--- title: "Separately excited elevator motor with field flux reduced" subject: "Electrical Machines I" chapter: 4 key: "semester_4/electrical_machines_i:elevator-motor-300nm-flux-reduced" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/elevator-motor-300nm-flux-reduced site: BCT Engineering --- # Separately excited elevator motor with field flux reduced Electrical Machines I · Chapter 4: DC Motor - Asked in: 2080 Chaitra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/elevator-motor-300nm-flux-reduced - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4 ## Question A separately excited dc motor drives an elevator which requires a constant torque of 300 Nm. The motor is connected to a 600 V dc supply and rotates at 1500 rpm. The armature resistance is 0.5 Ω. Determine the armature current. If the field flux is reduced by 10 %, determine the armature current and motor speed. ## Short answer At 1500 rpm, ω=157.08 rad/s and required power is 47,124 W. Solving (600−0.5Ia)Ia=47,124 gives Ia1≈84.5 A. With flux reduced to 0.9φ, torque constant gives Ia2≈93.9 A and speed N2≈1653 rpm. ## Answer In a separately excited motor $T = k\phi I_a$ and $E_b = k\phi\omega$. The developed power $E_b I_a = T\omega$. **Given:** $T = 300$ N·m (constant), $V = 600$ V, $N_1 = 1500$ rpm, $R_a = 0.5\ \Omega$. #### Part 1: Armature current $$ \omega_1 = \frac{2\pi \times 1500}{60} = 157.08\ \text{rad/s} $$ $$ E_{b1} I_{a1} = T \omega_1 = 300 \times 157.08 = 47\,124\ \text{W} $$ With $E_{b1} = 600 - 0.5 I_{a1}$: $$ \begin{aligned} (600 - 0.5 I_{a1}) I_{a1} &= 47\,124 \\ 0.5 I_{a1}^2 - 600 I_{a1} + 47\,124 &= 0 \\ I_{a1} &= \frac{600 - \sqrt{600^2 - 4(0.5)(47\,124)}}{2(0.5)} = 84.49\ \text{A} \end{aligned} $$ (The other root, about 1115 A, is not a practical operating point.) $$ E_{b1} = 600 - 0.5 \times 84.49 = 557.76\ \text{V} $$ #### Part 2: Flux reduced by 10% ($\phi_2 = 0.9\phi_1$) Torque is constant, so $\phi_1 I_{a1} = \phi_2 I_{a2}$: $$ I_{a2} = \frac{84.49}{0.9} = 93.88\ \text{A} $$ $$ E_{b2} = 600 - 0.5 \times 93.88 = 553.06\ \text{V} $$ Speed: $$ \begin{aligned} N_2 &= N_1 \times \frac{E_{b2}}{E_{b1}} \times \frac{\phi_1}{\phi_2} \\ &= 1500 \times \frac{553.06}{557.76} \times \frac{1}{0.9} = 1652.6\ \text{rpm} \end{aligned} $$ **Answer:** Initially $I_a \approx 84.5$ A. With 10% less flux, $I_a \approx 93.9$ A and speed $\approx 1653$ rpm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the initial armature current found? A. Solve Eb·Ia = T·ω with Eb = V - Ia·Ra B. Ia = V/Ra C. Ia = T/(Ra) D. Ia = T·ω/V directly Answer: A. Eb Ia = 300×157.08 = 47124 W gives a quadratic with Ia = 84.49 A. (Key point: Power method) 2. Why is the 1115 A root rejected? A. It gives negative speed B. It is impractical; it is far above a normal operating current C. It gives negative torque D. It exceeds the supply voltage Answer: B. The practical root 84.5 A gives Eb of about 558 V. (Key point: Root choice) 3. Torque is constant and flux is cut by 10%. What happens to Ia? A. It falls to 0.9 × Ia1 B. It stays at 84.5 A C. It rises to Ia1/0.9 = 93.9 A D. It rises to 1.1 × Ia1 exactly Answer: C. T ∝ φIa, so φ1Ia1 = φ2Ia2. (Key point: Current with weaker field) 4. What is the new speed? A. About 1350 rpm B. About 1500 rpm C. About 1667 rpm D. About 1653 rpm Answer: D. N2 = 1500×(553.06/557.76)/0.9 = 1652.6 rpm. 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