--- title: "Armature size, 10 kW 900 rpm motor" subject: "Electric Machine Design" chapter: 5 key: "semester_5/electric_machine_design:dc-motor-10kw-900rpm" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-motor-10kw-900rpm site: BCT Engineering --- # Armature size, 10 kW 900 rpm motor Electric Machine Design · Chapter 5: DC Machine Design - Asked in: 2082 Baishakh, 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-motor-10kw-900rpm - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-5 ## Question Calculate the diameter (D) and length (L) of armature for a 10 kW, 4 pole, 900 rpm, 220 V shunt motor. Given full load efficiency = 0.80, maximum gap flux density = 0.85 Wb/m², specific electrical loading = 30,000 ampere conductor/meter, field form factor (Kf) = 0.7. Assume that maximum efficiency occurs at full load and the field current is 3% of rated current. The pole face is square. ## Short answer Armature power Pa found by loss split at max efficiency: Pa=10.875 kW. Output coefficient Co=π²Bav·ac·10⁻³=176.2; D²L=Pa/(Co·n)=4.115×10⁻³ m³. For square pole face, L=0.5498D, so D=0.196 m, L=0.108 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). #### Step 1: Armature power (losses split at maximum efficiency) $$ \begin{aligned} \text{Input} &= \frac{10}{0.80} = 12.5\ \text{kW} \\ I_L &= \frac{12500}{220} = 56.82\ \text{A} \\ I_f &= 0.03 \times 56.82 = 1.705\ \text{A} \\ I_a &= I_L - I_f = 56.82 - 1.705 = 55.11\ \text{A} \\ \text{Total losses} &= 12500 - 10000 = 2500\ \text{W} \end{aligned} $$ At maximum efficiency, variable loss (armature copper loss) = constant loss, so armature copper loss $= 2500/2 = 1250$ W. $$ \begin{aligned} R_a &= \frac{1250}{55.11^2} = 0.4115\ \Omega \\ E_b &= V - I_aR_a = 220 - 55.11\times0.4115 = 197.3\ \text{V} \\ P_a &= E_b I_a = 197.3 \times 55.11 = 10\,875\ \text{W} = 10.875\ \text{kW} \end{aligned} $$ #### Step 2: Specific magnetic loading and output coefficient The field form factor $K_f = B_{av}/B_m = 0.7$; Sawhney takes $\psi$ (pole arc/pole pitch) $= K_f = 0.7$. $$ \begin{aligned} B_{av} &= K_f B_m = 0.7\times0.85 = 0.595\ \text{Wb/m}^2 \\ C_0 &= \pi^2 B_{av}\, ac\times10^{-3} = \pi^2\times0.595\times30000\times10^{-3} = 176.2 \\ n &= 900/60 = 15\ \text{rps} \end{aligned} $$ $$ D^2L = \frac{P_a}{C_0 n} = \frac{10.875}{176.2\times15} = 4.115\times10^{-3}\ \text{m}^3 $$ #### Step 3: Square pole face For a square pole face, core length = pole arc: $$ L = \psi\,\tau = \psi\frac{\pi D}{p} = \frac{0.7\pi D}{4} = 0.5498D $$ $$ \begin{aligned} D^3 &= \frac{4.115\times10^{-3}}{0.5498} = 7.485\times10^{-3}\ \text{m}^3 \\ D &= 0.1956\ \text{m} \\ L &= 0.5498 \times 0.1956 = 0.1075\ \text{m} \end{aligned} $$ #### Check - Pole pitch $\tau = \pi D/4 = 0.154$ m; pole arc $= 0.7\tau = 0.1075$ m $= L$ (square). - Peripheral speed $= \pi D n = \pi\times0.1956\times15 = 9.2$ m/s (safe). - Frequency $= pn/2 = 4\times15/2 = 30$ Hz (below 50 Hz). **Answer: Armature diameter $D \approx 0.196\ \text{m}\ (196\ \text{mm})$, armature core length $L \approx 0.108\ \text{m}\ (108\ \text{mm})$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is armature copper loss obtained? A. All 2500 W B. One-third of 2500 W C. 10 kW x 0.2 D. Half of total loss 2500 W = 1250 W Answer: D. At max efficiency, variable = constant loss. (Key point: Losses) 2. How is Ia found? A. Ia = line current B. Ia = line current - field current (3% of it) = 55.11 A C. Ia = line + field D. Ia = output/220 Answer: B. IL = 56.82, If = 1.705. (Key point: Field current) 3. What is Bav? A. 0.85 B. 0.85/0.7 C. Kf x Bm = 0.7 x 0.85 = 0.595 D. 0.7 Answer: C. Used in C0 = 176.2. (Key point: Bav) 4. What are D and L? A. D about 196 mm, L about 108 mm B. D about 138 mm, L about 76 mm C. D about 108 mm, L about 196 mm D. D about 168 mm, L about 92 mm Answer: A. D^3 = 7.485e-3, L = 0.5498 D. 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