--- title: "Armature size, 7.5 kW 1800 rpm motor" subject: "Electric Machine Design" chapter: 5 key: "semester_5/electric_machine_design:dc-motor-7-5kw-1800rpm" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-motor-7-5kw-1800rpm site: BCT Engineering --- # Armature size, 7.5 kW 1800 rpm motor Electric Machine Design · Chapter 5: DC Machine Design - Asked in: 2074 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-motor-7-5kw-1800rpm - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-5 ## Question Calculate the diameter and length of armature of 7.5 kW, 4 pole, 1800 rpm, and 220 V shunt motor. Given: full load efficiency = 0.83, maximum gap flux density = 0.9 Wb/m², specific electric loading = 30,000 ampere conductor per meter, field form factor = 0.7. Assume that the maximum efficiency occurs at full load and the field current is 2.5% of rated current. The pole is square face and consider all the possible losses in the machine. ## Short answer Armature power 8.04 kW, C₀ = 186.5, D²L = 1.437×10⁻³ m³; with square pole face L = 0.5498D, D ≈ 0.138 m (138 mm), L ≈ 0.076 m (76 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). #### Step 1: Armature power (losses split at maximum efficiency) $$ \begin{aligned} \text{Input} &= \frac{7.5}{0.83} = 9.036\ \text{kW} \\ I_L &= \frac{9036}{220} = 41.07\ \text{A} \\ I_f &= 0.025 \times 41.07 = 1.027\ \text{A} \\ I_a &= I_L - I_f = 41.07 - 1.027 = 40.05\ \text{A} \\ \text{Total losses} &= 9036 - 7500 = 1536\ \text{W} \end{aligned} $$ At maximum efficiency, variable loss = constant loss, so armature copper loss $= 1536/2 = 768$ W. (Field copper, iron, friction and brush losses form the constant part, so all losses are counted.) $$ \begin{aligned} R_a &= \frac{768}{40.05^2} = 0.479\ \Omega \\ E_b &= V - I_aR_a = 220 - 40.05\times0.479 = 200.8\ \text{V} \\ P_a &= E_b I_a = 200.8 \times 40.05 = 8042\ \text{W} = 8.04\ \text{kW} \end{aligned} $$ #### 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) $= K_f = 0.7$. $$ \begin{aligned} B_{av} &= K_f B_m = 0.7\times0.9 = 0.63\ \text{Wb/m}^2 \\ C_0 &= \pi^2 B_{av}\, ac\times10^{-3} = \pi^2\times0.63\times30000\times10^{-3} = 186.5 \\ n &= 1800/60 = 30\ \text{rps} \end{aligned} $$ $$ D^2L = \frac{P_a}{C_0 n} = \frac{8.04}{186.5\times30} = 1.437\times10^{-3}\ \text{m}^3 $$ #### Step 3: Square pole face For a square pole face, core length = pole arc: $$ L = \psi\,\tau = \psi\frac{\pi D}{p} = \frac{0.7\pi D}{4} = 0.5498D $$ $$ \begin{aligned} D^3 &= \frac{1.437\times10^{-3}}{0.5498} = 2.614\times10^{-3}\ \text{m}^3 \\ D &= 0.1378\ \text{m} \\ L &= 0.5498 \times 0.1378 = 0.0757\ \text{m} \end{aligned} $$ #### Check - Pole pitch $\tau = \pi D/4 = 0.108$ m; pole arc $= 0.7\tau = 0.0757$ m $= L$ (square). - Peripheral speed $= \pi D n = \pi\times0.1378\times30 = 13.0$ m/s (well below 30 m/s). - Frequency $= pn/2 = 4\times30/2 = 60$ Hz (acceptable for a small machine). **Answer: Armature diameter $D \approx 0.138\ \text{m}\ (138\ \text{mm})$, armature core length $L \approx 0.076\ \text{m}\ (76\ \text{mm})$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is armature power Pa obtained? A. Pa equals output B. Pa = input C. Pa = output x efficiency D. Armature copper loss is half total loss at max efficiency, then Pa = Eb Ia Answer: D. Ra = 0.479, Eb = 200.8 V, Pa = 8.04 kW. (Key point: Armature power) 2. What is D^2L? A. Pa x C0 x n B. Pa/C0 C. C0 n / Pa D. Pa/(C0 n) = 1.437e-3 m^3 Answer: D. 8.04/(186.5 x 30). (Key point: D^2L) 3. What are D and L? A. D about 138 mm, L about 76 mm B. D about 196 mm, L about 108 mm C. D about 168 mm, L about 92 mm D. D about 76 mm, L about 138 mm Answer: A. L = 0.5498 D. 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