Elementos finitos

PROBLEMAS

FLUJO DE CALOR EN ESTADO ESTACIONARIO PARA 1 DIRECCIÓN

* Matriz de rigidez elemento “e”

K(e)=KXAL(e)1-1-11Conducción+hPL(e)62112Convección+hA0001Convección final

* Vector fuerza elemento “e”

f(e)=QAL(e)211Conducción+hT∞PL(e)611Convección+hT∞A01Convección final

1. For the one-dimensional composite bar shown in Figure P13-1, determine the interfacetemperatures. For element 1,'let ^ = 200 \V/(m-°C); for element 2. !et Ka, = !00 W/(m • °C); and for element 3, let K^ = 50 W/(m • °C). Let A = 0.1 m2. The left end has a constant temperature of 100 °C and the right end has a constant temperature of 300 °C.

DATOS
Número de Elementos: 3
Número de Nodos: 4
A= 0,1 m2
KXX1= 200 W/m°C
KXX2= 100 W/m°C
KXX3= 50 W/m°CK(1)=200Wm2°C∙0,1m22m1-1-11=10-10-1010Conducción

K(2)=100Wm2°C∙0,1m22m1-1-11=10-10-1010Conducción

K(3)=50Wm2°C∙0,1m22m1-1-11=10-10-1010Conducción

f(1)=f(2)=f(3)=00

MATRIZ GLOBAL

10 -10 0 0 -10 0 0 20-10 0 -10 20 -10 0-10 10T1T2T3T4=0000

T1=100 °C

20-10 0-10 20-10 0-1010T2T3T4=100000

T4=300 °C

20-10-10 20T2T3=10003000

T2=166.67 °C
T3=233.33 °CRESULTADO

T=100166.67233.33300 °C

2. For the ene-dimensional rod shown in Figure PI3-2 (;asulated except at the ends), determine the temperatures at 173, 2L/3, and L. Let Kr_-, = 3 Btu/(h.-in.-°F), h - 1.0 Btu/(h-in2-°F), and TK = 0°F. The temperature at the left end is 200 °F.

DATOS
Número de Elementos: 3
Número de Nodos: 4
KXX= 3 Btu/h in°F =36 Btu/h ft °F
h= 1 Btu/h in2 °F = 144Btu/h ft2 °F
T∞= 0 °F
T1= 200 °F
L1= L2= L3= 3 in = 0.25 ft
R= 2 in= 0.17

K(1)=K(2)=36Btuhft°F∙π∙0,172ft20,25ft1-1-11= 13,07-13,07-13,07 13,07Conducción

K(3)=36Btuhft°F∙π∙0,172ft20,25ft1-1-11Conducción+36Btuhft2°F π∙0,172ft20001Convección final

K(3)= 13,07-13,07-13,07 13,07+00013,07= 13,07-13,07-13,07 26,14

f(1)=f(2)=00

f(3)=hT∞A01Convección final=36Btuhft2°F ∙0°F∙π∙0,172ft201=00

MATRIZ GLOBAL

13,07-13,07 0 0 -13,07 0 0 26,14 -13,07 0 -13,07 26,14 -13,07 0-13,07 26,14T1T2T3T4=0000

T1=200 °F

26,14-13,07 0-13,07 26,14-13,07 0-13,0726,14T2T3T4=261400

T2=150 °F
T3=100 °F
T4=50 °F

RESULTADO

T=20015010050 °F

3. A rod with uniform cross-sectional area of 2 in2 and thermalconductivity of 3 Btu/ (h-in.-°F) has heat flow in the x direction only (Figure). The right end is insulated. The left end is maintained at 50 °F, and the system has the linearly distributed heat flux shown. Use a two-element model and estimate the temperature at the node points and the heat flow at the left boundary.

DATOS
Número de Elementos: 2
Número de Nodos: 3
A= 2 in2
KXX= 3 Btu/h in°F
T1= 50°f
L1= L2= 30 in
q1= 0
q2= 3
q3= 6

K(1)=K(2)=3Btuh in °F∙2in230 in 1-1-11= 0,2-0,2-0,2 0,2Conducción

f1=Q∙A∙L211=3Btuh211=1,51,5Btuh

f2=Q∙A∙L211=6Btuh211=33Btuh

MATRIZ GLOBAL

0,2-0,2 0-0,2 0,4-0,2 0-0,2 0,2T1T2T3=1,54,53

T1= 50 °F

0,4 -0,2-0,2 0,2T2T3=14,53

T2=87.5 °F
T3=102.5 °F

RESULTADO

T=5087,5102,5 °F

4. The rod of 1-m. radiusshown in Figure generals heat initially at the rate of uniform Q - 10,000 Btu/(h-ft3} throughout the rod. The left edge and perimeter of the rod are insulated, and the right edge is exposed to an environment of T∞=100°F. The convection heat-transfer coefficient between the wall 'and the environment is h - 100 Btu/(h-ft2-°F). The thermal conductivity of the rod is K^ = 12 Btu/fh-ft-T). The length ofthe rod is 3 in. Calculate the temperature distribution in the rod. Use at least three elements in your finite element model

DATOS
Número de Elementos: 3
Número de Nodos: 4
KXX= 12 Btu/h ft °F
h= 100 Btu/h ft2 °F
Q= 100 Btu/h ft3
T∞= 100 °F
L1= L2= L3= 1 in = 0.083 ft
R= 1 in= 0.083 ft

K(1)=K(2)=12Btuhft°F∙π∙0,0832ft20,083ft 1-1-11= 3,13-3,13-3,13 3,13Conducción...