h2kAP
_1/3
(Tb T) (1.83)
The same process can be applied to triangular fins (which will require information on
Bessel
functions that will be presented in the following chapter). Under the assumption of
b2/L2 1
which is consistent with the 1D heat transfe

Bib = (10)(.01)/400 = 0.0025 which is small enough to validate the 1D
approximation of the
fin.
1.3.2 Simple fins of uniform cross section
The most simple type of fin has a constant cross sectional area, The fin equation
reduces to
T N2T = 0 (1.58)
where

5. Consider now a pin fin with a square cross section, with b denoting the thickness
of the fin
and L the length. Fins of this type are to be mounted in a rectangular array on a
circuit
board, subject to the constraint that the gap between adjacent fin su

final result is
u=
r
a3
_
2 aear +
2
Bi
1 ea_
_a
Bi 1
_
(a + 2) ea
_
+
2
a3
1 ear_
(1.54)
20 CHAPTER 1. PRELIMINARIES AND REVIEW
The dimensionless temperature is given from T = u/r. Obtaining the center
temperature takes
some mathematical maneuvers, becau

2T
x2 (3.4)
T
x
_
0
= 0, T(x = 1, t) = 0 (3.5)
T(x, t = 0) = 1 (3.6)
The solution procedure can be stated as follows: we assume that the solution T can
be expressed
as a product of two functions, each of which depends only on a single variable. In
other w

in the appendix.
Recognize also that the integral in the above equation would be zero, for n 6= m,
whenever n
and m satisfy any homogeneous boundary condition at the boundaries. That is, if
the BCs were
3.3. ORTHOGONAL FUNCTIONS AND ORTHOGONALITY 59
in th

Te=50 C instable
Te=40 C instable
Te=30 C instable
Te=50 C stable
Te=40 C stable
Te=30 C stable
ZONE STABLE
ZONE INSTABLE
0.1
1
10
100
1000
10000
10 100 1000
Re
Ri
Te=50 C instable
Te=40 C instable
Te=30 C instable
Te=50 C stable
Te=40 C stable
Te=30 C st

qz
=
kxx kxy kxz
kxy kyy kyz
kxz kyz kzz
T/x
T/y
T/z
(1.8)
Material such as crystals can posses a highly anisotropic structure, and accordingly
the thermal
conductivity of these materials can be equally anisotropic: heat can be transferred
more effective

(2.3)
This obviously gives T = 1 at r = 1. If we use the fact that I0 = I1 and K0 = K1, we
find that
the adiabatic BC is satisfied at r = a.
The heat flux from the fin follows from
qfin = kAB
dT
dr
_
ri
=
2kbri(TB T)
ro
dT
dr
_
a
= 2akb(TB T)N
K1(Na)I1(N)

Rconv
(1.68)
i.e., N2 is proportional to the ratio of axial conduction resistance to surface
convection resistance.
Once N exceeds a certain value (around 3) the resistance to heat transfer in the fin
is dominated
by axial conduction. As mentioned above,

typically posed for spherical coordinates as finite T at r 0 needs to be given in a
more precise
mathematical form. This is done by imposing
r2 dT
dr
_
r0
=0
which is essentially the same as saying that the total heat transfer rate at the
center is zero.

=
r
a2 ear
2
a3 ear + c1r + c2
1.2. ONEDIMENSIONAL STEADY CONDUCTION 19
The integrations in the above were performed using integration by parts which is
an extremely
useful formula for integration of a product of two functions. For the indefinite
integra

75
100
125
50 75 100 125 150
s
= 280 W/K
= 830 W/K
= 1400 W/K
qtc
qtc
qtc
qth (W/K)
Figure 6.3. Influence of the flow rates on the time constant (temperature
step on the hot or on the cold fluid).
0
5
10
15
20
25
30
0 500 1000 1500
(s)
= 210 W/K
= 628 W/K

Recognize that if cosh(x) is a solution to the DE, then cosh(a+x) (where a is a
constant) is also a
solution. This amounts to a shifting of the system origin. The BC at x = 0 gives the
final solution:
T=
cosh[N(1 x)]
cosh(N)
(1.63)
1.3.3 Measures of fin p

40.5
41.0
41.5
42.0
0 2 4 6 8 10
Rayon ( mm )
Temprature ( C )
t=0s
t = 20 s
t = 24 s
t = 30 s
t = 50 s
Figure 4.7. Velocity and temperature profiles at z = 300 mm ; qv = + 5.10-3
m3/h, T = + 2 C, qv o = 3.10-2 m3/h, Teo = 40 C, h a = 10 W/m2.K.
0
1
2
3
4

4 74 77 +4% 71,8 -3%
Table 5.2. Convective coefficient in (W m-2 K-1) for 5 s excitation.
Fluxmtre Model h cst
c0
c0
h
h
Model h(t)
c0
c0
h
h
1 37 49 +32% 42,3 +14%
2 50 65 +30% 56 +12%
3 62 70,6 +14% 67 +8%
4 74 80 +8% 80 +8%
Heat Exchangers Under Transi

can now be viewed as a linearized radiation heat transfer coefficient, denoted
hrad.
1.2 OneDimensional Steady Conduction
1.2.1 The Thermal Resistance
The most simple conduction situation consists of one dimension, steady heat
transfer without
sources or

t=4s
t=7s
t = 11 s
t = 26 s
Figure 4.2. Velocity and temperature profiles for z = 200 mm, T = + 10 C.
Transient Convective Heat Transfer
J. of the Braz. Soc. of Mech. Sci. & Eng. Copyright 2005 by ABCM January-March 2005, Vol. XXVII, No.1 / 87
0
0.2
0.4
0

z Pe r r
VT
r
rUT
tr
T
The initial and boundary conditions are as follows:
t 0 , 0 z 1 , 0 r 1: T 1 (uniform temperature)
t+ 0 ,
; r ; T T
r
z:r41
2
1
2
01
2
where
( T
TT
T
T
ea
e>
0 or < 0)
0 0 0
;
rr
T
r:
w
*
w
Bi T
r
T
r 1 : 0 ;
where
f
* ha R
Bi
(gen

generated within its volume at a rate
q = q 0 ear/R (1.48)
18 CHAPTER 1. PRELIMINARIES AND REVIEW
where q0 and a are constants. The above function is chosen as representative of
heat generation
by nuclear decay (fission). Heat is convected from the surfac

possible, however, to represent E as a sum of energies of small volume elements
within the system
with each element assumed to be in thermodynamic equilibrium at any instant. As
the volume
of the elements go to zero the sum can be expressed as an integra

dAs
dAs
dx
_x + . . . = P(x)_x
in which P is the fin perimeter (again a function of x). The two previous equations
are combined
and divided by _x, and the limit of _x 0 is taken. This gives
k
d
dx
Ac
dT
dx hP(T T) = 0 (1.56)
which is known as the fin equ

M. Hadidi, M. Guellal, M. Lachi, J. Padet Loi de rponse dun
changeur thermique soumis des chelons de temprature aux deux
entres. Int. Communications in Heat and Mass Transfer, 22, N1, p.145-154
(1995)
M. Lachi, N. El Wakil, J. Padet - The time constant of

r2 T
r
+
1
r2 sin
sin
T
+
1
r2 sin
2T
2 (1.14)
1.1.3 Boundary conditions
Much of the course will focus on methods of solving Eq. (1.10). Two principal
elements go into
the solution method: 1) application of mathematical techniques to obtain a general
s

carrying current. Again, the circuit analysis will not be valid under these conditions,
and we are
forced to formally solve the conduction equation, for the given boundary conditions,
to obtain the
temperature profile within the object.
Start again with t

G. Polidori, J. Padet Heating and partial cooling problems in unsteady
forced convection. European Phys. J. Appl. Phys. 4, p.235-238 (1998)
M. Rebay, G. Polidori, J. Padet - Complment danalyse sur les
coulements de type couche limite en convection laminai

hot stream
0.1
0.12
0.14
0.16
0.18
0.2
0.22
0 50 100 150 200 250 300
Time (s)
Effectiveness
0.153
Figure 6.10. Temporal evolution of the effectiveness on the hot fluid side.
Numerical Model [64]
In order to get a better understanding of transient states i

Again, the choice A = 0 leads to T = 0 as the solution which is obviously not
correct. Rather,
the outcome should be
cos() = 0 (3.18)
This represents an transcendental equation for ; meaning that an infinite number
of roots exist
for the equation. For thi

hand, put it in the form of Eq. (2.30), and deduce the constants A, B, C, D, and n
from inspection.
44 CHAPTER 2. ADVANCED 1D ANALYTICAL METHODS
To give an example, start again with the annular fin DE, which is
r2T + rT r2N2T = 0 (2.32)
Recognize that r h

Z
1
0
cos[
1
2
(2m 1)x] dx
=
Z1
0
cos[
1
2
(2m 1)x]
X
n=1
An cos[
1
2
(2n 1)x]
!
dx
=
X
n=1
An
Z1
0
cos[
1
2
(2m 1)x] cos[
1
2
(2n 1)x] dx (3.23)
The orders of integration and summation can be switched in the above equation
because it is
assumed that the

(1)nBi
n(2
n + Bi2)1/2
(3.55)
Similarly, the second integral is
Z1
0
cos2(nx) dx =
1
2
_
1+
cos(n) sin(n)
n
_
=
1
2
_
1+
Bi
2
n + Bi2
_
Again, the eigencondition was used to eliminate the trigonometric functions.
Combining the previous two equations, the

= Am
1
2
or
Am =
4(1)m1
(2m 1)
=
2(1)m1
m
(3.25)
This is the sought explicit equation for Am. Of course, the index m can be set to any
number
or symbol we choose, i.e. An. The final, complete solution to the temperature field is
then
T(x, t) = 2
X
n=1
(1