Department Of Polymer And Petrochemical Engineering Heat And Mass .

Department of polymer And Petrochemical EngineeringHeat And Mass TransferAssistant lecture:Qusai A.MahdiCONDUCTION-CONVECTION SYSTEMSThe heat that is conducted through a body must frequently be removed (or delivered)by some convection process. For example, the heat lost by conduction through afurnace wall must be dissipated to the surroundings through convection. In heatexchanger applications a finned-tube arrangement might be used to remove heat froma hot liquid. The heat transfer from the liquid to the finned tube is by convection. Theheat is conducted through the material and finally dissipated to the surroundings byconvection. Obviously, an analysis of combined conduction-convection systems isvery important from a practical standpoint.Fin equation:Consider the one-dimensional fin exposed to a surrounding fluid at a temperature T as shown in figure . The temperature of the base of the fin is Tb. We approach theproblem by making an energy balance on an element of the fin of thickness dx asshown in the figure. The defining equation for the convection heat-transfer coefficientis recalled as q hA(Tw T ) where the area in this equation is the surface area forconvection. Let the cross-sectional area of the fin be (Ac) and the perimeter be (P).dTThen the energy quantities are Energy in left face q x kA c,dxEnergy out right face q x dx kAcdTdT d 2T]x dx kAc ( 2 dx)dxdx dxEnergy lost by convection hPdx(T T )Energy in left face Energy out right face Energy lost by convectiond 2T1dx hp(T T )dx] kA* dxdx 22d Thp (T T )2kAdxkA

Department of polymer And Petrochemical EngineeringHeat And Mass TransferAssistant lecture:Qusai A.Mahdilet [θ T T ] and [m 2 hp]kAdT d(θ T ) dθ dxdxdx22dT dθ dx 2 dx 2d 2θ m 2θ (governor equation)2dxd DD 2θ m 2θ ( D 2 m 2 )θ 0 D mdxθ Ce Dxθ C1 e mx C 2e mx [ general solution of governor equation ]Can be applied to three types of fins, depending on physical situationCASE 1 The fin is very long, and the temperature at the end of the fin isessentially that of the surrounding fluid.CASE 2 The end of the fin is insulatedCASE 3 The fin is of finite length and loses heat by convection from its end.For case1 :- The boundary conditions areθ(0) θ b Tb T at at [x 0]x L Tfin tip T i.e (TL T ) then,x Lθ(L) TL T 0cond convx 0 θ θbAppling this boundary conditions in general solution equation, yieldsθ(0) C1 e-m(0) C2em(0) C1 C2 θ bat x L θ(L) 0 C1 em( ) C2e m( ) C1 (0) C2 ( )C2 0θ b C1e m(0) C1 θ bθ (x) θ b e mxAnd the solution becomesθ (x) xT T e mx eθ b Tb T hpkAcFor case2:- negligible heat loss from the fin tip (insulated fin tip ,qfin tip 0)the boundary conditions areat x 0 θ(0) θ b

Department of polymer And Petrochemical EngineeringHeat And Mass TransferAssistant lecture:Qusai A.Mahdiθ(0) C1 e-m(0) C2em(0) C1 C2 θ b .(a)at x Lqfin(tip) 00 mC1 e-mL θ 0 x L x mC2em(L) ] m.(b)Solving equations (a and b) for C1 and C2 the solution is obtained asθ (x)e mxe mx θ b 1 e 2mL 1 e 2mLθbθbC2 then 2mL1 e1 e 2mLe m(x -L)e m(x -L)e m(x -L) e m(x -L) mL e e mL e mL e mLe mL e mLwhere C1 θ (x)θbθ (x)e m(L -x) e -m(L -x) Orθbe mL e mLθ (x) T(x) T cosh[m(L x)] θbTb T coshmLa-Actual fin with convectionat the tipqfinAcpqfinCase 3 the boundary conditions areat x 0 θ (0) θ bqconduction(x L) qconvection(x L)Lc kA c k θ x θ xx LT(x) T Tb T x L hAc θ x L h(TL T ) x Lcoshm(L c x)coshmL cAll of the heat lost by the fin must be conducted into the base at x 0.Using the equations for the temperature distribution, we can compute theheat loss from q kAdT]x 0dxAn alternative method of integrating the convection heat loss could beused:LLq hP(T T )dx hP dx In most cases, however, the first equation is00easier to apply.

Department of polymer And Petrochemical EngineeringHeat And Mass TransferAssistant lecture:Qusai A.Mahdi m(0)) hPkAc θ bThe heat flow for case 1 is q kA( mθ b e11 ) hPkAc θ b tanhmLFor case 2, q kAθ b m( 2mL1 e1 e 2mLFor case 3 q hPkA cθ b tanhmLcFin efficiency: Actual heat transferredHeat which would be transferred(if entire fin area were at base temp)qηf fin the fin efficiency for case1 infinite long finq finmaxηf hPkA c (Tb T )ηf(verylongfin) ηf(insulatedtipfin) hA fin (Tb T )q finq fin,max 1 hp1 L kA c mLhPkAθ b tanhmL tanhmL hPLθ bmLwhere (z)is the depth of the fin and (t) is the thickness. Now, if the fin issufficiently deep, the term (2z) will be large compared with (2t ), andmL 2hz2hL LktzktMultiplying numerator and denominator by (L1/2 ) gives mL 2h 3/2LkLt(Lt) is the profile area of the fin, which we define as [Am Lt] So thatmL 2h 3/2LkA mA corrected length Lc is then used in all the equations which apply for the caseof the fin with an insulated tip.Lc L AcpAc z * tp 2(z t) if z tLc L t,The error which results from this approximation will be less than 82ht 1/2 1πd 2 /4 L d/4percent when ( ) , Lc L πd2k2Lc L where Af in is the total surface area of the fin. Since[Afin pL] for fins withconstant cross section.z*t2z

Department of polymer And Petrochemical EngineeringHeat And Mass TransferAssistant lecture:Qusai A.MahdiExample (1) An aluminum fin [k 200 W/m· ] 3.0 mm thick and 7.5 cm longprotrudes from a wall. The base is maintained at 300 , and the ambienttemperature is 50 with h 10 W/m2· . Calculate the heat loss from the finper unit depth of material.Solution:- We may use the approximate method of solution by extending thefin a fictitious length t/2 and then computing the heat transfer from a fin withinsulated tip.hPh(2z 2t) 5.774kAktz, A (1)(3 10 3 ) 3 10 3 m2 [4.65in 2 ],m Lc L t/2 7.5 0.15 7.65cmq (tanhmLc ) hPkAθ bAnd q (5.774)(200)(3*10-3)(300-50)tanh[(5.774)(0.0765)] 359 W/m[373.5 Btu/h·ft]Circumferential fins of rectangular profileL r2 r1 Lc L t/2 r2c r1 Lc Am tLcExample (2) Aluminum fins 1.5 cm wide and 1.0 mm thick are placed on a 2.5cm-diameter tube to dissipate the heat. The tube surface temperature is 170 ,and the ambient-fluid temperature is 25 .Calculate the heat loss per fin for h 130 W/m2· .assume k 200W/m.oC for aluminum.Solution. For this example we can compute the heat transfer by using the finefficiency curves . The parameters needed areLc L t / 2 1.5 0.05 1.55 cm , r1 2.5/2 1.25 ,r2c r1 Lc 1.25 1.55 2.80 cmr2c/ r1 2.80 / 1.25 2.24, Am t(r2c - r1 ) (0.001)(2.8 – 1.25)(10-2) 1.55*10-5m2L3c / 2 (h 1/ 2130) (0.0155) 3 / 2 0.396kAm(200)(1.55 10 5 )From figηf 82 percent.qf hAf(Tb-T ) ,Af surface areaAf 2π(rc22-r1)q max 2 (r r )h(T0 T ) 2 (2.8 1.252 )(10 4 )(130)(170 25) 74.35W[253.7Btu/h]22c212

Department of polymer And Petrochemical EngineeringHeat And Mass TransferAssistant lecture:Qusai A.Mahdiqact (0.82)(74.35) 60.97W [208Btu / h]

Department of polymer And Petrochemical EngineeringHeat And Mass TransferAssistant lecture:Qusai A.MahdiFIN EFFECTIVENESS :-TbTbAbqfinqfinAbThe performance of the fins is judged on the basic of the enhancement in heattransfer relative to the no fin case ε fin q withfinη A hθ fin fin b For the insulated-tipq withoutfinhA b θ bfin. Where Afin pL is the total surface area of the fin and Ab A is the base area.ɛfin 1 indicated that the addition of fins to the surface does not affect heat transfer.ɛfin 1 indicated that the fin actually acts as insulation slowing down the heat transferfrom the surface. This situation can occur when fins made of low thermal conductivitymaterials are used .ɛfin 1 indicated that the fins are enhancing heat transfer from the surface.THERMAL CONTRCT RESISTANCET TT TT Tq k A A 1 2 A 2 A 2 B k B A 2 B 3 or x A1 / hC A xBT1 T3q x A / k A A 1 / hc A xB / k B AABqqXA(a)XATT1T2A1/hc A is called the thermal contactT2Bresistance and hc is called the 3304stainless-steel bars, 10 cm long, have ground surfaces and are exposed to airwith a surface roughness of about 1 μm. If the surfaces are pressed togetherwith a pressure of 50 atm and the two-bar combination is exposed to an overalltemperature difference of 100 , calculate the axial heat flow and temperaturedrop across the contact surface.Solution:- The overall heat flow is subject to three thermal resistances, oneconduction resistance for each bar, and the contact resistance. For the bars,Rth x(0.1)(4) 8.679 /WkA (16.3) (3 10 2 ) 2X

Department of polymer And Petrochemical EngineeringHeat And Mass TransferAssistant lecture:Qusai A.MahdiFrom Table the contact resistance is Rc 1(5.28 10 4 )(4) 0.747 /Whc A (3 10 2 ) 2The total thermal resistance is thereforeΣRth (2)(8.679) 0.747 18.105and the overall heat flow is q T100 5.52W [18.83Btu / h] Rth 18.105The temperature drop across the contact is found by taking the ratio of thecontact resistance to the total thermal resistance: Tc Rc(0.747)(100) T 4.13 Rth18.105[39.43 ]In this problem the contact resistance represents about 4 percent of the totalresistance.ProblemsQ1)Water and air are separated by a mild-steel plane wall. It is proposed toincrease the heat-transfer rate between these fluids by adding straightrectangular fins of 1.27-mm thickness and 2.5-cm length spaced 1.27 cm apart.The air-side and water-side heat-transfer coefficients may be assumed constantwith values of 11.4 and 256 W/m 2.K respectively. Determine the percentchange in total heat transfer when fins are placed on (a) the water side, (b) theair side, and (c) both sidesQ2)Straight rectangular fin ,made of aluminum(k 208 W/m.0c)4mmthickness(t)and 100mm width(w).The environment condition is h 60 W/m2. 0Cand T2 3000C .Its found that the efficiency of the fin is(0.85)and the basetemperature Tb 12000C? 1-calculate the heat loss from the base of the fin.2Detrmine the performance of the fin. Note Neglect the convection at the tipof the finQ3)To increase the heat dissipation from a 2.5 cm O.D tube ,circumferentialfins made of aluminum (k 200 W/m.K) are soldered to the outer surface .Thefins are 0.1 cm thick and have an outer diameter of 5.5 cm as shown in fig .Ifthe tube temperature is 100 0C , the environmental temperature is 25 0C,and theheat transfer coefficient between the fin and , the environmental is 25W/m2.K,calculate the rate of heat loss from the fin.

Department of polymer And Petrochemical EngineeringHeat And Mass TransferAssistant lecture:Qusai A.MahdiQ4)An experimental device that produces excess heat is passively cooled. Theadditional of pin fins to the casing of this device is being considered toaugment the rate of cooling .Consider a copper pin fin 0.25cm in a diameterthat protrudes from a wall at 95 0C into ambient air temperature as shown infigure .The heat transfer is mainly by natural convection with a coefficientequal to 10W/m2.k .Calculate the heat loss, assuming (a)the fin is "infinitelylong"(b)the fin is 2.5 cm long and the coefficient at the end is the same asaround circumferences .Finally(c)how long would the fin have to be for theinfinitely long solution to be correct within 5percent?Q5)Q6)Q7)The figure below shows part of a set of radial aluminumfins(k 180W/m.K)that are to be fitted to a small air compressor .The devicedissipates 1 Kw by convection to surrounding air which is at 20 0C .each fin is100mm long ,30mm high and 5mm thick .The tip of each fin may be assumedto be adiabatic and heat transfer coefficient of h 15 W/m2.K acts over the