EXTENDED SURFACES / FINS Convection: Heat transer etween a solid surace and a moving luid is governed y the Newton s cooling law: q = ha(t s -T ). Thereore, to increase the convective heat transer, one can Increase the temperature dierence (T s -T ) etween the surace and the luid. Increase the convection coeicient h. This can e accomplished y increasing the luid low over the surace since h is a unction o the low velocity and the higher the velocity, the higher the h. Example: a cooling an. Increase the contact surace area A. Example: a heat sink with ins.
Extended Surace Analysis T P: the in perimeter A c : the in cross-sectional area q x x = ka dt qx+ dx= qx + C dx dqx dx dx A C is the cross-sectional area dqconv = h( das )( T T ), where da S is the surace area o the element dqx Energy Balance: qx = qx+ dx+ dqconv = qx + dx + hdas ( T T ) dx ka dt 2 C dx + hp( T T ) dx = 0, 2 dx i k, A are all constants. C
Extended Surace Analysis (contd.) 2 dt hp ( dx ka T T 2 ) = 0, A second - order, ordinary dierential equation C Deine a new variale θ( x) = T( x) T, so that 2 d θ 2 2 hp 2 2 m θ = 0, where m =, ( D m ) θ = 0 2 dx ka Characteristics equation with two real roots: + m & - m The general solution is o the orm mx θ( x) = C e + C e 1 2 mx To evaluate the two constants C and C, we need to speciy two oundary conditions: C The irst one is ovious: the ase temperature is known as T(0) = T The second condition will depend on the end condition o the tip 1 2
Extended Surace Analysis (contd...) For example: assume the tip is insulated and no heat transer dθ/dx(x=l)=0 The temperature distriution is given y Tx ( )- T θ cosh ml ( x) = = T T θ cosh ml The in heat transer rate is q ka dt = C ( x = 0) = hpkac tanh ml = Mtanh ml dx These results and other solutions using dierent end conditions are taulated in Tale 3.4 in HT textook, p. 118.
Temperature distriution or ins o dierent conigurations Case Tip Condition Temp. Distriution Fin heat transer A Convection heat cosh m( L x) + ( h ) sinh m( L x) sinh ml + ( h ) cosh ml transer: mk Mθ mk o hθ(l)=-k(dθ/dx) h x=l cosh ml + ( ) sinh ml cosh ml + ( h ) sinh ml mk mk B C Adiaatic (dθ/dx) x=l =0 Given temperature: θ(l)= θ L cosh m( L x) Mθ 0 tanh ml cosh ml ( θ L )sinh m( L x) + sinh m( L x) (cosh ml θ L ) θ θ Mθ0 sinh ml sinh ml D θ T T Ininitely long in θ(l)=0 θ = θ (0), = T m 2 T, hp ka C M mx e M θ 0 = hpka C θ
Example An Aluminum pot is used to oil water as shown elow. The handle o the pot is 20-cm long, 3-cm wide, and 0.5-cm thick. The pot is exposed to room air at 25 C, and the convection coeicient is 5 W/m 2 C. Question: can you touch the handle when the water is oiling? (k or aluminum is 237 W/m C) T = 25 C h = 5 W/ m 2 C x 100 C
Example (contd...) We can model the pot handle as an extended surace. Assume that there is no heat transer at the ree end o the handle. The condition matches that speciied in the ins Tale, case B. h=5 W/ m 2 C, P=2W+2t=2(0.03+0.005)=0.07(m), k=237 W/m C, A C =Wt=0.00015(m 2 ), L=0.2(m) Thereore, m=(hp/ka C ) 1/2 =3.138, M= (hpka C )(T -T )=0.111θ =0.111(100-25)=8.325(W) T( x) - T θ cosh m( L x) = = T T θ cosh ml T 25 cosh[ 3138. ( 0. 2 x)] =, 100 25 cosh( 3138. * 0. 2) Tx ( ) = 25 + 62. 32 *cosh[ 3138. ( 0. 2 x)]
Example (contd ) Plot the temperature distriution along the pot handle 100 T( x) 95 90 85 0 0.05 0.1 0.15 0.2 x As shown, temperature drops o very quickly. At the midpoint T(0.1)=90.4 C. At the end T(0.2)=87.3 C. Thereore, it should not e sae to touch the end o the handle
Example (contd...) The total heat transer through the handle can e calculated also. q =Mtanh(mL)=8.325*tanh(3.138*0.2)=4.632(W) Very small amount: latent heat o evaporation or water: 2257 kj/kg. Thereore, the amount o heat loss is just enough to vaporize 0.007 kg o water in one hour. I a stainless steel handle is used instead, what will happen: For a stainless steel, the thermal conductivity k=15 W/m C. Use the same parameter as eore: m hp kac 1/ 2 = 12.47, M = C = hpka = 0.0281
Example (contd...) T ( x) T T T cosh m( L = cosh ml x) T ( x) = 25 + 12.3cosh[12.47( L 100 x)] 75 Tx () 50 25 0 0 0.05 0.1 0.15 0.2 x Temperature at the handle (x=0.2 m) is only 37.3 C, not hot at all. This example illustrates the important role played y the thermal conductivity o the material in terms o conductive heat transer.
Fins-2 I the pot rom previous lecture is made o other materials other than the aluminum, what will e the temperature distriution? Try stainless steel (k=15 W/m.K) and copper (385 W/m.K). Recall: h=5w/m 2 C, P=2W+2t=2(0.03+0.005)=0.07(m) A C =Wt=0.00015(m 2 ), L=0.2(m) Thereore, m ss =(hp/ka C ) 1/2 =12.47, m cu =2.46 M ss = (hpk ss A C ) (T -T )=0.028(100-25)=2.1(W) M cu = (hpk ss A C ) θ =0.142(100-25)=10.66(W) For stainless steel, Tss ( x) - T θ cosh m( L x) = = T T θ cosh ml Tss 25 cosh[ 12.47( 0. 2 x)] =, 100 25 cosh( 12.47 * 0. 2) T ( x) = 25 + 12. 3*cosh[ 12.47( 0. 2 x)] ss
Fins-2 (contd...) For copper, T cu Tcu ( x) - T θ cosh m( L x) = = T T θ cosh ml cosh[ 2.46( 02. x)] =, cosh( 2.46 * 02. ) T ( x) = 25 + 66. 76 * cosh[ 2.46( 0. 2 x)] cu 25 100 25 100 T( x) T ss ( x) T cu ( x) 95 90 85 80 75 copper aluminum stainless steel 0 0.04 0.08 0.12 0.16 0.2 x
Fins-2 (contd...) Inside the handle o the stainless steel pot, temperature drops quickly. Temperature at the end o the handle is 37.3 C. This is ecause the stainless steel has low thermal conductivity and heat can not penetrate easily into the handle. Copper has the highest k and, correspondingly, the temperature inside the copper handle distriutes more uniormly. Heat easily transers into the copper handle. Question? Which material is most suitale to e used in a heat sink?
Fins-2 (contd...) How do we know the adiaatic tip assumption is good? Try using the convection heat transer condition at the tip (case A in ins tale) We will use the aluminum pot as the example. h=5 W/m 2.K, k=237 W/m.K, m=3.138, M=8.325W Long equation T ( x) T θ cosh[ ml ( x)] + ( h/ mk)sinh[ ml ( x)] = = T T θ cosh ml + ( h / mk)sinh ml - cosh[ 3138. ( 0. 2 x)] + 0. 00672sinh[ 3138. ( 0. 2 x)] = cosh( 0. 6276) + 0. 00672sinh( 0. 6276) Tx ( ) = 25 + 62. 09{cosh( 0. 6276 3138. x) + 0. 00672sinh( 0. 6276 3138. x)}
Fins-2 (contd.) T( x) T c ( x) 100 96.25 92.5 T: adiaatic tip T c : convective tip 88.75 85 0 0.04 0.08 0.12 0.16 0.2 x T(0.2)=87.32 C T c (0.2)=87.09 C Note 1: Convective tip case has a slightly lower tip temperature as expected since there is additional heat transer at the tip. Note 2: There is no signiicant dierence etween these two solutions, thereore, correct choice o oundary condition is not that important here. However, sometimes correction might e needed to compensate the eect o convective heat transer at the end. (especially or thick ins)
Fins-2 (contd...) In some situations, it might e necessary to include the convective heat transer at the tip. However, one would like to avoid using the long equation as descried in case A, ins tale. The alternative is to use case B instead and accounts or the convective heat transer at the tip y extending the in length L to L C =L+(t/2). With convection L C =L+t/2 Insulation t Original in length L t/2 L Then apply the adiaatic condition at the tip o the extended in as shown aove.
Fins-2 (contd...) Use the same example: aluminum pot handle, m=3.138, the length will need to e corrected to L C =l+(t/2)=0.2+0.0025=0.2025(m) Tcorr ( x) - T θ cosh ml ( c x) = = T T θ cosh ml T corr cosh[ 3138. ( 0. 2025 x)] =, cosh( 3138. * 0. 2025) T ( x) = 25 + 62. 05*cosh[ 3138. ( 0. 2025 x)] corr 25 100 25 c
Fins-2 (contd...) T( x) T c ( x) 100 96.25 92.5 T(0.2)=87.32 C T c (0.2)=87.09 C T corr (0.2025)=87.05 C T corr ( x) 88.75 85 0 0.04 0.08 0.12 0.16 0.2 slight improvement over the uncorrected solution
Correction Length The correction length can e determined y using the ormula: L c =L+(A c /P), where A c is the cross-sectional area and P is the perimeter o the in at the tip. Thin rectangular in: A c =Wt, P=2(W+t) 2W, since t << W L c =L+(A c /P)=L+(Wt/2W)=L+(t/2) Cylindrical in: A c =(π/4)d 2, P= πd, L c =L+(A c /P)=L+(D/4) Square in: A c =W 2, P=4W, L c =L+(A c /P)=L+(W 2 /4W)=L+(W/4)
R( ml) Optimal Length o a Fin In general, the longer the in, the higher the heat transer. However, a long in means more material and increased size and cost. Question: how do we determine the optimal in length? Use the rectangular in as an example: q = M tanh ml, or an adiaatic tip in 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 ml ( q ) = M, or an ininitely long in Their ratio: R(mL)= q ( q ) = tanh ml Note: heat transer increases with ml as expected. Initially the rate o change is large and slows down drastically when ml> 2. R(1)=0.762, means any increase eyond ml=1 will increase no more than 23.8% o the in heat transer.
Temperature Distriution For an adiaatic tip in case: R θ T T cosh m( L x) = = T T cosh ml High ΔT, good in heat transer Use m=5, and L=0.2 as an example: Low ΔT, poor in heat transer 1 1 R θ ( x) 0.8 0.648054 0.6 0 0.05 0.1 0.15 0.2 0 x 0.2
Correction Length or a Fin with a Non-adiaatic Tip The correction length can e determined y using the ormula: L c =L+(A c /P), where A c is the cross-sectional area and P is the perimeter o the in at the tip. Thin rectangular in: A c =Wt, P=2(W+t) 2W, since t << W L c =L+(A c /P)=L+(Wt/2W)=L+(t/2) Cylindrical in: A c =(π/4)d 2, P= πd, L c =L+(A c /P)=L+(D/4) Square in: A c =W 2, P=4W, L c =L+(A c /P)=L+(W 2 /4W)=L+(W/4)
Fin Design T T Total heat loss: q =Mtanh(mL) or an adiaatic in, or q =Mtanh(mL C ) i there is convective heat transer at the tip hp where m=, and M= hpkacθ = hpka C( T T ) ka Use the thermal resistance concept: q = hpka tanh( ml)( T T ) = t, C c ( T T ) R where R is the thermal resistance o the in. For a in with an adiaatic tip, the in resistance can e expressed as R t, ( T T ) 1 = = q hpka [tanh( ml)] C t,
Fin Eectiveness How eective a in can enhance heat transer is characterized y the in eectiveness ε : Ratio o in heat transer and the heat transer without the in. For an adiaatic in: ε q q hpka tanh( ml) kp q ha ( T T ) ha ha C = = = = C C C I the in is long enough, ml>2, tanh(ml) 1, it can e considered an ininite in (case D o tale3.4) kp k P ε = hac h AC In order to enhance heat transer, ε > 1. However, I 2 will e considered justiiale ε <1 then we have an insulator instead o a heat in ε tanh( ml)
Fin Eectiveness (contd...) ε kp k P = hac h AC To increase ε, the in s material should have higher thermal conductivity, k. It seems to e counterintuitive that the lower convection coeicient, h, the higher ε. But it is not ecause i h is very high, it is not necessary to enhance heat transer y adding heat ins. Thereore, heat ins are more eective i h is low. Oservation: I ins are to e used on suraces separating gas and liquid. Fins are usually placed on the gas side. (Why?)
Fin Eectiveness (contd...) P/AC should e as high as possile. Use a square in with a dimension o W y W as an example: P=4W, AC=W2, P/AC=(4/W). The smaller W, the higher the P/AC, and the higher ε. Conclusion: It is preerred to use thin and closely spaced (to increase the total numer) ins.
Fin Eectiveness (contd...) The eectiveness o a in can also e characterized as ε q q ( T T )/ R R = = = = q ha ( T T ) ( T T )/ R R t, th, C t, h t, It is a ratio o the thermal resistance due to convection to the thermal resistance o a in. In order to enhance heat transer, the in's resistance should e lower than that o the resistance due only to convection.
Fin Eiciency Deine Fin eiciency: η = where q represents an idealized situation such that the in is made up max o material with ininite thermal conductivity. Thereore, the in should e at the same temperature as the temperature o the ase. qmax = ha ( T T ) q q max
Fin Eiciency (contd ) T(x)<T or heat transer to take place For ininite k T(x)=T, the heat transer is maximum T x x Total in heat transer q Real situation Ideal heat transer q max Ideal situation
Fin Eiciency (cont.) Use an adiaatic rectangular in as an example: η q M tanh ml hpkac ( T T ) tanh ml = = = q ha ( T T ) hpl( T T ) max tanh ml = = hp L ka c tanh ml ml max t, 1/( η ha ) Rt, (see Tale 3.5 or η o common ins) The in heat transer: q = η q = η ha ( T T ) q T T T T = =, where Thermal resistance or a single in. R t, t, 1 ha 1 As compared to convective heat transer: R t, = ha In order to have a low er resistance as that is required to enhance heat transer: R > R or A < η A = η
Overall Fin Eiciency Overall in eiciency or an array o ins: q q Deine terms: A : ase area exposed to coolant A : surace area o a single in A t : total area including ase area and total inned surace, A t =A +NA N: total numer o ins
Overall Fin Eiciency (contd ) q = q + Nq = ha( T T ) + Nη ha ( T T ) t = h[( A NA ) + Nη A ]( T T ) = h[ A NA (1 η )]( T T ) t t NA = hat[1 (1 η)]( T T) = ηohat( T T ) A t NA Deine overall in eiciency: ηo = 1 (1 η) A t
Heat Transer rom a Fin Array T T q = haη ( T T ) = where R = t t O t, O RtO, hatη O Compare to heat transer without ins 1 q= ha( T T) = h( A + NA, )( T T) = ha where A is the ase area (unexposed) or the in, To enhance heat transer Aη That is, to increase the eective area η. t O >> A OA t 1
L 1 t Thermal Resistance Concept A=A +NA, R =t/(k A) T 1 T T 1 T 2 T T R 1 =L 1 /(k 1 A) R = 1/( haη ) t, O t O T 2 T q T1 T T1 T = = R R + R + R 1 t, O