CHAPTER 2
ONE-DIMENSIONAL STEADY STATE
CONDUCTION
2.1 Examples of One-dimensional
Conduction:
2.1.1 Plate with Energy Generation and
Variable Conductivity
1
Example 2.1: Plate with internal energy generation
q  and a variable k
k  k o (1   T )
0o C
Find temperature distribution.
0
(1) Observations
• Variable k
• Symmetry
• Energy generation
• Rectangular system
• Specified temperature at boundaries
q
0o C
x
L
Fig. 2.1
2
(2) Origin and Coordinates
Use a rectangular coordinate system
(3) Formulation
(i) Assumptions
• One-dimensional
• Steady
• Isotropic
• Stationary
• Uniform energy generation
3
(ii) Governing Equation
Eq. (1.7):
d  dT 
k
  q   0
dx  dx 
k  k o (1   T )
(2.1)
(a)
(a) into eq. (2.1)
d
dx

 (1 

T)
d T  q 
0

dx  ko
(b)
(iii) Boundary Conditions.
Two BC are needed:
4
T (0)  0
(c)
T ( L)  0
(d)
(4) Solution
Integrate (b) twice
T 

T
2

2
q 
2ko
2
x  C1 x  C 2
(e)
BC (c) and (d)
C1 
q L
2ko
,
C2  0
(f)
5
(f) into (e)
q L x 
x
T  T 
1   0


 ko 
L
2
2
(g)
Solving for T
T 
1


1

2
q L x 
x

1 

 ko 
L
(h)
q L x 
x

1 

 ko 
L
(i)
Take the negative sign
T 
1


1

2
6
(5) Checking
• Dimensional check
• Boundary conditions check
• Limiting check: q   0, T
• Symmetry Check:

1 1
q Lx x
  

(  1)
2
dx
2 
 ko L

dT
0

1
2
(
q L
 ko
)(
2x
 1 ) (j)
L
Setting x = L/2 in (j) gives dT/dx = 0
7
• Quantitative Check
Conservation of energy and symmetry:
q(0)  
q AL
(k)
2
q( L) 
q AL
(l)
2
Fourier’s law at x = 0 and x = L
q ( 0 )   k o 1   T ( 0 ) 
dT ( 0 )
dx

q AL
(m)
2
8
q ( L )   k o 1   T ( L ) 
dT ( L )
dx

q AL
(n)
2
(6) Comments
Solution to the special case:
k = constant: Set   0
2.1.2 Radial Conduction in a Composite
Cylinder with Interface Friction
9
Example 2.2: Rotating shaft in sleeve, frictional
heat at interface, convection on
outside. Conduction in radial direction.
Determine the temperature
distribution in shaft and sleeve.
h, T 
T1
Rs
r
Ro
0
(1) Observations
• Composite cylindrical wall
• Cylindrical coordinates
• Radial conduction only
q i
T2
sleeve
F ig. 2 .2
10
• Steady state:
Energy generated = heat conducted through the
sleeve
• No heat is conducted through the shaft
• Specified flux at inner radius of sleeve, convection
at outer radius
(2) Origin and Coordinates
Shown in Fig. 2.2
11
(3) Formulation
(i) Assumptions
• One-dimensional radial conduction
• Steady
• Isotropic
• Constant conductivities
• No energy generation
• Perfect interface contact
• Uniform frictional energy flux
• Stationary
12
(ii) Governing Equation
Shaft temperature is uniform. For sleeve: Eq. (1.11)
d  dT 1 
r
0


dr  dr 
(2.2)
(iii) Boundary Conditions
Specified flux at R s :
q i   k 1
dT 1 ( R s )
(a)
dr
13
Convection at R o :
 k1
dT 1 ( R o )
dr
 h [ T1 ( R o )  T  ]
(b)
(4) Solution
Integrate eq. (2.2) twice
T1  C 1 ln r  C 2
(c)
BC give C1 and C2
C1  
q i R s
k1
(d)
14
and
C 2  T
q i R s 
k1 

 ln R o 

k1 
hR o 
(e)
(d) and (e) into (c)
T1 ( r )  T 
q i R s  R o
k1 


 ln

k1 
r
hR o 
(f)
hR o / k = Biot number
Shaft temperature T2: Use interface boundary condition
T 2 ( r )  T 2 ( R s )  T1 ( R s )
(g)
15
Evaluate (f) at r = Rs and use (g)
T2 ( r )  T
q i R s  R o
k1 


 ln

k1  R s
hR o 
(5) Checking
• Dimensional check
• Boundary conditions check
• Limiting check: q i  0
(h)
h, T 
T1
Rs
r
Ro
0
q i
T2
sleeve
F ig. 2 .2
(6) Comments
• Conductivity of shaft does not play a role
16
• Problem can also be treated formally as a
composite cylinder. Need 2 equations and 4 BC.
2.1.1 Composite Wall with Energy Generation
q  . Plate 1
is sandwiched between two plates.
Outer surfaces of two plates at T o .
Example 2.1: Plate 1 generates heat at
To
x
L2
k2
q

L1
0
L2
To
k1
k2
Find the temperature distribution in
the three plates.
Fig. 2.3
17
(1) Observations
• Composite wall
• Use rectangular
coordinates
To
x
L2
k2
q

L1
0
L2
• Symmetry: Insulated
center plane
k1
k2
To
Fig. 2.3
• Heat flows normal to plates
• Symmetry and steady state:
Energy generated = Energy conducted out
(2) Origin and Coordinates
Shown in Fig. 2.3
18
(3) Formulation
(i) Assumptions
• Steady
• One-dimensional
• Isotropic
• Constant conductivities
• Perfect interface contact
• Stationary
(ii) Governing Equations
Two equations:
19
2
d T1
dx

2
q 
2
0
d T2
(a)
dx
k
2
0
(b)
(iii) Boundary Conditions
Four BC:
Symmetry:
dT 1 ( 0 )
0
(c)
dx
Interface:
k1
dT 1 ( L1 / 2 )
dx
 k2
dT 2 ( L1 / 2 )
(d)
dx
20
T1 ( L1 / 2 )  T 2 ( L1 / 2 )
(e)
Outer surface:
T 2 ( L1 / 2  L 2 )  T o
(f)
(4) Solution
Integrate (a) twice
T1 ( x )  
q 
2
x  Ax  B
2 k1
(g)
Integrate (b)
T 2 ( x )  Cx  D
(h)
21
Four BC give 4 constants: Solutions (g) and (h) become
2 
2



q L1 1 k 1 L 2 x
T1 ( x )  T o 

 

2
2 k 1  4 k 2 L1 L1 
(i)
q L1  1 L 2
x 
T 2 ( x )  To 

 

2 k 2  2 L1 L1 
(j)
2
(5) Checking
• Dimensional check: units of
q L
2
:
k
22
q  ( W/m
3
2
2
)L (m )
o
k ( W/m  C )

o
C
• Boundary conditions check
• Quantitative check:
1/2 the energy generated in center plate = Heat
conducted at x  L1 / 2
L1
2
q    k 1
dT 1 ( L1 / 2 )
(k)
dx
23
(i) into (k)
 k1
dT 1 ( L1 / 2 )
dx

L1
q 
2
Similarly, 1/2 the energy generated in center plate
= Heat conducted out
L1
2
q    k 2
dT 2 ( L1 / 2  L 2 )
(l)
dx
(j) into (l) shows that this condition is satisfied.
• Limiting check:
(i) If q   0 , then T1 ( x )  T 2 ( x )  T o .
(ii) If L1  0 then T1 ( x )  T o .
24
(6) Comments
Alternate approach: Outer plate with a specified
flux at x  L1 / 2 and a specified temperature at
x  L1 / 2  L 2 .
2.2 Extended Surfaces - Fins
2.2.1 The Function of Fins
Newton's law of cooling:
q s  hA s ( T s  T  )
(2.3)
25
Options for increasing q s
:
• Increase h
• Lower T 
• Increase A s
Examples of Extended Surfaces (Fins):
• Thin rods on condenser in back of refrigerator
• Honeycomb surface of a car radiator
• Corrugated surface of a motorcycle engine
• Disks or plates used in baseboard radiators
26
2.2.2 Types of Fins
(a) constant area
straight fin
(b) variable area
straight fin
(c) pin fin
(d) annular fin
Fig. 2.5
27
Terminology and types
• Fin base
• Fin tip
• Straight fin
• Variable cross-sectional area fin
• Spine or pin fin
• Annular or cylindrical fin
2.2.3 Heat Transfer and Temperature
Distribution in Fins
• Heat flows axially and laterally (two-dimensional)
• Temperature distribution is two-dimensional
28
2.2.4 The Fin Approximation
Neglect lateral temperature
variation
T  T (x)
r
d
T
x
T
h,T
Criterion:
Biot number = Bi
Fig. 2.6
Bi = hd /k << 1
Bi 
d /k
1/ h

(2.4)
Internal resis tan ce
external resis tan ce
29
2.2.5 The Fin Heat Equation: Convection at
Surface
(1) Objective:
Determine fin heat transfer rate.
Need temperature distribution.
(2) Procedure:
Formulate the fin heat equation.
Apply conservation of energy.
• Select an origin and coordinate axis x.
• Assume Bi  0 . 1 ,  T  T ( x )
• Stationary material, steady state
30
ds
C
y
d
y
s
0
x
,T
h

d
x
q

d
x
d
q
x
x
d
d
a
(
)
d
d
c
b
(
)
(c
)
F
.2
.7
Conservation of energy for the element dx:
E in  E g = E out
E in

qx
dq x

E out  q x 
dx  dq c
dx
(a)
(b)
(c)
31
dA s
C
qx 
qx
dq x
dx
dx
dx
dy
ds
dq c
(b)
(c )
(b) and (c) into (a)
dq x

Eg 
dx  dq c
dx
Fourier's law and Newton’s law
dT
q x   kA c
dx
dq c  h ( T  T  ) dA s
Energy generation
(d)
(e)
(f)
32
E g  q A c ( x ) dx
(g)
(e), (f) and (g) into (d)
dT 
d 
 kA c ( x ) dx  dx  h (T  T  ) dA s  q Ac ( x ) dx  0
dx 

(2.5a)
Assume constant k
2
d T
dx
2

1
dA c dT
A c ( x ) dx dx

h
kA c ( x )
(T  T )
dA s
dx

q 
0
k
(2.5b)
• (2.5b) is the heat equation for fins
• Assumptions:
(1) Steady state
(2) Stationary
33
(3) Isotropic
(4) Constant k
(5) No radiation
(6) Bi << 1
• Ac , dA c / dx , and dA s / dx are determined from
the geometry of fin.
2.2.6 Determination of dAs /dx
From Fig. 2.7b
dA s  C ( x ) ds
C ( x ) = circumference
ds = slanted length of the element
(a)
34
For a right triangle
ds  [ dx
2
2 1/2
 dy s ]
(b)
(b) into (a)

 C ( x ) 1 
dx

dA s
2
 dy s 

 
 dx  
1/2
(2.6a)
For dy s / dx << 1
dA s
 C (x)
(2.6b)
dx
2.2.7 Boundary Conditions
Need two BC
35
2.2.8 Determination of Fin Heat Transfer
Rate q f :
h
T
,

0
q
(
0
)
q
s
q
s
x
h
,T

q
s
F
ig
.2
.8
Conservation of energy for q   0 :
q f  q(0)  q s
(a)
Two methods to determine q f :
36
(1) Conduction at base.
Fourier's law at x = 0
q f  q ( 0 )   kA c ( 0 )
dT ( 0 )
(2.7)
dx
(2) Convection at the fin surface.
Newton's law applied at the fin surface
q f  qs 
A s h[T ( x )  T ]dA s
(2.8)
• Fin attached at both ends: Modify eq. (2.7)
accordingly
• Fin with convection at the tip: Integral in eq. (2.8)
includes tip
37
• Convection and radiation at surface: Apply eq. (2.7).
Modify eq. (2.8) to include heat exchange by
radiation.
2.2.9 Applications: Constant Area Fins with
Surface Convection
h
,T

0
T
o
C
x
h
T
,

A
c
F
i
g
.2
.9
38
A. Governing Equation
Use eq. (2.5b). Set
dA c / dx  0
h
,T

(a)
0
x
T
o
y s = constant
C
h
T
,

A
c
F
ig
.2
.9
dy s / dx  0
Eq. (2.6a)
dA s / dx  C
(b)
(a) and (b) into eq. (2.5b)
2
d T
dx
2

hC
kA c
(T  T )  0
(2.9)
39
Rewrite eq. (2.9)
  T  T
m
2

hC
(c)
(d)
kA c
Assume T  = constant, (c) and (d) into (2.9)
2
d 
dx
2
2
m  0
Valid for:
(1) Steady state
(2) constant k, A c and T 
(2.10)
40
(3) No energy generation
(4) No radiation
(5) Bi 1
(6) Stationary fin
B. Solution
Assume: h = constant
 ( x )  A1 exp( mx )  A 2 exp(  mx )
(2.11a)
 ( x )  B 1 sinh mx  B 2 cosh mx
(2.11b)
41
C. Special Case (i):
• Finite length
• Specified temperature at base, convection at tip
Boundary conditions:
h, T 
x
0
C
ht
h, T 
To
Ac
F ig. 2.10
T ( 0 )  To
k
dT ( L )
dx
 h t [T ( L )  T  ]
 (0)   o
(e)
(f)
(h)
42
k
d ( L )
dx
 ht  ( L )
(i)
Two BC give B1 and B2
 (x)
o


T ( x )  T
To  T 
(2.12)
cosh m  L  x    h t mk  sinh m  L  x 
cosh mL   h t mk  sinh mL
Eq. (2.7) gives q f
q f  [ k Ac C h ]
1 / 2
( T o  T  )[sinh mL  ( h t / mk )cosh mL ]
cosh mL  ( h t / mk ) sinh mL
43
(2.13)
C. Special Case (ii):
• Finite length
• Specified temperature at base, insulated tip
BC at tip:
d ( L )
0
(j)
dx
Set h t  0 eq. (2.12)
 (x)
o

T ( x )  T
To  T 
Set h t  0 eq. (2.13)
q f  kA c Ch 
1 2

cosh m ( L - x )
(2.14)
cosh mL
To  T   tanh
mL
(2.15)
44
2.2.10 Corrected Length Lc
• Insulated tip: simpler solution
• Simplified model: Assume insulated tip, compensate
by increasing length by  L c
• The corrected length is L c
Lc  L   Lc
(2.16)
• The correction increment Lc depends on the
geometry of the fin:
Increase in surface area due to  L c = tip area
Circular fin:

2
ro
 2 ro  L c
45
 L c  ro / 2
Square bar of side t
 Lc  t / 4
2.2.11 Fin Efficiency 
f
Definition

f

qf
(2.17)
q max
q max  hA s To  T 
A s = total surface area
46
Eq. (2.17) becomes

f

qf
(2.18)
h A s ( To  T  )
2.1.12 Moving Fins
Examples:
• Extrusion of plastics
T sur
h, T 
x
To
dx
(a)
• Drawing of wires and
dq r
sheets
• Flow of liquids
U
(b )
m hˆ
qx
dx
dq c
d hˆ
ˆ

m (h 
dx )
dx
dq x
qx 
dx
dx
F ig. 2.11
47
dq r
Heat equation:
dq c
d hˆ
ˆ
m ( h 
dx )
dx
m hˆ
Assume:
q
• Steady state
• Constant area
• Constant velocity U
• Surface convection and radiation
qx 
x
dx
dq x
dx
dx
(b)
F ig. 2.11
Conservation of energy for element dx
q x  m hˆ  q x 
dq x
dx  m hˆ  m
dx
d hˆ
dx
 dq c  dq r
(a)
m   UA c
(b)
48
d hˆ  c p dT
(c)
Fourier’s and Newton’s laws
q x   kA c
dT
(d)
dx
dq c  h ( T  T  ) dA s
(e)
dA s  C dx
(f)
4
(g)
dq r    ( T
4
 T sur ) dA s
(b)-(g) into (a) assume constant k
2
d T
dx
2

 c p U dT
k
dx

hC
kA c
(T  T ) 

k
(T
4
4
 T sur )  0
49
(2.19)
Assumptions leading to eq. (2.19):
(1) Steady state
(2) Constant U, k, P and  ,
(3) Isotropic
(4) Gray body
(5) Small surface enclosed by a much larger surface and
(6) Bi << 1
50
2.2.13 Application of Moving Fins
h
,T

xU
Example 2.4
Plastic sheet leaves furnace
at T o .
W
t
fu
rn
a
c
e
in
s
u
la
te
db
o
tto
m
T
o
F
ig
.2
.1
2
Sheet is cooled at top by convection.
Assume:
(1) Steady state
(2) Bi < 0.1
(3) No radiation
(4) No heat transfer from bottom
Determine the temperature distribution in the sheet.
51
Solution
h
,T

xU
W
(1) Observations
t
fu
rn
a
c
e
• Constant area fin
• Temperature is one-dimensional
• Convection at surface
• Specified furnace temperature
• Fin is semi-infinite
• Constant velocity
in
s
u
la
te
db
o
tto
m
T
o
F
ig
.2
.1
2
(2) Origin and Coordinates
52
(3) Formulation
(i) Assumptions
(1) One-dimensional
(2) Steady state
(3) Isotropic
(4) Constant pressure
(5) Constant U, k, P and  ,
(6) Negligible radiation
(ii) Governing Equation
Eq. (2.19)
2
d T
dx
2

 c p U dT
k
dx

hC
kA c
(T  T )  0
(2.20)
53
A c  Wt
(a)
C  W  2t
(b)
(a) and (b) into eq. (2.20)
2
d T
dx
2
 2b
dT
m T c
2
(c)
dx
where
b
 c pU
2k
,
m
2

h (W  2 t )
, c
kWt
h (W  2 t )
kW t
T
(d)
(iii) Boundary Conditions
T ( 0 )  To
(e)
T (  )  finite
(f)
54
Eq. (c) is:
• Linear
• Second order
• Constant coefficients
(4) Solution:
Eq. (A-6b), Appendix A
T  C 1 exp(  bx 
C 2 exp(  bx 
2
2
2
2
b  m x) 
b  m x) 
(2.21)
c
m
2
B.C. (f)
C1  0
(g)
55
B.C. (e)
C 2  T0 
c
m
(h)
2
(d), (g) and (h) into (2.21)
T ( x )  T
To  T
  c pU
 exp 

 2 k
(
 c pU 2
2k
) 
h (W  2 t ) 
x
kW t 

(2.22)
(5) Checking
Dimensional check: Each term in the exponential in
eq. (2.22) is dimensionless
56
Boundary conditions check: Eq.(2.22) satisfies (e) and (f).
Limiting checks:
(i) If h  0 :
(ii) If U   :
T ( x )  To
T ( x )  To
(6) Comments
(i) Temperature decays exponentially
(ii) Motion slows decay
57
2.2.14 Variable Area Fins
Ac  Ac ( x )
Example: Cylindrical or annular fin
Governing equation:
Usually has variable coefficients
h , T
Case (i) : The Annular Fin
h , T
dr
Fig. 2.13
2
d T
dr
2

1
dA c dT
A c ( r ) dr
dr

h
kA c ( r )
(T  T )
dA s
dr
0
58
(2.23)
r
A c ( r )  2 r t
(a)
dA c / dr  2  t
(b)
h , T
h , T
Eq.(2.6a) gives dA s / dr
Fig. 2.13
2

 dy s 
 C ( r ) 1  
 
dr
 dr  

dA s
dr
1/2
For y s = constant:
dy s / dr  0
C ( r )  2 ( 2 r )
(c)
59
r
Eq. (2.6a):
dA s
 2 ( 2 r )
(d)
dr
(a), (b) and (d) into eq. (2.23)
2
d T 1 dT

 ( 2 h / kt )( T  T  )  0
2 r dr
dr
(2.24)
Case (ii): Triangular straight fin
Fin equation: Constant k,
eq. (2.5b)
Ac  2 W y s ( x )
y
L
h
,T

y
s
x
(e)
0
h,T

t
d
F 2
60
y
L
h
,T

y
s
x
0
h,T

t
d
x
F 2
i
.
y s  ( x / L )( t / 2 )
(f)
A c  (W t / L ) x
(g)
dA c
W t/L
(h)
dx
Eq. (2.6a):
dA s
dx
2 1/2
 2W [1  ( dy s / dx ) ]
2 1/2
 2W [ 1  ( t / 2 L ) ]
(i)
61
(g), (h) and (i) into eq. (2.5b)
2
d T
dx
2

1 dT
x dx

  2 hL / kt  1   t / 2 L 

2 1/2
1 / x ( T
 T )  0
(2.25)
Equations (2.24) and (2.25) are:
•Linear
•Second order
•Variable coefficients
62
2.3 Bessel Differential Equations and Bessel
Functions
2.3.1 General form of Bessel Equations

2
x
2

d y
dx
2
2
 (1  2 A ) x  2 B x
2
 C D x
2C
2
2

dy
(2.26)
dx
2
2
2
 B x  B (1  2 A ) x  A  C n
2
y0
Note the following:
(1) Eq. (2.26) is linear, second order with variable
coefficients
(2) A, B, C, D, and n are constants
63
(3) n is called the order of the differential equation
(4) D can be real or imaginary
2.3.2 Solution: Bessel Functions
• Form: Infinite power series solutions
• General solution: depends on the constants n and D
(1) n is zero or integer, D is real
y( x )  x
A

C
C
exp( B x ) C 1 J n ( D x )  C 2 Y n ( D x )

(2.27)
where
C 1 , C 2 = constant of integration
64
J n ( Dx
Y n ( Dx
C
C
) = Bessel function of order n of the first kind
) = Bessel function of order n of the second kind
Note the following:
(i) The term ( D x C ) is the argument of the Bessel
function
(ii) Values of Bessel functions are tabulated
(2) n is neither zero nor a positive integer, D is real
y( x )  x
A

C
C
exp( B x ) C 1 J n ( D x )  C 2 J  n ( D x )

(2.28)
65
(3) n is zero or integer, D is imaginary
y( x )  x
A

C
C
exp( B x ) C 1 I n ( p x )  C 2 K n ( p x )

(2.29)
where
p
In
D
, i is imaginary =
1
i
= modified Bessel function of order n of the first kind
K n = modified Bessel function of order n of the second
kind
66
(4) n is neither zero nor a positive integer, D is imaginary
y( x )  x
A

C
C
exp( B x ) C 1 I n ( p x )  C 2 I  n ( p x )

(2.30)
2.3.3 Form of Bessel Functions
J n , Y n , J  n , I n , I  n and K n :
• Symbols for infinite power series
• The form of each series depends on n
Example: n 
2 , Bessel function J 2 ( x )

J2(x) 

k 0
k
(  1) ( x / 2 )
2k 2
k !  k  2 !
(2.31)
67
2.3.4 Special Closed-form Bessel Functions:
n
odd integer
2
For n  1 / 2
J1 / 2 ( x) 
2
x
sin x
(2.32)
cos x
(2.33)
and
J 1 / 2 ( x ) 
2
x
68
For n = 3/2, 5/2, 7/2, … use Eq. (2.32) or eq. (2.33) and
the recurrence equation:
J k 1 / 2 ( x ) 
2k  1
x
J k 1 / 2 ( x )  J k  2 / 3 ( x )
k = 1, 2, 3, … (2.34)
For the modified Bessel functions, n = 1/2
I1 / 2 ( x ) 
2
x
sinh x
(2.35)
cosh x
(2.36)
and
I 1 / 2 ( x ) 
2
x
69
For n = 3/2, 5/2, 7/2, …. use eq. (2.35) or eq. (2.36) and
the recurrence equation
I k 1 / 2 ( x ) 
2k  1
x
I k 1 / 2 ( x )  I k  2 / 3 ( x )
k = 1, 2, 3, … (2.37)
2.3.5 Special Relations for n = 0, 1, 2, …
n
(2.38a)
n
Y n ( x )  (  1) Yn ( x )
(2.38b)
In( x)  In( x)
(2.38c)
K n(x)  K n(x)
(2.38d)
70
J  n ( x )  (  1) J n ( x )
2.3.6 Derivatives and Integrals of Bessel
Functions

x
dx
d

x
dx
d
n
Z n  mx
n
Z n  mx

 mx
 
 mx

n
Z n  1  mx
n

Z n  1  mx
Z  J , Y , I (2.39)

  mx  n Z
n  1  mx
 
n
 mx
Z n  1  mx 
Z  K
Z
(2.40)
 J , Y , K (2.41)
Z  I
(2.42)
71
n

mZ n  1  mx   Z n  mx 

d

x
 Z n  mx   
dx
n
  mZ

mx   Z n  mx
n

1

x
n

 mZ n  1  mx  
Z n  mx

d

x
 Z n  mx   
dx
n
 mZ

mx  
Z n  mx 
n

1

x
x
n
n
Z n  1 ( mx ) dx  ( 1 / m ) x Z n ( mx )
Z  J ,Y , I
(2.43)

Z  K
(2.44)

Z  J ,Y , K
(2.45)
Z  I
(2.46)
Z  J ,Y , I
(2.47)
72

x
n
Z n  1  mx dx   1 m  x
n
Z n  mx

Z  J ,Y , K
(2.48)
2.3.7 Tabulation and Graphical Representation
of Selected Bessel Functions
Table 2.1
x
J0(x)
Jn(x)
I0(x)
In(x)
0

1
0
0
0
1

0

Yn(x) Kn(x)
-
0

0
73
F ig . 2 .1 5 G raphs of selected B essel fu nctio ns
74
2.4 Equidimensional (Euler) Equation
2
x
2
d y
dx
2
 a1 x
dy
dx
 a0 y  0
(2.49)
Solution:Depends on roots r1 and r2
r1 , 2 
 ( a1  1) 
2
( a1  1)  4 a 0
(2.50)
2
Three possibilities:
(1) Roots are distinct
y( x )  C1 x
r1
 C2x
r2
(2.51)
75
(2) Roots are imaginary
y( x )  x
a
C 1 cos( b log
x )  C 2 sin( b log x ) 
(2.52)
(3) One root only
r
y ( x )  x ( C  C log x )
1
2
(2.53)
2.5 Graphically Presented Fin Solutions to Fin
Heat Transfer Rate q f
Fin efficiency 
:
f

f

qf
h A s ( To  T  )
(2.18)
76
Fig. 2.16 Fin efficiency of three types of straight fins [5]
77
Fig. 2.17 Fin efficiency of annular fins of constant thickness [5]
78