Elements of Heat Transfer Analysis

Transcription

1 !H-ref,mmc,rev/5/0prt/5/0 Elements of Heat ransfer Analyss (A Bref Introducton to Heat ransfer) Mchael M. Chen Unversty of Mchgan. Introducton A. General dscussons hese lectures consttute a compact presentaton of elementary heat transfer analyss, ntended to accompany the study of thermodynamcs n a frst course on thermal scences n mechancal engneerng. Students wth a good understandng of these pages and wllng to use some deductve reasonng should e ale to perform effectve analyses of heat transfer for many engneerng purposes. Occasonal reference to more extensve property tales, charts and formulas n a dedcated textook on heat transfer or handook may e helpful for some prolems [Incropera & DeWtt 996, Mlls 99, Ozsk 985]. Most of the methodologes llustrated here are algerac. Calculus and dfferental equatons are used sparngly, prmarly to help students gan famlarty and eneft from the clarty and physcal nsght they rng to analyss, and to ndcate what more advanced analyss mght ental. Heat transfer s concerned wth the study of the transport of energy due to temperature dfferences. As engneers, we are more nterested n the rate of heat transfer, denoted n these notes y Q & [unt watts, or W, whch s J/s], rather than n the amount of heat energy transferred 3. In addton, we wll also fnd t convenent to work wth the heat transfer rate per unt area q&, [W/m ], whch we wll call the heat flux. Heat flux s clearly a vector, snce t has a magntude and a drecton. For many purposes the drecton s ovous y context, so we wll fnd t adequate to work wth the scalar verson of heat flux, denoted y q&. Student should make an effort to dstngush these from the amount of heat transfer Q [J] and heat transfer per unt mass q [J/kg] used n the thermodynamc dscussons of ths course. Heat transfer takes place n one of three modes: conducton, convecton and radaton. We wll refly dscuss each of these modes n ths secton. Practcal prolems of heat transfer frequently nvolve more than one component and more than one mode. he asc sklls for dealng wth mult-component, mult-mode heat transfer wll e dscussed n Secton. More detaled dscussons for asc calculatons for each of the three modes wll e dscussed n the Square rackets[ ] wth name(s) and year are keys to lst of references. Otherwse [ ] are used ndcate unts and dmensons, as ndcated n the next footnote. Another arevated ntroducton to heat transfer, wth dfferent emphass, can also e found n Chapter of Sonntag and Borgnakke, Introducton to Engneerng hermodyancs, Wley, 00, the current textook for ME30. 3 In some heat transfer ooks Q s used to denote the heat transfer rate [W] and q s used to denote heat flux [W/m ].

2 !H-ref,mmc,rev/5/0prt/5/0 next three sectons, to e followed y other sectons that wll develop our sklls further and ntroduce a few addtonal concepts. B. Conducton Heat ransfer Conducton s heat transfer due to temperature non-unformtes n a contnuous sustance, ether statonary or n moton. Consder the plane sla shown n Fg.. Oservatons supplemented y common sense reasonng suggest that the heat transfer rate Q & [W] must e proportonal to the temperature dfference [K], cross-sectonal area A [m ], and the nverse of the thckness B [m]. Q& = Ak () B Area = A Q & B Fg. Heat conducton through a plane sla he coeffcent of proportonalty k s a materal property, called the thermal conductvty. Unt consstency requres that the unt of k e [W/mK]. More careful analyss would ndcate that a more general way to wrte ths equaton s q& = k () cond where denotes the temperature gradent. hs s a vector whose drecton and magntude are the drecton and magntude of the steepest slope of temperature ncrease. he negatve sgn s necessary snce heat transfer only proceed from n the drecton of decreasng temperature. he x- component of oth sdes of Eq. () s q& x = k (3) x Smlar equatons can e otaned for the y- and z- components. Eq. () s known as the Fourer s Law of Heat Conducton. he algerac form, Eq. () can e used as an approxmaton for heat transfer through plane layers or thn non-planar layers n steady state f k s relatvely unform n the layer.

3 !H-ref,mmc,rev/5/0prt/5/0 3 Snce n general k s a materal property 4, t s a functon of and p. In most cases, however, the pressure dependence s small and k depends prmarly on temperature. ypcal thermal conductvtes for several classes of materal are shown n ale. It s advsale for students to e famlar wth the relatve magntudes of thermal conductvtes of dfferent classes of materal or representatve values, to facltate quck decsons n analyses or desgn. Other propertes n n ale wll e dscussed n later sectons. Materal, at temperature(s) Damond, at 300, 400K Alumnum oxde, at 300, 600 K Metals Pure copper, at 300 K Brass, a copper alloy, at 300, 600 Alumnum alloy 04-6, at 300, 600 K Caron steel, AISI 00, at 300, 000 K Stanless steel, AISI 30, at 300, 000 K Common Non-metallc Materals Plate glass, at 300 K Plastcs akelte, at 300 K Plastcs eflon, at 300 K Wood, common types, at 300 K Glass fer nsulaton, at 300 K Urethane foam nsulaton, at 300 K Lquds Water, at 300, 400 K Engne ol at 300 K Gases Helum, at 300, 600 K Ar, at 300, 600 K ale Selected hermal Propertes s shown W/mK 300, , , 49 77, , , , , , K 0 6 J/m α k/ρc 300 K 0 6 m /s If etter sources are not avalale, approxmate conductvtes at ntermedate temperatures for solds and lquds can e otaned y lnear nterpolaton. Interpolaton, and modest extrapolaton, for gas conductvtes can est e done y assumng logk-versus-log as lnear. Note that k s for common engneerng metals vary etween 5 00 W/mK, wth stanless steel near the ottom and alumnum alloys near the top. Pure metals have much hgher conductvtes than ther own alloys, ut are of no engneerng sgnfcance. Example Basc Conducton 4 he exceptons are very dlute gases or very thn (nanometer scale) layers, for whch molecular-scale effects ecome mportant.

4 !H-ref,mmc,rev/5/0prt/5/0 4 A hot water heater has an exposed surface area of.8 m. After t s nsulated wth nch of ferglass nsulaton, t s found that the nsde surface of the nsulaton s at 0 F, and outsde surface s at 90 F. Determne the heat loss through the nsulaton. Soluton: = 0 F 90 F = 30 F = 30*5/9=6.67 K A =.8 m B = n = *0.054=0.054 m k = W/mK (est nformaton convenently avalale, from ale ) Hence Q & = Ak /B =.8*0.046*6.67/0.054 [m ][W/mK][K/m] = W Students are strongly recommended to pay attenton to the followng n dong homework:. All quanttes not n SI unts are frst converted to SI unts to nsure unt consstency.. Arthmetc expresson s frst wrtten n symolc form, as n the case of = Ak to assst gradng and future checkng, and to nsure partal credt ncase of numercal error. hs s then followed y numercal susttutons n the same order so the grader can determne the value used for each parameter. 3. he numercal susttutons s accompaned y a unt consstency check n wrtten form, as n [m ][W/mK][K/m]. hs can e great help n spottng errors. hs may e omtted as you gan expertse and confdence, or when the prolem s smple and ovous. C. Convecton and the Convectve Heat ransfer Coeffcent he word convecton refers to the conveyance of energy y movng materal, usually a flud. he detaled study of convecton would nclude the study of the flow feld and ts effect on temperature dstruton. It s a suject of nterest not only to mechancal engneers ut also to other professons, for example chemcal and aeronautcal engneers and atmospherc scentsts. In elementary heat transfer analyses, however, the study of convecton prmarly refers to the calculaton of rate of heat transfer etween a sold surface and a ody of flud, due to the comned effects of flow and conducton. he consequence of the comned effects of flow and conducton n the flud s to concentrate the temperature varatons n the flud to a small layer adjacent to the surface, as llustrated y the shaded regons n Fg.. hs evokes the mage that the flud at large s well mxed, wth nearly unform temperature, and there exsts a ncreasngly stagnant flm near the wall whch consttutes the prmary resstance to heat transfer. hs s the thermal oundary layer, also called conducton oundary layer.

5 !H-ref,mmc,rev/5/0prt/5/0 5 For quanttatve calculatons, t s convenent to defne the convectve heat transfer coeffcent 5 h, attruted to Issac Newton: q& = h ; where w flud (4) As was the case n Eq. (), s customarly employed as a postve numer. he user should keep track of the drecton of heat transfer y oservng the sense of the temperature drop. he suscrpt w (wall) refers to the sold-flud nterface. As mght e expected, h vares wth the flow confguraton, the flud velocty, the dmenson of the mmersed oject or contanng vessel, and the flud propertes. It should e recognzed that ths s not a fundamental equaton, ut s merely a defnton of h for convenent computaton of the convectve heat flux. he actual value of h must e provded separately y detaled analyss or emprcal correlatons ased on accumulaton of data. Furthermore, there s also an amguty n assgnng flud. By conventon, for external flows, llustrated n Fg. a and c, flud s taken to e the undstured flud temperature away from the wall, usually denoted as. For nternal flows, llustrated n Fg., flud vares wth axal poston and s taken to e the average temperature over the cross secton wth the flow rate as the weghtng factor. hs wll e dscussed n greater detal n a later secton. f = w V V w V a w f = ave c Fg. Common confguratons of convecton. a. External flow Forced Convecton,. Internal flow Forced Convecton, c. Free or Natural Convecton 5 he heat transfer coeffcent s almost unversally denoted y h, the same symol used for specfc enthalpy. Students should e alert to dstngush the two y context.

6 !H-ref,mmc,rev/5/0prt/5/0 6 Forced convecton refers to convecton due to an mposed velocty, as shown n a and of Fg., whereas natural or free convecton refers to flows generated naturally y the uoyancy of heated ar, as shown n c of Fg.. As an example for forced convecton, consder a crcular rod or ppe wth surface temperature w, sujected to a cross wnd of ar at velocty U and temperature. hs s an external flow, forced convecton prolem, for whch s computed as w. For ranges of condtons shown y the nequaltes, an approxmate formula wth aout 5% accuracy s V h For ar, 80 K < 30 K, 0.5 VD mf < < <. D ( h n W/m K, V n m/s, D n m) (5) hs formula has een derved from a more elaorate and more general formula (see Secton 4) y usng propertes of ar at 300 K. If the mean flm temperature mf ( w + ) / strays too far from 300 K, the assumed propertes wll no longer e vald, and results wll not e accurate. As an example for free convecton, consder a heated horzontal ppe located n stll ar. In ths case uoyancy would nduce a local ar flow just around the ppes, n the manner of Fg. c. For ranges of condtons shown, an approxmate formula wth aout 5% accuracy s h =. 55 ( ) D For ar, 0.00 < D 80 K < 3 mf < 00, < 30 K, < 60, ( h n W/m K, n K, D n m) Both of these formulas are approxmatons of the more elaorate and more general formulas to e dscussed n Secton 4. Note that these specalzed formulas constructed for restrcted applcatons must e used wth the specfed set of unts, and must e converted carefully when parameters n other unts are nvolved. In contrast, fundamental physcal laws and asc defntons, such as Eqs. (), (3) and (4), are general and can e used wth any set of selfconsstent unts. he exponents n these formulas are emprcal and wll change wth condtons. Wthn the specfed range of valdty, the exponents gve a concse ndcaton of the dependence of h on parameters U and D. For example, t can e shown from Eq. (5) that for ths forced convecton confguraton, a one percent ncrease of U wll lead to a 0.65% ncrease n h, and each percent ncrease of D wll lead to a 0.35% decrease n h. Smlarly, Eq. (6) tells us that h for free convecton s approxmately proportonal to /3-power of the temperature dfference, and nversely proportonal to the one-tenth power of D. hs type of understandng of the trends s very valuale n engneerng desgn and analyss. Representatve values of h are gven n ale. he forgong dscusson clearly shows that the convecton heat transfer coeffcent represents a form of conducton modfed y flud moton. Hence we cannot smultaneously (6)

7 !H-ref,mmc,rev/5/0prt/5/0 7 consder the conducton and convecton through the same ody of flud, or doule countng would result. ale. ypcal Values of Heat ransfer Coeffcents (n W/m K) (From varous sources) Free Convecton Gases 5 0 (Some dependence) Lquds Forced Convecton Gases 5 50 (Weak dependence) Lquds 50 0,000 Bolng and Condensaton (Strong-Weak dependence) h for radaton near room temperature h for radaton around 600 K,500 00, ~50 Example. Forced Convecton An ar heater conssts of electrc heatng elements, 0.5 n dameter and 3 ft long, each dsspatng 00 W. A fan lows ar, at 5 C across the heaters at 5000 ft/mn. Determne the heat transfer coeffcent and the surface temperature of the heater. Soluton: From Eq. (4), and the defnton of, the total heat transfer rate s ( ) Q& = Ah w whch can e solved to yeld w + Q& / Ah = V = 5000**0.054/60=5.4 m/s D = 0.5*0.054 = 0.07 m h = 4.68*5.4^0.65/0.07^0.035 = 44.6 W/m K A = 3**0.054*p()*0.07= m (End surfaces neglected) w = Q& + / Ah = 5+00/(44.6* ) [W]/[m ][W/m K] = 7.8 C A retrospectve check shows that the computed mean flm temperature would e aout 76 C, consderaly aove the allowed lmt. hs means that the results are lkely to e qute naccurate, wth errors greater than 5%, whch corresponds to a error n greater than 5 K. If more accurate results are desred, the more elaorate formulas of Secton 4, or a dedcated text ook on heat transfer, should e consulted.

8 !H-ref,mmc,rev/5/0prt/5/0 8 Comments on Accuracy Although the result of the last example s not very accurate, t nevertheless can serve very useful engneerng functons. For example, an estmated temperature of 8 C s proaly suffcent to show that the surface s unlkely to start a fre f t s n contact wth a pece of wood or paper. On the other hand, t s proaly hot enough to present a urn hazard to humans and pets, and should e desgned wth safeguards to prevent drect contact. Example 3 Free Convecton A horzontal heatng element, 0.5 dameter and 3 feet long, s located n a room wth stll ar at 0 C. he surface temperature of the heatng element s mantaned at 40 C. Determne the free convecton heat transfer coeffcent and the total heat transfer rate. Soluton hs rod has the same dmensons as the rod n Example. From Eq. (6), wth = 50 0 = 30 K, ( ) 0. 3 h =. 55 =.55*30^0.3/0.07^0.= W/m K 0. D Q& fr c = Ah = *6.504*30 [m ][W/m K][K] = 7.8 W D. Radaton Heat ransfer It s apparent to most oservers that very hot surfaces such as stoves and freplaces, as well as the sun, can transmt heat wthout the help of an ntervenng materal, and that the heat can e asored y ojects whch are n lne-of-sght of such hot ojects. hs s thermal radaton, emtted y all ojects not at asolute zero temperature. We now know that thermal radaton s a form of electromagnetc wave. All electromagnetc waves carry energy, and represent transmtted power. ypcal thermal radaton dffers from waves ntended for sgnal transmsson n that t s spread over a wde wavelength range (or equvalently, frequency range). he sun s radaton s thermal radaton wth the ulk of ts power spread over a wavelength range of µm. Evoluton has made our eyes adapt to ths ountful source of radaton. So we now call ths range the vsle lght range. Ojects at lower temperatures generally emt at longer wavelengths. Most earthound radaton heat transfer occurs n the nfrared range, wth wavelengths aove µm. hermal radaton has essentally the same propertes as lght, travelng n straght lnes n unform, transparent meda and can e transmtted, asored, or reflected from surfaces. Ar at moderate temperatures s essentally transparent to nfrared radaton, whereas most lquds and solds are not. Hence an mportant class of radatve heat transfer prolems s concerned wth heat exchange among surfaces, ether n vacuum or mmersed n ar. he general prncples and methodology of such analyses wll e ntroduced n later sectons. o provde an example for the relatve magntude and mathematcs of radatve heat transfer, we wll cte wthout dervaton one smple formula for an mportance case of radatve transfer. hs s the case when a small ody s completely enclosed y a larger enclosure e, as shown n Fg. 3a. We wll later show that as long as e s at least two- or three-fold as large as,

9 !H-ref,mmc,rev/5/0prt/5/0 9 essentally all of the radaton emtted from wll e asored y e after ether sngle or multple reflecton n the cavty. On the other hand, most of the radaton emtted y e, n whchever drecton, s lkely to e reflected y other parts of e and n the process joned y addtonal emssons, so that when t fnally reaches t wll nearly equal the deal theoretcal ntensty correspondng to e s surface temperature, ndependent of the surface propertes. Under such condtons the net radatve heat transfer etween and e can e satsfactorly calculaton y the approxmate expresson r, e 4 4 ( ) Q& = A ε σ (7) e In the aove equaton, σ s a unversal physcal constant called the Stephan-Boltzmann constant, equal to 5.67(0) 8 W/m K 4. ε s a dmensonless surface property called the emssvty []. Note that heat transfer s proportonal to the dfference of the 4 th power of temperatures 6, ut depends only on the surface area and emssvty ε of ody, and not on those of e. More detaled explanaton for ths apparent asymmetry n the formula wll e dscussed later. he emssvty s the rato of the emssve power of a surface compared to that of a perfect emtter, called a lack ody. For all ut hghly polshed metals, the emssvty s typcally etween 0.7 and. Sample emssvtes for selected surface are shown n ale 3. e a Fg. 3 a. A small ody completely enclosed n a larger ody.. A wndow pane wth one sde exchangng heat radatvely wth nteror walls, furnture and occupants, and another sde exchangng heat wth the exteror radatve envronment, ncludng ground, trees, clouds, mst and mosture n the atmosphere, and deep space. he aove equaton can e generalzed to apply to the radatve exchange etween any small surface exchangng heat radatvely wth a comnaton of odes whch enclose t optcally. Examples nclude the nsde surface of the wndow pane shown n Fg. 3, exchangng heat radatvely wth a comnaton of surfaces whch completely occupy ts vew, ncludng all the nteror walls and the furnture. Note that common glass s opaque to nfrared radaton. he 6 For essentally lnear processes such as convecton and conducton, t s mmateral whether temperature s Celsus or Kelvn. For a non-lnear process such as radaton, t s essental to use the asolute temperature.

10 !H-ref,mmc,rev/5/0prt/5/0 0 nsde surface of the glass therefore cannot communcate radatvely wth the outsde. hermal radaton nsde the room does not pass through the wndow. Smlarly, the exteror surface of the wndowpane exchanges heat radatvely wth the outsde ground, the trees, cloud, droplet-laden atmosphere, and the deep space ackground. In ether case e n the aove equaton s replaced, a sutaly defned mean radatve envronment temperature: y mre r, e 4 4 ( ) Q& = A ε σ (8) mre For certan calculatons t s more convenent to recast these results n terms the more famlar heat transfer coeffcent. We may then defne a radaton heat transfer coeffcent n a manner smlar to the convectve heat transfer coeffcent: r, e r ( ) Q & A h (9) mre Solvng for h from these two equatons, we have, after re-arrangement 3 3 hr = 4ε σ ave + 4 ave f 0. 3 ave ε σ (0) ave where ave ( + mre ) /, mre. he approxmate form s usually adequate snce rarely s greater than 0. 3ave. ypcal values of the radatve heat transfer coeffcent are also lsted n ale, for comparson wth convectve coeffcents. ale 3 Sample Ranges of Emssvtes and Asorptvtes Clean, hghly polshed metals Common metal surfaces Oxdzed metal surfaces Clean Al O 3 (whte), at room temperature 0.7 At C Most surfaces at room temperature, ncludng some apparently whte surfaces Example 4 Radatve Heat ransfer and Comned Radaton and Convecton For the prolem of Example 3, f the mean radatve envronment temperature s also 0 C, determne the radatve heat loss and the total heat loss due to radaton and convecton. Assume the emssvty of heatng element to e 0.9. Soluton

11 !H-ref,mmc,rev/5/0prt/5/0 It s mportant to convert all s to Kelvn: = = =33.5 K, Smlarly, mre =83.5 K, ave =98.5 K. Q& r, e = A ε σ 4 4 ( ) mre w = *0.9*5.67E-8*(33.5^4-83.5^4) [m ][][W/m K 4 ][K 4 ] = W We can next try to solve the prolem usng the radatve heat transfer coeffcent, and see f we get essentally the same result. Snce the nequalty s satsfed, we may use the smple approxmate equaton for h: h Q& r r, e 3 4εσave A h r = 4*0.9*5.67E-8*98.5^3 [] [W/m K 4 ][K 3 ]= 5.40 W/m K ( ) mre a = *5.40*30 [m ][W/m K][K]= 5.9 W he dfference etween ths and the prevous result of W s small. Ar s transparent. Hence convecton and radaton can proceed ndependently and smultaneously. Hence the total heat transfer rate s the sum of the two, or Q & = Q& + Q& = = 3.05 W total free c rad In ths prolem, the ament condtons for oth convecton and radaton are oth equal to 0 C. herefore we can also comne the two heat transfer coeffcents and use a total heat transfer coeffcent equal to =.94 W/m K. Usng ths value Q& total ( + h ) = A hfr c rad = *.94*30 [m ][W/m K][K]= 3.04 W Example 5 Comned Radaton and Convecton -- Inverse Prolem If the heatng element of Examples 3 and 4 s known to dsspate 0 W y convecton and radaton, what s s the surface temperature? Comments: Comnng radaton and convectve heat transfer, we have Q& total 4 4 ( ) + Ah( ) = Aε w σ w mre w o solve for the unknown w nvolves a messy, hghly non-lnear algerac prolem. he dffculty s not allevated y usng the radatve heat transfer coeffcent formulaton, snce wthout a known surface temperature, the radatve heat transfer coeffcent cannot e determned. Hence such a prolem must e solved usng one of the teraton methods. hs example wll e completed n Secton.F.

12 !H-ref,mmc,rev/5/0prt/5/0. Formulaton and Soluton Methodologes A. Conservaton Laws Heat transfer analyss usually requres two types of equatons. he relatonshps dscussed aove, relatng the conductve, convectve and radatve heat fluxes to varous forms of temperature non-unformtes, are called consttutve relatonshps. he second type of equatons are the conservaton laws. he prmary conservaton law used n heat transfer s the conservaton of energy, though sometmes t s supplemented y the conservaton of mass and conservaton of momentum. Beyond the smplest of prolems, at least some of the varales of nterest, ncludng temperature, heat flux, etc., are unknown. he conservaton of energy s then used to te them together n algerac and/or dfferental equatons, whch are then solved to permt calculaton of the desred quanttes. Examples wll e gven n su-secton C elow and n later sectons. In tme-dependent heat transfer prolems and n convecton prolems, changes n nternal energy and enthalpy are often expressed n a per-volume ass as ρ d and ρc p d respectvely. he specfc heats c v and c p always appear n comnaton wth ρ, never can thus e vewed as a sngle property Sample values are shown n ale. It s alone. ρcp useful to note that for solds and lquds, ther values cluster around.5(0) 6 J/m 3 K, and s almost always wthn the range.7 4.(0) 6 J/m 3 K. hs means crude ut relale estmates can often e made wthout detaled knowledge of the two separate propertes ρ and c p. he alty to do ths can e of consderale value n analyss and desgn when precse property values are not avalale. B. Unt and Dmensons Unts play mportant roles n heat transfer analyss. hs s partcularly true n the present transton perod etween tradtonal Amercan unts and the new standard, SI. Whle SI unts are self-consstent, so that no converson s necessary etween work unt [N-m] and the unt of thermal energy [J], ths s not true for the tradtonal Amercan unts. Careless use of nconsstent unts s a frequent cause of errors. Mathematcal expressons ased on physcal laws must also have unt consstency each term n the expresson must e reducle to the same unts. hs s a valuale tool for checkng the correctness of dervatons and calculatons for correctness. Students should develop the hat of checkng unts constantly as he/she s performng the analyss. Once the hat s developed, t can e effortless and s not tme-consumng, ut can often save a lot of the tme wasted on erroneous computatons. Self-consstent unts are derved from just a few asc quanttes, called dmensons. For example, the SI unt of force, N, s really [kg-m/s ], derved through the equaton F = ma. Hence the dmenson for force s [ML/ ], where M, L and denote the dmensons of mass, length and tme. he checkng of dmensonal consstency s clearly a requste for unt consstency. A smple and effectve way to nsure dmensonal consstency s to render equatons dmensonless y makng all varales and parameters n the equaton dmensonless. Examples of such an approach can e found n Secton 4. c v

13 !H-ref,mmc,rev/5/0prt/5/0 3 Another common source of error s the careless treatment of prefxes k, M, m (klo, mega, mll), etc. Whle kg s part of the asc, self-consstent SI system, other unts, such as kj, mm, are not. For example, to determne the velocty correspondng to a gven knetc energy, usng kj n the equaton K.E. = mv / would lead to erroneous answers. If energy s gven n kj, t s necessary to convert t to J efore usng ths equaton to calculate the velocty. C. Resstances, Conductances and Comnaton Rules o hghlght the value of the concepts to e dscussed elow, we shall egn y consderng a smple composte wall consstng of two layers of dfferent materals onded together, an nner layer of thckness B a wth conductvty k a, and a outer layer of thckness B wth conductvty k. he nsde and outsde temperatures are, 3 respectvely. What s the heat transfer rate per unt area? he soluton to ths prolem s not straghtforward, ecause the temperature at the nterface etween the two layers s not known. he concept of resstance wll greatly smplfy the analyss of ths type of prolems. he conducton and convecton expressons, and the approxmate expresson for radaton dscussed n Secton all show the heat transfer rates to e proportonal to the temperature dfference. hs s analogous to the flow of electrc current, whch s proportonal to the voltage dfference across a component. he analogy permts us the defne the heat transfer resstance, denoted y R, as the rato of the temperature dfference and the heat transfer rate, wth unt [K/W]: R [K/W], or Q& () Q& R For example, the conducton resstance for the plane sla, the convectve resstance for an area A wth coeffcent h, and radatve resstance for a small oject n a large enclosure, accordng to Eqs. ()(4) and (0) respectvely, are B Rk, sla = ; Rv = ; Rr, a = = () Ak Ah Ah 4Aε σ where the suscrpts k, v, r denote conducton, convecton, and radaton respectvely. he great eneft of orrowng the resstance concept from electrcal crcutry s that the well-estalshed comnaton rules for electrcal resstances also apply here for heat transfer resstances. hese rules are for the calculaton of the equvalent resultant resstance for comnatons of the ndvdual resstances, accordng to whether they are n seres or parallel: For Seres Resstances: Equvalent resstance equals to sum of ndvdual resstances n R e = R = r a 3 ave for seres resstances (3) For Parallel Resstances: Inverse of equvalent resstance equals sum of nverses of ndvdual resstances

14 !H-ref,mmc,rev/5/0prt/5/0 4 n = R e R = for parallel resstances (4a) An alternatve statement for Eq. (4) s to make use of the concept of conductance, whch s just the nverse of the resstance G (5) R he comnaton rule for parallel resstance can now e stated as Equvalent Conductance s equal to the sum of conductances: n R e = R = for parallel resstances (4) Shown on the rght of the equatons are standard schematc representatons for resstance networks, whch we have also orrowed. hs can e very valuale for constructng schematcs to ad n vsualzng the relatonshp etween dfferent heat transfer components n an overall system. Dervaton of the comnaton rules he orgn of the comnaton rules s the conservaton of energy. he comnaton rule for parallel resstances smply states that the total heat transfer s equal to the sum of heat transfer through all resstances. hs s ecause for each resstance, Q & = / R. Snce s the same for all parallel resstances, the total heat transfer s smply the product of and the sum of all (/R) s. o derve the comnaton rule for seres resstances, we consder the composte wall of two slas, descred n the lead paragraph of ths secton. he prolem conssts of two components (slas), a and, n seres, wth known overall = 3. he temperature of the medan nterface s not known. o solve ths prolem, we oserve that n steady state, conservaton of energy requres oth slas to have the same heat transfer rate Q &. Referrng to the defnton of resstance, we can wrte, for layers a and : a : : 3 = R Q& a = R Q& We have two equatons for the two unknowns Q & and. he algera of elmnatng the s partcularly easy -- we smply add oth sdes of the equatons. he result s a formula for Q & n terms of the overall and ( R + R ), whch s clearly the equvalent resstance: = ( R + R )Q & 3 a hs proves the comnaton rule for two resstances n seres. he proof can e easly extended to any numer of resstances n seres.

15 !H-ref,mmc,rev/5/0prt/5/0 5 Applcaton of the Comnaton Rules he comnaton rules can e used systematcally to reduce a complex system wth multple components to a smple system wth only one equvalent resstance or conductance. hs s llustrated y the followng example. Example 6 An ndustral oven, wth nsde surface temperature at 0 C, conssts of a 0. m thck ceramc wall of conductvty k =.6 W/mK, covered y a layer of nsulaton 0. m thck wth k = 0.05 W/mK. he outsde wall s exposed to ar at 5 C wth a convectve heat transfer coeffcent of 30 W/m K and to a mean radatve envronment at 5 C, wth a radatve heat transfer coeffcent of 7 W/m K. (a) Determne the heat loss per m of the wall. () Determne the nterface temperature f and the surface temperature s. Soluton: he crcut representaton for ths prolem s shown n Rcer Rns Rconv out Rrad We shall frst comne the two parallel components -- the convectve and radatve heat transfer coeffcents. We may use ether Eq. (4a) or (4), though latter s much smpler and smply states that radaton and convecton n ar can proceed ndependently and are addtve. Hence the equvalent resstance R vr for comned convecton and radaton s R vr = = A h =/(30+7) = K/W, for m ( + h ) conv rad hs reduced the prolem to three resstances n seres. he total resstance, for m of area, s hence Bcer Bns R tot = Rcer + Rns + Rvr = + + R Ak Ak cer = 0./ / = =.490 K/W (a) For heat transfer rate Q& = = (0 5)/.490 = 38.5 W per m. R tot tot? f? s () o determne the temperatures, we make use of the relatonshp etween temperature dfference and resstance. Frst, for the ceramc wall ns vr

16 !H-ref,mmc,rev/5/0prt/5/0 6 Q& = R f cer cer n n f =. Hence R cer cer = Q& R = *0.065 = = 7.6 C. Next, for the comned convecton/radaton coeffcents Q& = R s vr vr out = s R vr out vr. Hence = + Q& R = * = = 6.3 C. Unt Area Resstance and the R-Value In Example 6, we computed the heat transfer rate Q & and resstance R for unt area of the wall, ut retaned the unts for the overall heat transfer and resstance, [W] and [K/W] respectvely. If we desre, we can also express the results n terms of unt-area symols and unts. For heat transfer per unt area, ths s the heat flux q& [W/m ]. Snce resstance s nversely proportonal to the area, the unt s [m K/W]. Note that we can not call ths the resstance per unt area or per m ecause of recprocal relatonshp. We shall refer to t as the unt area resstance. he R-value often cted n connecton wth uldng products n the US s the unt area resstance n [ft hr F/Btu]. For example, consder a 3.5 thck layer of ferglass nsulaton. Wth k = W/mK, the unt area resstance value would e B / k = 3.5*.054/0.04 =. m K/W. Converted to tradtonal Amercan unts, we have R-Value =.*3.808^*.8/3.43 =.6 ft hr F/Btu. hs s close to, ut not necessarly equal to the marked R-value of roll nsulaton eng sold n uldng materal stores ecause of varatons of product and manufacturng processes. D. Domnant and Non-domnant Components Most real-lfe prolems n engneerng nvolve more than one component. In these prolems t mmensely mportant to have an apprecaton for what s domnant and what s not. Knowng what s domnant s valuale n desgn teratons, as well as n allocatng tme and effort for analyss. Clearly effort expended n detaled, accurate analyss of a domnant component pays far greater dvdends than effort expended on an unmportant component. For most people the alty and hat to assess quckly what s domnant does not come naturally, ut s developed after repeated practce. Consder the prolem of Example 6. For ths prolem, the overall resstance, for m of area, was found to e.49 K/W. Of ths, the nsulaton accounts for.4 K/W, or aout 96% of total. hs s clearly the domnant resstance. In contrast, the ceramc wall accounts for aout.5%, and the comnaton of convecton and radaton coeffcents accounts for only.%. Suppose some one ponts out that the accuracy of the convecton heat transfer coeffcent s very low, only 30%. Is t worthwhle for us to re-evaluate these coeffcents? Proaly not. hs s ecause a 30% error of.% of the total resstance s only 0.33%, whch s clearly too low to

17 !H-ref,mmc,rev/5/0prt/5/0 7 e of concern. If we wsh to mprove the accuracy of the overall computaton, we must focus on the thckness of the nsulaton and the accuracy of the value of thermal conductvty. Smlar consderatons may also suggest that we smply neglect the radatve contruton to heat transfer. Snce t s much smaller than convecton, and snce even convecton s not too mportant, the neglect of radaton would not ntroduce sgnfcant errors n ths case. Suppose we wsh to reduce the total heat loss. Clearly ncreasng the thckness of the ceramc wall s not cost-effectve, snce t accounts for only.5% of the total resstance. A small ncrease of the thckness of the nsulaton would accomplsh the goal much more economcally. Conversely, we may desre to change the desgn to ncrease the effcency of heat transfer, n order to reduce the nsde surface temperature for the same heat transfer rate. Clearly ncreasng the ar coolng would not e effectve, snce convecton only accounts for a neglgle fracton of the resstance. Decreasng the thckness of nsulaton s the most effectve measure for ths purpose. E. General Method for Lnear Networks Not all networks can e smplfed y comnaton rules. For example, the smple network n Fg. 4. s a non-comnale network. Fortunately, there s a general method for solvng all lnear resstance networks. We wll frst descre the method for the smple network shown n Fg. 4.. Afterwards the general procedure wll e descred. Let and and all the fve resstances e known. We wsh to determne the total heat transfer from node to node 4, along wth the temperatures at nodes and 3. Conservaton of energy requres that n steady state, the total heat transfer toward any node must vansh. Applyng ths prncple to nodes and 3, we have R a R R c R c R + 4 d R e 3 = 0 = 0 hs consttute a set of two lnear algerac equatons, wth two unknowns and. hese of course can e solved n a straghtforward manner, y any of the many methods for lnear algerac equatons. Once and are known, the heat transfer can e otaned. (6) (7) R 3 Re Rc 4 Ra Rd Fg. 4. Example of a non-comnale resstance network

18 !H-ref,mmc,rev/5/0prt/5/0 8 hs procedure can e extended to any network wth any numer of unknown nodal temperatures. Invokng the conservaton of energy wll usually lead to one equaton for each node wth an unknown temperature. Soluton of these equatons y Gaussan elmnaton or other matrx nverson schemes s straghtforward. Iteraton methods may also e used. Once the unknown temperatures have een found, the computaton of heat transfer rates follows drectly. F. emperature-dependent Coeffcents, Non-lneartes here s a potental catch- stuaton that occur frequently n heat transfer analyss. Most resstances and conductances are temperature-dependent. he temperature-dependence of the coeffcents for free convecton and radaton s qute ovous. Even conducton and forced convecton are at least somewhat temperature-dependent, ecause the conductvty and other physcal propertes are temperature dependent. When we egn to solve a heat transfer prolem, the temperature dstruton s frequently not known. How can we proceed? he dffculty s a symptom of the non-lnearty of the prolem. Wth temperaturedependent resstances and conductances, nether the resstance-comnaton method nor the smultaneous equaton method descred n su-sectons C and E apples. he answer to ths dffculty s teraton. Rather than attemptng to solve the prolem drectly, many teraton methods are far more effcent n ths type of stuatons. For example, the Newton-Raphson method would e qute effectve. For most heat transfer prolem, however, a specal form of the successve susttuton method s oth effcent and ntutvely easy to understand. For smple prolems, the method can e easly mplemented wth a programmale computer. For ether smple or complex prolems, t can also e easly mplemented wth a spreadsheet program such as Excel. o start the teraton process, reasonale guesses (these are called frst terates ) are frst assumed for each of the unknown temperatures requred for the evaluaton of the coeffcents. hese wll permt the resstances to e determned. Wth known tentatve resstances, the crcut can e constructed and solved, as f t were a lnear prolem. he temperatures from ths soluton are often mprovements over the frst terates, and can e used to as second terates to evaluate mproved coeffcents and resstances. he processes can e repeated untl the temperatures no longer change. he valty of the method depends on the fact that frequently, the coeffcents are only weakly dependent on temperature. hus a gven relatve error n temperature would lead to a smaller relatve error n the coeffcents. When these coeffcents are used to compute new temperatures, the relatve error would e smaller than those orgnally assumed. As n all teraton methods, occasonally condtons are such that succeedng terates do not represent mproved values, ut may ether oscllate or contnue to change wthout settlng down to a converged value. In ths case the teraton s sad to dverge and another teraton method must e found. Fortunately, the successve susttuton method descred here wll converge for most heat transfer prolems. We wll llustrate ths method y completng the soluton of Example 5. Example 5 (contnued from Secton.D)

19 !H-ref,mmc,rev/5/0prt/5/0 9 hs s the same prolem as Example 4, except that the surface temperature s unknown and that the heatng element s dsspatng 0 W. Wth mre = ( )( h h ) Q & + = A s fc r D = 0.07 m, A = m (Prevously computed. End surfaces neglected) For frst terate, assume surface temperature to e 33.5 K (40 C). he free convecton and radaton heat transfer coeffcents would then e W/m K and 5.40 W/m K respectvely. R = s ( + ) A h fc h r = /( *( )) =.30 K/W = + Q& R = *.30=39.7 K total Wth ths surface temperature, we can re-evaluate the free convecton and radaton heat transfer coeffcents h fr c =. 55 ( ) D ( + )/ =.55*( )^0.3/0.07^0.= W/m K ave = s = ( )/ = K h ra 3 4ε a σave = 4*0.9*5.67E-8*306.6^3 = W/m K hese can the used to calculate R, s n turn, startng a new cycle of teraton. In sx teratons 7, the converged value for the surface temperature s found to e 35.3 K. Iteraton numer s Assumed hfr-c ave hr (Exact) R s Computed * It s nterestng to note that should the gven power of the heatng element e 400 W or hgher nstead of 0 W, the teraton process descred here would dverge. hs s ecause at the hgh temperatures correspondng to these hgh power levels, radaton domnates. he radaton heat transfer coeffcent at thse hgh temperatures s very strongly temperature-dependent.

20 !H-ref,mmc,rev/5/0prt/5/ *Iteraton can stop at ths pont wth less than 0.3% error. 3. Steady Conducton through Non-planer Geometres A. he Dfferental Equaton for Heat Conducton for Constant Densty Meda (hs susecton s for your enrchment only. horough understandng not requred.) Consder a sold or a statonary flud wth non-unform ut steady temperature ( x, y, z). We wll see that ( x, y, z) oeys a dfferental equaton that can e otaned from the conservaton of energy. he dfferental equaton can then e solved to determne the heat transfer ehavor of complex ody shapes. We wll conser the conservaton of energy for a very small volume element dxdydz center at locaton (x,y,z). Snce qx denotes the heat flux n the x-drecton, ( dx)( qx / x) denotes the dfference etween the heat flux on the rght surface and that on the left surface. Hence the net rate of outward heat transfer due to q x for ths element s qx ( dydz) dx x where the product n the parenthess represents the cross-sectonal area. Smlarly, the net rate of outward heat transfer due to qy and q z for ths element are q y qz ( dzdx) dy, ( dxdy) dz y z he sum of these three terms s the total rate of outward heat transfer for the element dxdydz. Conservaton of energy states that, n the asence of volume change (whch leads to work), the total rate of outward heat transfer must e equal to the rate of decrease of nternal energy, equal to the mass ( dxdydz )ρ tmes the rate of decrease of specfc nternal energy ( u / t). Puttng these together, we have u t q x q x y ( dxdydz) = ( dydz) dx + ( dzdx) dy + ( dxdy) y q dz z Recognzng that ( u / t) = c( / t), and makng use of Eq. (3) and ts counterparts n y- and z- drectons, the dfferental equaton for heat conducton ecomes ρ c = k + k + k (8) t x x y y z z z

21 !H-ref,mmc,rev/5/0prt/5/0 Frequently smpler forms of ths equaton are satsfactory. For example, for constant thermal conductvty, k can e moved n front of the dfferentaton sgn and wth ρc to form α = k/ρc: = α + + (9) t x y z In cylndrcal coordnates wth ax-symmetry (.e., ndependent of θ ), the same equaton takes the form = α r + r (0) t r r z For -D ax-symmetrc prolems n polar coordnates, the z-dervatve term s omtted. For steady state prolems, the tme dervatves vansh, the reduced equatons just consst of contents of the parentheses set equal to zero. B. Steady Conducton through a Cylndrcal Shell We now consder a cylndrcal shell of length L, and nner and outer rad R and R o, at temperatures and o respectvely. Although ths prolem can e solved from the more general partal dfferental equaton dscussed n the last secton, the smple geometry permts a straghtforward formulaton n ordnary dfferental equatons, whch wll e llustrated here. he cross secton of the shell s shown n Fg. 5. he total heat transfer rate Q at any radus r s gven y Eq. (3) to e Q d = () () r L( πr) dr In ths case we expect the heat flux qr to decrease as r ncreases, ecause the same total heat transfer s spread over a larger cylndrcal area. So the dfferental equaton requred s otaned y requred Q to e constant and unchangng: d d d Q() r = 0 or r = 0 f k = unform 8 () dr dr dr he oundary condtons are ( R ) = ( Ro ) o, = (3) r 8 hs equaton can also e otaned from Eq. (0) y settng the tme and z-dervatves equal to zero. R R o

22 !H-ref,mmc,rev/5/0prt/5/0 Fg. 5. he Cross-secton of a cylndrcal shell hs system can e solved easly y two successve ntegratons and evaluatng the constants wth the help of oundary condtons. he results s πl Q = R ln R o k C. More Complex Shapes, Conducton Shape Factors For more complex ody shapes than the plane sla and the cylndrcal shell, the prncple of the method of analyss s the same. he only dfference s that n most cases the conservaton of energy leads to a partal dfferental equaton 9 + x y = 0 wth oundary condtons gven at all exteror surfaces. Methods for solvng such equatons are eyond the scope of ths ref ntroducton. hey can e found n mathematcal texts, especally those on numercal methods. For frequently encountered ody shapes, however, such solutons have already een otaned y engneers and taulated n reference sources. We wll refly descre the use of such a resource. It can e shown from dmensonal consderatons that all solutons of steady heat condton prolems wth an mposed are n the followng format: Q = Sk (6) he factor S, called the Conducton Shape factor, contans all the mportant nformaton on the shape and sze of the ody. It has the dmenson [L], whch s a consequence of the fact that heat transfer rate s proportonal to the cross secton area, of dmenson [L ], and nversely proportonal to the length of the heat conducton path, of dmenson [L]. aulatons of the conducton shape factor can e found n many heat transfer reference sources, ncludng most (4) (5) 9 hs can e otaned from Eq. (9) y settng the parenthess = 0 and omttng the z-dervatve term.

23 !H-ref,mmc,rev/5/0prt/5/0 3 text ooks. An arevated lst of some of the most useful shape factors s shown n ale 4. for llustratve purposes and to provde for smple calculatons. D. emperature-dependent Conductvty and the Average Conductvty hermal conductvtes are strong functons of temperature. For prolems spannng a small enough temperature range, ths varaton may e nconsequental. Assumng the conductvty to e constant and equal to any representatve value wthn the range would lead to lttle error. On the other hand, f the varaton of conductvty s large, one may ask f there s a technque to avod the complcatons of solvng non-lnear equatons? he answer s yes, at least for steady state conducton prolems. For steady conducton prolems, t can e shown that the average thermal conductvty defned n the usual sense wll yeld the correct heat transfer calculatons: k kd (7) ale 4. A Short Lst of Selected Conducton Shape Factors S Q Sk Descrpton Schematc Condtons Shape Factor Eccentrc cylndrcal shell of length L, dameters D & D o, and eccentrcty e. Rectangular shell of length L, wth square cross sectons of sdes D & D o. Addtve correcton for edge of ntersectng walls, when walls have een computed usng plane sla formula Cylnder, of dam. D & length L, surface at, ured n semnfnte medum normal to plane D o B L e D o D o > D and L >> D Do + D 4e o cosh DoD D o / D <.4 D o / D >.4 π L π L ln D o D ( / ) π L 0. 93ln D o D L / B > L ( / ) π L L >> D ln( 4L / D)

24 !H-ref,mmc,rev/5/0prt/5/0 4 surface, at Example 7. Determne the heat transfer rate per meter for a.5 cm dam. ppe at 00 C, nsulated wth cm of ferglass nsulaton wth the outer surface temperature equal to 30 C. Soluton: We wll solve ths prolem wth two methods and compare the results. Method. We wll use Eq. (4) for the cylndrcal shell. πl Q = R ln R o k =*π*l*k*(00 30)/ln(4.5/.5) =*π**0.046*70/ln(4.5/.5)=34.4 W Method, We wll use the entry for eccentrc cylnders n the shape factor tale, and settng the eccentrcty equal to zero. π L π L Q = Sk = k = k cosh Do + D e cosh + * 4. 5 *. 5 DoD π L π L π = k = k = * 70 = W cosh (. 777) 4. Convecton Heat ransfer Coeffcents A. Correlatons of h n terms of Nusselt Numers We have prevously ntroduced the heat transfer coeffcent h for the computaton of convectve heat transfer: q& = h ; where w flud (8) Only two sample formulas for determnng h were gven. We shall now dscuss how to determne h for wde ranges of condtons. In prncple, h can e evaluated from solutons of the partal dfferental equatons of heat transfer and flud flow. hs can e qute nvolved and s the concern of more advanced studes of convecton. For smple and frequently encountered flow confguratons, use can made of complatons of formulas and correlatons already estalshed y analyss and experments. hese can e found for n handooks and textooks of elementary heat transfer. hese formulas are typcally gven n the form of a dmensonless parameter called Nusselt numer, defned as hl/k, where L s a characterstc length, ether length, heght or dameter as s approprate for the prolem. For forced convecton when the velocty s mposed, the Nusselt numer s represented as a functon of the Reynolds numer Re UL/ν, a dmensonless velocty for vscous flow, and the Prandtl numer Pr ν /α, a flud property equal to the rato of momentum

25 !H-ref,mmc,rev/5/0prt/5/0 5 and heat dffusvtes. ν denotes the knematc vscosty [m /s], equal to µ /ρ, the dynamc vscosty [N-s/m or kg/s-m] dvded y densty [kg/m 3 ]. One physcal nterpretaton of ν s the dffusvty for momentum. α s the heat dffusvty [m /s], k/ρcp. For free convecton, when convecton s due to uoyancy assocated wth the temperature dstruton, the velocty s unknown. In ths case the Nusselt numer s typcally gven as a functon of the Raylegh numer Ra gβ L 3 /να. o avod amguty n the choce of the length parameter, frequently Nu, Re and Ra are wrtten wth the proper length parameter as a suscrpt. For example, Nu D, Re D, Ra D use the dameter D for the length parameter L n ther defntons. As a general rule, flud propertes can vary sgnfcantly etween the surface and the flud nteror. A short lst of the most mportant correlatons are gven elow as examples and for the convenence of students. More complete lsts can e found n dedcated textooks. hese are often dvded nto groups or chapters for nternal forced convecton, external forced convecton, and free convecton. Sample flud propertes requred for these calculatons are gven n ale 5. a. urulent Flow nsde Crcular ues hs s an example of nternal forced convecton. A frequently cted correlaton s Nu D = 0.03 ReD. Pr / for ReD > 0, L / D > 0, 0.7 < Pr < 60 (9) In ths equaton, the Nusselt and Reynolds numers are ased on the nternal ppe dameter, and the velocty n the Reynolds numer s the average flud velocty (suscrpt af) over the cross-sectonal area of the ppe. hd UaveD Nu D ; ReD (30) k ν U af c A c 4 Volumetrc flow rate Mass flow rate uda = = (3) A A ρa c he defnton of the heat transfer coeffcent, and the computaton of heat transfer, also make use of the average flud (suscrpt af) temperature: q w af ( ) h (3) U w A af c af A c uda For many prolems the flud propertes may change sgnfcantly etween the wall temperature and the flud temperature. Some correlatons gve specfc remedes to account for ths change. A general procedure, sutale for modest varatons of property or when etter nformaton s not avalale, s to evaluate the propertes at the flm temperature, taken to e the md-pont etween the wall temperature and the flud temperature: c (33)