--- title: "Equivalent inertia, speed profile and quadrants (Fig 2b)" subject: "Utilization of Electrical Energy" chapter: 2 key: "semester_7/utilization_of_electrical_energy:equivalent-inertia-speed-profile-quadrants" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/equivalent-inertia-speed-profile-quadrants site: BCT Engineering --- # Equivalent inertia, speed profile and quadrants (Fig 2b) Utilization of Electrical Energy · Chapter 2: Electric Drive System - Asked in: 2071 Chaitra - Video: [Equivalent Value of Drive parameters ( Moment of inertia and torque)](https://www.youtube.com/watch?v=PAk4iVE1-vY) — Electrical point - 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/equivalent-inertia-speed-profile-quadrants - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-2 ## Question [Figure 1a: a motor (electric torque Te, speed ωm) is directly coupled to Load 1 (load torque Tl1); the motor and Load 1 together have inertia J1. Load 1's shaft drives Load 2 (load torque Tl2, inertia J2, speed ωm2) through a gear pair with n1 teeth on the motor side and n2 teeth on the load side.] [Figure 2b: electric torque Te (N-m) against time t (s): Section A, 0 to 0.1 s: +41; Section B, 0.1 to 0.2 s: +5; Sections C and D, 0.2 to 0.3 s (C from 0.2 to 0.25, D from 0.25 to 0.3): −67; Section E, 0.3 to 0.4 s: +5; Section F, 0.4 to 0.5 s: +41; after 0.5 s: +5.] Consider a dc motor driving system (Figure 1a) with following parameter; J2 = 0.01 kg-m², n1/n2 = 2, J1 = 0.08. Determine the equivalent 'Jeq' of the system. If electric torque profile of figure 2b is applied to motor, construct the profile of the speed (Assume Load torque TL = 5 NM). Using the plots (T vs. t and ω vs. t), discuss the quadrant of operation in each section A to F as shown. ## Short answer J_eq = 0.12 kg·m². Speed rises 0→30 rad/s (A), holds 30 (B), falls 30→0 (C), reverses 0→-30 (D), holds -30 (E), rises -30→0 (F). Quadrants: I, I, II, III, IV, IV. ## Answer **Assumptions:** the drive starts from rest at $t = 0$; the load torque $T_L = 5$ N-m is referred to the motor shaft and stays constant (as given) in every section. #### Equivalent inertia Load 2 runs at $\omega_{m2} = \dfrac{n_1}{n_2}\omega_m = 2\omega_m$. Equating kinetic energies: $$ \begin{aligned} J_{eq} &= J_1 + \left(\frac{n_1}{n_2}\right)^2J_2 = 0.08 + (2)^2(0.01) \\ &= 0.08 + 0.04 = 0.12\ \text{kg-m}^2 \end{aligned} $$ #### Speed profile From $T_e - T_L = J_{eq}\dfrac{d\omega_m}{dt}$, in each section the acceleration is constant: $$ \alpha = \frac{T_e - 5}{0.12} $$ | Section | Time (s) | $T_e$ (N-m) | $T_e - T_L$ | $\alpha$ (rad/s²) | $\Delta\omega$ | $\omega$ at end (rad/s) | |---|---|---|---|---|---|---| | A | 0 to 0.1 | 41 | 36 | 300 | +30 | 30 | | B | 0.1 to 0.2 | 5 | 0 | 0 | 0 | 30 | | C | 0.2 to 0.25 | -67 | -72 | -600 | -30 | 0 | | D | 0.25 to 0.3 | -67 | -72 | -600 | -30 | -30 | | E | 0.3 to 0.4 | 5 | 0 | 0 | 0 | -30 | | F | 0.4 to 0.5 | 41 | 36 | 300 | +30 | 0 | | after | > 0.5 | 5 | 0 | 0 | 0 | 0 | Sample working, section A: $\omega = 0 + 300\times0.1 = 30$ rad/s; section C: $\omega = 30 - 600\times0.05 = 0$. ```text Te (N-m) 41 |___ ___ 5 | |____ ____| |_____ 0 +---+----+----+----+---+---+-----> t (s) -67 | |____|____| 0 .1 .2 .25 .3 .4 .5 A B C D E F w (rad/s) 30 | /------\ | / \ 0 +--/----------\-----------/------> t | \ / -30 | \-------/ 0 .1 .2 .25 .3 .4 .5 ``` #### Quadrant of operation | Section | $T_e$ | $\omega$ | Quadrant | Operation | |---|---|---|---|---| | A | + | + (rising) | I | Forward motoring, accelerating | | B | + | + (constant) | I | Forward motoring, steady speed | | C | - | + (falling) | II | Forward braking (regenerative) | | D | - | - (rising in reverse) | III | Reverse motoring, accelerating | | E | + | - (constant) | IV | Reverse braking (motor torque opposes reverse motion, steady) | | F | + | - (falling to 0) | IV | Reverse braking, decelerating to rest | **Answer:** $J_{eq} = 0.12$ kg-m²; speed rises to 30 rad/s (A), holds (B), reverses to -30 rad/s (C, D), holds (E) and returns to zero at 0.5 s (F); quadrants I, I, II, III, IV, IV for A to F. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is Jeq for J1 = 0.08, J2 = 0.01 and n1/n2 = 2? A. 0.09 B. 0.12 C. 0.48 D. 0.10 Answer: B. Jeq = J1 + (n1/n2)^2 J2 = 0.08 + 4(0.01) = 0.12 kg-m^2. (Key point: Jeq calculation) 2. In section A (Te = 41, TL = 5), what is the acceleration? A. 300 rad/s^2 B. 340 rad/s^2 C. 36 rad/s^2 D. 600 rad/s^2 Answer: A. alpha = (41 - 5)/0.12 = 300 rad/s^2, giving 30 rad/s in 0.1 s. (Key point: Acceleration in section A) 3. In section C, Te is negative and speed is positive but falling. Which quadrant? A. I B. II C. III D. IV Answer: B. Negative torque with positive speed is forward braking, quadrant II. (Key point: Quadrant of section C) 4. In section D, Te is negative and speed is negative and increasing in magnitude. Which operation? A. Forward motoring B. Forward braking C. Reverse motoring (quadrant III) D. Reverse braking Answer: C. Torque and speed both negative is reverse motoring, quadrant III. 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