--- title: "Main dimensions of 30 kW 220 V shunt generator" subject: "Electric Machine Design" chapter: 5 key: "semester_5/electric_machine_design:dc-generator-30kw-220v" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-generator-30kw-220v site: BCT Engineering --- # Main dimensions of 30 kW 220 V shunt generator Electric Machine Design · Chapter 5: DC Machine Design - Asked in: 2080 Bhadra, 6 marks - Video: [DC Machine Main Dimension numerical problem 2 #electricalengineering #electricmotors](https://www.youtube.com/watch?v=u_HbAS-skR8) — EEE, SJBIT - 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-220v - 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 220 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. ## Short answer Ia=138.2 A, E=226.6 V, Pa=31.32 kW. Co=117.4, D²L=0.01778 m³. Square pole face yields 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 question does not state a length condition, so the usual assumption is made: **square pole face** (core length = pole arc). The full-load armature voltage **drop** is taken as 3% of terminal voltage. #### Step 1: Armature power from currents and emf $$ \begin{aligned} I_L &= \frac{30000}{220} = 136.36\ \text{A} \\ I_f &= \frac{220}{120} = 1.833\ \text{A} \\ I_a &= I_L + I_f = 138.20\ \text{A} \\ E &= V + I_aR_a = 220 + 0.03\times220 = 226.6\ \text{V} \\ P_a &= E I_a = 226.6\times138.20 = 31.32\ \text{kW} \end{aligned} $$ #### Step 2: Loadings and $D^2L$ $$ \begin{aligned} B_{av} &= \psi B_m = 0.7\times0.85 = 0.595\ \text{Wb/m}^2 \\ C_0 &= \pi^2\times0.595\times20000\times10^{-3} = 117.4 \\ n &= 900/60 = 15\ \text{rps} \\ D^2L &= \frac{31.32}{117.4\times15} = 0.01778\ \text{m}^3 \end{aligned} $$ #### Step 3: Square pole face $$ L = \psi\frac{\pi D}{p} = \frac{0.7\pi D}{4} = 0.5498D $$ $$ \begin{aligned} D^3 &= \frac{0.01778}{0.5498} = 0.03233\ \text{m}^3 \\ D &= 0.3186\ \text{m} \\ L &= 0.5498\times0.3186 = 0.1751\ \text{m} \end{aligned} $$ #### Check - Pole pitch $\tau = \pi D/4 = 0.250$ m; pole arc $= 0.7\tau = 0.1751$ 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. How is the armature current of the shunt generator found? A. Ia = IL - If B. Ia = IL C. Ia = IL x 1.03 D. Ia = IL + If, with If = V/Rf Answer: D. In a shunt generator the armature supplies both the load and the field: Ia = 136.36 + 1.833 = 138.2 A. (Key point: Armature current of shunt generator) 2. With the armature drop 3% of terminal voltage, what is the armature power used in the output equation? A. Pa = E Ia, with E = 220 + 0.03 x 220 = 226.6 V, giving 31.32 kW B. Pa = V Ia = 30.4 kW C. Pa = 30 kW exactly D. Pa = E Ia with E = 220 - 6.6 V Answer: A. The generated emf is E = V + IaRa = 226.6 V, so Pa = 226.6 x 138.2 = 31.32 kW. (Key point: Armature emf and power) 3. No length condition is given. What is assumed to find L? A. L = D B. L = pole pitch C. L = 2D D. A square pole face: L = pole arc = 0.7πD/4 = 0.5498D Answer: D. The usual assumption is a square pole face, so L = ψπD/p = 0.5498D, giving D = 0.319 m and L = 0.175 m. (Key point: Assumption for length) ## Close variants set in other exams - [30 kW 240 V generator, square pole](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-30kw-240v-square) — Electric Machine Design, chapter 5, asked 2× - [30 kW 240 V generator, 3% drop](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-30kw-240v-3pct) — Electric Machine Design, chapter 5, asked 3× - [30 kW 240 V generator, 2.5% drop](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-30kw-240v-2-5pct) — Electric Machine Design, chapter 5, asked 3× ## Related in Electric Machine Design - [Armature dimensions of 50 kW shunt generator](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-50kw-armature) — Electric Machine Design, chapter 5 - [Main dimensions of 600 kW square-pole generator](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-600kw-square) — Electric Machine Design, chapter 5 - [Main dimensions of 200 kW shunt generator](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-200kw) — Electric Machine Design, chapter 5 - [Armature size, 10 kW 900 rpm motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-motor-10kw-900rpm) — Electric Machine Design, chapter 5, asked 5× - [Main dimensions of 5 kW generator (Bav 0.42)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-5kw-bav-042) — Electric Machine Design, chapter 5, asked 4× - [Main dimensions of 5 kW generator (Bav 0.40)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-5kw-bav-040) — Electric Machine Design, chapter 5, asked 4× ## Related in other subjects - [Short shunt to long shunt compound generator (10 kW, 220 V)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashort-shunt-cumulative-10kw-220v-long) — Electrical Machines I, chapter 3 - [30 kW shunt generator: emf, copper loss, efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Agen-shunt-30kw-loss-1000w) — Electrical Machines, chapter 3, asked 2× - [30 kW shunt generator: emf, copper loss, efficiency (100 W)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Agen-shunt-30kw-loss-100w) — Electrical Machines, chapter 3, asked 2× - [EMF, copper loss and efficiency of 30 kW shunt generator](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-gen-30kw-200v-emf-losses) — Electrical Machines I, chapter 3