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
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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.
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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:
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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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32
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98
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