Transferencia de calor-problema
Enviado por Edgard Mendoza • 3 de Noviembre de 2022 • Tarea • 332 Palabras (2 Páginas) • 72 Visitas
TRANSFERENCIA DE CALOR |
TRANSFERENCIA DE CALOR PI143-A
PRACTICA CALIFICADA N° 3
El cuerpo bidimensional mostrado en la figura tiene la temperatura inicial en la superficie y la temperatura del interior como las calculadas. En un instante dado, considerado tiempo cero la cara a 500 C disminuye, de forma rápida, su temperatura hasta 30 C. Tomando Δx= Δy 15 cm y α=1.29*105 m2/ calcúlense las temperaturas de los nodos 1,2,3 y 4 al cabo de 30 min. Realice el cálculo utilizando el método explícito. Tómese K=45 W/mxC.
[pic 1]
[pic 2][pic 3]
[pic 4] | [pic 5] | [pic 6] |
[pic 7] | [pic 8][pic 9][pic 10] | |
[pic 11] |
[pic 12]
SOLUCION. –
Hallamos las temperaturas de los nodos internos 1,2,3 y 4 con la siguiente ecuación
[pic 13]
NODO 1: 100+ T2+ T3+500-4* T1=0 ………………………………...(1)
NODO 2: T1+100+ T4+500-4* T2=0 …………………………………(2)
NODO 3: 100+ T4+ 100+T1-4* T3=0 …………………………………(3)
NODO 4: T3+ 100+100+T2-4* T4=0 …………………………………(4)
Observamos del grafico que T1=T2 y T3=T4 . Por lo tanto, solo tendríamos un sistema de 2 ecuaciones con 2 incógnitas.
Usando las ecuaciones (2) y (3) obtenemos:
T1=T2 =250 °C
T3=T4 =150°C
[pic 14]
[pic 15][pic 16]
[pic 17][pic 18] | ||
[pic 19] | [pic 20] | |
[pic 21]
Utilizando el método explicito hallamos las nuevas temperaturas:
Α*Δt/Δx=1/4
p | T1 | T2 | T3 | T4 |
0 | 250 | 250 | 150 | 150 |
1 | 132.5 | 132.5 | 150 | 150 |
2 | 103.125 | 103.125 | 120.625 | 120.625 |
3 | 88.4375 | 88.4375 | 105.9375 | 105.9375 |
4 | 81.09375 | 81.09375 | 98.59375 | 98.59375 |
5 | 77.421875 | 77.421875 | 94.921875 | 94.921875 |
6 | 75.5859375 | 75.5859375 | 93.0859375 | 93.0859375 |
7 | 74.6679688 | 74.6679688 | 92.1679688 | 92.1679688 |
8 | 74.2089844 | 74.2089844 | 91.7089844 | 91.7089844 |
9 | 73.9794922 | 73.9794922 | 91.4794922 | 91.4794922 |
10 | 73.8647461 | 73.8647461 | 91.3647461 | 91.3647461 |
11 | 73.807373 | 73.807373 | 91.307373 | 91.307373 |
12 | 73.7786865 | 73.7786865 | 91.2786865 | 91.2786865 |
13 | 73.7643433 | 73.7643433 | 91.2643433 | 91.2643433 |
14 | 73.7571716 | 73.7571716 | 91.2571716 | 91.2571716 |
15 | 73.7535858 | 73.7535858 | 91.2535858 | 91.2535858 |
16 | 73.7517929 | 73.7517929 | 91.2517929 | 91.2517929 |
17 | 73.7508965 | 73.7508965 | 91.2508965 | 91.2508965 |
18 | 73.7504482 | 73.7504482 | 91.2504482 | 91.2504482 |
19 | 73.7502241 | 73.7502241 | 91.2502241 | 91.2502241 |
20 | 73.7501121 | 73.7501121 | 91.2501121 | 91.2501121 |
21 | 73.750056 | 73.750056 | 91.250056 | 91.250056 |
22 | 73.750028 | 73.750028 | 91.250028 | 91.250028 |
23 | 73.750014 | 73.750014 | 91.250014 | 91.250014 |
24 | 73.750007 | 73.750007 | 91.250007 | 91.250007 |
25 | 73.7500035 | 73.7500035 | 91.2500035 | 91.2500035 |
26 | 73.7500018 | 73.7500018 | 91.2500018 | 91.2500018 |
27 | 73.7500009 | 73.7500009 | 91.2500009 | 91.2500009 |
28 | 73.7500004 | 73.7500004 | 91.2500004 | 91.2500004 |
29 | 73.7500002 | 73.7500002 | 91.2500002 | 91.2500002 |
30 | 73.7500001 | 73.7500001 | 91.2500001 | 91.2500001 |
31 | 73.7500001 | 73.7500001 | 91.2500001 | 91.2500001 |
32 | 73.75 | 73.75 | 91.25 | 91.25 |
33 | 73.75 | 73.75 | 91.25 | 91.25 |
34 | 73.75 | 73.75 | 91.25 | 91.25 |
35 | 73.75 | 73.75 | 91.25 | 91.25 |
36 | 73.75 | 73.75 | 91.25 | 91.25 |
37 | 73.75 | 73.75 | 91.25 | 91.25 |
38 | 73.75 | 73.75 | 91.25 | 91.25 |
...