Thermal model, brake disc 2

[rulmismo]

Hi,

I`ve developed a "lumped capacities" thermal model of a brake disc.

----------|--R1-|---------------
^ | | |
(I) ---C1 ----C2 -R2
| --- ---- -
| | | |
--------------------------------

The input is "brake power" represented as a current generator.

I've tweaked the model values so it represents in the voltage of C1 capacitor the brake disc Tª.

The thing is that I know the brake force in the disc (nu_dynamc * caliper force) but I am not sure of how is the dimension correct relation from "brake force on disc [Newton]" to "brake power [in Watts]".
I am a little confused between to use:
- Brake force * instant vehicle velocity (m/s)
- Brake force * instant disc rotation velocity (m/s)

I know it is easy, but I am an electronics guy!!
I someone can help I will appreciate it

Thanks!

 

[gruntguru]

Use the velocity at the point where your "brake force" is measured. If you mean the force at the surface of the brake rotor, you must multiply by rotor velocity (at the same radius you measure the force). If you mean the force acting on the car to slow it down, multiply by the velocity of the car.

 

[zekeman]

"I know it is easy, but I am an electronics guy!!"

If you don't know, it is NOT easy.

Transient problems are never easy.


If you are looking for a temperature solution for a given brake force and coefficient of friction, then the problem is not so simple since the heat flux is a function of r, the radius of the disc.

The input heat flux to the disc is stress*coefficient of sliding friction * velocity.

So,since velocity= V(r), you see, it is not a 1 dimensional problem as you assumed.

The solution can be obtained classically and involves Bessel functions. Or you can use one of the many numerical.

Personally, I like the former.

 

[gruntguru]

Althought Zekeman is right, I think you can get a respectable result by assuming a constant heat flux across the friction surface of the disc - after all, the brake manufacturer will be designing towards that solution. The centroid of the pad is probably a reasonable assumption for the equivalent radius for force application and disc velocity. This will probably coincide with the radius of gyration of the disc-swept-area which is a [(R1^2 + R2^2)/2]^0.5

 

[rulmismo]

Thanks guys, your detail is far more that expected!

Really I `ve a simulation of the brake discs manufacturer; after tuning it, my model reproduces almost the same results for Tª (over capacitor C1) with same acceleration/decceleration profile used in their simulation.

I understand that in the simulation they show "maximun Tª on disc" (voltage over C1 in my model) and do not give details about Tª distribution on disk. Maybe voltage over C2 can be seen as a "mean Tª" of the disk (but don´t know really).

Just for curisity they state that 350ºC is the limit of this disc temperature.

I don´t know what model they use, but it seems that i.e. for R2 they assume a constant value, because initially I did it with R2=f(vehicle speed,Nusselt number) representig convection to ambient, and my results in that case differ from manufacturer in the high speed acceleration zones, however tuning to a R2 constant I can got similar results.

Regarding to my first question I now see, as gruntgurut told, that is no difference if you transform the force to ground or disc application point, and take the speed at same point

 

[zekeman]

"Really I `ve a simulation of the brake discs manufacturer; after tuning it, my model reproduces almost the same results for Tª (over capacitor C1) with same acceleration/decceleration profile used in their simulation."

They probably used a single lumped mass and their R1 included radiation which is higher than convection.


I also looked over your model; your current source should be in parallel with R1. The model below ( incidentally very clever way of representing it; I shall use it in the future)corrects it and I show C1 to be 1/2 of the disc and c2 the other 1/2 with R2 the resistance of the condutive path between them. R3 is the radiative/convective resistance
at the far surface.


------ ----|--R2---------------
^ ' | | |
(I) ' ---C1 ----C2 -R3
| R1 --- ---- -
| ' | | |
----'---------------------------

If you want to use the single mass, assuming very small resistance, R2 then I get


------ ---------------
^ ' | |
(I) ' ---C1' R3
| R1 --- -
| ' | |
----'------------------
if you combine R3 in parallel with R1 it further degenerates to
------ --|
^ ' |
(I) R1' ---C1'
| ---
| ' |
----'-------
R1'=R1 parallel with R3
C1'=C1+C2

 

[rulmismo]

Thanks zekeman for your detailed reply,

I have not R1 but still get a 90%fit to their model,

I interpret C1 as the outer part of the disk in contact with the pad, and C2 as the inner part or zone not in contact with pad where the heat flux goes after the initial "heat up".

I have calculated h(convection) and h(radiation) for a disc of similar dimensions at 210ºC at I get
h(convection)=5,2 W/m2K
h(radiation)=0,4 W/m2K
so it seems (if I did it well) that radiation is not very important at least up to 300ºC or more

I am afraid now that will be difficult to adjust the model because I only can measure easily in field the pad Tª and not the disk (and the pad Tª probably is no so similar to disk max. Temp.)

I adjoint the excel sheet as an example, sorry but includes some spanish but all the units should be dimension-correct

Regards!

 

[rulmismo]

After reviewing my calculation I was wrong about radiation (I used ºC instead of ºK for the formulae!)

So for a disc of similar dimensions at 210ºC I get:
h(convection)=4,5 W/m2K
h(radiation)=10 W/m2K

With my very simplistic assumptions seems that radiation is as important as convection at standstill and with Tª as low as 50ºC.
If T>350ºC the radiation even makes that the influence of forced convection will be low.

I updated the excel with correct radiation values: