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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) To calculate the braking time, we can use the formula: t = (final velocity - initial velocity) / deceleration First, we need to convert the initial velocity (95 km/h) to meters per second (m/s): Initial velocity = 95 km/h * (1000 m/km) * (1 h/3600 s) = 26.39 m/s Since the vehicle comes to a stop, the final velocity = 0 m/s. Then, we can find braking time: t = (0 - 26.39 m/s) / (-6 m/s²) = 4.4 s (ii) To calculate the braking distance, we can use the formula: Braking distance = initial velocity * time + 0.5 * deceleration * time² Braking distance = 26.39 m/s * 4.4 s + 0.5 * (-6 m/s²) * (4.4 s)² = 57.85 m (iii) To find the total energy absorbed by each brake, we need to find the kinetic energy (KE) of the vehicle first: KE = 0.5 * mass * initial_velocity² Given that an automobile vehicle weighs 13.5 kN, we first convert it to mass (kg): Mass = 13.5 kN * (1000 N/kN) / 9.81 m/s² = 1375.38 kg Now, we find the kinetic energy of the vehicle: KE = 0.5 * 1375.38 kg * (26.39 m/s)² = 476528.05 J As mentioned in the problem, the kinetic energy of the rotating parts is 10% of the kinetic energy of the moving vehicle: Rotating parts KE = 476528.05 J * 10% = 47652.80 J Then, subtract the rotating parts' kinetic energy from the total kinetic energy to get the energy absorbed by the brakes: Energy absorbed by brakes = 476528.05 J - 47652.80 J = 428875.25 J Since there are 4 brakes, the total energy absorbed by each brake is: Energy absorbed by each brake = 428875.25 J / 4 = 107218.81 J (iv) To find the torque capacity of each brake, we can first find the angular deceleration (α) using the formula: α = deceleration / radius Given the tire diameter of 750 mm, the radius is 375 mm or 0.375 m. Now, we can find the angular deceleration: α = 6 m/s² / 0.375 m = 16 rad/s² Now, we can find the torque capacity using the formula: Torque capacity = moment of inertia (I) * α The moment of inertia (I) of each brake drum assembly can be found using the formula: I = 0.5 * m * r², where m is the mass and r is the radius With a mass of 10 kg for each brake drum assembly, the moment of inertia is: I = 0.5 * 10 kg * (0.375 m)² = 0.703 kg·m² Now, we can find the torque capacity: Torque capacity = 0.703 kg·m² * 16 rad/s² = 11.25 N·m (v) To find the temperature rise of the brake drum assembly, we can use the formula: Temperature rise = (Energy absorbed by each brake) / (mass of brake drum assembly * specific heat capacity) Temperature rise = 107218.81 J / (10 kg * 460 J/kg°C) = 23.3°C
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