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Heat and Mass Transer Pro. Pradip Dutta Module 3, Learning Ojetives: Students should reognize the in euation. Students should know the general solutions to the in euation.

1 Heat and Mass Transer Pro. Pradip Dutta Module 3, Learning Ojetives: Students should reognize the in euation. Students should know the general solutions to the in euation. Students should e ale to write oundary onditions or (a) very long ins, () insulated tip ins, () onvetive tip ins and (d) ins with a speiied tip temperature. Students should e ale to apply the oundary onditions to the in euation and otain a temperature proile. Students should e ale to apply the temperature proile to the Fourier Law to otain a heat low through the in. Students should e ale to apply the onept o in eiieny to deine an euivalent thermal resistane or a in. Students should e ale to inorporate ins into an overall eletrial network to solve -D, SS prolems with no soures. Indian Institute o Siene Bangalore

2 Heat and Mass Transer Pro. Pradip Dutta MODULE 3 Extended Surae Heat Transer 3. Introdution: Convetion: Heat transer etween a solid surae and a moving luid is governed y the Newton s ooling law: = ha(t s -T ), where T s is the surae temperature and T is the luid temperature. Thereore, to inrease the onvetive heat transer, one an Inrease the temperature dierene (T s -T ) etween the surae and the luid. Inrease the onvetion oeiient h. This an e aomplished y inreasing the luid low over the surae sine h is a untion o the low veloity and the higher the veloity, the higher the h. Example: a ooling an. Inrease the ontat surae area A. Example: a heat sink with ins. Many times, when the irst option is not in our ontrol and the seond option (i.e. inreasing h) is already strethed to its limit, we are let with the only alternative o inreasing the eetive surae area y using ins or extended suraes. Fins are protrusions rom the ase surae into the ooling luid, so that the extra surae o the protrusions is also in ontat with the luid. Most o you have enountered ooling ins on air-ooled engines (motoryles, portale generators, et.), eletroni euipment (CPUs), automoile radiators, air onditioning euipment (ondensers) and elsewhere. 3. Extended surae analysis: In this module, onsideration will e limited to steady state analysis o retangular or pin ins o onstant ross setional area. Annular ins or ins involving a tapered ross setion may e analyzed y similar methods, ut will involve solution o more ompliated euations whih result. Numerial methods o integration or omputer programs an e used to advantage in suh ases. We start with the General Condution Euation: dt α dτ system = T + k () Ater making the assumptions o Steady State, One-Dimensional Condution, this euation redues to the orm: dt + = dx k 0 () This is a seond order, ordinary dierential euation and will reuire oundary onditions to evaluate the two onstants o integration that will arise. Consider the ooling in shown elow: Indian Institute o Siene Bangalore

3 Heat and Mass Transer Pro. Pradip Dutta h, T y T s, A L The in is situated on the surae o a hot surae at T s and surrounded y a oolant at temperature T, whih ools with onvetive oeiient, h. The in has a ross setional area, A, (This is the area through with heat is onduted.) and an overall length, L. Note that as energy is onduted down the length o the in, some portion is lost, y onvetion, rom the sides. Thus the heat low varies along the length o the in. We urther note that the arrows indiating the diretion o heat low point in oth the x and y diretions. This is an indiation that this is truly a two- or three-dimensional heat low, depending on the geometry o the in. However, uite oten, it is onvenient to analyse a in y examining an euivalent one dimensional system. The euivalent system will involve the introdution o heat sinks (negative heat soures), whih remove an amount o energy euivalent to what would e lost through the sides y onvetion. Consider a dierential length o the in. h, T T 0, Δx Aross this segment the heat loss will e h (P Δx) (T-T ), where P is the perimeter around A Δ x. the in. The euivalent heat sink would e ( ) Euating the heat soure to the onvetive loss: Indian Institute o Siene Bangalore

4 Heat and Mass Transer Pro. Pradip Dutta ( ) h P T T = A (3) Sustitute this value into the General Condution Euation as simpliied or One- Dimension, Steady State Condution with Soures: dt h P ( ) dx k A T T = 0 (4) whih is the euation or a in with a onstant ross setional area. This is the Seond Order Dierential Euation that we will solve or eah in analysis. Prior to solving, a ouple o simpliiations should e noted. First, we see that h, P, k and A are all independent o x in the deined system (They may not e onstant i a more general analysis is desired.). We replae this ratio with a onstant. Let then: m = h P k A (5) dt m ( T T ) = 0 (6) dx Next we notie that the euation is non-homogeneous (due to the T term). Reall that nonhomogeneous dierential euations reuire oth a general and a partiular solution. We an make this euation homogeneous y introduing the temperature relative to the surroundings: Dierentiating this euation we ind: θ T - T (7) Dierentiate a seond time: Sustitute into the Fin Euation: dθ dt = + 0 (8) dx dx d θ dt = (9) dx dx d θ m θ = 0 (0) dx This euation is a Seond Order, Homogeneous Dierential Euation. 3.3 Solution o the Fin Euation We apply a standard tehniue or solving a seond order homogeneous linear dierential euation. Try θ = e α x. Dierentiate this expression twie: 3 Indian Institute o Siene Bangalore

5 Heat and Mass Transer Pro. Pradip Dutta dθ dx = α α e x d θ = α dx α e x () () Sustitute this trial solution into the dierential euation: Euation (3) provides the ollowing relation: α e α x m e α x = 0 (3) α = ± m (4) We now have two solutions to the euation. The general solution to the aove dierential euation will e a linear omination o eah o the independent solutions. Then: θ = A e m x + B e -m x (5) where A and B are aritrary onstants whih need to e determined rom the oundary onditions. Note that it is a nd order dierential euation, and hene we need two oundary onditions to determine the two onstants o integration. An alternative solution an e otained as ollows: Note that the hyperoli sin, sinh, the hyperoli osine, osh, are deined as: sinh( mx ) = e e mx mx osh( mx ) = e + e mx mx (6) We may write: C m x D m x C e mx e m x D e mx e m x + C + D C D mx osh( ) + sinh( ) = + = e + e We see that i (C+D)/ replaes A and (C-D)/ replaes B then the two solutions are euivalent. θ = C osh( m x) + D sinh( m x ) (8) Generally the exponential solution is used or very long ins, the hyperoli solutions or other ases. Boundary Conditions: Sine the solution results in onstants o integration we reuire oundary onditions. The irst one is ovious, as one end o the in will e attahed to a hot surae and will ome into thermal euilirium with that surae. Hene, at the in ase, mx (7 ) 4 Indian Institute o Siene Bangalore

6 Heat and Mass Transer Pro. Pradip Dutta θ(0) = T 0 - T θ 0 (9) The seond oundary ondition depends on the ondition imposed at the other end o the in. There are various possiilities, as desried elow. Very long ins: For very long ins, the end loated a long distane rom the heat soure will approah the temperature o the surroundings. Hene, θ( ) = 0 (0) Sustitute the seond ondition into the exponential solution o the in euation: 0 θ( ) = 0 = A e m + B e -m () The irst exponential term is ininite and the seond is eual to zero. The only way that this euation an e valid is i A = 0. Now apply the seond oundary ondition. θ(0) = θ 0 = B e -m 0 B = θ 0 () The general temperature proile or a very long in is then: θ(x) = θ 0 e -m x (3) I we wish to ind the heat low through the in, we may apply Fourier Law: k A dt k A d θ = = dx dx (4) Dierentiate the temperature proile: So that: where dθ dx = θ me o mx (5) h P mx mx = k A θ0 e = h P k A e θ0 k A = mx Mθ e (6) 0 M = hpka. Oten we wish to know the total heat low through the in, i.e. the heat low entering at the ase (x=0). = h P k A θ 0 = Mθ 0 (7) The insulated tip in Assume that the tip is insulated and hene there is no heat transer: 5 Indian Institute o Siene Bangalore

7 Heat and Mass Transer Pro. Pradip Dutta dθ dx x=l = 0 (8) The solution to the in euation is known to e: Dierentiate this expression. θ = C osh( m x) + D sinh( m x ) (9) dθ = Cm sinh( mx ) + Dm osh( mx ) (30) dx Apply the irst oundary ondition at the ase: 0 θ ( 0) = θ0 = C sinh( m 0) + D osh( m 0) (3) So that D = θ 0. Now apply the seond oundary ondition at the tip to ind the value o C: dθ ( L) = 0 = Cm sinh( m L) + θ0m osh( m L) dx (3) whih reuires that osh( ml) C = θ 0 (33) sinh( ml) This leads to the general temperature proile: osh m( L x) θ ( x) = θ0 (34) osh( ml) We may ind the heat low at any value o x y dierentiating the temperature proile and sustituting it into the Fourier Law: k A dt k A d θ = = dx dx (35) So that the energy lowing through the ase o the in is: = hpka θ 0 tanh( ml) = Mθ0 tanh( ml) (36) I we ompare this result with that or the very long in, we see that the primary dierene in orm is in the hyperoli tangent term. That term, whih always results in a numer eual to or less than one, represents the redued heat loss due to the shortening o the in. 6 Indian Institute o Siene Bangalore

8 Heat and Mass Transer Pro. Pradip Dutta Other tip onditions: We have already seen two tip onditions, one eing the long in and the other eing the insulated tip. Two other possiilities are usually onsidered or in analysis: (i) a tip sujeted to onvetive heat transer, and (ii) a tip with a presried temperature. The expressions or temperature distriution and in heat transer or all the our ases are summarized in the tale elow. Tale 3. Case Tip Condition Temp. Distriution Fin heat transer A Convetion heat osh m( L x) + ( h )sinh m( L x) sinh ml + ( h ) osh ml transer: mk Mθ mk o hθ(l)=-k(dθ/dx) h x=l osh ml + ( )sinh ml osh ml + ( h )sinh ml mk mk B C D Adiaati (dθ/dx) x=l =0 Given temperature: θ(l)= θ L Ininitely long in θ(l)=0 oshm( L x) Mθ 0 tanh ml oshml ( θ L θ )sinhm( L x) + sinhm( L x) (oshml L ) θ θ Mθ0 sinhml sinh ml mx e M θ Fin Eetiveness How eetive a in an enhane heat transer is haraterized y the in eetiveness, ε, whih is as the ratio o in heat transer and the heat transer without the in. For an adiaati in: hpkac tanh( ml) kp ε = = = = tanh( ml) (37) ha ( T T ) ha ha C C I the in is long enough, ml>, tanh(ml), and hene it an e onsidered as ininite in (ase D in Tale 3.). Hene, or long ins, In order to enhane heat transer, C C C kp k P ε = (38) ha h A ε should e greater than (In ase ε <, the in would have no purpose as it would serve as an insulator instead). However ε is onsidered unjustiiale eause o diminishing returns as in length inreases. 7 Indian Institute o Siene Bangalore

9 Heat and Mass Transer Pro. Pradip Dutta To inrease ε, the in s material should have higher thermal ondutivity, k. It seems to e ounterintuitive that the lower onvetion oeiient, h, the higher ε. Well, i h is very high, it is not neessary to enhane heat transer y adding heat ins. Thereore, heat ins are more eetive i h is low. Oservations: I ins are to e used on suraes separating gas and liuid, ins are usually plaed on the gas side. (Why?) P/AC should e as high as possile. Use a suare in with a dimension o W y W as an example: P=4W, AC=W, P/AC=(4/W). The smaller the W, the higher is the P/AC, and the higher the ε.conlusion: It is preerred to use thin and losely spaed (to inrease the total numer) ins. The eetiveness o a in an also e haraterized y ( T T ) / R t, t, h ε = = = = (39) hac ( T T ) ( T T ) / Rt, h Rt, It is a ratio o the thermal resistane due to onvetion to the thermal resistane o a in. In order to enhane heat transer, the in s resistane should e lower than the resistane due only to onvetion. R 3.5 Fin Eiieny The in eiieny is deined as the ratio o the energy transerred through a real in to that transerred through an ideal in. An ideal in is thought to e one made o a peret or ininite ondutor material. A peret ondutor has an ininite thermal ondutivity so that the entire in is at the ase material temperature. real η = = ideal h P k A θ tanh( m L) L h ( P L) θ L (40) T x Real situation T x Ideal situation Simpliying euation (40): 8 Indian Institute o Siene Bangalore

10 Heat and Mass Transer Pro. Pradip Dutta η = k A h P θ tanh( m L) tanh( m L) L = L θ m L L (4) The heat transer through any in an now e written as:. = ( T T (4) η. h. A The aove euation provides us with the onept o in thermal resistane (using eletrial analogy) as Rt, = (43) η. h. A Overall Fin Eiieny: Overall in eiieny or an array o ins Deine terms: A: ase area exposed to oolant A: surae area o a single in At: total area inluding ase area and total inned surae, At=A+NA N: total numer o ins Heat Transer rom a Fin Array: = + N = ha ( T T ) + Nη ha ( T T ) t = h[( A NA ) + Nη A ]( T T ) = h[ A NA ( η )]( T T ) t t NA = hat[ ( η)]( T T ) = ηohat( T T ) A t NA Deine overall in eiieny: ηo = ( η) A t 9 Indian Institute o Siene Bangalore

11 Heat and Mass Transer Pro. Pradip Dutta T T = haη ( T T ) = where R = t t O t, O RtO, haη t O Compare to heat transer without ins = ha( T T ) = h( A + NA, )( T T ) = ha where A is the ase area (unexposed) or the in, To enhane heat transer Aη That is, to inrease the eetive area η O A t. Thermal Resistane Conept: t O >> A L t A=A+NA, T T R= t/(ka) T T 0 Indian Institute o Siene Bangalore

12 Heat and Mass Transer Pro. Pradip Dutta T T T T R =L /(k A) R = /( haη ) t, O t O T T T T = = R R + R + R t, O Indian Institute o Siene Bangalore

13 Heat and Mass Transer Pro. Pradip Dutta Prolem : A long, irular aluminium rod attahed at one end to the heated wall and transers heat through onvetion to a old luid. (a) I the diameter o the rod is triples, y how muh would the rate o heat removal hange? () I a opper rod o the same diameter is used in plae o aluminium, y how muh would the rate o heat removal hange? Known: long, aluminum ylinder ats as an extended surae. Find: (a) inrease in heat transer i diameter is tripled and () inrease in heat transer i opper is used in plae o aluminum. Shemati: T, h Aluminum Or opper D Assumptions: () steady-state onditions, () one-dimensional ondution, (3) onstant properties, (4) uniorm onvetion oeiient, (5)rod is ininitely long. Properties: aluminum (pure): k=40w/m. K; opper (pure): k=400w/m. K Analysis: (a) or an ininitely long in, the in rate is r = M = ( hpka ) θ = ( hπdkπd / 4) θ π = ( hk) D 3 θ Where P=πD and A =πd /4 or the irular ross-setion. Note that D 3 /. Hene, i the diameter is tripled, (3D) ( D) = 3 3 = 5. And there is a 50 % inrease in heat transer. Indian Institute o Siene Bangalore

14 Heat and Mass Transer Pro. Pradip Dutta () in hanging rom aluminum to opper, sine k /, it ollows that ( ) Cu k 400 Cu = = ( ) Al k Al 40 =.9 And there is a 9 % inrease in the heat transer rate. Comments: () eause in eetiveness is enhaned y maximum P/A = 4/D. the use o a larger numer o small diameter ins is preerred to a single large diameter in. () From the standpoint o ost and weight, aluminum is preerred over opper. Indian Institute o Siene Bangalore

15 Heat and Mass Transer Pro. Pradip Dutta Prolem : Two long opper rods o diameter D= m are soldered together end to end, with solder having melting point o C. The rods are in the air at 50C with a onvetion oeiient o 0W/m. K. What is the minimum power input needed to eet the soldering? Known: Melting point o solder used to join two long opper rods. Find: Minimum power needed to solder the rods. Shemati: Copper D=m Juntion T =650 0 C Air T = 5 0 C H=0W/m. K Assumptions: () steady-state onditions, () one-dimensional ondution along the rods, (3) onstant properties, (4) no internal heat generation, (5) negligile radiation exhange with surroundings, (6) uniorm, h and (7) ininitely long rods. Properties: opper T = (650+5) 0 C 600K: k=379 W/m.K Analysis: the juntion must e maintained at C while energy is transerred y ondution rom the juntion (along oth rods). The minimum power is twie the in heat rate or an ininitely long in, Indian Institute o Siene Bangalore

16 Heat and Mass Transer Pro. Pradip Dutta = = (hpka ) (T T ) min sustituting numerial values, W = 0 (π 0.0m) 379 min m.k thereore, min = 0.9W W m.k π (0.0m) 4 (650 5) 0 C. Comments: radiation losses rom the rod are signiiant, partiularly near the juntion, therey reuiring a larger power input to maintain the juntion at C. Indian Institute o Siene Bangalore

17 Heat and Mass Transer Pro. Pradip Dutta Prolem 3: Determine the perentage inrease in heat transer assoiated with attahing aluminium ins o retangular proile to a plane wall. The ins are 50mm long, 0.5mm thik, and are eually spaed at a distane o 4mm (50ins/m). The onvetion oeiient aeted assoiated with the are wall is 40W/m. K, while that resulting rom attahment o the ins is 30W/m. K. Known: Dimensions and numer o retangular aluminum ins. Convetion oeiient with and without ins. Find: perentage inrease in heat transer resulting rom use o ins. Shemati: L=50mm N=50m - W=width h w = 30W/m.K(with ins) h wo =40W/m.K(without ins) Aluminium Assumptions: () steady-state onditions, () one-dimensional ondution, (3) onstant properties, (4) negligile in ontat resistane, (6) uniorm onvetion oeiient. Properties: Aluminum, pure: k 40W/m. K Analysis: evaluate the in parameters L = L + A L L 3/ p 3/ = L t = m (h (h w w / KA / KA t = m max / P ) / P ) it ollows that, η = = η = 0.7h (0.0505m) = hene w wlθ 3/ m = = W / m.k 0.05m (wθ 3 30W / m.k 40W / m.k ) = 6 m.6w / m / 6 m K(wθ / ) Indian Institute o Siene Bangalore

18 Heat and Mass Transer Pro. Pradip Dutta w w w With the ins, the heat transer rom the walls is = N + ( Nt)wh w θ W 4 = 50.6 (wθ ) + (m m) 30W / m.k(wθ m.k W = ( ) (wθ ) = 566wθ m.k = h m wθ = 40wθ Without the ins,. wo wo ) Hene the perentage inreases in heat transer is w wo 566wθ = 40wθ = 4.6 = 46% Comments: I the inite in approximation is made, it ollows that = / (hpka ) θ = [h w wkwt] / θ = (30**40*5*0-4 ) / wθ =.68w θ. Hene is overestimated. Indian Institute o Siene Bangalore

19 Heat and Mass Transer Pro. Pradip Dutta Module 3: Short uestions. Is the heat low through a in truly one-dimensional? Justiy the one-dimensional assumption made in the analysis o ins.. Insulated tip ondition is oten used in in analysis eause (hoose one answer) A. Fins are usually delierately insulated at their tips B. There is no heat loss rom the tip C. The heat loss rom the tip is usually insigniiant ompared to the rest o the in, and the insulated tip ondition makes the prolem mathematially simple 3. What is the dierene etween in eetiveness and in eiieny? 4. The ins attahed to a surae are determined to have an eetiveness o 0.9. Do you think the rate o heat transer rom the surae has inreased or dereased as a result o addition o ins? 5. Fins are normally meant to enhane heat transer. Under what irumstanes the addition o ins may atually derease heat transer? 6. Hot water is to e ooled as it lows through the tues exposed to atmospheri air. Fins are to e attahed in order to enhane heat transer. Would you reommend adding ins to the inside or outside the tues? Why? 7. Hot air is to e ooled as it is ored through the tues exposed to atmospheri air. Fins are to e attahed in order to enhane heat transer. Would you reommend adding ins to the inside or outside the tues? Why? When would you reommend adding ins oth inside and outside the tues? 8. Consider two inned surae whih are idential exept that the ins on the ist surae are ormed y asting or extrusion, whereas they are attahed to the seond surae aterwards y welding or tight itting. For whih ase do you think the ins will provide greater enhanement in heat transer? Explain. 9. Does the (a) eiieny and () eetiveness o a in inrease or derease as the in length is inreased? 0. Two pin ins are idential, exept that the diameter o one o them is twie the diameter o the other? For whih in will the a) eiieny and () eetiveness e higher? Explain.. Two plate ins o onstant retangular ross setion are idential, exept that the thikness o one o them is twie the thikness o the other? For whih in will the a) eiieny and () eetiveness e higher? Explain.. Two inned surae are idential, exept that the onvetion heat transer oeiient o one o them is twie that o the other? For whih in will the a) eiieny and () eetiveness e higher? Explain. Indian Institute o Siene Bangalore

Known: long, aluminum cylinder acts as an extended surface.

Known: long, aluminum cylinder acts as an extended surface. Prolem : A long, irular aluminium rod attahed at one end to the heated all and transers heat through onvetion to a old luid. (a) I the diameter o the rod is triples, y ho muh ould the rate o heat removal