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

Typical finned-tube heat exchangers

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

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

Ac(x) dAs(x) T(x) x dx T∞, h

When k = constant,

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

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

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

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

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

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

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

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

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

T(r) T∞, h

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

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

present case: Thus, boundary conditions:

Heat loss from the fin T(r) T∞, h

Fin Performance fin effectiveness fin resistance fin efficiency

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.

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

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.

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

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

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

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

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

Efficiency of annular fins of rectangular profile

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

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

let

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

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

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

0.15 0.95

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

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

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

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

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.

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.

+ 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 = 11.25 W

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

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

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

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

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

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

Properties: Table A.4. air

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