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I'm brazilian and I'm new to using Mathematica software. I need to solve a problem involving The Heat Equation 3D transiente for a break disk. I have the equations and boundaries conditions below and I need the Temperature T[r,z,t]. Is it possible to solve in Mathematica? I did something to solve this problem, but the code didn't work well. Could someone help me?

enter image description here

The Code is below:

(* Geometria do Disco - Unidade no SI *)

r1 = 0.05; (*Raio do furo do disco*)
r2 = 0.1; (*Raio até o inicio da pinça*)
r3 = 0.17; (*Raio externo do disco*)
l = 0.015;(*Expessura do disco*)
d = l/2;(*Metade da espessura do disco*)
TC = 30; (*Temperatura inicial do disco em °C*)
Ti = TC - 273;(*Temperatura inicial do disco em K*)
Sd = 0.8;(*Área da Superfície de contato do disco*)

(* Propriedades do Disco - Unidade no SI *)

cd = 0.5;(*Calor específicio à pressão constante do material do disco*)
kd = 47;(*Coeficiente de condutividade térmica do disco*)
ρd = 7860;(*Massa específica do disco*)
ξd = Sqrt[cd*kd*ρd];(*Efusividade do disco*)
Sp = 0.8;(*Área da superfície de contato do disco*)

(* Propriedades da Pastilha - Unidade no SI *)

angulo = 60; (*Ângulo de contato da pastilha*)
∅ = (angulo*2*π)/
  180;(*Ângulo de contato da pastilha em radianos*)
μ = 0.6;(*Constante de atrito pastilha-disco*)
ρp = 20;(*Massa específica da pastilha*)
kp = 85;(*Coeficiente de condutividade térmica da pastilha*)
cp = 0.7;(*Calor específico da pastilha*)
ξp = Sqrt[cp*kp*ρp](*Efusividade da pastilha*)
Sd = 0.45 (*Área da surpefície de contato da pastilha*)

(* Propriedades do Fluido - Unidade no SI *)

h = 100;(*Coeficiente de conveção do ar*)
T∞ = 35;(*Temperatura ambiente do ar em °C*)
T∞k = 
  T∞ - 273;(*Temperatura ambiente do ar em K*)
ρ = 1.21;(*Massa específica do ar*)
H = h/kd;(*Razão utilizada para simplificação dos cálculos*)

(* Calculos - Unidade no SI *)

σ = (ξd*
   Sd)/(ξd*Sd - ξp*Sp);(*Coeficiente de partição de calor*)
Subscript[P, max] = 25;(*Pressão máxima aplicada na pastilha*)
Subscript[ω, o] = 20;(*Velocidade angular do disco de freio*)
Subscript[t, b] = 5.2;(*Tempo de frenagem*)
α = (kd/(ρd*
    cd));(*Difusividade Térmica do material do disco*)
heatflux = q[t_] == ∅/(2*π)*μ*σ*Subscript[P, 
    max]*r2*Subscript[ω, 
    o]*(1 - t/Subscript[t, b]);(*Fluxo de calor pastilha-disco*)

solution = 
 NDSolveValue[{D[T[r, z, t], r, r] + (1/r)*D[T[r, z, t], r] + 
     D[T[r, z, t], z, z] - (1/\[Alpha])*D[T[r, z, t], t] == 
    NeumannValue[H*(T[r, z, t] - T\[Infinity]k), 
      r == r1 && 0 <= z <= d] + 
     NeumannValue[H*(T\[Infinity]k - T[r, z, t]), 
      r == r3 && 0 <= z <= d] + NeumannValue[0, z == 0] + 
     NeumannValue[H*(T\[Infinity]k - T[r, z, t]), 
      z == d && r1 <= r <= r2] + 
     NeumannValue[heatflux, z == d && r2 <= r <= r3], 
   T[r, z, 0] == Ti}, T, {r, r1, r3}, {z, 0, l}, {t, 0, 10}]

error: NDSolveValue::ndnum: Encountered non-numerical value for a derivative at t == 0.`.

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  • $\begingroup$ I can't read the text in the image. $\endgroup$ – andre314 Sep 26 '17 at 15:33
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    $\begingroup$ A good rule of thumb on stack exchange is to care that question are usefull for others. Here I don't think anybody will take time to decode all the code to see if eventualy, the question would be usefull to him. $\endgroup$ – andre314 Sep 26 '17 at 15:33
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    $\begingroup$ Can you tell me the source of the problem description? Is this from a book? $\endgroup$ – user21 Sep 26 '17 at 22:52
  • $\begingroup$ I am from a university competition team in which the project is called Baja SAE. I have to make a code that describes how the heat is transferred from the brake pad to the brake disc. However, I am not being able to specify the boundary conditions in the problem. Could you help me with that? The source is from a published article called: Analysis of heat conduction in a disk brake system $\endgroup$ – Gabriel Nunes Sep 26 '17 at 23:05
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Not a answer. I have rewritten the code of the OP for those who are fluent in Mathematica, and in a minimal form :

NDSolve[{
  Derivative[0, 2, 0][T][r, z, t] + 
    Derivative[1, 0, 0][T][r, z, t]/r + 
    Derivative[2, 0, 0][T][r, z, t] == 
   83.61*Derivative[0, 0, 1][T][r, z, t], 
    {T[r, z, t] == 
    If[r == 0.05, -238 + Derivative[1, 0, 0][T][r, z, t]/2.127], 
        T[r, z, t] == 
    If[r == 0.17, -238 - Derivative[1, 0, 0][T][r, z, t]/2.127],
     Derivative[0, 1, 0][T][r, z, t] == If[z == 0, 0],
     Derivative[0, 1, 0][T][r, z, t] == 
    If[0.05 <= r <= 0.1 && z == 0.0075, (100/47)*(-238 - T[r, z, t]), 
     q[t_] == 11.66*(1 - 0.1923*t)]
    },
    {T[r, z, 0] == -243}
    },
 T,
 {r, 0.05, 0.17},
 {z, 0, 0.0075},
 {t, 0, 10}]

It seems that there are several problems :
- If[z == 0, 0] ???
- q[t_] == 11.66*(1 - 0.1923*t)???
- ... (eventually)

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Thanks @andre for code.

This works 

NDSolve[{Derivative[0, 2, 0][T][r, z, t] + Derivative[1, 0, 0][T][r, z, t]/r    
   + Derivative[2, 0, 0][T][r, z, t] ==83.61*Derivative[0, 0, 1][T][r, z, t], 
  T[r, z, t] == -238 + Derivative[1, 0, 0][T][r, z, t]/2.127 /. r -> 0.05, 
  T[r, z, t] == -238 - Derivative[1, 0, 0][T][r, z, t]/2.127 /. r -> 0.17,
  Derivative[0, 1, 0][T][r, z, t] == 0 /. z -> 0, 
  Derivative[0, 1, 0][T][r, z, t] == (100/47)*(-238 - T[r, z, t]) /.z ->0.0075, 
  T[r, z, 0] == -243}, 
  T, {r, 0.05, 0.17}, {z, 0, 0.0075}, {t, 0, 10}]

but gives out some warnings and doesn't calculate till the final moment $t=10$. Perhaps playing with options like 

MaxSteps -> 1000000, StartingStepSize -> 0.0001, MaxStepSize -> 0.01

could help.

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