1. Heat Transfer from Extended Surfaces Heat Transfer Enhancement by Fins
Bare surface
Finned surface

2. Typical finned-tube heat exchangers

3. Straight fin of uniform cross section
Straight fin of nonuniform cross section
Annular fin
Pin fin

4. Equation for Extended Surfaces
Ac(x) dx
dAs(x) dx Tb
T∞, h

5. Ac(x)
dAs(x)
T(x) x dx
T∞, h

6. When k = constant,

7. Fins of Uniform Cross-Sectional Area
P: fin perimeter Tb P dx
Ac(x) = constant,
and dAs = Pdx
dAs Ac x

8. excess temperature : q(x) = T(x) - T∞L Tb
T(x) dx x
where
boundary conditions
at x = 0:

9. L Tb
T(x)
at x = L: 3 cases x dx
1) very long fin (L → ∞):
2) convection tip:
3) negligible heat loss: adiabatic tip

10. Temperature distribution
1) long fin:
2) convection tip:
3) adiabatic tip:

11. Total heat loss by the finor L
1) long fin: Tb P Ac
2) convection tip: dx
dAs x
3) adiabatic tip:

12. Example 3.9
air
Find:
Temperature distribution T(x) and heat loss qf when the fin is constructed from: a) pure copper, b) 2024 aluminum alloy, and c) type AISI 316 stainless steel.
Estimate how long the rods must be for the assumption of infinite length to yield an accurate estimate of the heat loss.
Assumption:
very long fin

13. qf 1) For a very long fin heat loss: Copper: k = 398 W/m.K
Air
heat loss:
Copper: k = 398 W/m.K
Aluminum alloy: k = 180 W/m.K
Stainless steel: k = 14 W/m.K
conductivity at
Copper: 8.3 W
Aluminum alloy: 5.6 W
Stainless steel: 1.6 W qf

15. qf = MtanhmL (long fin: qf = M)2)
For the adiabatic condition at the tip
qf = MtanhmL (long fin: qf = M)
To get an accuracy over 99% or
Copper: 0.19 m
Aluminum alloy: 0.13 m
Stainless steel: 0.04 m

16. Fins of Nonuniform Cross-Sectional Area : Annular Fin
T∞, h

17. T(r)
T∞, h

18. When k = const,
in terms of excess temperature
where
boundary conditions: when an adiabatic tip is presumed

19. Solutions to generalized Bessel equations
General Solution:
where
Particular solutions
g : real
n : real
n : zero or integer
and
g : imaginary
n : fractional

20. present case:
Thus,
boundary conditions:

23. Heat loss from the fin
T(r)
T∞, h

24. Fin Performance
fin effectiveness
fin resistance
fin efficiency

25. 1. Fin effectiveness
Ac,b: fin cross-sectional
area at the base Tb
heat loss without fin
Ac,b
T∞, h
fin effectiveness:
(rule of a thumb)
design criteria:
Assume hs are the same for with or without
fin.

26. Ex) long straight fin with uniform cross-
sectional area
In order to get high fin performance
large k material
installation of fins at the lower h side
thin shape

27. provides upper limit of ef, which is reached as L approaches infinity.
long fin
adia. tip
Practically qf for the adiabatic tip reaches 98% of heat transfer when mL = 2.3.
Thus, the fin length longer than L = 2.3/m is not effective.

28. 2. Fin resistance Tb
T∞, h
Ac,b
qb: driving potential
Thermal resistance due to convection at the exposed base: Rt,b

29. qmax: heat loss when the whole fin is assumed at Tb
3. Fin efficiency x Ac
dAs Tb L dx P
qmax: heat loss when the
whole fin is assumed at Tb
Ex) straight fin of uniform cross-sectional
area with an adiabatic tip

30. Errors can be negligible if
For an active tip, the above relation can be used with fin length correction.
rectangular fin: Tb x L t D
pin fin:
Errors can be negligible if or

31. When w >> t, P ~ 2w, w t Lc Ap
Corrected fin profile area

32. Efficiency of straight fins
(rectangular, triangular, and parabolic profile)

33. Efficiency of annular fins of rectangular profile

34. Overall Surface Efficiency
single fin efficiency:
overall efficiency of array of fins:
At : area of fins + exposed portion of the base

36. In the case of press fit:
thermal contact resistance Tb Tc T∞

38. let

39. Example 3.10 Annual fins Engine cylinder Cross-section
(2024 T6 Al alloy)
air
Find:
Increase in heat transfer, Dq = qt – qwo, associated
with using fins

40. 1)
Heat transfer rate
H = 0.15 m
t = 6 mm
hf: known

41. k = 186 at 400 K To get hf, use Fig. 3.19. Parameters: H = 0.15 m
t = 6 mm
Parameters:
2024 T6 Al
k = 186 at 400 K

42. 0.15
0.95

43. H = 0.15 m
t = 6 mm
Without fins
The amount of increase in heat transfer

44. Comments:
Fixed fin thickness : 6 mm, Minimum fin gap : 4 mm  Nmax= H/S =15

45. Fixed fin gap : 4 mm, Minimum fin thickness : 2 mm  Nmax= H/S =25

46. Example 3.11 Hydrogen-air Proton Exchange Membrane (PEM) fuel cell
Without finned heat sink
With finned heat sink
Known:
Dimensions of a fuel cell and finned heat sink
Fuel cell operating temperature
Rate of thermal energy generation: W
Power production: P = 9 W
Relationship between the convection coefficient and the air channel dimensions

47. Find:
Net power of the fuel cell-fan system for no heat sink, Pnet = P – Pf
# of fins N needed to reduce the fan power consumption by 50%
Assumptions:
Steady-state conditions
Negligible heat transfer from the edges of the fuel cell, as well as from the front and back faces of the finned heat sink
1D heat transfer through the heat sink
Adiabatic fin tips
Negligible radiation when the heat sink is in place.

48. Fan power consumption:
P = 9 W
Fuel cell 1.
Fan power consumption:
Volumetric flow rate of cooling air:
The fan consumes more power than is generated by the fuel cell, and the system cannot produce net power.

49. + exposed base of finned side
2. To reduce the fan power consumption by 50%,
Aluminum fined sink:
k = 200 W/m.K
Lc = 50 mm
contact joint
(contact)
+ finned sink base
(conduction)
+ exposed base of finned side
(convection)
+ fins
(conv+cond)
Thermal circuit
= W

50. Lc = 50 mm Aluminum fined sink: k = 200 W/m.K
: 2 sides of the heat sink assembly

51. Lc = 50 mm Aluminum fined sink: k = 200 W/m.K
 N, h are needed to evaluate Rt,b

52. Lc = 50 mm Aluminum fined sink: k = 200 W/m.K For a single fin,
For a fin with an insulated fin tip,

53. Aluminum fined sink:
k = 200 W/m.K
Lc = 50 mm

54. Aluminum fined sink:
k = 200 W/m.K
For N fins,
 N, h are also needed to evaluate Rt,f

55. The total thermal resistance, Rtot
The equivalent fin resistance, Reqiuv

57. Properties: Table A.4. air

58. For N = 22,
11 on top and 11 on the bottom,
For N = 20,
For N = 24,