Heat Transfer with Fins

Outline • • • • • Heat Transfer with Fins Review previous material What is a fin and why do we use them? Examples of fins Analysis of fins Fin effectiveness 2 Review Parallel Resistances 1 Rtotal T∞ = • Tsurr 1 Rtotal = 1 Rconv + Review Slab with Convection 1 htotal = = As hconv + As hrad 1 = hconv + hrad As Rtotal Q& Figure 3-26 from Çengel, Heat and Mass Transfer T −T First find heat flux from h1, h2, q& = = ∞1 ∞ 2 A 1 +L+ 1 L, k, T∞1, and T∞2 h1 k T2 = T∞ 2 + q& h2 h2 Once heat flux is known, find T1 and T2 T1 = T∞1 − 3 Review Composite Cylindrical Shell Figure 3-6 from Çengel, Heat and Mass Transfer L Rrad Define total heat transfer coefficient, htotal Figure 3-5 from Çengel, Heat and Mass Transfer A is area normal to heat flow q& h1 4 Effect of Insulation Thickness 2500 Review: Adding insulation to a cylinder (or sphere) can actually increase heat transfer! 2000 1500 physical radius Heat loss (Btu/hr) 1000 Maximum heat transfer is at rc = k/h, which may be less than physical radius 500 T∞1 − T∞2 Q& = ⎛r ⎞ ⎛ ⎞ ⎛r ⎞ 1 1 1 1 r 1 + ln⎜ 2 ⎟ + ln⎜ 3 ⎟ + ln⎜ 4 ⎟ + h1 2πr1L 2πk1L ⎜⎝ r1 ⎟⎠ 2πk2 L ⎜⎝ r2 ⎟⎠ 2πk3 L ⎜⎝ r3 ⎟⎠ h2 2πr4 L 5 0 0 1 2 3 4 5 6 7 8 9 10 Radius (in) 6 1 Resistances for Pipe Insulation Review Conduction Shape Factor 12 10 • Simplified analysis – for multidimensional geometries with each surface at a uniform temperature – Use shape factor, S, whose equation is found from tables like Çengel Table 3-7 & = kS(T – T ) – Basic equation: Q 1 2 – S must have dimensions of length 8 Resistance 6 (hr·ft·F/Btu) Insultation Resistance Convection Resistance Total Resistance 4 2 0 0 0.5 1 1.5 2 2.5 3 • Equations for S depend on parameters in the different geometries Insulation Thickness (in) 7 What is a Fin? • A fin is a extended surface to increase area for convection heat transfer Q& = hAs (Ts − T∞ ) • Goal is to increase & As to increase Q 8 Innovative Fin Designs • Important application is cooling of power electronics • How do we analyze fin effectiveness? • Automobile radiator is example of fin – Start with simple 1D fin Figure 3-33 from Çengel, Heat Transfer 9 Figure 3-35 from Çengel, Heat Transfer • S T0 = Tb • F Analysis • Conduction in xdirection • Convection from fin surface • Heat balance over differential volume element – p = perimeter dT ⎤ dT ⎤ ⎡ ⎡ + hpΔx(T − T∞ ) ⎢− kAc dx ⎥ = ⎢− kAc dx ⎥ ⎣ ⎦x ⎣ ⎦ x + Δx 11 Figure 3-34 from Çengel, Heat Transfer 10 What is Fin Shape? • Analysis does not consider shape of cross section, only area and perimeter – Derivation of fin equation assumes constant cross section • Fin may be in the shape of a uniform cylinder, but not a cone • Is fin two dimensional? – Yes, but one-dimensional analysis is accurate if hδ/k < 0.2, where δ = D for circular fins, thickness for rectangular fins 12 2 Fin Equation and Solution Boundary Conditions for C1, C2 θ = C1e mx + C2e − mx • T0 = Tb = base temperature at x = 0 • Define θ = T – T∞ so θb = Tb – T∞ • The fin equation for constant k and Ac is d 2θ hp − m 2θ = 0 m2 = 2 kAc dx • First condition: base temperature is Tb so θ = θb = Tb – T∞ at x = 0 • Alternatives for second condition – Infinitely long fin (requires C1= 0) – Negligible heat transfer from end at x = L – Convection and radiation heat transfer coefficient for end at x = L • The solution to this differential equation is: θ = C1e mx + C2e − mx 13 14 Infinitely Long Fin Zero Heat Transfer at x = L θ = C1e mx + C2 e −mx θ = C1e mx + C2e − mx • C1 = 0 to keep solution finite • At x = 0, θ = θb = requires C2 = θb θ = θb e −mx q& = − k ⇒ T − T∞ = (Tb − T∞ )e dT −x = − k (Tb − T∞ )e dx hp kAc • At x = 0, θ = θb; at x = L, dθ/dx = 0 • Algebraic details at end θ = θb cosh m( L − x) − x hp kAc (− dθ d (T − Tb ) dT = = = mC1e mx − mC2e −mx dx dx dx hp kAc ) Q& x =0 = Ac q& x =0 = kAc (Tb − T∞ ) hp kAc = kAc hp (Tb − T∞ ) cosh mL dθ − m sinh m( L − x) kθb m sinh m( L − x) q& = −k = = −kθb dx cosh mL cosh mL kA θ m sinh mL( L − x) Q& x=0 = Ac q& x =0 = c b = kAc hp (Tb − T∞ ) tanh mL cosh mL 15 16 Hyperbolic Tangent Convection 1 0.9 0.8 tanh x 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.5 1 1.5 2 2.5 3 x 17 • For convection (+ radiation) at x = L use approximation of extra length • Use adjusted length, Lc, to give area Ac • Modified length, Lc = L + Ac/p, will give convection from end Figure 3-39 from Çengel, Heat Transfer 18 3 Problem Solution • A pin (cylindrical) fin has a diameter of 4 mm, a length of 5 cm, and a thermal conductivity of 200 W/m·K. If the heat transfer coefficient is 70 W/m2·K, with a surface temperature of 50oC and an air temperature of 20oC, what is the heat transfer with and without the fin? • Given: k = 200 W/m·K, h = 70 W/m2·K, D = 0.004 m, L = 0.05 m, Ts = 50oC, T∞ & with and without fin = 20oC, Find: Q • Use equation for finite length fin with end correction for convection Q& x=0 = kAc hp (Tb − T∞ ) tanh mLc m = hp kAc Lc = L + Ac p p = πD = π(0.004 m ) = 0.01257 m Ac = πD 2 4 = π(0.004 m )2 4 = 0.00001257 m 2 ( ) Lc = L + Ac p = 0.05 m + 0.00001257m2 (0.01257 m) = 0.051 m hp 70 W m ⋅ K 0.01257 m = = 18.71 m−1 kAc m2 ⋅ K 200 W 0.00001257m2 m= 19 20 Solution II Increase of 40 times? Q& x =0 = kAc hp (Tb − T∞ ) tanh mLc = (50 − 20)K • 70 W m2 ⋅ K (0.01257 m )(0.00001257 m 2 ) 200 W m⋅ K Q& fin = 1.05 W ⎡18.71 (0.051 m )⎤⎥ tanh ⎢ ⎦ ⎣ m • Without fin available area is Ac Q& no fin = Ac h(Tb − T∞ ) = 70 W m2 ⋅ K (0.00001257 m )(30 K ) = 0.0264 W 2 • Increase by factor of 39.64 (effectiveness, ε) • Factor of 40 increase in heat transfer is just for area of fin – A practical installation of small fins like this one would have several fins on a surface – Consider two areas on original surface • Area where pins occur will have increase by factor shown in fin analysis • Area without fins has usual flat surface result – Will consider this difference in determining fin effectiveness 21 Fin Effectiveness Constant Ac Fin Effectiveness • Effectiveness, εfin, is ratio of heat transfer with fin to heat transfer with no fin ε fin = 22 • Infinitely long fin • Fin with end convection Q& fin Q& fin = & Qno fin hAb (Tb − T∞ ) • Ab is fin area at base, same as Ac for fin with constant cross sectional area • Want fin effectiveness greater than one to get additional heat transfer 23 ε ∞fin = ε convfin = Q& x =0 = kAc hp (Tb − T∞ ) Q& x =0 = kAc hp (Tb − T∞ ) tanh mL kAc hp (Tb − T∞ ) Q& x =0 kp = = hAc (Tb − T∞ ) hAc (Tb − T∞ ) hAc kAc hp (Tb − T∞ ) tanh mL kp = tanh mL hAc (Tb − T∞ ) hAc 24 4 What makes a fin effective? ε ∞fin = kp hAc ε convfin = What makes a fin effective? II ⎛ hp ⎞ kp kp L ⎟⎟ tanh mL = tanh⎜⎜ hAc hAc ⎝ kAc ⎠ ε convfin = ⎛ hp ⎞ kp kp tanh mL = tanh ⎜⎜ L ⎟⎟ hAc hAc ⎝ kAc ⎠ • Choose a base case and vary each parameter from 0.1 to 10 times base value • For infinite fin high k, low h and high ratio of p/Ac make a fin effective • For pin fin, p = πD and Ac = πD2/4, so p/Ac = 4/D; small D gives better ε • Effect of variables on fin with convection at end is not clear – Rectangular fin with width, w, and thickness, t, so that Ac = tw and p = 2(t + w) • Base case has k = 200 W/m·K, w = 0.03 m, t = 0.005 m, h = 70 W/m2·K, L = 0.08 m – Base mLc = 1.05 and ε = 28.5 25 26 Effect of Parameters on Rectangular Fin Effectiveness Overall Fin Effectiveness Fin Effectiveness 110 100 Base Case 90 k = 200 W/m·K t = 0.005 m w = 0.03 m 80 70 60 thermal conductivity 2 h = 70 W/m ·K L = 0.08 m mL = 1.05 fin thickness, t fin width, w heat transfer coefficient ( 40 30 no fin 20 10 0 0.1 1 10 Ratio of parameter to base value 27 Fin Efficiency for uniform cross section A fin = Lc p Figure 3-39 from Çengel, Heat Transfer • Compare actual heat transfer to ideal case where entire fin is at base temperature Q& fin η fin = & = Q fin,max Q& fin hA fin (Tb − T∞ ) ) Q& fin h ε fin A fin + Aunfin (Tb − T∞ ) = Q& hA(T − T ) fin length 50 • Original area, A = (area with fins, Afin) + (area without fins, Aunfin) 29 b ∞ Q& fin A fin Aunfin ⎞ ⎛ ⎟ εtotal = & = ⎜⎜ ε fin + Qno fin ⎝ A A ⎟⎠ Figure 3-45 from Çengel, Heat Transfer 28 Fin Efficiency II • Relation to effectiveness Q& fin ε fin hAc (Tb − T∞ ) ε fin Ac η fin = = = & Q fin,max hA fin (Tb − T∞ ) A fin • Recall pin fin problem with D = 0.004 m and Lc = 0.051 m so Afin = pLc = πDLc = π(0.004 m)(0.051 m) = 0.0006409 m2 – Previously showed ε = 39.64 so η = (39.64) (pD2/4)/(0.0006409 m2) = 0.777 30 5 Fin Efficiency Data Fin Efficiency Data II Figure 3-43 from Çengel, Heat Transfer A p = Lc t Figure 3-42 from Çengel, Heat Transfer 31 32 From Çengel, Heat Transfer Electronic Heat Sinks • Designed to protect equipment like power transistors from overheating • Characterized by thermal resistance, R, & = power dissipation = such that Q (Tdevice - T∞)/R • See Table 3-6 in Çengel for examples with R values – First part of table shown on next chart 33 34 Problem Solution • A power transistor that dissipates 120 W has a maximum operating temperature of 70oC. Cooling air is available at 25oC. Are any of the heat sinks on the previous page suitable for this transistor? & = 120 W, T = 70oC, T = • Given: Q surf ∞ 25oC • Find: Do devices have sufficient R Q& = Tsurf − T∞ R ⇒ R= Tsurf − T∞ 70o C − 25o C 1.125o C = = 40 W Q& W • Only the HS 5030 mounted vertically (with R = 0.9oC/W) will satisfy this cooling requirement • The R values for the other heat sinks are too large (1.2oC/W and 5oC/W) 35 36 6 Extra Charts Zero Heat Transfer at x = L • The following charts show the details of the fin equation solution for no convection (dT/dx = 0) at x = L θ = C1e mx + C2e −mx dθ d (T − Tb ) dT = = = mC1e mx − mC2e − mx dx dx dx • At x = 0, θ = θb; at x = L, dθ/dx = dT/dx= 0 dθ dT = dx x = L dx θb = C1 + C2 θb = C1 + C1e C1 = θb 1 + e 2mL x= L = mC1e mL − mC2e −mL = 0 C2 = C1e 2mL 2 mL C2 = C1e 2mL = θb e 2 mL 1 + e 2mL 37 Zero Heat Transfer at x = L II • Combine results from last chart θ = C1e θ= mx + C2e θb 1 + e 2mL − mx e mx + C1 = θb e 2 mL 1 + e 2mL θb θb e 2 mL C2 = 1 + e 2 mL 1 + e 2mL θ e mx + θb e 2mL e − mx e −mx = b 1 + e 2mL • Multiply fraction top and bottom by e-mL θ= θb 1 + e 2mL e mx + θb e 2mL 1 + e 2mL e −mx = θb e − mL e mx + θb e mL e − mx e −mL + e mL 39 38 Zero Heat Transfer at x = L III • Rearrange exponential terms on last chart and introduce hyperbolic cosine, cosh(x) = (ez + e-z)/2 θ= θb e − mL e mx + θb e mL e − mx = θb e m( x− L ) + e − m( x − L ) e −mL + e mL e −mL + e mL 2θ cosh m( L − x) θb cosh m( L − x) = b = 2 cosh mL cosh mL • Final result: θ = θb cosh m( L − x) cosh mL 40 7