View
240
Download
4
Category
Preview:
Citation preview
Steady Heat Conduction (Chapter 3) Zan Wu zan.wu@energy.lth.se Room: 5123
Agenda
• Steady-state heat conduction - without internal heat generation
- with internal heat generation
• Fins, extended surfaces - Rectangular fin
- Triangular fin
- Circular fin
Heat conduction equation isotropic material
If λ constant ⇒ homogeneous and isotropic material
)181( −′+
∂∂
∂∂
+
∂∂
∂∂
+
∂∂
∂∂
=∂∂
Qzt
z
yt
yxt
xtc
λ
λλτ
ρ
2 2 2
2 2 2 (1 19)t t t t Qax y z cτ ρ
′ ∂ ∂ ∂ ∂= + + + − ∂ ∂ ∂ ∂
a thermal diffusivitycλρ
=
Simple Plane Wall
Steady heat conduction
One dimensional case No internal heat generation, Q’ = 0; Constant λ (1-19) Solution: t = c1x + c2
BC: x = 0, t = t1; x = b, t = t2
0=τ∂∂
⇒
0zy≡
∂∂
=∂∂
0dx
td2
2=
Simple Plane Wall
Heat flow
Alternate formulation
dxdtAQ λ−=
)33(xb
tttt 121 −
−+=
)43()tt(b1AQ 21 −−λ=
1 2
currentpotential resistance
bt t QAλ
− = ⋅
Composite Wall
“Serial circuit”
Convection resistance
1 4 ( thermal resistence)t t Q− = ⋅∑
QA
bA
bA
btt3
3
2
2
1
141 ⋅
λ
+λ
+λ
=−
)63(
Ab
Ab
Ab
ttQ
3
3
2
2
1
141 −
λ+
λ+
λ
−=
1( ) Convective resistance: f wQ A t tA
αα
= − ⇒
Composite Wall, convective BC
tf2
tf1
b1
λ1 λ2 λ3
b2 b3
α1
α2
)73(
A1
Ab
A1
ttQ
2
3
li i
i
1
2ff1 −
α+
λ+
α
−=
∑=
Composite Circular Wall (Shell)
i of f
32
1 1 1 2 2 3 0
1 1 1 1ln ln2 2 2 2i
t tQ rr
r L L r L r r Lπ α π λ π λ π α
−=
+ + +
Fouling
The accumulation and formation of unwanted materials on the surfaces of processing equipment One of the major unsolved problems in heat transfer
Governing equation
)181( −′+
∂∂
∂∂
+
∂∂
∂∂
+
∂∂
∂∂
=∂∂
Qzt
z
yt
yxt
xtc
λ
λλτ
ρ
2
2 0d t Qdx λ
′+ =
21 2
'2Qt x c x cλ
= − + +General solution
Boundary conditions
0=x 0dtdx
=
bx = ( )wall fdt t tdx
λ α− = −
At the plane of symmetry
Adiabatic or insulated BC
Fin
• A fin is a surface that extends from an object to increase the rate of heat transfer to or from the environment
• Typically, the fin material has a relatively high thermal conductivity
• Used in various applications to increase the heat transfer from surfaces.
• Increase surface area
• Microfin
Rectangular fin
Boundary conditions:
long and thin fin, the heat transferred at the fin tip is negligible
)313(0AC
dxd
2
2−=ϑ
λα
−ϑ
)tt( f−=ϑb
2bZ
Z2ACm2
λα
=λ⋅α
≈λα
=
)tt(dxdt:Lx fLx −α′=λ−= =
0dxdt
Lx=
=
f111 tttt:0x −=ϑ=ϑ⇒==
x dx
L
t1
Q1
.
tf
b Z
Rectangular fin
Solution:
(cosh2
sinh )2
mx mx
mx mx
e emx
e emx
−
−
+=
−=
1 2
3 4cosh sinh
mx mxC e C e
C mx C mx
ϑ −= + =
= +
Hyperbolic functions
For x = L ϑ = ϑ2
1 1
cosh ( ) (3 38)cosh
f
f
t t m L xt t mL
ϑϑ
− −= = −
−
2
1
1cosh mL
ϑϑ
=
heat transfer from the fin ?Q
1 10
sinh ( )coshx
d mLQ A A mdx mL
CmA
ϑλ λ ϑ
αλ
=
= − = − ⋅ ⋅ −
= ⇒
1 1 1tanh 2 tanh (3 40)Q C A mL b Z mLα λ ϑ α λ ϑ= ⋅ = −
Rectangular fin
Rectangular fin
If the condition below is used
one has
and
and
LxLxdx
d=
=ϑα′=
ϑ
λ−
)413(mLsinh
mmLcosh
)xL(msinhm
)xL(mcosh
1−
λα′
+
−λα′
+−=
ϑϑ
)423(mLsinh
mmLcosh
11
2 −
λα′
+=
ϑϑ
)433(mLtanh
m1
mLtanhmAmQ 11 −
λα′
+
+λα′
ϑλ=
Rectangular fin
Fig. 3-13. Arrangement of rectangular fins
1
preferable if
0dQdL
>
1 ( )Q function L=L
.
Z
t1
b
1Q
Criterion for benefit:
0dLQd 1 >
mLtanhm
1
mLtanhmAmQ 11
λα′
+
+λα′
ϑλ=
2
2
2
21
1
N
mLtanhmmLcosh
mm
N
)mLtanhm
1(mLcosh
m
AmdLQd
+
λα′
λα′
−λα′
+⋅ϑλ=
⇒> 0dLQd 1
0mLtanhmm
mLtanhm
1 22
2>
λα′
−λ
α′−
λα′
+
0m
1 22
2>
λ
α′−
Rectangular fin
Assume α′ = α ⇒
or
⇒
rule of thumb:
b2
ACm2
λα
=λα
=
02
b1 2
2>
λ⋅α
λα−
λα
>2
b1
)463(1b
2−>
αλ
)473(5b
2−>
αλ
Fin effectiveness, fin efficiency
1:
2:
1
1
from the fin from the base area without the fin
η =
1
1
from the finfrom a similar fin but with λ
φ == ∞
Optimal fin
Criterion: maximum heat flow at a given weight M = ρ b L Z = ρ Z A1
A1 = b L, Z, ρ are given.
Find maximum for A1 = b⋅L, constant. (3-40): C ≈ 2Z , A = b⋅Z
mLtanhACQ 11 ϑλα=
b2
ACm2
λα
=λα
=
1Q
⋅λα
⋅ϑ⋅αλ=b
Ab
2tanhZb2Q 111
LZb
Optimal rectangular fin
Condition
1 0 gives optimum
after some algebra one obtains
21.419 (3 55)/ 2
dQdb
Lb b
λα
=
= −
Fin arrangement
After some algebra one finds:
)523(b
Ab
2tanhZb2
mLtanhAmQ
11
11
−
λα
ϑαλ=
=ϑλ=
)583(b
Ab
2u 1 −⋅λα
=
)543(ucosh
u3utanh 2 −=
{ 1(3 54) from the condition / 0}dQ db− =
)a613(4
1utanh
uZ1QA 233
3
1
11 −
λα⋅
ϑ
=
Weight of the fin
M = ρ b L Z = ρ Z A1 =
ρ/λ is the material parameter
see Table 3-1.
Aluminum is applied instead of copper. ρ/λ Aluminum: 11.8; copper: 23.0
Why not Magnesium? ρ/λ Magnesium: 10.2
λρ⋅
α⋅
ϑ
= 232
3
1
1
41
utanhu
Z1Q
Straight triangular fin
ϑ = t − tf
Heat balance ⇒
Solution:
K0 → ∞ as x → 0 ⇒ B = 0 because ϑ is finite for x = 0
x = L ϑ = ϑ1 ⇒
)623(0bL2
x1
dxd
x1
dxd
2
2−=ϑ
λα
−ϑ
+ϑ
bL2
λα
=β
( ) )x2(BKx2AI 00 β+β=ϑ
( )L2AI01 β=ϑ
LxbZA ⋅=
δ
L
dx
x
b t1
tf1Q
Bessel differential equation
I0 and K0 are the modified Bessel functions of order zero
Triangular fin
⇒
)L2(IA
0
1β
ϑ=
)653()L2(I)x2(I
0
0
1−
ββ
=ϑϑ
Lx1 dx
dtAQ=
λ−=
Lx
0
011 dx
)x2(dI)L2(I
1bZQ=
β
βϑλ−=
Triangular fin
Introduce 2
/
xd d d dxdx d dx d
ξ βξ β
ξ ξ
=
⇒ = ⋅ =
=
ξξ
β=
β
== Lx
0
Lx
0d
)(dIL/dx
)x2(dI
)L2(Ib
2)L2(IL/ 11 βλα
=ββ=
)663()L2(I)L2(Ib2ZQ
0
111 −
ββ
αλϑ=⇒
Table 3.2 for numerical values of I0 and I1
Summary of formulae for rectangular and triangular fins
optimal fin
optimal fin
)383(mLcosh
)xL(mcosh
ft1tftt
1−
−=
−−
=ϑϑ
b22mλα
=
)403(mLtanh1Zb21Q −ϑαλ=
mLtanhb
2αλ
=η
mLmLtanh
=ϕ
)553(b
2419.12/b
L−
αλ
=
)653()L2(0I)x2(0I
1−
ββ
=ϑϑ
bL2
λα
=β
)L2(0I)L2(1I1Zb21Q
ββ
⋅ϑ⋅αλ=
)L2(0I)L2(1I
b2
ββ
⋅αλ
=η
L)L2(0I/)L2(1I
βββ
=ϕ
)673(b
2309.12/b
L−
αλ
=
Formulas for fin efficiencies
Some simple calculations give:
Rectangular fin
Triangular fin
mLtanhb
2αλ
=η
)L2(I)L2(I
b2
0
1ββ
αλ
=η
L)L2(I/)L2(I 01
βββ
=ϕ
mLmLtanh
=ϕ
Fin effectiveness
Fin efficiency
42
Circular or annular fins
Heat conducting area
A = 2πr b
Convective perimeter
C = 2 ⋅ 2πr = 4πr
r1 r2
b
Recommended