--- title: "Time of reversal of a drive" subject: "Utilization of Electrical Energy" chapter: 2 key: "semester_7/utilization_of_electrical_energy:drive-reversal-time" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/drive-reversal-time site: BCT Engineering --- # Time of reversal of a drive Utilization of Electrical Energy · Chapter 2: Electric Drive System - Asked in: 2079 Baisakh, 8 marks - 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/drive-reversal-time - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-2 ## Question A drive has following parameters: J = 10 kg-m², T = 100 − 0.1N, N-m, Passive load torque T1 = 0.05N, N-m, where N is the speed in rpm. Initially the drive is operating in steady-state. Now it is to be reversed. For this motor characteristic is changed to T = −100 − 0.1N, N-m. Calculate the time of reversal. ## Short answer Initial steady speed is 666.67 rpm and final reverse speed is -666.67 rpm. Using J dω/dt = T - Tl gives time constant 6.981 s. Reversal to 95% of final speed takes t = 6.981 ln 40 = 25.75 s. ## Answer **Initial steady state.** Motor torque equals load torque: $$ 100 - 0.1N = 0.05N \;\Rightarrow\; N_0 = \frac{100}{0.15} = 666.67\ \text{rpm} $$ **After the change.** Motor characteristic $T = -100 - 0.1N$ (valid for positive and negative speed). The passive load torque $0.05N$ always opposes motion, so the same expression holds for negative speed. New steady state: $$ -100 - 0.1N = 0.05N \;\Rightarrow\; N_f = -666.67\ \text{rpm} $$ **Dynamic equation** ($\omega = \frac{2\pi}{60}N$): $$ J\frac{2\pi}{60}\frac{dN}{dt} = T - T_l = -100 - 0.15N $$ $$ \frac{dN}{dt} = -\frac{0.15}{J(2\pi/60)}\left(N + 666.67\right) $$ Time constant: $$ \tau = \frac{J(2\pi/60)}{0.15} = \frac{10\times0.10472}{0.15} = 6.981\ \text{s} $$ Solution: $$ N(t) = N_f + (N_0 - N_f)e^{-t/\tau} = -666.67 + 1333.33\,e^{-t/6.981} $$ Since the speed approaches $-666.67$ rpm only asymptotically, **reversal is taken as complete when the speed reaches 95 % of the final value**, i.e. $N = -633.33$ rpm (usual textbook assumption, Dubey). $$ \begin{aligned} t_r &= \tau\ln\frac{N_0 - N_f}{N - N_f} = 6.981\ln\frac{666.67+666.67}{-633.33+666.67} \\ &= 6.981\ln\frac{1333.33}{33.33} = 6.981\ln 40 = 25.75\ \text{s} \end{aligned} $$ (Braking to zero speed alone takes $6.981\ln 2 = 4.84$ s.) **Answer:** reversal time $\approx 25.75$ s (to 95 % of the reverse steady-state speed of $-666.67$ rpm). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the initial steady-state speed? A. 666.67 rpm, from 100 - 0.1N = 0.05N B. 1000 rpm C. 500 rpm D. -666.67 rpm Answer: A. Setting motor torque equal to load torque: 100 = 0.15N, so N = 666.67 rpm. (Key point: Initial speed) 2. What is the final steady speed after the characteristic changes? A. 666.67 rpm B. -666.67 rpm C. -1000 rpm D. 0 rpm Answer: B. -100 - 0.1N = 0.05N gives N = -666.67 rpm. (Key point: Final speed) 3. What is the time constant of the drive? A. J(2 pi/60)/0.15 = 6.98 s B. J/0.15 = 66.7 s C. 0.15/J = 0.015 s D. 10 s Answer: A. Torque slope total is 0.15, so tau = 10 x 0.10472 / 0.15 = 6.981 s. (Key point: Time constant) 4. How is reversal time defined, and what is its value? A. Time to reach 0 rpm, 4.84 s B. Time to reach 95 % of final speed, 25.75 s C. Time to reach 100 % of final speed, infinite D. Time equal to tau, 6.98 s Answer: B. Speed approaches final value asymptotically, so 95 % is used: t = 6.981 ln 40 = 25.75 s. 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