EXTENDED SURFACES / FINS
Convection: Heat transfer between a solid surface and a moving fluid is governed by the Newton’s cooling law: q = hA(Ts -T). Therefore, to increase the convective heat transfer, one can
Increase the temperature difference (Ts -T) between the surface and the fluid.
Increase the convection coefficient h. This can be accomplished by increasing the fluid flow over the surface since h is a function of the flow velocity and the higher the velocity, the higher the h. Example: a cooling fan.
Increase the contact surface area A. Example: a heat sink with fins. Extended Surface Analysis
Tb P: the fin perimeter Ac: the fin cross-sectional area
x
dT dqx qkA qqxdx x dx xCdx dx
AC is the cross-sectional area dqconv h( dAS )(T T ), where dAS is the surface area of the element dq Energy Balance: qq dqq x dx hdA() T T x xdx conv x dx S dT2 kA dx hP(), T T dx 0 if k, A are all constants. C dx 2 C Extended Surface Analysis (contd….) dT2 hP 2 (),TT 0 A second - order, ordinary differential equation dx kAC
Define a new variable ()xTxT = () , so that 2 d 2222 hP 2 m 00,,() where m Dm dx kAC Characteristics equation with two real roots: + m & - m The general solution is of the form mx mx ()xCeCe12
To evaluate the two constants C12 and C, we need to specify two boundary conditions:
The first one is obvious: the base temperature is known as T(0) = Tb The second condition will depend on the end condition of the tip Extended Surface Analysis (contd...)
For example: assume the tip is insulated and no heat transfer d/dx(x=L)=0
The temperature distribution is given by )( TxT xLm )(cosh b TT b mL)cosh( The fin heat transfer rate is given by dT kAq x )0( hPkA tanh tan mLMmL f C dx C Temperature distribution for fins of different configurations
Case Tip Condition Temp. Distribution Fin heat transfer A Convection heat xLm h xLm )(sinh)()(cosh mL h cosh)(sinh mL mk mk transfer: Mo mL h sinh)(cosh mL mL h sinh)(cosh mL h(L)=-k(d/dx)x=L mk mk
B Adiabatic xLm )(cosh M tanh mL 0 (d/dx)x=L=0 cosh mL C Given temperature: L xLmxLm )(sinh)(sinh)( (cosh mL L ) (L)=L b M b sinh mL 0 sinh mL
mx D Infinitely long fin e M 0 (L)=0
2 hP , mTT kAC
b b ,)0( hPkAMTT bC Example
An Aluminum pot is used to boil water as shown below. The handle of the pot is 20-cm long, 3-cm wide, and 0.5-cm thick. The pot is exposed to room air at 25C, and the convection coefficient is 5 W/m2 C. Question: can you touch the handle when the water is boiling? (k for aluminum is 237 W/m C)
T = 25 C h = 5 W/ m2 C
x
100 C Example (contd...)
We can model the pot handle as an extended surface. Assume that there is no heat transfer at the free end of the handle. The condition matches that specified in the fins Table, case B. Use the following data: h=5 W/ m2 C, P=2W+2t=2(0.03+0.005)=0.07(m), k=237 W/mC, 2 AC=Wt=0.00015(m ), L=0.2(m) 1/2 Therefore, m=(hP/kAC) =3.138, M=(hPkAC)(T b -T)=0.111b=0.111(100-25)=8.325(W) T()x - T coshm (L x ) TTbb cosh mL Tx 25 cosh[3138 . ( 0 . 2 )] , 100 25 cosh(3138 . * 02 . ) Tx()25 62 . 32 *cosh[. 3138 (. 0 2 x )] Example (contd…)
Plot the temperature distribution along the pot handle
100
95 T(x )
90
85 0 0.05 0.1 0.15 0.2 x As shown in the figure, temperature drops off but not very steeply. This is because k of aluminium is fairly high. At the midpoint, T(0.1)=90.4C. At the end T(0.2)=87.3C. Therefore, it should not be safe to touch the end of the handle. The end condition is insulated, hence the gradient is zero. Example (contd...)
The total heat transfer through the handle can be calculated also. qf =Mtanh(mL)=8.325tanh[(3.138)(0.2)]=4.632 W
If a stainless steel handle is used instead, what will happen? For a stainless steel, the thermal conductivity k=15 W/m°C, which is much less compared to aluminium. Using the same process parameter as before:
2/1 hP m ,47.12 hPkAM C 0281.0 kAC Example (contd...) )( TxT xLm )(cosh b TT cosh mL xT xL )](47.12cosh[3.1225)(
100
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 of the thermal conductivity of the material in the temperature distribution in a fin. Fin Design
T Total heat loss: qf =Mtanh(mL) for an Tb adiabatic fin, or qf =Mtanh(mLC) if there is convective heat transfer at the tip
hP where mT= , and M= hPkACCbb hPkA (T ) kAc Use the thermal resistance concept:
()TTb qmfbhPkAC tanh(L )(TT ) Rtf,
where Rtf, is the thermal resistance of the fin. For a fin with an adiabatic tip, the fin resistance can be expressed as
()TTb 1 Rtf, q f hPkAC [tanh(mL )] Fin Effectiveness
How effective a fin can enhance heat transfer is characterized by the fin effectiveness f : Ratio of fin heat transfer and the heat transfer without the fin. For an adiabatic fin:
qqff hPkAC tanh(mL ) kP f tanh(mL ) qhATTCb() hA C hA C If the fin is long enough, mL>2, tanh(mL) 1, it can be considered an infinite fin (case D of table3.4) kP k P f hACC h A
In order to enhance heat transfer, f 1.
However, f 2 will be considered justifiable
If f <1 then we have an insulator instead of a heat fin Fin Effectiveness (contd...)
kP k P f hACC h A
To increase f , the fin’s material should have higher thermal conductivity, k. It seems to be counterintuitive that the lower convection coefficient, h, the higher f . But it is not because if h is very high, it is not necessary to enhance heat transfer by adding heat fins. Therefore, heat fins are more effective if h is low. Observation: If fins are to be used on surfaces separating gas and liquid. Fins are usually placed on the gas side. (Why?) Fin Effectiveness (contd...)
P/AC should be as high as possible. Use a square fin with a dimension of W by W as an example: P=4W, AC=W2, P/AC=(4/W). The smaller W, the higher the P/AC, and the higher f.
Conclusion: It is preferred to use thin and closely spaced (to increase the total number) fins. Fin Effectiveness (contd...)
The effectiveness of a fin can also be characterized as
qqff()/ TTR bt , fRth, f qhATTCb()()/ TTRb th,, R tf It is a ratio of the thermal resistance due to convection to the thermal resistance of a fin. In order to enhance heat transfer, the fin's resistance should be lower than that of the resistance due only to convection. Fin Efficiency
q f Define Fin efficiency: f qmax where qmax represents an idealized situation such that the fin is made up of material with infinite thermal conductivity. Therefore, the fin should be at the same temperature as the temperature of the base. qhATTmax f ()b Fin Efficiency (contd…)
For infinite k T(x) T b x x Total fin heat transfer qf Ideal heat transfer qmax Real situation Ideal situation Fin Efficiency (cont.) Use an adiabatic rectangular fin as an example: q f MmLtanh hPkAcb()tanh T T mL f qhATTmax fb() hPLTT () b tanhmL tanh mL (see Table 3.5 for of common fins) hP mL f L kAc The fin heat transfer: qqffmax ffb hATT ( ) TTbb TT 1 qRftf, where , 1/( ffhA ) R tf, ffhA Thermal resistance for a single fin. 1 As compared to convective heat transfer: R tb, hAb In order to have a low er resistance as that is required to enhance heat transfer: RRtb,, t f or A b f A f Overall Fin Efficiency Overall fin efficiency for an array of fins: Define terms: Ab : base area exposed to coolant qf Af : surface area of a single fin qb At: total area including base area and total finned surface, At =Ab+NA f N: total number of fins Overall Fin Efficiency (contd…) qqNqhATTtb f bb() NhATT ffb () h[( Atf NA ) N ffb A ]( T T ) h [ A tffb NA (1 )]( T T ) NAf hAtfbOtb[1(1)]()() T T hA T T At NAf Define overall fin efficiency: Of 1 (1 ) At Heat Transfer from a Fin Array TTb 1 qhATTttOb ( ) where RtO, RhAtO, t O Compare to heat transfer without fins 1 qhATT()( hANAT )() T bbb, fbhA where Abf, is the base area (unexposed) for the fin To enhance heat transfer AAtO That is, to increase the effective area O At . Thermal Resistance Concept L1 t A=Ab +NAb,f Rb=t/(k b A) T 1 T1 T2 Tb T T hAR )/(1 R1=L 1/(k 1 A) ,Ot Ot TT11 TT T2 Tb q R RRR1,btO