Heat Transfer from Extended Surfaces

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

5. T(x) T∞, h Ac(x) dAs(x) x dx

6. When k = constant,

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

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

9. L Tb T(x) at x = L: 3 cases dx x 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. or Total heat loss by the fin L 1) long fin: Tb P Ac 2) convection tip: dAs dx x 3) adiabatic tip:

12. air Example 3.9 • Find: • Temperature distribution T(x) and heat loss qfwhen 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. Air conductivity at qf Copper: 8.3 W Aluminum alloy: 5.6 W Stainless steel: 1.6 W 1) For a very long fin heat loss: Copper: k = 398 W/m.K Aluminum alloy: k = 180 W/m.K Stainless steel: k = 14 W/m.K

15. or 2) For the adiabatic condition at the tip qf = MtanhmL (long fin: qf = M) To get an accuracy over 99% 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. in terms of excess temperature where When k = const, boundary conditions: when an adiabatic tip is presumed

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

20. Thus, present case: boundary conditions:

23. Heat loss from the fin T(r) T∞, h

24. Fin Performance • fin effectiveness • fin resistance • fin efficiency

25. (rule of a thumb) 1. Fin effectiveness Ac,b: fin cross-sectional area at the base Tb heat loss without fin Ac,b T∞, h fin effectiveness: 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. long fin adia. tip provides upper limit of ef, which is reached as L approaches infinity. 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 Ac,b T∞, h qb: driving potential Thermal resistance due to convection at the exposed base: Rt,b

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

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

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

38. let

39. air Example 3.10 Annual fins Engine cylinder Cross-section (2024 T6 Al alloy) 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. H = 0.15 m t = 6 mm To get hf, use Fig. 3.19. 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: 11.25 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. 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. contact joint (contact) + finned sink base (conduction) + exposed base of finned side (convection) + fins (conv+cond) 2. To reduce the fan power consumption by 50%, Aluminum fined sink: k = 200 W/m.K Lc = 50 mm Thermal circuit = 11.25 W

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