Heat Due to Friction In Brake Pads

[Jobistober]

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. I am not looking for a definite answer at this point, just a general equation I can use.

 

[so-crates]

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.

 

[Jobistober]

??

thanks 'crates

when you have some time, can you explain in more detail

 

[russ_watters]

Its actually pretty simple: the brakes dissipate all of the kinetic energy of the car when it stops.

 

[Chronos]

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.

 

[Jobistober]

Lets not Get off subject here...

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

 

[russ_watters]

Its just as I said, the kinetic energy equation: e=1/2mv^2

 

[Jobistober]

So

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?

 

[Chronos]

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.

 

[russ_watters]

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.

 

[Jobistober]

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.

 

[Chronos]

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.

 

[Gokul43201]

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.

 

[Jobistober]

Ok

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?

 

[Gokul43201]

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.

 

[Chronos]

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.

 

[Jobistober]

how can slope be a factor in KE, would not slope only affect velocity, which is already a variable in the KE equation?

 

[Gokul43201]

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)

 

[Chronos]

how can slope be a factor in KE, would not slope only affect velocity, which is already a variable in the KE equation?

It increases [downhill] or decreases [uphill] the potential energy of the system mass due to gravity. Picture the bicycle sitting on a table attached to a counterweight suspended off the end of the table across a pulley. How much counterweight do you need to add to keep the bicycle from moving when the table is perfectly level? None. How much counterweight do you need to keep it from moving when you tilt the table? It depends on the sine of the angle of inclination. If you tilt it 90 degrees, the counterweight must be equal in mass to that of the bicycle. So basically you just add the counterweight mass required to offset the inclination angle to the mass of the bicycle. So, you just substitute m = [m + m*sine(theta)], where theta is the angle of inclination, when calculating KE.

 

[wd40]

I can presume you want the equations in order to determine the maximum heat generated in order to try and accommodate it in your design.
There seem to be an agreement that this is changing with many parameters. Unless you want to conduct a parametric study, I advise you to conduct an experiment at an extreme condition and extrapolate the data from there.
This might be a solution.

Luck.

 

[Jobistober]

It increases [downhill] or decreases [uphill] the potential energy of the system mass due to gravity. Picture the bicycle sitting on a table attached to a counterweight suspended off the end of the table across a pulley. How much counterweight do you need to add to keep the bicycle from moving when the table is perfectly level? None. How much counterweight do you need to keep it from moving when you tilt the table? It depends on the sine of the angle of inclination. If you tilt it 90 degrees, the counterweight must be equal in mass to that of the bicycle. So basically you just add the counterweight mass required to offset the inclination angle to the mass of the bicycle. So, you just substitute m = [m + m*sine(theta)], where theta is the angle of inclination, when calculating KE.

Ok, but I have been calculating the heat while the bike is decelerating on a relatively horizontal surface. Would I still have to factor in the slope to get the max initial kinetic energy?

 

[Chronos]

As you are aware, at any given initial velocity, stopping distance on a steep slope is much greater than on a level surface [which translates into more work done by the brake system]. The easiest thing to do is calculate heat rise on a level surface at maximum weight and velocity, and double it. Just guessing, I would say downhill braking is where the problem you are tackling usually occurs.

 

[saroshmumtaz]

the ans is
heat=fpV,
where f is friction coef, p is pressure, and v is velocity.

 

[ezskater]

Hi all,
I'd like to add a question re the "skidding" case. Basically, let's assume a brake pad is pressed against a surface that quickly moves along (can be the brake pad on the rim or, in the case of an inline skate with a heel brake, a brake pad and the road itself). Let's look at a quasi-stationary case: rolling down a long decline with constant slope and speed. So we have to dissipate some constant power (in the order of a few hundred Watts [10% slope - 10m/s speed - 1 m/s vertical speed - 80 kg mass - 800 N weight - about maybe 50% power dissipated into air friction - makes 400 W]).
In a first order approach I would say that the brake pad would not heat up higher than the temperature of the road. Because that surface moves along quickly, it does not heat up significantly while the pad glides along. The temperature at the interface road/pad is constant. This is where the heat is generated. So the road basically acts like an infinite and "zero-resistive" heat sink.
I think experience does not support this approximation. In particular, if you have an inline skate with a heel brake, the heel brake DOES heat up. How much? What's wrong with the first order approximation? What would be a more realistic model?
Thanks, Hanno

 

[ezskater]

It's me again ...
I just opened a new thread in the General Engineering Section to further discuss this. I think my problem has not so much to do with Materials and Chemical. So if you're interested in what I brought up, please go to the General Engineering Section.
Hanno

 

[jaap de vries]

Interested in actual temperature?

If you are interested in the actual temperature then remember that the heat transfer (per unit area) is equal to the temperature gradient times the thermal conductivity. Obviously the thermal conductivity of rubber is a lot smaller. For small geometries a finite difference approach can be suprisingly accurate.


\dot{q}'' = k \cdot \frac{dT}{dx} \approx k \cdot \frac{T_2-T_1}{x_2-x_1}

 

[Thenymphadora]

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?

urm..if you convert 18.89 K to degree celcius..the answer will be in negative value..can u expain this to me?

 

[russ_watters]

That's a delta-T. A temperature change.