Extended Surfaces/FinsExtended Surface AnalysisExtended Surface Analysis (cont.)Extended Surface Analysis (cont.)Temperature distribution for fins of different configurationsExampleExample (cont.)Slide 8Slide 9Slide 10Extended Surfaces/FinsConvection: 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 AnalysisxTbq kAdTdxx Cq qdqdxdxx dx xx dq h dA T Tconv S ( )( ), where dA is the surface area of the elementSAC is the cross-sectional areaEnergy Balance: if k, A are all constants.xCq q dq qdqdxdx hdA T TkAd Tdxdx hP T T dxx dxconv xxSC        ( )( ) ,220P: the fin perimeterAc: the fin cross-sectional areaExtended Surface Analysis (cont.) d TdxhPkAT Tx T x TddxmhPkAD mx C e C eCCmx mx22222 2 21 200 0       ( ) ,( ) ( ) ,, , ( )( ), A second - order, ordinary differential equationDefine a new variable = so that where mCharacteristics equation with two real roots: + m & - mThe general solution is of the formTo evaluate the two constants C and C we need to specify two boundary conditions: The first one is obvious: the base temperature is known as T(0) = TThe second condition will depend on the end condition of the tip21 2b Extended Surface Analysis (cont.)For example: assume the tip is insulated and no heat transferd/dx(x=L)=0The temperature distribution is given by-The fin heat transfer rate isThese results and other solutions using different end conditions aretabulated in Table 3.4 in HT textbook, p. 118.T x TT Tm L xmLq kAdTdxx hPkA mL M mLb bf C C( )cosh ( )cosh( ) tanh tanh    0the following fins tableTemperature distribution for fins of different configurationsCase Tip Condition Temp. Distribution Fin heat transfer A Convection heat transfer: h(L)=-k(d/dx)x=L mLmkhmLxLmmkhxLmsinh)(cosh)(sinh)()(cosh MmLmkhmLmLmkhmLsinh)(coshcosh)(sinh B Adiabatic (d/dx)x=L=0 mLxLmcosh)(cosh  mLMtanh C Given temperature: (L)=L mLxLmxLmbLsinh)(sinh)(sinh)(  mLmLMbLsinh)(cosh D Infinitely long fin (L)=0 mxe M bCbbChPkAMTTkAhPmTT,)0(,2Note: This table is adopted from Introduction to Heat Transfer by Frank Incropera and David DeWittExampleAn 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 25C, 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) 100 CT = 25 Ch = 5 W/ m2 CxExample (cont.)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. h=5 W/ m2 C, P=2W+2t=2(0.03+0.005)=0.07(m), k=237 W/m C, AC=Wt=0.00015(m2), L=0.2(m)Therefore, m=(hP/kAC)1/2=3.138, M=(hPkAC)(Tb-T)=0.111b=0.111(100-25)=8.325(W)T x TT Tm L xmLT xT x xb b( )cosh ( )coshcosh[ . ( . )]cosh( . * . ),( ) . * cosh[ . ( . )]-   25100 253138 0 23138 0 225 62 32 3138 0 2Example (cont.)Plot the temperature distribution along the pot handle0 0.05 0.1 0.15 0.2859095100T( )xxAs shown, temperature drops off very quickly. At the midpointT(0.1)=90.4C. At the end T(0.2)=87.3C.Therefore, it should not be safe to touch the end of the handleExample (cont.)The total heat transfer through the handle can be calculated also. qf=Mtanh(mL)=8.325*tanh(3.138*0.2)=4.632(W)Very small amount: latent heat of evaporation for water: 2257 kJ/kg. Therefore, the amount of heat loss is just enough to vaporize 0.007 kg of water in one hour.If 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 before:0281.0,47.122/1CChPkAMkAhPmExample (cont.))](47.12cosh[3.1225)(cosh)(cosh)(xLxTmLxLmTTTxTb0 0.05 0.1 0.15 0.20255075100T x( )xTemperature at the handle (x=0.2 m) is only 37.3 °C, not hot at all. This example illustrates the important role played by the thermal conductivity of the material in terms of conductive heat