trabajo 2 calor

8/22/2019 Trabajo 2 Calor

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Nombre: Andres Gamboa

Codigo: 244627

4,60Consider the two-dimensional tube of a noncircular cross section formed by rectangular

and semicylindrical subdomains patched at the common dashed control surfaces in a mannersimilar to that described in Problem 4.59. Note that, along the dashed control surfaces,

temperatures in the two subdomains are identical and local conduction heat fluxes to the

semicylindrical subdomain are identical to local conduction heat fluxes from the rectangular

subdomain. The bottom of the domain is held at Ts 100C by condensing steam, while the

flowing fluid is characterized by the temperature and convection coefficient shown in the

sketch. The remaining surfaces are insulated, and the thermal conductivity is k 15 W/mK.

Find the heat transfer rate per unit length of tube, q, using . Hint: Take advantage of the symmetry of the problem by considering only half of theentire domain.

Solucin:

Dividiendo la seccin transversal en las regiones infinitesimales que se van a analizar:


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Numerando los nodos:

Para los nodos 12, 15, 18, 21.

Los balances de energia son los siguientes

Nodo 12

Desarrollando:


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De igual forma con los nodos del mismo tipo:

Nodo 15

Nodo 18

Nodo 21

Para los nodos 13, 16, 19 y 22.

Nodo 13

Desarrollando:

( ) Nodo 16

( )


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Nodo 19 ( )

Nodo 22 ( )

Nodo 9

Desarrollando

Nodo 10


5/13

Desarrollando

Nodo 8

Desarrollando

Nodo 7


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Desarrollando

Nodo 6

Para los nodos 11, 14, 17, 20

Nodo 11

( )

Nodo 14

( ) Nodo 17

( )

Nodo 20

( )

Nodo 1, 2, 3, 4 y 5.


7/13

Resolviendo el sistema de ecuaciones:

El flujo de calor prima es igual a:

[ ]

NODO T (K) T(C)

1 373,15 100

2 373,15 100

3 373,15 100

4 373,15 100

5 373,15 100

12 351,41 78,2557

15 343,72 70,5674

18 339,25 66,0952

21 337,78 64,6317

13 352,21 79,063

16 344,47 71,3155

19 339,95 66,7984

22 338,47 65,3176

10 363,45 90,2962

9 362,37 89,2213

8 359,6 86,4489

7 361,46 88,313

6 361,81 88,6634

11 347,44 74,292

14 339,89 66,7394

17 335,66 62,5094

20 334,29 61,1407


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4.77 The diagonal of a long triangular bar is well insulated,while sides of equivalent length are

maintained at uniform temperatures Ta and Tb.

a) Establish a nodal network consisting of five nodes along each of the sides. For one ofthe nodes on the diagonal surface, define a suitable control volume and derive the

corresponding finite-difference equation. Using this form for the diagonal nodes and

appropriate equations for the interior nodes, find the temperature distribution for the

bar. On a scale drawing of the shape, show the 25, 50, and 75C isotherms.

Solucin

Poniendo los modos y numerndolos:

Como no se conoce la medida de los lados del tringulo, entonces haremos como primera

suposicin que .Para los nodos 2, 3, 4, 5 y 6, 10, 13, 15:


9/13

Para los nodos 11, 7 y 8:

Nodo 7

Haciendo el balance de energa:

Desarrollando

De forma anloga para los otros dos nodos.

Nodo 8

Nodo 11

Para los nodos 9, 12 y 14.

Nodo 9

Haciendo el balance de energia:


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Desarrollando

Analoga con los otros dos nodos.

Nodo 12

Nodo 14

Desarrollando el sistema de ecuaciones:

El perfil de temperaturas de 25 C(anaranjado), 50 C(morado) y 75 C(verde), se presenta acontinuacin:

Nodo T (K) T (C)

2 273,15 0,00

3 273,15 0,00

4 273,15 0,00

5 273,15 0,00

6 373,15 100,00

7 323,15 50,00

8 301,72 28,57

9 287,44 14,29

10 373,15 100,00

11 344,58 71,43

12 323,15 50,00

13 373,15 100,00

14 358,86 85,71

15 373,15 100,00


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b) An alternate and simpler procedure to obtain the finite-difference equations for thediagonal nodes follows from recognizing that the insulated diagonal surface is a

symmetry plane. Consider a square 5x5 nodal network, and represent its diagonal as a

symmetry line. Recognize which nodes on either side of the diagonal have identical

temperatures. Show that you can treat the diagonal nodes as interior nodes and

write the finite-difference equations by inspection.

Solucin

Desarrollando la hiptesis para el nodo 9:

Haciendo el balance de energa:

Desarrollando


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Esta ecuacin es igual a la hallada en el punto (a), por tanto la hiptesis es correcta.

4.59 Calculate the heat transfer per unit depth into the page, q, using . The base of the rectangular subdomain is held at T h=20 C, while thevertical surface of the cylindrical subdomain and the surface at outer radius r0are at Tc=0 C.

The remaining surfaces are adiabatic, and the thermal conductivity is k=10 W/m*K.

Solucin

Poniendo los nodos y numerndolos:


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1

4

32

5

7

6

1110

98

12

14

13

15

1624

23

22

17

18

21

20

19

trabajo 2 calor

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