Department of Metallurgical Engineering Heat Transfer Phenomena Prof. Dr.: Kadhim F. Alsultani CONDUCTION-CONVECTION SYSTEMS The 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 a furnace wall must be dissipated to the surroundings through convection. In heatexchanger applications a finned-tube arrangement might be used to remove heat from a hot liquid. The heat transfer from the liquid to the finned tube is by convection. The heat is conducted through the material and finally dissipated to the surroundings by convection. Obviously, an analysis of combined conduction-convection systems is very 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 the problem by making an energy balance on an element of the fin of thickness dx as shown in the figure. The defining equation for the convection heat-transfer coefficient is recalled as q hA(T w T ) where the area in this equation is the surface area for convection. Let the cross-sectional area of the fin be (Ac) and the perimeter be (P). dT Then the energy quantities are Energy in left face = q x kA c , dx Energy out right face = q x dx kA c dT dT d 2T ]x dx kA c ( 2 dx) dx dx dx Energy lost by convection = hPdx(T T ) Energy in left face= Energy out right face+ Energy lost by convection d 2T 1 kA dx hp(T T)dx ] 2 kA* dx dx 2 d T hp (T T) 2 kA dx Department of Metallurgical Engineering Heat Transfer Phenomena Prof. Dr.: Kadhim F. Alsultani let [θ T T ] and [m 2 hp ] kA dT d(θ T ) dθ dx dx dx d 2 T d 2θ dx 2 dx 2 d 2θ m 2θ (governor equation) 2 dx d D D 2θ m 2θ ( D 2 m 2 )θ 0 D m dx θ CeDx θ C1 e mx C 2e mx [ general solution of governor equation ] Can be applied to three types of fins, depending on physical situation CASE 1 The fin is very long, and the temperature at the end of the fin is essentially that of the surrounding fluid. CASE 2 The end of the fin is insulated CASE 3 The fin is of finite length and loses heat by convection from its end. For case1 :- The boundary conditions are θ(0) θ b Tb T at at [x = 0] x =L=∞ Tfin tip=T∞ i.e (TL =T∞) then, x=L θ(L) TL T 0 cond=conv x=0 θ=θb Appling this boundary conditions in general solution equation, yields θ(0) C1 e-m(0) C2em(0) C1 C2 θ b at x =L=∞ θ(L) 0 C1 em( ) C2em( ) C1 (0) C2 () C2 0 θ b C1e m(0) C1 θ b θ (x) θ b e mx And the solution becomes θ (x) x T T e mx e θ b Tb T hp kA c For case2:- negligible heat loss from the fin tip (insulated fin tip ,qfin tip=0 )the boundary conditions are Department of Metallurgical Engineering Heat Transfer Phenomena Prof. Dr.: Kadhim F. Alsultani at x = 0 θ(0) θ b θ(0) C1 e-m(0) C2em(0) C1 C2 θb ....(a) at x L qfin(tip) 0 0 mC1 e-mL θ 0 x L x mC 2em(L) ] m.......(b) Solving equations (a and b) for C1 and C2 the solution is obtained as θ (x) e mx e mx θ b 1 e 2mL 1 e 2mL where C1 θ (x) θb θb 1 e 2mL θb then 1 e 2mL e m(x -L) e m(x -L) e mL e mL C2 e m(x -L) e m(x -L) mL e e mL e mL e mL θ (x) e m(L -x) e -m(L -x) Or θb e mL e mL θ (x) T(x) T cosh[m(L x)] θb Tb T coshmL a-Actual fin with convection at the tip qfin Ac p qfin Case 3 the boundary conditions are at x = 0 θ (0) θ b qconduction(x=L)= qconvection(x=L) Lc kA c k θ x θ x x L T(x) T Tb T x L hAc θ x L h(TL T ) x L coshm(L c x) coshmL c All 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 the heat loss from q kA dT ]x 0 dx An alternative method of integrating the convection heat loss could be used: Department of Metallurgical Engineering Heat Transfer Phenomena Prof. Dr.: Kadhim F. Alsultani L L 0 0 q hP(T T )dx hPdx In most cases, however, the first equation is easier to apply. m(0) ) hPkA c θ b The heat flow for case 1 is q kA(mθ b e 1 1 ) hPkA c θ b tanhmL For case 2, q kAθ b m( 2mL 1 e 1 e 2mL For case 3 q hPkA cθ b tanhmL c Fin efficiency: Actual heat transferre d Heat which would be transferre d(if entire fin area were at base temp) q η f fin the fin efficiency for case1 infinite long fin q finmax ηf hPkA c (Tb T ) ηf(verylongfin) ηf(insulatedtipfin) hA fin (Tb T ) q fin q fin, max 1 hp 1 L kA c mL hPkA θ b tanhmL tanhmL hPLθ b mL where (z)is the depth of the fin and (t) is the thickness. Now, if the fin is sufficiently deep, the term (2z) will be large compared with (2t ), and mL 2hz 2h L L ktz kt Multiplying numerator and denominator by (L1/2 ) gives mL 2h 3/2 L kLt (Lt) is the profile area of the fin, which we define as [Am = Lt] So that mL 2h 3/2 L kA m A corrected length Lc is then used in all the equations which apply for the case of the fin with an insulated tip. Lc L Ac p Ac z * t p 2(z t) if z t Lc L t ,The error which results from this approximation will be less than 8 2 ht 1 πd 2 /4 percent when ( )1/2 , Lc L L d/4 2k 2 πd Lc L z*t 2z Department of Metallurgical Engineering Heat Transfer Phenomena Prof. Dr.: Kadhim F. Alsultani where Af in is the total surface area of the fin. Since[Afin =pL] for fins with constant cross section. Example (1) An aluminum fin [k = 200 W/m·℃] 3.0 mm thick and 7.5 cm long protrudes from a wall. The base is maintained at 300℃, and the ambient temperature is 50℃ with h = 10 W/m2·℃. Calculate the heat loss from the fin per unit depth of material. Solution:- We may use the approximate method of solution by extending the fin a fictitious length t/2 and then computing the heat transfer from a fin with insulated tip. hP h(2z 2t) 5.774 kA ktz , A (1)(3 10 3 ) 3 10 3 m 2 [4.65in 2 ] L c L t/2 7.5 0.15 7.65cm q (tanhmL c ) hPkA θ b ,m And 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 profile L= r2 − r1 Lc = L +t/2 r2c= r1 + Lc Am= tLc Example (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 are Lc = 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 cm r2c/ r1 = 2.80 / 1.25 = 2.24, Am = t(r2c - r1 ) = (0.001)(2.8 – 1.25)(10-2) = 1.55*10-5 m2 L3c / 2 ( h 1/ 2 130 ) (0.0155) 3 / 2 0.396 kAm (200)(1.55 10 5 ) Department of Metallurgical Engineering Heat Transfer Phenomena Prof. Dr.: Kadhim F. Alsultani From figηf = 82 percent. qf=hAf(Tb-T∞) ,Af=surface area q max 2 (r 2 2c Af=2π(rc22-r1) r )h(T0 T ) 2 (2.8 1.252 )(10 4 )(130)(170 25) 74.35W[253.7Btu/h] 2 1 qact (0.82)(74.35) 60.97W [208Btu / h] 2 Department of Metallurgical Engineering Heat Transfer Phenomena Prof. Dr.: Kadhim F. Alsultani Tb FIN EFFECTIVENESS :- Tb Ab qfin qfin Ab The performance of the fins is judged on the basic of the enhancement in heat transfer relative to the no fin case ε fin q withfin η A hθ fin fin b For the insulated-tip q withoutfin hA b θ b fin. 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 transfer from the surface. This situation can occur when fins made of low thermal conductivity materials are used . ɛfin>1 indicated that the fins are enhancing heat transfer from the surface. THERMAL CONTRCT RESISTANCE T T T T T T q k A A 1 2 A 2 A 2 B k B A 2 B 3 or x A 1 / hC A xB T1 T3 q x A / k A A 1 / hc A xB / k B A A B q q XA (a) XA T T1 T2A 1/hc A is called the thermal contact T2B resistance and hc is called the contact T3 coefficient. Example (b) :Two 3.0cm diameter 1 2 3 304 stainless-steel bars, 10 cm long, have ground surfaces and are exposed to air with a surface roughness of about 1 μm. If the surfaces are pressed together with a pressure of 50 atm and the two-bar combination is exposed to an overall temperature difference of 100℃, calculate the axial heat flow and temperature drop across the contact surface. Solution:- The overall heat flow is subject to three thermal resistances, one X Department of Metallurgical Engineering Heat Transfer Phenomena Prof. Dr.: Kadhim F. Alsultani conduction resistance for each bar, and the contact resistance. For the bars, Rth x (0.1)( 4) 8.679 ℃/W kA (16.3) (3 10 2 ) 2 From Table the contact resistance is Rc 1 (5.28 10 4 )( 4) 0.747 ℃/W hc A (3 10 2 ) 2 The total thermal resistance is thereforeΣRth = (2)(8.679) + 0.747 = 18.105 and the overall heat flow is q T 100 5.52W [18.83Btu / h] Rth 18.105 The temperature drop across the contact is found by taking the ratio of the contact resistance to the total thermal resistance: Tc Rc (0.747)(100) T 4.13 ℃ Rth 18.105 [39.43℉] In this problem the contact resistance represents about 4 percent of the total resistance.
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