1. CHAPTER 2 ONE-DIMENSIONAL STEADY STATE CONDUCTION
2.1 Examples of One-dimensional Conduction:
2.1.1 Plate with Energy Generation and Variable Conductivity

2. ¢ (1) Observations q x L and a variable k
Example 2.1: Plate with internal energy generation
and a variable k L q C o
Fig. 2.1 x
Find temperature distribution.
(1) Observations
• Variable k
• Symmetry
• Energy generation
• Rectangular system
• Specified temperature at boundaries

3. (2) Origin and Coordinates
Use a rectangular coordinate system
(3) Formulation
(i) Assumptions
• One-dimensional
• Steady
• Isotropic
• Stationary
• Uniform energy generation

4. (ii) Governing Equation
(2.1)
(a)
(a) into eq. (2.1)
(b)
(iii) Boundary Conditions.
Two BC are needed:

5. (c)
(d)
(4) Solution
Integrate (b) twice
(e)
BC (c) and (d) ,
(f)

6. (f) into (e)
(g)
Solving for T
(h)
Take the negative sign
(i)

7. (5) Checking • Dimensional check • Boundary conditions check
• Limiting check:
• Symmetry Check:
(j)
Setting x = L/2 in (j) gives dT/dx = 0

8. • Quantitative Check
Conservation of energy and symmetry:
(k)
(l)
Fourier’s law at x = 0 and x = L
(m)

9. (n)
(6) Comments
Solution to the special case:
k = constant: Set
2.1.2 Radial Conduction in a Composite Cylinder with Interface Friction

10. (1) Observations outside. Conduction in radial direction.
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.
(1) Observations
• Composite cylindrical wall
• Cylindrical coordinates
• Radial conduction only

11. (2) Origin and Coordinates
• 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

12. (3) Formulation (i) Assumptions • One-dimensional radial conduction
• Steady
• Isotropic
• Constant conductivities
• No energy generation
• Perfect interface contact
• Uniform frictional energy flux
• Stationary

13. (ii) Governing Equation
Shaft temperature is uniform. For sleeve: Eq. (1.11)
(2.2)
(iii) Boundary Conditions
Specified flux at
(a)

14. (4) Solution Convection at (b) Integrate eq. (2.2) twice (c)
BC give C1 and C2
(d)

15. and
(e)
(d) and (e) into (c)
(f)
= Biot number
Shaft temperature T2: Use interface boundary condition
(g)

16. (5) Checking (6) Comments Evaluate (f) at r = Rs and use (g) (h)
• Dimensional check
• Boundary conditions check
• Limiting check:
(6) Comments
• Conductivity of shaft does not play a role

17. 2.1.1 Composite Wall with Energy Generation
• Problem can also be treated formally as a composite cylinder. Need 2 equations and 4 BC.
2.1.1 Composite Wall with Energy Generation
Example 2.1: Plate 1 generates heat at
is sandwiched between two plates. Outer surfaces of two plates at
Plate 1
Find the temperature distribution in the three plates.

18. (2) Origin and Coordinates
(1) Observations
• Composite wall
• Use rectangular coordinates
• Symmetry: Insulated center plane
• Heat flows normal to plates
• Symmetry and steady state:
Energy generated = Energy conducted out
(2) Origin and Coordinates
Shown in Fig. 2.3

19. (3) Formulation (i) Assumptions • Steady • One-dimensional • Isotropic
• Constant conductivities
• Perfect interface contact
• Stationary
(ii) Governing Equations
Two equations:

20. (b)
(a)
(iii) Boundary Conditions
Four BC:
Symmetry:
(c)
Interface:
(d)

21. (4) Solution (e) Outer surface: (f) Integrate (a) twice (g)
Integrate (b)
(h)

22. (5) Checking Four BC give 4 constants: Solutions (g) and (h) become
(j)
(5) Checking
• Dimensional check: units of

23. • Boundary conditions check
• Quantitative check:
1/2 the energy generated in center plate = Heat conducted at
(k)

24. (i) into (k)
Similarly, 1/2 the energy generated in center plate
= Heat conducted out
(l)
(j) into (l) shows that this condition is satisfied.
• Limiting check:
then
(i) If
(ii) If
then

25. 2.2 Extended Surfaces - Fins 2.2.1 The Function of Fins
(6) Comments
Alternate approach: Outer plate with a specified flux at
and a specified temperature at
2.2 Extended Surfaces - Fins
2.2.1 The Function of Fins
Newton's law of cooling:
(2.3)

26. Options for increasing
• Increase h
• Lower
• Increase
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

27. 2.2.2 Types of Fins 2.5 Fig. straight fin constant area (a)
variable area
(b)
2.5
Fig.
annular fin
(d)
pin fin
(c)

28. 2.2.3 Heat Transfer and Temperature Distribution in Fins
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

29. 2.2.4 The Fin Approximation x T h , r Criterion: Bi = h /k << 1
Neglect lateral temperature variation x T h , d r 6 . 2
Fig
Criterion:
Biot number = Bi
(2.4)
Bi = h /k << 1

30. 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
• Stationary material, steady state

31. Conservation of energy for the element dx:
(b)
(c)

32. (b) and (c) into (a)
(d)
Fourier's law and Newton’s law
(e)
(f)
Energy generation

33. • Assumptions: (g) (e), (f) and (g) into (d) Assume constant k (2.5b)
• (2.5b) is the heat equation for fins
• Assumptions:
(1) Steady state
(2) Stationary

34. 2.2.6 Determination of dAs /dx
(3) Isotropic
(4) Constant k
(5) No radiation
(6) Bi << 1 •
and
are determined from
the geometry of fin.
2.2.6 Determination of dAs /dx
From Fig. 2.7b
(a)
= circumference
ds = slanted length of the element

35. 2.2.7 Boundary Conditions << 1 For a right triangle (b)
(b) into (a)
(2.6a)
For
<< 1
(2.6b)
2.2.7 Boundary Conditions
Need two BC

36. 2.2.8 Determination of Fin Heat Transfer Rate
Conservation of energy for
(a)
Two methods to determine

37. (1) Conduction at base.
Fourier's law at x = 0
(2.7)
(2) Convection at the fin surface.
Newton's law applied at the fin surface
(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

38. 2.2.9 Applications: Constant Area Fins with Surface Convection
• 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

39. A. Governing Equation
Use eq. (2.5b). Set
(a)
= constant
Eq. (2.6a)
(b)
(a) and (b) into eq. (2.5b)
(2.9)

40. Rewrite eq. (2.9)
(c)
(d)
= constant, (c) and (d) into (2.9)
Assume
(2.10)
Valid for:
(1) Steady state
(2) constant k,
and

41. B. Solution (3) No energy generation (4) No radiation (5) Bi 1
(6) Stationary fin
B. Solution
Assume: h = constant
(2.11a)
(2.11b)

42. C. Special Case (i): • Finite length
• Specified temperature at base, convection at tip
Boundary conditions:
(e)
(f)
(h)

43. (i)
Two BC give B1 and B2
(2.12)
Eq. (2.7) gives
(2.13)

44. C. Special Case (ii): • Finite length
• Specified temperature at base, insulated tip
BC at tip:
(j)
Set
eq. (2.12)
(2.14)
Set
eq. (2.13)
(2.15)

45. 2.2.10 Corrected Length Lc • Insulated tip: simpler solution
• Simplified model: Assume insulated tip, compensate by increasing length by
• The corrected length is
(2.16)
• The correction increment Lc depends on the geometry of the fin:
Increase in surface area due to
= tip area
Circular fin:

46. 2.2.11 Fin Efficiency Square bar of side t Definition (2.17)
= total surface area

47. 2.1.12 Moving Fins Eq. (2.17) becomes (2.18) • Extrusion of plastics
Examples:
• Extrusion of plastics
• Drawing of wires and sheets
• Flow of liquids

48. • Steady state Heat equation: Assume: • Constant area
• Constant velocity U
• Surface convection and radiation
Conservation of energy for element dx
(a)
(b)

49. (c)
Fourier’s and Newton’s laws
(d)
(e)
(f)
(g)
(b)-(g) into (a) assume constant k
(2.19)

50. 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

51. 2.2.13 Application of Moving Fins
Example 2.4
Plastic sheet leaves furnace at
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.

52. (2) Origin and Coordinates
Solution
(1) Observations
• Constant area fin
• Temperature is one-dimensional
• Convection at surface
• Specified furnace temperature
• Fin is semi-infinite
• Constant velocity
(2) Origin and Coordinates

53. (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)
(2.20)

54. (a)
(b)
(a) and (b) into eq. (2.20)
(c)
where
(d)
(iii) Boundary Conditions
(e)
finite
(f)

55. Eq. (c) is: (4) Solution: • Linear • Second order
• Constant coefficients
(4) Solution:
Eq. (A-6b), Appendix A
(2.21)
B.C. (f)
(g)

56. (5) Checking B.C. (e) (h) (d), (g) and (h) into (2.21) (2.22)
Dimensional check: Each term in the exponential in
eq. (2.22) is dimensionless

57. Boundary conditions check: Eq.(2.22) satisfies (e) and (f).
Limiting checks:
(i) If
(ii) If
(6) Comments
(i) Temperature decays exponentially
(ii) Motion slows decay

58. Example: Cylindrical or annular fin
Variable Area Fins
Example: Cylindrical or annular fin
Governing equation:
Usually has variable coefficients
Case (i) : The Annular Fin
(2.23)

59. (a)
(b)
Eq.(2.6a) gives
For
= constant:
(c)

60. Case (ii): Triangular straight fin
Eq. (2.6a):
(d)
(a), (b) and (d) into eq. (2.23)
(2.24)
Case (ii): Triangular straight fin
Fin equation: Constant k,
eq. (2.5b)
(e)

61. (f)
(g)
(h)
Eq. (2.6a):
(i)

62. (g), (h) and (i) into eq. (2.5b)
(2.25)
Equations (2.24) and (2.25) are:
•Linear
•Second order
•Variable coefficients

63. 2.3 Bessel Differential Equations and Bessel Functions
2.3.1 General form of Bessel Equations
(2.26)
Note the following:
(1) Eq. (2.26) is linear, second order with variable coefficients
(2) A, B, C, D, and n are constants

64. 2.3.2 Solution: Bessel Functions
(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
(2.27)
where
= constant of integration

65. = Bessel function of order n of the first kind
= Bessel function of order n of the second kind
Note the following:
(i) The term
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
(2.28)

66. (3) n is zero or integer, D is imaginary
(2.29)
where
i is imaginary =
= modified Bessel function of order n of the first kind
= modified Bessel function of order n of the second
kind

67. 2.3.3 Form of Bessel Functions
(4) n is neither zero nor a positive integer, D is imaginary
(2.30)
2.3.3 Form of Bessel Functions
and
• Symbols for infinite power series
• The form of each series depends on n
Example:
Bessel function
(2.31)

68. 2.3.4 Special Closed-form Bessel Functions:
(2.32)
and
(2.33)

69. For n = 3/2, 5/2, 7/2, … use Eq. (2. 32) or eq. (2. 33) and
For n = 3/2, 5/2, 7/2, … use Eq. (2.32) or eq. (2.33) and the recurrence equation:
k = 1, 2, 3, …
(2.34)
For the modified Bessel functions, n = 1/2
(2.35)
and
(2.36)

70. 2.3.5 Special Relations for n = 0, 1, 2, …
For n = 3/2, 5/2, 7/2, …. use eq. (2.35) or eq. (2.36) and the recurrence equation
(2.37)
k = 1, 2, 3, …
2.3.5 Special Relations for n = 0, 1, 2, …
(2.38a)
(2.38b)
(2.38c)
(2.38d)

71. 2.3.6 Derivatives and Integrals of Bessel Functions
(2.39)
(2.40)
(2.41)
(2.42)

72. (2.43)
(2.44)
(2.45)
(2.46)
(2.47)

73. (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)
Yn(x)
Kn(x) 1
-  

75. 2.4 Equidimensional (Euler) Equation
(2.49)
Solution:Depends on roots r1 and r2
(2.50)
Three possibilities:
(1) Roots are distinct
(2.51)

76. 2.5 Graphically Presented Fin Solutions to Fin Heat Transfer Rate
(2) Roots are imaginary
(2.52)
(3) One root only
(2.53)
2.5 Graphically Presented Fin Solutions to Fin Heat Transfer Rate
Fin efficiency
(2.18)

77. Fig. 2.16 Fin efficiency of three types of straight fins [5]

78. Fig. 2.17 Fin efficiency of annular fins of constant thickness [5]