This document summarizes the analysis of heat transfer through a long thin fin and through a solder wire melting upon contact with a hot surface. For the long fin problem, averaging the governing equation over the fin's cross-section allows reducing the problem to 1D with temperature dependent only on the axial coordinate. This yields a dimensionless equation and boundary conditions that can be solved analytically. For the solder wire, a 1D analysis is valid when the Biot number is small. Non-dimensionalizing and averaging the energy equation yields an ordinary differential equation for the dimensionless cross-sectional average temperature. Solving this provides the temperature profile, from which the heat flux from the surface into the wire can be

1. Fundamentals of Transport Phenomena
ChE 715
Lecture 17
Ch 3 cont’d
• Complete Fin Problem
E l b li• Example prob. on scaling
• Transient prob-similarity method
Spring 2011

2. Long Fin Problem
Constant temperature (T0)
2W
L
2W
x
Heat transfer coefficient h, ambient temperature T∞.
L
z
“Thin Fin” approximation: assume γ = L/W >> 1.
Develop the dimensionless transport problem in terms of γ and .W
hW
Bi ≡p p p γ W
k
How does this approach simplify the problem when BiW << 1?
What does scaling tell us about the temperature profile in the x and z-direction?

3. Long Fin Problem
For Bi << 1, temp variation in x direction negligible! So T=T(z) good approx. (LAST CLASS)
Replace local value with cross-sectional average (at const. z):
0
1
( ) ( , )
W
T z T x z dx
W
≡ ∫
Remember energy eq.:gy q
2 2
2 2
0
d T d T
dx dz
+ =
Averaging each term over the cross-sectional average in the energy eq.:
2
1 1
x WW
T T h
=
∂ ∂ 2 2 2
1 1
W W
T T⎛ ⎞∂ ∂ ∂
2
00
1 1
( )
x
T T h
dx T T
W x W x Wk
∞
=
∂ ∂
= = − −
∂ ∂∫
2 2 2
2 2 2
0 0
1 1T T
dx Tdx
W z z W z
⎛ ⎞∂ ∂ ∂
= =⎜ ⎟
∂ ∂ ∂⎝ ⎠
∫ ∫
Energy eq. is now:
2
2
( ) 0
T h
T T
z Wk
∞
∂
− − =
∂

4. Long Fin Problem
Averaging BCs involving z:
0(0)T T=
( ) ( )
dT h
L T L T⎡ ⎤= − −⎣ ⎦
Define dimensionless temperature and z:
( ) ( )L T L T
dz k
∞⎡ ⎤⎣ ⎦
0
( )
T T
T T
ζ ∞
∞
−
Θ =
−
z
W
ζ =
W more representative
than length for thin
objects.
Temperature
is scaled.
Dimensionless Energy eq:
2
Bi
∂ Θ
Θ
Dimensionless BCs
(0) 1Θ = ( ) ( )Biγ γ
∂Θ
= Θ L
2
( ) 0
T h
T T
∂
2
Bi
ζ
= Θ
∂
(0) 1Θ = ( ) ( )Biγ γ
ζ
= − Θ
∂
Are eqns. properly scaled?
L
W
γ =2
( ) 0T T
z Wk
∞− − =
∂

5. Long Fin Problem
Scaling z coord:
m is
undetermined
constant
m
Z Bi ζ=
Dimensionless Energy Eq. becomes:
constant
2
∂ Θ
2
∂ Θ
Choose m = ½ then:
2
2
0m
Bi Bi
Z
∂ Θ
− Θ =
∂
2
Bi
ζ
∂ Θ
= Θ
∂
Choose m = ½, then:
Scale for z is1/ 2
1/2 h
Z Bi z
Wk
ζ
⎛ ⎞
= = ⎜ ⎟
⎝ ⎠
1/2
h
Wk
⎛ ⎞
⎜ ⎟
⎝ ⎠
Wk⎝ ⎠ Competition between heat
conduction and heat loss

6. Long Fin Problem
BC at top with scaled coord
is then:
∂Θ 1/2
( ) ( )Bi
Z
∂Θ
Λ = − Θ Λ
∂
1/2
h⎛ ⎞h
L
Wk
⎛ ⎞
Λ ≡ ⎜ ⎟
⎝ ⎠
where
If Bi 0 : And BCs are:If Bi 0 :
2
2
0
d
dZ
Θ
− Θ = ( ) 0
d
dZ
Θ
Λ =(0) 1Θ =
Solving:
( ) cosh tanh sinhZ Z ZΘ = Λ( ) cosh tanh sinhZ Z ZΘ = − Λ

7. Long Fin Problem
( ) cosh tanh sinhZ Z ZΘ = − Λ
( ) 1ZΘ →
Limiting cases:
“Short” fin (Λ << 1 or ) :
1/2
Wk
L
⎛ ⎞
<< ⎜ ⎟ ( 0)Λ →( ) 1ZΘ →Short fin (Λ << 1 or ) :L
h
<< ⎜ ⎟
⎝ ⎠
( 0)Λ →
Short fin essentially isothermal
“Long” fin (Λ >> 1) : ( ) Z
Z e−
Θ → ( )Λ → ∞
Long fin reaches ambient temperatureLong fin reaches ambient temperature
well before the tip

8. Example Problem—heat transfer to solder
Solder wire melting upon contact
with hot surface at constant rate
Solder Wire
Air with hot surface at constant rate
2R
h
T
– Steady state
l h f f i λV
w
Hot Surface
Tm
T∞
z
– latent heat of fusion, λ
– Choose z-coord as shown
Under what conditions a one-dimensional analysis is valid?
Determine the temperature profile in the wire, T(z)
Determine the heat cond ction rate from the s rface to the ireDetermine the heat conduction rate from the surface to the wire

9. Example Problem—heat transfer to solder
When is T ≈ T(z) only?
[ ( , ) ]
R
T
k h T R z T
r
∞
∂
− = −
∂
z
[ ]
[ ]
(0, ) ( , )
~
( , )
T z T R z hR
Bi
T R z T k∞
−
=
−
z
W
hR
Bi
k
≡For << 1, T(0,z)-T(R,z)~0
Similar to fin
p bl mproblem
To determine T(z):
Energy Eq. applied to
^ ^
2
v
DT T
C C T k T Hρ ρ
∂⎛ ⎞
= + ∇ = ∇ +⎜ ⎟igy q pp
solder:
vp p vC C T k T H
Dt t
ρ ρ= + ∇ = ∇ +⎜ ⎟
∂⎝ ⎠
i
0 0zwv e−
( )v v vzw w
T
T e T
z
∂
∇ = − ∇ = −
∂
i i

10. Example Problem—heat transfer to solder
2
2
2
1 T T
T r
∂ ∂ ∂⎛ ⎞
∇ = +⎜ ⎟
⎝ ⎠
Cylindrical
Coord
z
2
r r r z
⎜ ⎟
∂ ∂ ∂⎝ ⎠
Coord.
Replacing the terms in the Energy Eq.z p g gy q
we obtain:
2
2
v1
0wT T T
r
∂ ∂ ∂ ∂⎛ ⎞
+ + =⎜ ⎟
∂ ∂ ∂ ∂⎝ ⎠ ˆ
k
C
α
⎛ ⎞
≡⎜ ⎟
⎜ ⎟
Where
BC’s:
2
r r r z zα⎜ ⎟
∂ ∂ ∂ ∂⎝ ⎠ pCρ⎜ ⎟
⎝ ⎠
T(r,0) = Tm
[ ]( )
T h
T R T
∂
T(r, ∞) = T∞
(melting temp) (air temp)
( ti BC)[ ]( , )
r R
T R z T
r k
∞
=
= − −
∂
(convection BC)

11. Example Problem—heat transfer to solder
Non-dimensionalize using:
T T
z
⎛ ⎞
m
T T
T T
∞
∞
−
Θ ≡
−
r
R
η =
z
R
ζ =
z
wv R
Pe
α
⎛ ⎞
≡⎜ ⎟
⎝ ⎠
Peclet’s Number
The PDE for
Θ(η,ζ) is:
2
2
1
0Peη
η η η ζ ζ
⎛ ⎞∂ ∂Θ ∂ Θ ∂Θ
+ + =⎜ ⎟
∂ ∂ ∂ ∂⎝ ⎠
BC’s:
⎝ ⎠
Θ(η,0) = 1 (1, )Bi ζ
∂Θ
= − ΘΘ(η ∞) = 0(η, )
1
(1, )Bi
η
ζ
η =
Θ
∂
Θ(η, ∞) 0

12. Example Problem—heat transfer to solder
Defining cross-sectional average temp:
1
z
1
0
1
0
( , )
( ) 2 ( , )
d
d
d
η ζ η η
ζ η ζ η η
η η
Θ
Θ = = Θ
∫
∫
∫z
Integrate each term of PDE:
0
∫
2
1
0Peη
⎛ ⎞∂ ∂Θ ∂ Θ ∂Θ
+ + =⎜ ⎟
over η to get ODE for Θ(ζ)
11
1
( ) ( )d
η=
⎛ ⎞∂ ∂Θ ∂Θ
∫
2
0Peη
η η η ζ ζ
+ +⎜ ⎟
∂ ∂ ∂ ∂⎝ ⎠
0 0
1
(1, ) ( )d Bi Bi
η
η η η η ζ ζ
η η η η =
⎛ ⎞∂ ∂Θ ∂Θ
= = − Θ ≅ − Θ⎜ ⎟
∂ ∂ ∂⎝ ⎠
∫
1 12 2
1∂ Θ ∂ ∂ Θ
1
P∂Θ ∂Θ2 2
2 2 2
0 0
1
2
d dη η η η
ζ ζ ζ
∂ Θ ∂ ∂ Θ
= Θ =
∂ ∂ ∂∫ ∫ 0
2
Pe
Pe dη η
ζ ζ
∂Θ ∂Θ
=
∂ ∂∫

13. Example Problem—heat transfer to solder
The ODE is then:
2
1
0
Pe
Bi
∂ Θ ∂Θ
Θ + + =
Reordering:
2
0
2 2
Bi
ζ ζ
− Θ + + =
∂ ∂
z 2
2
2 0Pe Bi
ζ ζ
∂ Θ ∂Θ
+ − Θ =
∂ ∂
(0) 1Θ =
( ) 0Θ ∞ =( )
The solution will be of the form:
pζ
The solution that satisfies both BC’s is:
p
e ζ
Θ =
Where:
2
2 0p Pe p Bi+ ∗ − =
( )2
8
exp
2
Pe Pe Bi
ζ
⎡ ⎤− + +
⎢ ⎥Θ =
⎢ ⎥
2
8
2
Pe Pe Bi
p
− ± +
=or:
2⎢ ⎥
⎢ ⎥⎣ ⎦

14. Example Problem—heat transfer to solder
Rate of heat conduction from the surface to the
wire (qz
(s))
Enlarging region near surface:
C V
Enlarging region near surface
Wire
z Liq.
Hot Surface
Liq.
C.V.
Mass balance:
w w L LQ Qρ ρ=
Energy balance:
( )2 ( ) ( )ˆ ˆ( ) 0s w
w L z zQ H H R q qρ π− + − =
then:
Where:
2
vw wQ Rπ=
( )
at hot
surfaceˆλ−
At solid
wire
Volumetric
flow rate
( ) ( )ˆvs w
z w zq qρ λ= +

15. Example Problem—heat transfer to solder
Rate of heat conduction from the surface to the
wire (qz
(s)) ( ) ( )ˆs w
z w zq v qρ λ= +
Enlarging region near surface:
Wire
z
Liq.
Hot Surface
Liq.
C.V.
( ) ( )w mk T TdT d
q k ∞− Θ
We can estimate qz
(w) as heat flux @ z=0, using the
temperature prof. we calculated:
and:
( )
0 0
m
z
z
q k
dz R d ζ
ζ= =
= − = −
( )2
8Pe Pe Bid + +Θ
( )( ) 2( )ˆ 8
2
s m
z w
k T T
q v Pe Pe Bi
R
ρ λ ∞−
= + + +
( )
0
8
2
Pe Pe Bid
d ζ
ζ =
+ +Θ
= −