--- title: "Brake energy and drum temperature rise" subject: "Machine Design II" chapter: 7 key: "semester_7/machine_design_ii:brake-energy-absorbed-temperature-rise" source: https://www.bctengineer.com/q/semester_7/machine_design_ii/brake-energy-absorbed-temperature-rise site: BCT Engineering --- # Brake energy and drum temperature rise Machine Design II · Chapter 7: Clutches and brakes - Asked in: year not given, 6 marks - Video: [Thermal cosideration of brake with numerical,part-7](https://www.youtube.com/watch?v=A4vuNzGcmts) — EduGrown - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/brake-energy-absorbed-temperature-rise - Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-7 ## Question A car of mass 1400 kg moving at 90 km/h is brought to rest in 6 s with uniform retardation on a level road by four identical drum brakes. Each brake drum has a mass of 7 kg with specific heat 500 J/kg K. Find (a) the energy absorbed by each brake, the average retardation, the stopping distance and the average power per brake, and (b) the temperature rise of each drum, assuming that all heat goes into the drum. (c) The same car descends a slope of 1 in 20 at a steady 36 km/h. Find the heat dissipation needed per brake and the steady temperature rise of each drum if the effective product of heat transfer coefficient and area is hA = 18 W/K per drum. ## Short answer Kinetic energy gives 109.4 kJ per brake, retardation 4.17 m/s², stopping distance 75 m, average power 18.2 kW. Temperature rise per drum 31.3 °C. On slope, dissipation 1.72 kW per brake, steady temperature rise 95 °C. ## Answer #### Data $m = 1400$ kg, $v = 90$ km/h $= 25$ m/s, $t = 6$ s. #### (a) Energy and power Kinetic energy to be absorbed: $$ E = \tfrac12 m v^2 = \tfrac12\times1400\times25^2 = 437\,500\ \text{J} $$ Per brake: $E_1 = 437\,500/4 = 109\,375$ J $= 109.4$ kJ. Average retardation: $a = v/t = 25/6 = 4.17\ \text{m/s}^2$. Stopping distance: $s = \dfrac{v t}{2} = \dfrac{25\times6}{2} = 75$ m. Average power per brake: $P_1 = \dfrac{E_1}{t} = \dfrac{109\,375}{6} = 18.2$ kW (with uniform retardation the power at the start is twice this value). #### (b) Temperature rise (no heat loss) $$ E_1 = m_d\,c\,\Delta T \;\Rightarrow\; \Delta T = \frac{109\,375}{7\times500} = 31.3\ ^\circ\text{C} $$ For repeated stops, the cooling between applications must also be checked; a heavier or finned drum keeps the temperature lower. #### (c) Descending a slope A slope of 1 in 20 gives $\tan\theta = 1/20$, so $\sin\theta = 0.04994 \approx 0.05$; the speed is $v = 36$ km/h $= 10$ m/s. At steady speed the brakes absorb the loss of potential energy: $$ P = m g\,v\sin\theta = 1400\times9.81\times10\times0.04994 = 6859\ \text{W} $$ Per brake: $6859/4 = 1715$ W. At steady state, heat generated equals heat dissipated, $Q = hA\,\Delta T$: $$ \Delta T = \frac{1715}{18} = 95.3\ ^\circ\text{C} $$ This is acceptable for cast-iron drums, but on long descents the temperature may rise further and cause **brake fade** (the coefficient of friction falls as the temperature rises), so engine braking is used. **Answer:** (a) 109.4 kJ per brake, retardation 4.17 m/s², distance 75 m, 18.2 kW; (b) 31.3 °C; (c) 1.72 kW per brake, steady rise 95 °C. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Energy absorbed by each of the four brakes when stopping from 25 m/s is A. 437.5 kJ B. 54.7 kJ C. 218.8 kJ D. 109.4 kJ Answer: D. E = (1/2)(1400)(25^2) = 437.5 kJ, shared by 4 brakes is 109.4 kJ. (Key point: Energy per brake) 2. With all heat in the drum, delta T is found from A. E_1 = h A delta T B. E_1 = m_d c delta T = 109375/(7 x 500) = 31.3 C C. E = m c delta T with the car mass D. E_1 = P t / c Answer: B. The drum absorbs E_1 so delta T = E_1/(m_d c) = 31.3 degrees C. (Key point: Temperature rise) 3. On a steady downhill descent, the brakes absorb A. The kinetic energy of the car B. The potential energy lost, P = m g v sin(theta) C. Nothing, speed is constant D. Half of m g v Answer: B. At constant speed kinetic energy does not change, so the loss of potential energy is dissipated, 6859 W total. (Key point: Steady descent heat) 4. The steady temperature rise per drum with hA = 18 W/K is A. 95 C, from delta T = Q/(hA) = 1715/18 B. 31 C C. 381 C D. 6859/18 = 381 C per drum Answer: A. At steady state heat generated equals heat dissipated; 1715 W per brake gives 95.3 C. (Key point: Steady temperature rise) ## Related in Machine Design II - [Gear ratios of a four-speed car](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aautomobile-gear-ratio-selection) — Machine Design II, chapter 5 - [Internal expanding shoe brake analysis](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Ainternal-expanding-shoe-brake) — Machine Design II, chapter 7 - [Clutch versus brake and self-energising](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aclutch-brake-differences-self-energizing) — Machine Design II, chapter 7 - [Simple band brake absorbing power](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aband-brake-power-absorbed) — Machine Design II, chapter 7 ## Related in other subjects - [Stopping distance and brake energy](https://www.bctengineer.com/question.md?key=semester_5%2Fautomotive_technology%3Abraking-distance-energy-numerical) — Automotive Technology, chapter 7 - [Moving observer: car counts on 800 m road](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Amoving-observer-car-800m-20kmph) — Transportation Engineering II, chapter 1 - [Floating car method on 3.2 km road C-D](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Afloating-car-3-2km-cd-trips) — Transportation Engineering II, chapter 1 - [Disc brakes and drum brakes](https://www.bctengineer.com/question.md?key=semester_5%2Fautomotive_technology%3Adisc-brake-vs-drum-brake) — Automotive Technology, chapter 7