Aleta Tipo Disco

TERCER TRABAJO DE FENOMENOS DE TRANSPORTE
ALETA TIPO DISCO
Flujo de entrada – Flujo de salida = 0
qr/r *2πr*2B-qr/r+∆r*2π(r+∆r)*2B-2h*2πr∆r(Tp-Ta)=0
lim∆r→0(qr/r+∆r *2π(r+∆r)*2B-qr/r*2πr*2B2π*2B*∆r+4πhr∆r(Tp-Ta)2π*2B*∆r)=0
ddrrqr=-hrB(Tp-Ta)
Operando:
1rddrrqr=-hr(Tp-Ta)B
qr=-kdTdr
1rddr-krdTdr=-h(T-Ta)B
Finalmente:
d2Tdr2+1rdTdr=hkB(T-Ta)

Siendo las condiciones de frontera:
CF1:r=R1, T=Ta
CF2: dTdr=0, -kdTdr=h(T-Ta)
Dependiendo del caso se puede aplicar la condición de frontera CF2
Procedemos a discretizar la ecuación:
Ti+1-2Ti+Ti-1(∆r)2+1i∆rTi+1-Ti∆r=hkB(Ti-Ta)

Ordenando:
1+1iTi+1-2+1i+h∆r2kBTi+Ti-1=-h(∆r)2kBTa

i=1: 2T2-3+h∆r2kBT1+T0=-h(∆r)2kBTa
i=2: 32T3-52+h∆r2kBT2+T1=-h∆r2kBTa
i=3: 43T4-73+h∆r2kBT3+T2=-h∆r2kBTa
i=4:54T5-94+h∆r2kBT4+T3=-h∆r2kBTa
i=5: 65T6-115+h∆r2kBT5+T4=-h(∆r)2kBTa
Usando:
CF2: -kdTdr=h(T-Ta)

Donde:
Tdr=h(T-Ta)
Datos para la ejecución del programa:
h=80wm2°C k=50Wm°C R1=0,1m B=0.0005m
R2=0,18m Tp=200°C Ta=20°C

PROGRAMACIÓN EN POLYMATH 6.0:

f(T1) = 2 * T2 - (3 + h * dr ^ 2 / (K * B)) * T1 + T0 + h * ((dr) ^ 2) * Ta / (B * K)f(T2) = (3 / 2) * T3 - (5 / 2 + h * dr ^ 2 / (K * B)) * T2 + T1 + h * ((dr) ^ 2) * Ta / (B * K)
f(T3) = (4 / 3) * T4 - (7 / 3 + h * dr ^ 2 / (K * B)) * T3 + T2 + h * ((dr) ^ 2) * Ta / (B * K)
f(T4) = (5 / 4) * T5 - (9 / 4 + h * dr ^ 2 / (K * B)) * T4 + T3 + h * ((dr) ^ 2) * Ta / (B * K)
f(T5) = (6 / 5) * T6 - (11 / 5 + h * dr ^ 2 / (K * B)) * T5 + T4 + h * ((dr) ^ 2) * Ta / (B * K)

T6 = T5- (1 - h * dr / K) + h * dr * Ta / K

T1(0) = 0
T2(0) = 0
T3(0) = 0
T4(0) = 0
T5(0) = 0

K = 50 # W/m°C
h = 120 # W/m2°C
B = 0.0005 # m
Tp = 300 # °C
T0 = Tp
Ta = 20 # °C
dr = 0.0056 # m

Hacemos:
β=h∆r2kB=80wm2°C*(0,016m)2238Wm°C *1m=8,605*10-5
Una solución:
r(m) | T °C |
0 | 200 |
0,016 | |
0,032 | |
0,048 | |
0,064 | |
0,08 | |

POLYMATH Report | |Nonlinear Equations | 02-Oct-2012 |

Calculated values of NLE variables
  | Variable | Value | f(x) | Initial Guess |
1 | T1 | 164.9715 | -1.119E-13 | 0 |
2 | T2 | 108.3684 | 3.02E-14 | 0 |
3 | T3 | 79.50092 | -2.665E-14 | 0 |
4 | T4 | 64.56774 | 1.599E-14 | 0 |
5 | T5 | 57.98815 | 1.776E-15 | 0 |

  | Variable | Value |
1 | dr | 0.0056 |
2| K | 50. |
3 | h | 120. |
4 | B | 0.0005 |
5 | Tp | 300. |
6 | T0 | 300. |
7 | Ta | 20. |
8 | T6 | 57.27039 |

Nonlinear equations
1 | f(T1) = 2 * T2 - (3 + h * dr ^ 2 / (K * B)) * T1 + T0 + h * ((dr) ^ 2) * Ta / (B * K) = 0 |
2 | f(T2) = (3 / 2) * T3 - (5 / 2 + h * dr ^ 2 / (K * B)) * T2 + T1 + h * ((dr) ^ 2) * Ta / (B * K) = 0 |
3 | f(T3) = (4 / 3)* T4 - (7 / 3 + h * dr ^ 2 / (K * B)) * T3 + T2 + h * ((dr) ^ 2) * Ta / (B * K) = 0 |
4 | f(T4) = (5 / 4) * T5 - (9 / 4 + h * dr ^ 2 / (K * B)) * T4 + T3 + h * ((dr) ^ 2) * Ta / (B * K) = 0 |
5 | f(T5) = (6 / 5) * T6 - (11 / 5 + h * dr ^ 2 / (K * B)) * T5 + T4 + h * ((dr) ^ 2) * Ta / (B * K) = 0 |

Explicit equations
1 | dr = 0.0056 |
  | m |
2 | K = 50 |
  | W/m°C |3 | h = 120 |
  | W/m2°C |
4 | B = 0.0005 |
  | m |
5 | Tp = 300 |
  | °C |
6 | T0 = Tp |
7 | Ta = 20 |
  | °C |
8 | T6 = T5 - (1 - h * dr / K) + h * dr * Ta / K |

General Settings
Total number of equations | 13 |
Number of implicit equations | 5 |
Number of explicit equations | 8 |
Elapsed time | 0.0000 sec |
Solution method | SAFENEWT|
Max iterations | 150 |
Tolerance F | 0.0000001 |
Tolerance X | 0.0000001 |
Tolerance min | 0.0000001 |

r(m) | T °C |
0 | 200 |
0,0056 | 164.9715 |
0,032 | 108.3684 |
0,048 | 79.50092 |
0,064 | 64.56774 |
0,08 | 57.98815 |

Se considera R1=0 y apartir de ese punto se toman las nuevas mediciones:
Usando CF2: -kdTdr=h(T-Ta)
i=1: 2T2-3-βT1=-Tp-βTa
i=2:...