.7; Field form factor = 0.7 ## Short answer Ia=127 A, E=247.2 V, Pa=31.39 kW. Co=117.4, D²L=0.01782 m³. Assuming 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). The full-load armature voltage **drop** $I_aR_a$ is 3% of terminal voltage. Field form factor $K_f = B_{av}/B_m = 0.7$ (equal to $\psi$). Since no length ratio is given, the standard assumption of a **square pole face** (L = pole arc) is used. #### 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. What role does the field form factor Kf = 0.7 play here? A. It is the ratio of field current to armature current B. It is the efficiency of the field winding C. It equals Bav/Bm, so Bav = 0.7 x 0.85 = 0.595 Wb/m^2 D. It is the stacking factor of the core Answer: C. Kf = Bav/Bm (equal to ψ here), so Bav = 0.595 Wb/m^2. (Key point: Field form factor) 2. With an armature drop of 3% of 240 V and Ia = 127 A, what is Pa? A. 30 kW B. 247.2 V x 127 A = 31.39 kW C. 240 x 127 = 30.48 kW D. 247.2 x 125 = 30.9 kW Answer: B. E = 240 + 0.03 x 240 = 247.2 V and Ia = 125 + 2 = 127 A, so Pa = E Ia = 31.39 kW. (Key point: Armature emf and power) 3. What is C0 for Bav = 0.595 and ac = 20000? A. C0 = 0.595 x 20000 = 11900 B. C0 = π x 0.595 x 20 = 37.4 C. C0 = π^2 x 0.85 x 20 = 167.8 D. C0 = π^2 x 0.595 x 20000 x 10^-3 = 117.4 Answer: D. C0 = π^2 Bav ac x 10^-3 = 9.87 x 0.595 x 20 = 117.4. (Key point: Output coefficient) 4. Since no length ratio is given, what assumption fixes L? A. Square pole face: L = pole arc = 0.5498D, giving D = 0.319 m and L = 0.175 m B. L equals the pole pitch C. L equals the armature diameter D. L is chosen from peripheral speed Answer: A. The standard assumption is L = pole arc = ψπD/p, which yields D = 0.319 m and L = 0.175 m. 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