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EXTENDED SURFACES - FINS The Function of Fins • Increase heat transfer rate for a fixed surface temperature, or • Lower surface temperature for a fixed heat transfer rate Newton's law of cooling Q& =aS (T − T ) (3.63) s s s ∞ • “Extending” S increases Q& for a fixed T s Rewrite (3.63) s s T =T + s ∞ Q& s aS s • “Extending” surface at fixed Q& s lowers Ts Based on presentations by professor Latif M. Jiji, The City College of New York • A surface is “extended” by adding fins • Examples of fins: • Thin rods on the condenser in back of refrigerator. • Honeycomb surface of a car radiator. • Corrugated surface of a motorcycle engine. • Coolers of PC boards. Types of Fins (a) constant area (b) variable area straight fin straight fin Fig. 3.12 (c) pin fin (d) annular fin Fin terminology and types • Fin base (pata žebra) • Fin tip (konec žebra) • Straight fin: Fig. 3.12 (a) and (b). • Variable cross-sectional area fin: Fig. 3.12 (b), (c) and • • (d). Spine or a pin fin: Fig. 3.12 (c). Annular or cylindrical: Fig. 3.12 (d). Heat Transfer and Temperature r Distribution in Fins • Heat flow through a fin is axial and lateral (2-D) δ T∞ x T h,TT∞ α, ∞ Fig. 3.13 • Temperature distribution is 2-D r δ • Note temperature profile, Fig. 3.13 T∞ x T h,TT∞ α, ∞ Fig. 3.13 The Fin Approximation Neglect temperature variation in the lateral direction r and assume uniform temperature at any section: T ≈ T(x) • Criterion for justifying this approximation: Biot number = Bi = αδ /λ << 1 Bi < 0.1 (3.64) • Meaning of the Biot number: δ/λ internal resistance Bi = = 1/α external resistance The Fin Heat Equation • Objective: Determine the heat transfer rate from a fin. Need the temperature distribution • Procedure: Formulate the fin heat equation: Conservation of energy for element dx. • Select an origin and a coordinate axis x • Assume Bi < 0.1: T ≈ T(x) • Conservation of energy for the element for steady state and no energy generation: E& in = E& out Energy in by conduction = Q& x Energy out by conduction = Q& x + (dQ& x /dx)dx (3.65) Energy out by convection = dQ& conv Neglect radiation: = Q& (a) x Eout = Q& x + dQ& x dx +dQ& conv dx (b) (a) and (b) into (3.65) d Q& x dx dx + dQ& conv =0 dS=P.dx dA s C P qx &c dq dQ conv qx + dqx dx dx (b) (c) Fourier's law and Newton’s law: dT & Q = − λA x dx (d) dQ& conv = α (T − T )P.dx ∞ (e) α is a surface average heat transfer coefficient A = cross-sectional conduction area dS = surface area of element through which heat is convected (d) and (e) into (c) d ⎡ dT ⎤ λA ⎥dx − α(T − T∞ )Pdx = 0 ⎢ dx ⎣ dx ⎦ (f) Assume: constant λ d 2T αP − (T − T∞ ) = 0 2 λA dx (3.66) Or assuming T∞ constant and replacing T with T − T∞ d 2 (T - T ) αP ∞ − (T − T∞ ) = 0 2 λA dx (3.66) • Equation (3.66) is the heat equation for fins Boundary Conditions Two B.C. are needed Determination of Fin Heat Transfer Rate Q& o • Conservation of energy applied to a fin at steady state: qQ& fo = conduction at the base = convection at the surface • Two methods to determine Q& o (1) Conduction at the base: Fourier's law d (T − T∞ ) & Qo = − λA dx x =0 (3.68) (2) Convection at the fin surface: Newton's law Q& o = ∫ α[T(x) − T∞ ] P.dx S (3.69) • Modify (3.69) to include heat exchange by radiation • Introducing a new parameter fin parameter m αP mL = L λA Rewrite (3.66) d 2 (T - T ) 2 ∞ −m (T − T ) = 0 ∞ dx 2 (3.70) ` • Equation (3.73) is valid for: (1) Steady state (2) Constant λ (3) No energy generation (4) No radiation (5) Bi << 1 (6) Constant fin area (7) Constant ambient temperature T∞ Solution d 2 (T - T ) 2 ∞ −m (T − T ) = 0 ∞ dx 2 (3.70) Solution to (3.70) is T(x) − T∞ = C1e mx + C 2 e − mx (3.71) C1 and C are integration constants. They depend on: ` • Location of the origin • Direction of coordinate axis x ` • Two boundary conditions Special Cases Consider 3 cases of constant area fins (i) Specified temperature at base, semi-infinite fin (ii) Finite fin, specified temperature at the base, insulated tip (iii)Finite fin, specified temperature at the base, heat transfer at the tip • • d 2 (T - T ) 2 ∞ −m (T − T ) = 0 Fin equation: (3.70) ∞ dx 2 Temperature solution: (3.71) • Objective: To determine: T(x) − T∞ = C1e mx + C 2 e − mx (1) The temperature distribution in the fin T ≈ T(x) (2) The heat transfer rate (tepelný tokQ& o ) Case (i): Semi-infinite fin with specified temperature at the base α,h, TT∞ 0 To • The base is at temperature To • Ambient temperature is T∞ • Fin equation: (3.70) • Solution: (3.71) x ∞ α, h, T∞ ∞ Ac A Fig. 3.16 T(x) − T∞ = C1e mx + C 2 e − mx • B.C. are: CP (3.71) T(0) =To , T(0) − T∞ = T0 − T∞ (a) T(∞) =T∞ , T( ∞ ) − T∞ = 0 (b) B.C. (b) 0 = C1 ⋅ ∞ + C 2 ⋅ 0, B.C. (a) ∴ C1 = 0 C 2 = T0 − T∞ T(x) − T∞ = (To − T∞ )exp( −mx) Θ= T(x) − T T −T o ∞ ∞ = exp( −mx) (3.72) • Fin heat transfer Q& o : d (T − T∞ ) & Qo = − λA dx x =0 = λAm(T0 − T∞ ) Q& o = λAα P (T0 − T∞ ) (3.73) Q& o = (T0 − T∞ ) λAα P Obviously, no dependence on the fin length Case (ii): Finite length fin with specified temperature at the base and insulated tip Same as Case (i) except the tip is insulated. ∂T & Q x = L = − λS ∂x =0 x=L 0 To x α,T h, T∞ ∞ x=L h, T∞∞ α,T Fig. 3.17 =0 General solution: C ht Tip B.C. d (T − T∞ ) dx P T(x) − T∞ = C1e mx + C 2 e − mx Ac • Two B.C. - fin base (specified temperature) and tip zero heat transfer rate give C1 and C2 , About hyperbolic functions: sinhx e x − e − x ex + e−x ex − e−x = x ; tghx = ; coshx = sinhx = coshx e + e − x 2 2 Solution: T ( x ) − T∞ cosh m( L - x ) = Θ= cosh mL To − T∞ Tip temperature (x=L): Similarly heat transfer rate (tepelný tok): TL − T∞ 1 = ΘL ≡ T0 − T∞ cosh mL d (T − T∞ ) & Qo = − λA dx sinh mL & Qo = λAm(T0 − T∞ ) cosh ml Q& o = tgh (mL ) λAα P (T0 − T∞ ) x =0 Compare with the semi-infinite fin: Q& o = (T0 − T∞ ) λAα P ΘL ≡ TL − T∞ = 0,014 T0 − T∞ TL − T∞ = 0,014.(T0 − T∞ ) Case (iii): Finite length fin with specified temperature at the base and heat transfer at the tip 0 To x α,T h, T∞ ∞ α,T∞ P ht h, T∞∞ α,T Ac Fig. 3.17 T ( x ) − T∞ cosh[m ( L − x )] + (α L / mλ )sinh[m ( L − x )] = Θ≡ T0 − T∞ cosh mL + (α L / mλ )sinh mL C Corrected Length Lc • Fins with insulated tips have simpler solutions than fins with convection at the tip • Simplified model: Assume insulated tip and compensate by increasing the length by ΔLc • The corrected length Lc is Lc = L + ΔLc • The correction increment ΔLc = t/2 fin half width • Error negligible if αt ≤ 0,0625 λ Fin Effectiveness (efektivnost) εf Fin Efficiency (účinnost) ηf Fin performance is described by two parameters: Fin Effectiveness εf : Measures heat transfer enhancement due to fin addition. Q& o Q& o & o = λAm(T0 − T∞ ) sinh mL Q = Defined as ε f = cosh ml & αA(To − T∞ ) Qbase Ratio of heat transfered by the fin and heat transfered from the original area A of the fin Qbase = αA(To − T∞ ) A For a semi-infinite fin: εf = λAαP (To − T∞ ) λP = αA(To − T∞ ) αA How to increase fin effectivness: • high conductivity material λ • high perimeter to area ratio P/A – thin fin and close each to other • use fin where heat transfer coefficient α is low εf must be >2, otherwise don’t use the fin. Fin Efficiency η f : Compares heat transfer from a fin with the maximum heat that the fin can transfer. Defined as Q& o (3.89) ηf = Q& o,max Q& o,max = heat transfer from fin if its entire surface is at the base temperature Q& o,max = αS f (To − T∞ ) S f = total fin surface area Introduce the above into (3.89) Q& o ηf = αS f (To − T∞ ) (3.90) Total Fin Efficiency (účinnost) ηtot ηTOT = Q& TOT Q& TOT,max Q& TOT = αSTOT (To − T∞ ) Sf So Total finned surface (including surface between individual fins) STOT = S f + S o STOT Total heat transfer rate from the total finned surface of an actual finned surface Q& TOT Q& TOT = αS f η f (To − T∞ ) + αSo (To − T∞ ) Q& TOT,max When the entire finned surface is at the base temperature (3.91) After some rearrangement of (3.91) Sf ⎡ ⎤ & ( QTOT = αSTOT ⎢1 − 1 − η f )⎥(To − T∞ ) ⎣ STOT ⎦ Taking the definition of the total efficiency Q& TOT ηTOT = ⇒ Q& TOT = αSTOT ηTOT (To − T∞ ) αSTOT (To − T∞ ) we come to the definition of a total efficiency ηTOT = 1 − Sf STOT (1 − η f ) Practical procedure how to calculate heat transfer rate from a finned surface 1. Specify efficiency of an individual fin ηf using graphs 2. STOT Calculate total surface of fins and= S f + So “free” space between fins 3. η TO 4. Calculate total S f efficiency ( ηTOT = 1 − 1 −η f ) STOT Calculate maximum heat transfer rate Q& TOT,max = αS TOT (To − T∞ ) 5. Calculate actual heat transfer rate Q& TOT = ηTOT Q& TOT,max So Sf Graphs of fin efficiency Fin efficiency for three types of straight fins Graphs of fin efficiency Fin efficiency for annular fins