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Question: An automobile vehicle weighing 13.5 kN is moving on a level road at a speed of 95 km/h. When the brakes are applied, it is subjected to a uniform deceleration of 6 m/s2. There are brakes on all four wheels. The tyre diameter is 750 mm. The kinetic energy of the rotating parts is 10% of the kinetic energy of the moving vehicle. The mass of each brake drum assembly is 10 kg and the specific heat capacity is 460 J/kg°C. Calculate (i) the braking time; (ii) the braking distance; (iii) the total energy absorbed by each brake; (iv) the torque capacity of each brake; and (v) the temperature rise of brake drum assembly.
(i) Braking Time: We'll first find out the braking time using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration/deceleration, and t is the time. Initial velocity (u) = 95 km/h = 95 * (1000/3600) m/s = 26.39 m/s (converted to m/s) Final velocity (v) = 0 m/s (as the car comes to a stop) Deceleration (a) = -6 m/s^2 (negative since its decelerating) Now, using the equation: 0 = 26.39 + (-6) * t t = 26.39 / 6 t = 4.40 s Braking Time (t) = 4.40 seconds (ii) Braking Distance: We can use the equation v^2 = u^2 + 2as, where s is the braking distance. 0 = 26.39^2 + 2 * (-6) * s s = 26.39^2 / (2 * 6) s = 60.87 m Braking Distance (s) = 60.87 meters (iii) Total Energy Absorbed Per Brake: Total potential energy (E_total) = (10/100) * 13.5 kN kinetic energy of rotating parts E_total = 1.35 kN = 1350 J (converted to Joule) Since there are 4 brakes, Energy absorbed per brake = 1350 / 4 Energy absorbed per brake = 337.5 J (iv) Torque Capacity Per Brake: First, we need to calculate the brake force (F): Force (F) = mass of vehicle * deceleration F = 13.5 kN * 1000 / 9.81 (converting kN to kg) * 6 m/s^2 F = 8229.36 N (Newton) Brake force per wheel = 8229.36 N / 4 Brake force per wheel = 2057.34 N Now, we can calculate torque capacity per brake (T) using the equation T = F * r, where r is the radius of the tire. Radius (r) = 750 mm / 2 = 375 mm = 0.375 m (converted to meters) Torque capacity per brake (T) = 2057.34 N * 0.375 m T = 771.51 Nm Torque Capacity Per Brake = 771.51 Nm (v) Temperature Rise of Brake Drum Assembly: Energy absorbed per brake drum assembly = 337.5 J Mass of brake drum assembly (m) = 10 kg Specific heat capacity (c) = 460 J/kg°C We can use the equation ∆Q = mc∆T or ∆T = ∆Q / (mc) to find the temperature rise (∆T). Temperature rise (∆T) = 337.5 / (10 * 460) ∆T = 0.0733°C Temperature Rise of Brake Drum Assembly = 0.0733°C
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