EML 4500 FINITE ELEMENT ANALYSIS AND DESIGN
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Transcript EML 4500 FINITE ELEMENT ANALYSIS AND DESIGN
CONVECTION BC
• Convection Boundary Condition
– Happens when a structure is surrounded by fluid
– Does not exist in structural problems
– BC includes unknown temperature (mixed BC)
Wall
qh
T
T
qh hS(T T)
Fluid Temperature
Convection Coefficient
– Heat flow is not prescribed. Rather, it is a function of temperature on
the boundary, which is unknown
• 1D Finite Element
– When both Nodes 1 and 2 are convection boundary
q1 hAT1 hAT1
q2 hAT2 hAT2
T1
T2
T1
T2
1
EXAMPLE (CONVECTION ON THE BOUNDARY)
• Element
equation
T 1Ґ
T1
h1
(1)
kA 1 1 T1 q1
(1)
L 1 1 T2 q2
T2
1
T3
2
h3
T 3Ґ
(2)
kA 1 1 T2 q2
(2)
L 1 1 T3 q3
• Balance of heat flow
(1)
– Node 1: q1 h1A(T1 T1 )
(1)
(2)
– Node 2: q2 q2 0
(2)
– Node 3: q3 h3 A(T3 T3 )
• Global matrix equation
1 1 0 T1 h1A(T1 T1 )
kA
1 2 1 T2
0
L
0 1 1 T3 h3 A(T3 T3 )
2
EXAMPLE cont.
• Move unknown nodal temperatures to LHS
kA
kA
h
A
1
L
L
2kA
kA
L
L
kA
0
L
T1 h1AT1
kA
T
0
2
L
T
h
AT
3 3 3
kA
h3 A
L
0
• The above matrix is P.D. because of additional positive terms
in diagonal
• How much heat flow through convection boundary?
– After solving for nodal temperature, use
q1(1) h1A(T1 T1 )
• This is convection at the end of an element
3
EXAMPLE 5.5: FURNACE WALL
• Firebrick
k1=1.2W/m/oC
hi=12W/m2/oC
• Insulating brick
k2=0.2W/m/oC
ho=2.0W/m2/oC
Insulating
brick
Firebrick
Ta = 20 C
Tf = 1,500
C
hi
x
0.12 m
0.25 m
0 T1 18,000
16.8 4.8
4.8 6.47 1.67 T 0
2
0
1.67 3.67 T3 40
{T } T {1,411 1,190 552}C Convection
No heat flow
boundary
1,500 C
ho
Convection
boundary
20 C
x
2
q(2)
h
(T
T
)
1054
W/m
3
0
a
3
Tf
T1
hi
T2
1
T3
2
ho
Ta
4
CONVECTION ALONG A ROD
• Long rod is submerged into a fluid
• Convection occurs across the entire surface
• Governing differential equation
d
dT
kA
AQ
hP
T
T 0, 0 x L
g
dx
dx
P 2(b h)
Convection
Fluid T
b
qi(e )
j
i
xi
h
q(je )
Convection
xj
5
CONVECTION ALONG A ROD cont.
• DE with approximate temperature
d
dT
kA
AQg hP T T R(x)
dx
dx
• Minimize the residual with interpolation function Ni(x)
d
dT
x dx kA dx AQg hP(T T) Ni (x)dx 0
i
xj
• Integration by parts
xj
xj
xj
xj
xj
dT
dT dNi
kA
Ni (x) kA
dx hPTNidx AQgNi (x)dx hPT Nidx
dx
dx dx
xi
xi
xi
xi
x
i
6
CONVECTION ALONG A ROD cont.
• Substitute interpolation scheme and rearrange
xj
x
j
dNj dNi
dNi
i i TN
j j )Ni dx
x kA Ti dx Tj dx dx dx x hP(TN
i
i
xj
(AQg hPT )Ni dx q(x j )Ni (x j ) q(x i )Ni (x i )
xi
• Perform integration and simplify
Tj
kA
(e) Ti
(e)
(e)
T
T
hpL
Q
q
i
j
i
i
L(e)
3
6
xj
Q(e)
(AQ
hPT
)Ni (x)dx
i
g
xi
• Repeat the same procedure with interpolation function Nj(x)
7
CONVECTION ALONG A ROD cont.
• Finite element equation with convection along the rod
kA
(e)
L
1 1 hPL(e)
1 1 6
(e)
(e)
2 1 Ti Qi qi
1 2 T Q(e) q(e)
j j
j
(e)
(e)
[k (e)
} {q(e) }
T ] [k h ] T {Q
• Equivalent conductance matrix due to convection
hPL(e)
k
6
(e)
h
2 1
1 2
• Thermal load vector
(e)
(e)
Q
AQ
L
hPL
T
i
g
(e)
{Q }
2
Q j
1
1
8
EXAMPLE: HEAT FLOW IN A COOLING FIN
• k = 0.2 W/mm/C, h = 2104 W/mm2/C
• Element conductance matrix
0.2 200 1 1 2 10 4 320 40 2 1
[k ] [k ]
1 2
40
6
1
1
(e)
T
(e)
h
• Thermal load vector
Convection
T = 30 C
2 10 320 40 30 1
{Q }
2
1
4
(e)
160 mm
• Element 1
330 C
x
1.25 mm
Insulated
120 mm
T1
T2
1
T3
2
T4
3
9
EXAMPLE: HEAT FLOW IN A COOLING FIN cont.
• Element conduction equation
1.8533 0.5733 T1 38.4 q1(1)
– Element 1 0.5733 1.8533 T2 38.4 q(1)
2
1.8533 0.5733 T2 38.4 q(2)
2
(2)
– Element 2
0.5733 1.8533 T3 38.4 q3
1.8533 0.5733 T3 38.4 q(3)
3
– Element 3
0.5733 1.8533 T 38.4 (3)
4
q4
• Balance of heat flow
– Node 1
q1(1) Q1
– Node 2
(2)
q(1)
2 q2 0
– Node 3
(2)
q(2)
q
3
3 0
– Node 4
q(3)
4 hA(T T4 )
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EXAMPLE: HEAT FLOW IN A COOLING FIN cont.
• Assembly
38.4 Q1
0
0 T1
1.853 .573
.573 3.706 .573
76.8
0 T2
76.8
0
.573 3.706 .573 T3
0
.573 1.853 T4 38.4 hA(T T4 )
0
• Move T4 to LHS and apply known T1 = 330
0
0 330 38.4 Q1
1.853 .573
76.8
.573 3.706 .573
T2
0
0
.573 3.706 .573 T3 76.8
0
.573 1.893
0
39.6
T4
• Move the first column to RHS after multiplying with T1=330
0 T2 265.89
3.706 .573
.573 3.706 .573 T 76.8
3
0
.573 1.893 T4 39.6
11
EXAMPLE: HEAT FLOW IN A COOLING FIN cont.
• Solve for temperature
T1 330C, T2 77.57C, T3 37.72C, T4 32.34C
350
300
250
200
150
100
50
0
0
40
80
120
12