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-   -   Heat dissipation between brake Pad and Disc (http://www.physicsforums.com/showthread.php?t=158582)

heiroglif Feb28-07 01:44 PM

Heat dissipation between brake Pad and Disc
 
I have begun modelling a basic braking system in Simulink/Matlab for my final year project. The model is designed from the time force is exerted to the pedal at a particular velocity, to when heat is produced from the frictional force between the brake pad and disc.

I have the modelling equation for the heat produced from the frictional force, yet im not sure if its correct;

dW/dt = F*V(dS/dt)

where W is the work done with respect to time.
F is the frictional force,
S is displacement,
and V is velocity

My question is. Knowing the initial temperatures of the disc and pad, and knowing the coefficient of friction of the pad material, how can i measure the rate of heat dissipation over time, until the car reaches the desired stopping time limit?

I need it in an equation that can be implemented into simulink.

Im guessing its a first order differenitial//

any suggestions?

Thanks, H

mybsaccownt Feb28-07 02:09 PM

it seems like the units don't work out

you want work, which is N*m/s or just joule/s,

ds/dt is also velocity, is it not?

so you have:

change in work per time = force*velocity*velocity?

so...W = N * m^2/s^2 = N*m/s <--nope


also it's a partial differential equation, but i suppose you could take (dw/dt)/(ds/dt) to get dw/ds

hmm...what's happening is that the kinetic energy...you know what, I just punched in 'rotational kinetic energy' into google and I think you would benefit from reading something like this:

http://theory.uwinnipeg.ca/mod_tech/node50.html

or this:

http://hyperphysics.phy-astr.gsu.edu/hbase/rke.html


I have yet to deal with R.K.E but it seems like...

the friction is going to cause a torque, which will do work, which will be enough work to transform the initial kinetic energy into heat

heiroglif Feb28-07 02:28 PM

Hi, thanks for your quick response...

ok...

Quote:

Quote by mybsaccownt (Post 1259524)
it seems like the units don't work out

you want work, which is N*m/s or just joule/s,

ds/dt is also velocity, is it not?

so you have:

change in work per time = force*velocity*velocity?

so...W = N * m^2/s^2 = N*m/s <--nope


I think mistyped the equation...you are right, so if we take out V,

and leave it as

dW/dt = F(dS/dt)

this should be ok now right?


Quote:

also it's a partial differential equation, but i suppose you could take (dw/dt)/(ds/dt) to get dw/ds
good point, if i use dw/ds, it should still contain time as a factor.

Quote:

hmm...what's happening is that the kinetic energy...you know what, I just punched in 'rotational kinetic energy' into google and I think you would benefit from reading something like this:

http://theory.uwinnipeg.ca/mod_tech/node50.html

or this:

http://hyperphysics.phy-astr.gsu.edu/hbase/rke.html


I have yet to deal with R.K.E but it seems like...
im checking it out now...

Quote:

the friction is going to cause a torque, which will do work, which will be enough work to transform the initial kinetic energy into heat
to my understanding....the frictional force causes a Torque on the wheels,
this is given by

Td = C*((Dd/2) - do)*δP

where

do=offset distance to front and rear calipers
Dd=diameter of disc
C=caliper coefficient.
δP=pressure generated at master cylinder

I dont want to get into modelling forces on the wheel unless i have to to get heat dissipation.

?

Stingray Feb28-07 04:07 PM

What exactly are you asking? Your English is very unclear. Is this for high school or university?

Anyway, you now have the correct expression for the rate at which the brakes are doing work. Unless the tires are slipping significantly, most of that energy is dissipated as heat at the pad-rotor interface.

Still, this completely ignores the details of the heat conduction through various parts. It also does not take into account the heat removed by the air. Both of these effects are extremely difficult to model in any interesting way. The best you could probably do is say that the heat distributes itself uniformly through the brake components, and that it disappears at a rate depending on some given function of temperature and velocity. That would be very crude, though. Anything better is far too difficult.

I don't see the point of including only the term you have. Its interpretation is pretty clear. Solving it by hand shows (unsurprisingly) that the amount of heat dissipated over a finite time interval would just be the change in kinetic energy of the car. No numerical solutions are necessary.

heiroglif Feb28-07 04:46 PM

...
 
Quote:

Quote by Stingray (Post 1259668)
What exactly are you asking? Your English is very unclear. Is this for high school or university?.

apologies if my writing is unclear. I will elaborate....I'd like to model a brake system using a variation of brake pads and brake discs, and find out how their material properties affect overall braking efficiency.

Knowing that, the friction material of the brake pad, and the type of metal composition of the disc have a limiting value, we can assume after the limiting value is reached, the brake performance becomes innaffective, or is drastically reduced.

I would like to analyse which combinations of material perform best, and i assume the best way to monitor this would be to look at the amount of heat produced, and the way it is dissipated, so that if the car stops in say 20metres at 30mph, and the limiting value is not reached, how does this change if we stop in 20m at 60mph (60mph would require more force, therefore more friction, therefore more heat).

Quote:

Anyway, you now have the correct expression for the rate at which the brakes are doing work. Unless the tires are slipping significantly, most of that energy is dissipated as heat at the pad-rotor interface.
by pad rotor interface we mean between disc and pad right?

if so, yes this is the bit i want to concentrate on.

Quote:

Still, this completely ignores the details of the heat conduction through various parts. It also does not take into account the heat removed by the air. Both of these effects are extremely difficult to model in any interesting way.
this is why if i base my project around disc and brake pad variations ONLY, i can make a project which analyses the best combinations of material.

I would presume, taking all the known affecting factors into account would most definately prove to be a big task, thats why, im trying to focus on one particular thing.

Quote:

The best you could probably do is say that the heat distributes itself uniformly through the brake components, and that it disappears at a rate depending on some given function of temperature and velocity. That would be very crude, though. Anything better is far too difficult.
this suggestion, seems like the proposal i made above. Would it still seem crude if the project intended to only take into account brake pad, and disc variations? (lol)

Quote:

I don't see the point of including only the term you have. Its interpretation is pretty clear. Solving it by hand shows (unsurprisingly) that the amount of heat dissipated over a finite time interval would just be the change in kinetic energy of the car. No numerical solutions are necessary.
what if i showed graphically how heat increases, or decreases, over time, with a given coefficient of friction, which represents a different brake pad material?

The project is a university project. The course is a 1/2 1/2 mix of computer science and mechanical engineering. Therefore, as you may guess, im not a physics whizz, im actually quite basic, but was more or less forced to do a stricly engineering project. Which is ok, but i am definately in need of some guidance.

Thanks for your response, H

Stingray Feb28-07 06:50 PM

A good brake pad will generally have a coefficient of friction which is approximately constant over some range of temperature. This range is typically a few hundred degrees. The friction coefficient usually drops very quickly outside of that range. Too cold can be as bad as too hot, at least from the pad's point of view (If things really get hot, the brake fluid will boil. Then your system won't be able to maintain any significant pressure.).

In any case, the actual value of the friction coefficient in the pad's operating range isn't very important. All cars are designed such that within the appropriate temperature range, braking performance is limited by the tires rather than the amount of force a driver can apply to the pedal.

The friction coefficient also doesn't affect heat generation directly. The driver has in mind a certain amount of deceleration every time he presses the brake pedal. If a particular pad doesn't grip as well, a driver would just press the pedal harder until he got the desired effect.

Given that, I'm not sure what you want to optimize, or how exactly you want to define "braking efficiency." I think you want to calculate the worst-case temperature of the disks as a function of some parameters. The simplest thing to think of would be the change in temperature after a single stop. But that's something that can be understood with a few lines of calculation on a piece of paper (given the approximations we're using).

In the real world, single stops are not difficult for any decent braking system. Problems come up when you try to brake hard over and over. This is something that might be good to use a computer for. The problem is that you have to come up with some model of how a car is driven. You also need to figure out how heat is taken out of the brakes. Otherwise the simulated temperature will keep rising without limit. Say that the heat Q in the braking system obeys
[tex]
d Q/dt = M a v - f(T,v) ~,
[/tex]
where M is the car's mass, a is its acceleration, v is its velocity, and f is the cooling function. The temperature of the brakes could then be
[tex]
T = T_{0} + Q/mc ~,
[/tex]
where T0 is ambient temperature, m is the mass of the braking system, and c is its specific heat (you could look this up for iron).

The problem here is f. Some googling seems to show that it's reasonable to take
[tex]
f(T,v) = k_{1}(T-T_{0}) (1+k_{2} v^{n}).
[/tex]
k1, k2, and n are constants. n should apparently be between 0.5 and 0.8. k1 can be reasonably estimated, but k2 can't. That depends on the details of the airflow around the brake components.

I hope that helps a bit.

heiroglif Mar5-07 02:49 PM

...
 
Hi Stingray,

thanks for your reply,

it did help, especially the equation for heat dissipation...

ive been away from the PC this weekend.

I just have a few comments, and questions, see if you can help me..

Quote:

Quote by Stingray (Post 1259793)
A good brake pad will generally have a coefficient of friction which is approximately constant over some range of temperature. This range is typically a few hundred degrees. The friction coefficient usually drops very quickly outside of that range. Too cold can be as bad as too hot, at least from the pad's point of view (If things really get hot, the brake fluid will boil. Then your system won't be able to maintain any significant pressure.).

In any case, the actual value of the friction coefficient in the pad's operating range isn't very important. All cars are designed such that within the appropriate temperature range, braking performance is limited by the tires rather than the amount of force a driver can apply to the pedal.

Indeed, the operating range is rarely crossed, yet by using a variety of friction coefficients id be representing a variety of brake pads, which is one of the aims outlined in the project.

Quote:

In the real world, single stops are not difficult for any decent braking system. Problems come up when you try to brake hard over and over. This is something that might be good to use a computer for.
Thanks, Yes this is what im trying to get at, repeated braking...usually more of a problem with drum brakes, yet, still a part of repeating braking with discs.

Quote:

The problem is that you have to come up with some model of how a car is driven. You also need to figure out how heat is taken out of the brakes. Otherwise the simulated temperature will keep rising without limit. Say that the heat Q in the braking system obeys
[tex]
d Q/dt = M a v - f(T,v) ~,
[/tex]
where M is the car's mass, a is its acceleration, v is its velocity, and f is the cooling function.
Ive taken some model equations from a dynamic mechanical systems book, which models the brake system in detail, from the time force is exerted to the pedal by the drivers foot. Since im not taking into account, anything after brake pad to disc action, its more important to model what happens when the car brakes rather than how the car is actually driven.

Ive been looking at the equation you have provided,

dQ/dt = Mav - f(T,v)

im not sure why we would need the cars mass, acc and velocity to calculate the rate of heat change over time?

could i substitute the cooling function part of the equation into my original equation for heat produced? making it:

dW/dt = Fd*(dS/dt) - f(T,v)

Where W = work
Fd = frictional force to disc
and S is displacement (or velocity of disc at point of brake pad contact)

?


Quote:

The problem here is f. Some googling seems to show that it's reasonable to take
[tex]
f(T,v) = k_{1}(T-T_{0}) (1+k_{2} v^{n}).
[/tex]
k1, k2, and n are constants. n should apparently be between 0.5 and 0.8. k1 can be reasonably estimated, but k2 can't. That depends on the details of the airflow around the brake components.
what does k2 represent?
can the k's be values which increases or decreases at each brake action, representing, less heat dissipation?

Stingray Mar5-07 04:08 PM

Quote:

Quote by heiroglif (Post 1264316)
Ive been looking at the equation you have provided,

dQ/dt = Mav - f(T,v)

im not sure why we would need the cars mass, acc and velocity to calculate the rate of heat change over time?

could i substitute the cooling function part of the equation into my original equation for heat produced? making it:

dW/dt = Fd*(dS/dt) - f(T,v)

Where W = work
Fd = frictional force to disc
and S is displacement (or velocity of disc at point of brake pad contact)

Yes. Those two equations are really the same. First of all, the rate at which work is being done must be (very close to) the rate at which heat is being generated: [itex]dQ/dt=dW/dt[/itex]. Also, the force on the disc and the force on the road are related by [itex]F_{d} = (R/r) F_{road}[/itex], where r is the radius to the caliper's effective point of action and R is the tire radius. Similarly, [itex]dS/dt = (r/R) v[/itex]. The consequence of this is that [itex]F_{d} dS/dt = F_{road} v[/itex], which is really a statement that energy is conserved.

Now if the braking force is the main (longitudinal) force acting on the car, Newton's 2nd law can be used to write [itex]F_{road} = Ma[/itex]. These substitutions translate your equation into mine. I think the variables I used are more useful because you usually conduct a braking test by specifying the velocity and acceleration of the car.



Quote:

what does k2 represent? can the k's be values which increases or decreases at each brake action, representing, less heat dissipation?
The k's shouldn't change. k1 is a coefficient which describes the rate at which heat is transferred due to conduction. See Fourier's law of heat conduction. k2 is just a parameter describing how much larger the heat transfer rate is due to convection (airflow). This is related to something called the Nusselt number, but I don't know much about it.

heiroglif Mar5-07 05:57 PM

...
 
:smile:
great, ill start modelling using these equations...thats helped me alot, once again, thanks for the quick and informative response!

H


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