Paul Kendall Alex Perez

 Virtually all of a cars linear energy is transferred to the brakes as thermal energy.  The faster the car stops, the less time the energy has to uniformly heat up the entire rotor. Significantly higher temperatures will appear at the surface than in the center of the rotor. The higher temperature causes the outside to expand more than the inside.  This uneven temperature gradient and the overall cycling temperature eventually warps the rotor which is the primary cause of pulsed braking.

 A car is traveling at 50 mph down the road. The traffic light turns red up ahead and the driver of the car the flexibility to stop slowly from that point to the light, or wait and brake rapidly right before the light. Assume that the minimum of sopping time is 3 seconds. How will stopping in 3 seconds vs. 10 seconds affect the stress in the front brakes? The car weighs a 2700 lbs. and has front brake rotors as described in the following diagram. 10" 5.5" 1.0" Heat from braking friction

 Weight of the car  Stopping time  Dimensions of the brake rotors  70% of the braking is done with the front brakes  Initial brake temperature is 15°C  Brakes are solid and made of plain carbon steel  Heat dissipation from radiation is neglected  Heat dissipation from convection (forced + natural) is constant  Final speed after braking is zero  Braking acceleration is constant

Velocity Position If there are two front brakes, and two sides to a brake Assuming a constant acceleration, the acceleration and stopping distance can be calculated from the stopping time. Using the mass of the car, acceleration, and dimensions of the disk rotor heat dissipation on each side of the rotor can be calculated.

 From this we learn: Stress is linearly proportional to the temperature gradient Higher temperature gradient = Higher stress Let:Where

 If the temperature profile in the center of the rotor doesn’t rise above the initial steady state temperature, the rotor can be approximated as a semi-infinite solid Eq  However, for a closer approximation, the non constant Cp and k (conductivity) properties of the metal will be included in the calculation using a numerical approach (see property plots below)

 Illustrated below is the temperature profile of the brake rotor using the semi- infinite solid approximation and numerical iterations using non constant properties. Note that the first and second iteration are virtually the same.

t=0.05 sec t=0.30 sec t=0.90 sec While the entire braking time and temperature can not be approximated using a semi infinite solid approach (the center rises above initial temperature), the initial braking temperatures are illustrated below. Note that the initial slopes of the 10 second brake profile are about 70% less than the 3 second brake profile. Therefore the stress will be about 70% less as well. t=0.05 sec t=0.30 sec t=0.90 sec

 An increase in linear velocity is almost linearly related to an increase in the average convective coefficient, h.  This means the faster someone is driving, the larger the heat transfer from hot rotors to the cooler ambient air.

Using Transient Conduction Theory, the time to cooldown to an arbitrary temperature can be calculated at the different convective coefficients. The chart shows that the faster someone drives after applying their brakes, it will exponentially decrease the time to cool the rotors down.

 Energy put into the brake rotor is linear with stopping time  Slope (and in turn stress) of the temperature profile is linearly proportional to stopping time (i.e. 3X faster stopping time is 3X the stress on the rotors)  Convective coefficient is nearly linear with velocity of the vehicle  An increase in velocity means an exponential decrease in the time it takes to cool the rotors  Suggested Driving Routine – Brake early and accelerate quickly to decrease the heat transfer to the rotors (when braking) and increase the heat transfer from the rotors (when accelerating).