--- title: "Hoist equilibrium speeds in four quadrants" subject: "Utilization of Electrical Energy" chapter: 2 key: "semester_7/utilization_of_electrical_energy:hoist-equilibrium-speeds-four-quadrants" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/hoist-equilibrium-speeds-four-quadrants site: BCT Engineering --- # Hoist equilibrium speeds in four quadrants Utilization of Electrical Energy · Chapter 2: Electric Drive System - Asked in: 2081 Baisakh, 8 marks; 2071 Shrawan (asked 2 times) - Video: [four quadrant operation of electric drive | electrical drive | dc drive | in hindi | animation |L-10](https://www.youtube.com/watch?v=ZD6eGKjmCmc) — Mind of Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/hoist-equilibrium-speeds-four-quadrants - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-2 ## Question A motor is used to drive a hoist. Motor characteristics are given by Quadrants I, II and IV: T = 200 − 0.2N, N-m Quadrants II, III and IV: T = −200 − 0.2N, N-m Where N is the speed in rpm. When hoist is loaded, the net load torque T1 = 100 N-m and when it is unloaded, net load torque T1 = −80 N-m. Obtain the equilibrium speeds for operation in all the four quadrants. ## Short answer Equate motor torque to load torque using the given characteristic for each quadrant. QI: 200-0.2N=100 → N=500 rpm; QII: 200-0.2N=-80 → N=1400 rpm; QIII: -200-0.2N=-80 → N=-600 rpm; QIV: -200-0.2N=100 → N=-1500 rpm. ## Answer Equilibrium (steady-state) speed is where the motor torque equals the load torque: $T = T_l$. **Data** - Characteristic 1 (quadrants I, II, IV): $T = 200 - 0.2N$ - Characteristic 2 (quadrants II, III, IV): $T = -200 - 0.2N$ - Loaded hoist: $T_l = 100$ N-m; unloaded hoist: $T_l = -80$ N-m (the counterweight is heavier than the empty cage, so the load torque is negative). A hoist load is an **active (gravity) load**, so $T_l$ keeps its sign when the speed reverses. **Quadrant I: forward motoring (hoisting the loaded cage)**, characteristic 1, $T_l = 100$: $$ \begin{aligned} 200 - 0.2N &= 100 \\ N &= \frac{200-100}{0.2} = 500\ \text{rpm} \end{aligned} $$ $T = +100$, $N = +500$: both positive, quadrant I. **Quadrant II: forward braking (hoisting the empty cage; counterweight pulls)**, characteristic 1, $T_l = -80$: $$ \begin{aligned} 200 - 0.2N &= -80 \\ N &= \frac{200+80}{0.2} = 1400\ \text{rpm} \end{aligned} $$ $T = -80$, $N = +1400$: torque negative, speed positive, quadrant II (regenerative braking; speed above no-load speed of 1000 rpm). **Quadrant III: reverse motoring (lowering the empty cage)**, characteristic 2, $T_l = -80$: $$ \begin{aligned} -200 - 0.2N &= -80 \\ N &= \frac{-200+80}{0.2} = -600\ \text{rpm} \end{aligned} $$ $T = -80$, $N = -600$: both negative, quadrant III. **Quadrant IV: reverse braking (lowering the loaded cage)**, characteristic 2, $T_l = 100$: $$ \begin{aligned} -200 - 0.2N &= 100 \\ N &= \frac{-200-100}{0.2} = -1500\ \text{rpm} \end{aligned} $$ $T = +100$, $N = -1500$: torque positive, speed negative, quadrant IV (regenerative braking during lowering). ```text N (rpm) II: 1400 | I: 500 (T=-80) | (T=+100) -----------+-----------> T III: -600 | IV: -1500 (T=-80) | (T=+100) ``` | Quadrant | Operation | Load torque (N-m) | Characteristic | Speed (rpm) | |---|---|---|---|---| | I | Forward motoring, hoisting loaded | 100 | $200-0.2N$ | 500 | | II | Forward braking, hoisting empty | -80 | $200-0.2N$ | 1400 | | III | Reverse motoring, lowering empty | -80 | $-200-0.2N$ | -600 | | IV | Reverse braking, lowering loaded | 100 | $-200-0.2N$ | -1500 | **Answer:** equilibrium speeds are 500 rpm (I), 1400 rpm (II), -600 rpm (III) and -1500 rpm (IV). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the condition for steady-state (equilibrium) speed? A. Motor torque equals load torque B. Motor torque is zero C. Speed equals zero D. Load torque is zero Answer: A. Equilibrium occurs where motor torque equals load torque, T = T_l. (Key point: Equilibrium condition) 2. For the loaded hoist (T_l = 100) on T = 200 - 0.2N, what is the speed? A. 1400 rpm B. 500 rpm C. -600 rpm D. 1000 rpm Answer: B. 200 - 0.2N = 100 gives N = 100/0.2 = 500 rpm. (Key point: Quadrant I speed) 3. For the empty hoist (T_l = -80) on T = 200 - 0.2N, what is the speed and mode? A. 1400 rpm, forward braking (regenerative) B. -600 rpm, reverse motoring C. 500 rpm, forward motoring D. -1500 rpm, reverse braking Answer: A. 200 - 0.2N = -80 gives N = 1400 rpm, above the no-load speed of 1000 rpm, so it is regenerative braking. (Key point: Quadrant II speed) 4. Which pair of speeds is correct for quadrants III and IV? A. -600 rpm (III, T_l = -80) and -1500 rpm (IV, T_l = 100) B. -1500 rpm (III) and -600 rpm (IV) C. 600 rpm (III) and 1500 rpm (IV) D. -200 rpm (III) and -300 rpm (IV) Answer: A. -200 - 0.2N = -80 gives -600 rpm; -200 - 0.2N = 100 gives -1500 rpm. 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