# Heat Due to Friction In Brake Pads

by Jobistober
 P: 10 I am designing a bike frame and I want to know how much heat will be generated between the rubber brake pads and the alloy rims. I have looked endlessly on the internet for formulas and I came here hoping someone could give me an answer. Im not looking for a definite answer at this point, just a general equation I can use.
 P: 309 A general equation hmmm, how about $$\Delta U = q + w$$ :-) No really you can use it, but the devil is in the details. The amount heat must depend on the coefficient of kinetic friction between the brake pads and the frame. Also, assuming the chemical constitution of the brake pads/frame doesn't change(would be a fair assumption), then q must equation work input - work output. Only problem is you also have to consider how quickly the brake pads will conduct that heat into the air, as the system is not isolated. I don't have time to go much further than that.
 P: 10 thanks 'crates when you have some time, can you explain in more detail
 Mentor P: 22,303 Heat Due to Friction In Brake Pads Its actually pretty simple: the brakes dissipate all of the kinetic energy of the car when it stops.
 Sci Advisor PF Gold P: 9,454 Friction is a real beast in the real world. It finds truly amazing ways to obey the laws of thermodynamics and energy conservancy. Heat is a major player, but other factors also come into play in a big way. In most mechanical systems, vibration [kinetic energy], not heat, is the preferred route to discharge energy. This is largely a curse for mechanical engineers. Heat is more easily managed than vibration.
 P: 10 thank you all for your input, but let me remind you that I started this thread looking for general equations that I can use to find the heat generated due to friction between bicycle brake pads and its rims, just as I said in the first post. Thanks
 Mentor P: 22,303 Its just as I said, the kinetic energy equation: e=1/2mv^2
 P: 10 kinetic energy in joules is equal to heat in joules created by friction? Must I assume coefficients of friction, heat dissipation into the air, material makeup, etc have nothing to do with it?
 Sci Advisor PF Gold P: 9,454 You can do a straight up calculation, but the temperature you end up with will be higher than what occurs in reality. Energy is bled off through in a number of other ways, such as rolling resistance. The brake will also normally seize the rim at some point, transfering all the friction to the tires as they skid against the road surface. Having said all that, this is the formula W = J x Q where W is the work [kinetic energy at initial velocity], J is Joules constant, and Q is the total amount of heat generated. What this does not tell you is how the heat is distributed between the brake pad and rim. To approximate that, you need to figure in the specific heat of the pad and rim material and their masses. The caloric energy is converted to temperature as follows Q = mcT where Q is the caloric energy, m is the mass, c is the specific heat of the material and T is the temperature rise. Calculate a temperature for both the rim and the pad assuming each will absorb the total heat energy. Of course neither will actually get that hot, but, it gives a starting point. You could split the difference, let the rim take half and the pad the other half, if you are curious what is more realistic. But, from a design perspective, I would be inclined to use the maximum possible temperature. I know for certain neither of them will get that hot, so my design has a nice fat safety margin. Hope that helps.
Mentor
P: 22,303
 Quote by Jobistober kinetic energy in joules is equal to heat in joules created by friction? Must I assume coefficients of friction, heat dissipation into the air, material makeup, etc have nothing to do with it?
Since you already know the brakes are capable of stopping the car, you don't need to do any friction calculations.

But Chronos is right - going from how much heat is dissipated to how, precisely, it is dissipated is not an easy task.
 P: 10 you input has helped Chronos, however could you specify what the value of joules constant is and what the units of T are in the equation Q=mcT please. I know I must sound like an idiot for asking such a dumb question and for that I apologize, but your help is greatly appreciated.
 Sci Advisor PF Gold P: 9,454 Joules constant is 4.184 Joules/calorie T is in units of C [degrees centrigrade] Suggestion: when you calculate the work done [removing the kinetic energy of the system] do it in Joules units. Those are not dumb questions. The only dumb questions are the ones you never ask.
 Emeritus Sci Advisor PF Gold P: 11,155 Let me do a rough calculation with some assumed numbers. You can change them to those you prefer. Mass of bike + rider = 220 lbs = 100 kg (I'm just using nice numbers...to make this easy for me ) Max. speed = 22 mph = 36 km/hr = 10 m/s So, KE = 0.5*100*10^2 = 5000 J Typically, most of this frictional loss comes from skidding. How tightly you squeeze the brakes, determines how much of the loss is at the brakes, and how much at the bottom of the tire. Assume all of this loss is from skidding. The heat generated there is divided about evenly between the ground and the bottom of the tires. So, about 2500J of heat goes into the tires, or about 1250 J per tire, if both wheels are skidding - if you lock both front and back brakes. Since rubber is a lousy conductor of heat, I'm going to assume that all the heat absorbed by the rubber stays in that portion of the tire that makes contact with the ground. I have absolutely no idea what the thickness, width and typical softness of tires, so I'm making an arbitrary guess that the volume of rubber involved is about 5cc, so I guess it mass is about 25 gm (I'm guessing that the density of this rubber is about 5 g/cc). Also, I've found the specific heat capacity of rubber to be between 1200 and 1700 J/K-kg. I'm going to use C = 1250 J/K-kg for niceness. So we have, Heat =1250 J = m*C*dT = .025*1250*dT. That gives dT = 40 K or about 70 F. Clearly this is an overestimate...but it's like a worst case number. Now this is the increase in temperature of the tire due to skidding. It is not the increase in temperature of the rim and brake padss due to the brakes. That can be done similarly, using the KE of the wheel (=mv^2, where m = mass of wheel). I'll get to it I find the time.
 P: 10 Now considering everything everyone else has said, I am going to use these equations and you people can back me up. However this time, I am going to use as close to real world numbers as possible. Everyone is free to correct me if I am wrong! Using actual numbers I have obtained on a run with my gravity bike: total mass (Bike and rider) = approx. 300 lbs = 136 Kg max speed = 47.7 mph = 21.32 m/s KE = (1/2)136(21.32)^2 = 30908.88 J Assuming the rims are made up of steel, mass and specific heat are: c= 452 J/Kg/K m= 4lbs = 1.81Kg Also assuming braking forces are divided half and half between the two rims: Q= 15454.44 J 15454.44= (1.81)(452)T 15454.44=818.12T T = 18.89 K = 32 F roughly 32 degrees ferenheit is the final temperature increase per rim, correct?
 Emeritus Sci Advisor PF Gold P: 11,155 This would be correct only if there's no skidding. And even so, you'll have to make the following adjustments. The heat would be divided between the rim and the brake pads, roughly equally. And stainless steel, being a fairly poor conductor, the entire rim does not get heated immediately - only the portion of the rim that gets to be in contact with the pad (affected volume=rim thickness*pad width*wheel circumference) This will give you an upper limit on the maximum instantaneous temperature reached by the rim. Radiation losses will be small and may be neglected. You could also test this with some kind of strip thermometer, to confirm that it's not grossly off.
 Sci Advisor PF Gold P: 9,454 Correct, Gokul. Which is what Job is looking for. He wants to know how hot the rims could get in the worst case. I should mention that the calculation does not take into account slope. You need both the slope and maximum initial velocity to get the right value for the total kinetic energy.
 P: 10 how can slope be a factor in KE, would not slope only affect velocity, which is already a variable in the KE equation?
 Emeritus Sci Advisor PF Gold P: 11,155 Realistic slopes will have a small effect, because when you use energy conservation, you have to include the PE too (which, in the absense of a slope, remains constant)

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