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# Fins - Heat Transfer from Extended Surfaces; 2D SS Conduction 3521

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This 64 page Class Notes was uploaded by Connor Barber on Friday January 30, 2015. The Class Notes belongs to 3521 at Ohio State University taught by Koelling in Spring2015. Since its upload, it has received 160 views. For similar materials see Transport 2 in Chemical Engineering at Ohio State University.

## Reviews for Fins - Heat Transfer from Extended Surfaces; 2D SS Conduction

Clutch. So clutch. Thank you sooo much Connor!!! Thanks so much for your help! Needed it bad lol

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Date Created: 01/30/15

Heat Transfer from Extended Surfaces Fins Driving force for heat transfer Fourier s Law Newton s Law of Cooling Heat transfer from extended surfaces OBJECTIVES Describe why fins are used and how they can enhance heat transfer Derive the governing equation for heat transfer in fins and summarize the four solutions for different boundary conditions Apply the appropriate fin equation to solve heat transfer problems in which fins are present Define fin effectiveness and fin efficiency and describe how these parameters are used to quantify the performance of fins HEAT TRANSFER FROM EXTENDED SURFACES Extended surface is commonly used to depict an important special case involving heat transfer by conduction within a solid and heat transfer by convection andor radiation from the boundaries of the solid Direction of heat transfer from the boundaries is perpendicular to the principal direction for heat transfer in the solid Extended surfaces used to enhance heat transfer between a solid and surrounding fluid Extended surface is called a fin EXAMPLES OF FINS EXAMPLES OF FINS TAPS A significant quantity of oil is transported across Alaska in the Trans Alaska Pipeline System TAPS About half of the BOOmile long TAPS is built above permafrost that needs to stay frozen to keep the pipeline stable Engineers have placed 124300 fins on the TAPS vertical support member that are on permafrost terrain two per suppon Fins transfer ground heat to the atmosphere whenever the ground temperature exceeds the air temperature EXAMPLES OF FINS TAPS Fins are actually heat pipes with anhydrous ammonia refrigerant or carbon dioxide that vaporizes below ground level rises and condenses above ground Heat is conducted from the ground into the pipes rises with the vapor and is conducted through the heat pipe walls and out into the cooler air through aluminum radiators at the top of each pipe EXAMPLES OF FINS TAPS Trams Alaska ipeiliine Passive Cunlimp Aluminum Fina an IEFL39 39 Ii 151151 FL I E quoth HIE 11 HE RH E M NHIM E39I H1 LEHLEE 39 Flu ljrm 1 1 IQrmwu 39339 561quot 113 th li r n mihimi H 3 h 1 1 IL erumn an FM 39i il39whml l warmtw rings I rigmvnn I Tia r11 TH 33913 my 1 11 rt h m 3 1 m y E IiFili IEHI IJ39LlL I L Erlr1iih i39lrlrlh39lvlh 2 la Him gum a wna m mm m 1m Inna E 1 in Jim 5 gm milk Enquot imn l 1 in L 5 Hal l H Hm i 1 w 1 A F k PiF Eli 39 Ser ce u COMBINED CONDUCTION AND CONVECTION With T1gtT2 temperature f gradients in the xdirection i w L sustain heat transfer by I conduction in the strut i 39i x With T1gtT2gtTOO there is concurrent heat transfer by T convection to the fluid qX and the magnitude of the temperature gradient decrease with increasing x HEAT TRANSFER FROM EXTENDED SURFACES q 2 Too Options Decrease TOO lncrease flowrate to increase h Plane wall at temperature Increase surface TS and area A area across Convection BC with TOO ltTS whlch convectlon Goal increase heat OCCUFS transfer from plane wall TIES it Thermal conductivity of fin has strong effect on the temperature distribution along the fin Directly impacts the enhancement of heat transfer rate ldeal fin characteristics High thermal conductivity of fin to minimize temperature variations from the base to the tip lnfinite fin k results in entire fin at temperature of base FIN CONFIGURATIONS Factors in selection of the fin configuration Space weight manufacturing method cost increase in pressure drop and the extent to which fins increase heat transfer Mathematical analysis of heat transfer enhancement depends on the crosssectional area variation and on the boundary conditions at the tip 4 general cases ltuOgt nZZmU Cwm Imgt mXOIgtZQmEm GENERAL CONDUCTION ANALYSIS Goal of analysis Calculation of extent to which particular extended surfaces or fin arrangements improves heat transfer from a surface to the surrounding fluid Must determine the temperature distribution along a fin GENERAL CONDUCTION ANALYSIS Conservation of energy qx qxAx dqwnv T Fourier39s Law qx kAC d dx Where AC is the cross sectional area which may vary with X Since the conduction heat rate at X dX may be expressed as dqx qxAx qx d dx x itfollowsthaLIT d dT diAC CZs Tw qxAx k14c k Ac jdx X X X dx dx dx dZT 1 1AC dT 1 h 1AS T T Convection heat transfer rate dxz AC dX dX AC F dX 00 O quOl lV TS Substitute rate equations into the Solution wnth appropriate BC provndes Tdlstrlbutlon Conservation of Energy Eqn Fourier s Law can calculate conduction rate at any x FINS WITH UNIFORM CROSSSECTIONAL AREA 2 d7239Jr 1 dAC dT 1 hdAS TTOOO dX AC dX dX Ack dX Twig Transform differential 39 quot equation by defining an excess temperature One BC at the base of the fin convection at tip adiabatic tip 39 fixed temperature Tb 1 infinite length fin gs Fz UCppgt P22w NM 4 1553quot AL DE FINS WITH UNIFORM CROSSSECTIONAL AREA d27 1 dA dT 1 h 0 1A ST TO 1X2 1X AC dX Cross sectional area is a constant ASP x dAC dX a AS dX d2ThPTTOOO a x2 kAC O P FINS WITH UNIFORM CROSSSECTIONAL AREA ija T gto Define m2 h P kA C d 26 dx2 9x Clem C2 BC16 OTb TOO 6b BC 2 at fin tip X L m26 O FINS WITH UNIFORM CROSSSECTIONAL AREA Case A Convection hx at tip 9X Clemquot Cze mx BC 1 90 Tb Too ab Humi BC 2 hACTL Too kAC H L Wmm BC 2 h6L k H I coshmL x ISsin m k ab coshmLn kjsinhmL Note slope decreases with increasing x reduction in conduction hx due to continuous convec onlosses FINS WITH UNIFORM CROSSSECTIONAL AREA To obtain the fin heat transfer Case A Convection hx at tip rate of apply Fourier s Law at the fin base Hump T1 m d T d6 tit tilt l Ami T MATCH m qf qb kAC I 1 H X X0 dX X0 sinhmL ooshmL qf l ooshmL sinhmL FINS WITH UNIFORM CROSSSECTIONAL AREA Case B Adiabatic tip 6x Clemx 026 BC 1 Tb Too 2 6b T cmnf39quoti quot 31 g r 39 d BC 2 I E Z 0 w I c llzi dx XZL dX H i coshmL x 6quot coshmL qf JthAC 6btanhmL 0 FINS WITH UNIFORM CROSSSECTIONAL AREA Case C Prescribed T at tip 6x Clemquot Cze mquot BC 1 Tb TOO 6b qmnuilr Tm BC 2 6L mu mfglL TTL l 6L sinhmx sinhmL x E 2 9b 6b SinhmL coshmL q thAC6b SinhmL b FINS WITH UNIFORM CROSSSECTIONAL AREA Case D Infinitely long fin 9X Clemx 026 BC 1 Tb Too 6b IL qmll L gtoo BC 2 9L 2 0 q 39 6 mX e 6b Qf FINS WITH UNIFORM CROSSSECTIONAL AREA T tllpt it39illlll39t Jli llt ihlllil il j and mat 1th I39mquot Iinre if Hirl ll l Pl r s srtrtinn Tip innfli mj FEtnpsErutllw Fin H t i i39EE L1 L Di att39i ltn lit i in T t 39 f l Hate ch meg mn mm CUSI 23rth I hfmk sinh mILL I 1 Ei h mL Miriam 435 mL Hf fll afar Ir hfm kd gdxlrt Liam ML Harm sinh mL 1 cash mL U m Si ll ML E ailiahaItia E Eh mL I M I h mL d fdllrd 339 1311511 mL C Flr gmfih d Ell lp r ll lfll ML L g sinh Hut Ei h HEEL I H fan39s3h NIL ELEM Jr smh mL quot Si 11h 3HL 3 In x t n L At 3930 E31731 U M 39 I M 2 T I m 2 Ifo g my in Tm M wxpmf b Hum 1 A qmnv I IF FINS PEFi lFOFi lMANCE Fins are used to increase the heat transfer from a surface by increasing the effective surface area Recall from the discussion about critical insulation thickness that an increase in area also means an increase in the conduction resistance from the original surface There is no assurance that the heat transfer will be enhanced through the use of fins Calculation of the fin effectiveness can determine whether the use of fins is worthwhile Ratio of fin heat transfer rate to the heat transfer rate that would exist without the f in FINS PEFlFOH 8f qf hACib b Acib fin cross sectional area at base Use of fins rarely justified unless sf 2 2 kP For case D 8 m Fin effectiveness enhanced by MANCE Wypitttl lir39illllt ii it tho rm wcr39limt lwal transfer m irFIitriiti i39l Pruitt fry irting Iii Free convection Gases 2 25 th S il Form cmwectim Gases EE EE U thit s ll E t Catt39ccti il with JillHEquot sittings Boiling iE liiEli 39fli ii i il Z Zl lil g i a choice of material with high thermal conductivity b increasing ratio of perimeter to crosssectional area thin fins c low convection coefficient need for fins greater with gas than liquid FIN EFFICIENCY An alternate means to assess fin thermal performance is quantified by the fin efficiency Maximum driving potential for convection is the temperature difference at the base Maximum rate fin could dissipate energy is the rate that would exist if the entire fin surface were at the base temperature qm ax hAf 6b 77f FIN EFFICIENCY Ef fi uy If immmm n lmingJ Straight Fina Emmwga w 11 ElkLt if L If A IL V Ealnhmi IIj mil Trimagaarimf quot EI IJLLE 1332 1 Mimi 1539 m1 guran P mh 131 Iran39Erlile EHgun LIL 13 E1 I ME A magi E mm 1 2 1 Hf Circular Fin RECIWIELIEW Air E FEE 2 3 HE I 1 4 Ff Fin Fina REFIWHIEHIIHIJI 5in w L fd I 32311 1L T f ii39iIEe I rb 2 I If 1 2351 L E DIE 3911 FIN EFFICIENCY Equla H Kim n m I JIM ch Iquot 1 f illi39fil 1E RIPE II 2 I ljnli39rlj mzi ng Humirijlfm r33 Emma F Fig I I 39 r Iat1IJHit III 1ch 7 E IEEIHL WHAT YOU NEED TO BE ABLE TO DO Describe situations in which fins can enhance heat transfer Identify conduction and convection processes that occur in a fin Describe the ideal material for a fin and identify the important properties for the ideal fin material Identify the appropriate boundary condition and use the equations for the temperature distribution and fin heat transfer rate to solve problems with fins Know how to quantify fin performance in terms of fin efficiency and fin effectiveness WORKSHEET 8 Heat Transfer In Extended Surfaces WORKSHEET 9 Temperature Distribution In Extended Surfaces 2D Steady State Conduction 2 D steadystate conduction solution Finite difference derivation and solution OBJECTIVES Identify key differences between 1D and 2D steadystate conduction Simplify the heat equation for 2D 88 conduction and introduce several solution methods for the resulting PDE Introduce the finite difference technique and demonstrate how this technique can be used to solve 2D and 3D 88 conduction problems 2D SS CONDUCTION In many cases a 1D analysis is not sufficient to accurately capture the temperature distribution Mathematics involved in solution of multi dimensional temperature distributions are more challenging Focus on two different approaches for 2D systems Exact solutions idealized conditions Numerical methods finite difference technique 2 D SS CONDUCTION Prismatic solid with 2D heat conduction Fourier s Law is applicable at every point Local heat flux is a vector that is everywhere perpendicular to isotherms Heat flow lines adiabats No heat can be conducted across an adiabat q at it in Iul 3951quot 1 2D SS CONDUCTION Two objectives in conduction analysis Determine the temperature distribution in the medium Solve the heat equation to determine Txy Determine the heat flux components Use Fourier s Law to determine qXA and qyA V 439 Ii quot let Jag A I In L 1391 Iatttherma JITF it t I ii Jt gt 39t t 39iI w w Hg t Hm t at Isuth arm 2D SS CONDUCTION Heat equation 82T02T82T zi 8X2 ay2 822 k a 81 82T 82T 0 8X2 8y2 ibk A ax Founer s Law irkg A 8y H 39 539 F n39 39 IE tHEFTTIEa Heat Hm Vi I th rm quotHES 2 1 3 2D ss CONDUCTION EXACT SOL N Thin rectangular plate 3 Assume negligible heat W transfer from surfaces of plate mm W lm Neglect Tgradients normal to x y plane L zo 2 D conduction I 9 Solution using Separation of 327 527 0 Variables 6X2 8 Define a TT1 2 1 2 2 aaaa0 8X2 8y2 2D ss CONDUCTION EXACT SOL N 826 826 Z 0 Transformation H rgu 1 8X2 8y2 results in 3 39 I301 905y0 homogeneeus BC Kirk W 3h BC 2 6L y O and normalized BC3 9XOO dependent a L4 5 variable 3 Lhasa BC4 9XW1 a TE Qw mw dgt En 1d2X 1 sz X a x2 a y2 Key LHS depends only on X RHS depends only on y equality can apply in general for any X or y only if both sides constant 2D SS CONDUCTION EXACT SOL N 6xyXltxgtYltygt 3 1de isz 12 1quot X dxz Y dyz X C1 cos x C2 sin1x Y 2 C362 C4ely DD lax 1119 6 C1 cos1x C2 sin xC3ely C462 6 TE BC1 9OyO C1O T7 2 1 BC3 6x 00 C3 C4 2D SS CONDUCTION EXACT SOL N 9 6xy W W 6 On sinsinh j L L TzTl Irishm T T1 HE39T l iTEd 1 j l I 39E39 L 6 2010 sinsinh j n1 L L BC4 6XW1 Cn n 123 n7rsinh n j M sinh E 32039 1 mm L a 7rn1 n L er smh L 3 2D ss CONDUCTION EXACT SOL N Exact solutions have been obtained for numerous geometries and boundary conditions Texts such as Conduction of Heat in Solids by Carslaw and Jaeger present exact solutions for numerous geometries boundary conditions and coordinate systems FINITE DIFFERENCE EQUATIONS Usually 2D problems have geometries andor boundary conditions that preclude analytic solutions Problems require numerical techniques Finite difference finite element or boundary element Numerical techniques provide the temperature at discrete points not at any point in the medium Use of numerical methods dominates the multi dimensional heat transfer analysis Numerous packages used for complex geometries FINITE DIFFEP points in the medium lENCE EQUATIONS First step in numerical analysis is definition of Subdivide medium of interest into a number of small regions Assign reference point to each region in its center node Aggregate of points nodal network grid or mesh T of node is average temperature in region Selection of nodes depends on geometric convenience and desired accuracy xquot gil r 1 quot 39W39 quotI l l39l F A I39 v51 FINITE DIFFERENCE FORM OF HEAT EQUATION Determination of a temperature distribution numerically requires that an approximate conservation equation be written for each node of unknown temperature Resulting set of equations must be solved simulataneously to obtain the temperature at each node Start with the heat equation and define a finite difference form quotMm u it h h r trig 31 1 Vi39ilf L E FINITE DIFFERENCE FORM OF HEAT EQUATION aZT aZT 8x2 8y2 5LT 5T 8x 8x z Ax O 2 T Hl l a T 399 8x2 1 1 m m m m 2 2 mm FINITE DIFFERENCE FORM OF HEAT EQUATION aZT aZT 2 2 0 8x 8y aZT Tm1n Tm 1n 2Tmn 8x2 m n Ax2 aZT Tmn1 Tmn 1 2Tmn 8W m At2 T mn1 T mn 1 T m1n Tm1n 4Tmn O Ax Ay For each mn node the heat equation exact differential equation is reduced to an approximate algebraic equation Approximate finite difference form of the heat equation can be applied at any interior node that is equidistant from its four neighboring nodes ENEHGY BALANCE METHOD Alternate derivation of finite difference equations can be done to permit this technique to be used on nodes other than interior nodes Can analyze phenomena such as problems involving multiple materials embedded heat sources and boundaries Finite difference equations obtained by applying conservation of energy to a control volume centered about the node Assume all heat flow is into the node Neglect accumulation unsteady state will be covered later in class 82T 82T O V riff 1 8X2 8y2 Jr i i 7 I E in E gen O rii fir swiftr 5 Eli all 4 Z qi gtmn 0 i1 ENERGY BALANCE METHOD Assume conduction process occurs through lanes that are oriented in the x or ydirection Assume a unit depth AE1 Tm n mn qm 1n gtmn O LAX Heat rate Finite difference approximation to T gradient at node boundary Area Thermal conductivity h L i u ENERGY BALANCE METHOD 4 Zqi gtmn 39 39 1 0 i1 qm 1n gtmn 39 m llnA m n If If 1 Eur qm1rn mrn m1lnAX m l Ii39quotP 1Ti mi E I Tm n Tm n qmn 1 gtmn I I my 1 li i Tm n Tm n qmn1 gtmn I I ENERGY BALANCE METHOD PROBLEM Using the energy balance method derive the finite difference equation for the mn node located on a plane insulated surface of a medium with uniform heat generation fir iiil 1 lineulated eurface ff 5 T quotJim if itIT I m 1 H I Unit depth normal 31 it Gil iquot i m gamer L 3 l fil39l T 1391 1ij m it n 1 it i u f ri ENERGY BALANCE METHOD PROBLEM Emil m H 1 Hngulatad EUFfI EE Ax Q1Q2Q3Q4Q739Ay39120 1 mt eri H in J quot quot T y 9 3 Unit depth Harmal i ta paper NH 395 5 J W J a I 39 3939 le 39I I H1 3n 32 m 43 FF Ax T T m H 1 2D SS CONDUCTION EQUATIONS 3111111112113 If ZILIIITZIIIJHII llilquotIEquot liff 39fillf ti Ellil lljil li ll Z Eulr gumtziazm Fil lit m i r i Equatim Err ix 2 l mi ii l at 1 LF il39i i F1 m 15 riquoti39i391ii39 Tm Elm J EarHim rm tilT I39 m WEN l 1quot Fur r u it Ir WEN1 Case hlt ftfi l39 nuclei 1r a Emmtl LEJI4 LI39I TEEl mli ha u I zfa mM 3 E a E E m C e H de at an illi izll39 l E t i 39t with cmwa i n 14 TW ll39 l39 EJI FJ Tam 1 k JET E ii I JI 235 Wild at 1 131mm SUff EE with EvJ J vEE39Ti l l 51 2 9 431 441 2D SS CONDUCTION EQUATIONS Slungnlary If 11min llrljit i liff HIP eq11e11ei11115 C n glll ti FiniteH Fferente Equmi n ml L LJ l is if st quotaquot r g E 1 Finlin 1 Tin E I 1 J Elli Case Nadia at an emner with enln vectin n J n it in 3 7 ME El n 14 T ll l Twin 1i E 439Em Lit 14 Case blade at a plan e 511 rf aee with L mifnrm heat u a b To obtain the finitedifference equation for an adiabatic insulated surface simply set h0 SOLUTION METHODS Matrix Inversion Expression of system of N finitedifference equations for N unknown nodal temperatures as A T C Coef Cient Solution Vector Righthand Side Vector of Constants Matnx NxN T1T2 TN C1C2CN I 1 Solution gt T A C Inverse of Coefficient Matrix GaussSeidel Iteration Each finitedifference equation is written in explicit form such that its unknown nodal temperature appears alone on the left hand side where i1 N and kis the level of iteration Iteration proceeds unti satisfactory convergence is achieved for all nodes Tik Tim I I lt8 FINITE DIFFERENCE SOLUTIONS What measures may be taken to insure that the results of a finitedifference solution provide an accurate prediction of the temperature field 1 A check should be made to verify that the results satisfy conservation of energy for the nodal network For steadystate conditions the rate of energy inflow must be balanced with the rate of outflow for a control surface about nodes where all adjacent temperatures have been evaluated 2 Perform grid studies to evaluate the results as a function of grid spacing Note that as Ax and Ay are reduced by a factor of two the number of finite difference equations is increased by a factor of four FINITE DIFFERENCE EXAMPLE Determine the steadystate temperatures of four equally spaced interior nodal points on the slab depicted below 300 0C Algorithm A B 1 Break up system into rid 4000c 200 C 9 C D 2 Derive finite difference equations 500 0C for each node 3 Solve simultaneous equations to obtain unknown temperatures FINITE DIFFERENCE EXAMPLE 14 24 34 44 Tm 1n 4quotTmn 0 23 33 43 13 T23 T21 732 T12 4T22 O T33 T31 T4 T22 4T3 0 12 22 32 42 T24 722 T33 T13 4723 Z 0 T14 732 T43 T23 4733 2 O 11 21 31 41 300 C T2 500 T32 400 47392 2 o T33 500 200 T22 4T32 0 A B 300 T22 T33 400 4T23 O 400 0C 200 0C 300 T32 200 T23 4T33 0 C D 500 C FINITE DIFFEl1ENCE EXAMPLE T23 500 T32 400 4T22 O T33 500 200 T22 4T32 O T2 2 7133 4T23 O 3 300 T32 200 T23 4T33 O 14 24 34 44 22 32 42 12 4T2 2 T2 3 T32 900 11 21 31 41 T22 4T32 T33 7OO T22 4T23 T33 7OO 0C T23 T32 4T3 3 500 A B 4 1 1 O 39 quotT2 2 900 400 0C 200 C 1 O 4 1 T23 700 C D 1 4 0 1 T32 700 O 1 1 4 3 3 0C FINITE DIFFERENCE EXAMPLE 14 24 34 44 1 1 O T22 900 0 4 1 T23 Z 700 13 23 33 43 4 O 1 T32 700 1 1 4 T33 500 12 22 32 42 400 11 21 31 41 350 350 300 C 300 A B 400 C 200 C 39 C D FINITE DIFFERENCE EXAMPLE Higher grid densities are more easily solved using finite difference packages or using programs like Matlab 300 3489 3264 3026 2638 3768 3508 3169 2667 3974 3842 3506 2891 4384 4341 4115 3491 500 FINITE DIFFERENCE EXAMPLE Consider a long bar of square crosssection 08m to a side and of thermal conductivity 2 WmK Three of the sides are maintained at a constant temperature of 300 C The fourth side is exposed to a fluid at 10000 for which the convection heat transfer coefficient is 10 Wm2K Use a grid spacing of 02m to determine the midpoint temperature and heat transfer rate between the bar and the fluid per unit length of the bar T 3 Ewart 183 m T it 2 Wifrnnlf Ell FINITE DIFFERENCE EXAMPLE it r pm a HIE ll h k a 5 39miaa l SEIiIIEMTIE L T3 3m at T2 T x y 12 m Symmetry line Bar LE 1 33 m E Wimrl i Ta T r T5 3m DIG 2quot quot T1GEDEJ C 5 qlal Gigi bl IE I 5 ASEUMPTIHES 1 Shadystat m di mami E ducti 1 C l l t l i pmpartieg ALYSIS a Ei mxidari g Ememr IE nancial EtW i is Shawn 311mm 11711 111mm nwemi m lmd f E ll l l l he ll IlJl ji39E T114 l liIEdilff l L i 311 L 33 5 4 3F 3 1 I t l li li Eq 429 by inspe i A130 can be i39ed 35 mtirim39 mmts E l l li l li g Emm emjr a plane with mlwecti n Emil 4421 rim ung that m meEK 222 12 MHEmeK 1 nd Nude T 1T5 31211 TE 2141 112 li jT Nude E 2T5 TIL Tar 1241 1E 112Tg if 3 39539 a m Tm H 4423 Zaral Ell l T ulL 2 TEEN 129352 Nada at 1 131am Surmsze with cxmn v c tim FINITE DIFFERENCE EXAMPLE 30m 30m 30m 1 50 20 Ea iiiii iiiHHLQH GMHLGEGG wa QQQQQHHL an H H me the E lll 11mm T nd the WdEp MI tmpemmre 33 T4 Hume 292 239 29r nzilz 2545 2401 1931 IN94 The heat rate by C v remen y ham and uid i3 given 33 li ntw 2W 39H 1 11 r I 132 NEW 39 f qt mm 39 E39i t b 41 mm 9 5392 13911 39 T39 2 2 11 x2391 3 Tm 11 xj7fi Tm 11y xf23m 71 In Symmetry lime T3 301 DE ux y y 2m E full 39K El2 mfl1 4 M 1 2191 mj3nu 1u K WHAT YOU NEED TO BE ABLE TO DO Simplify the heat equation for 2D 88 conduction Use the conduction shape factor and the dimensionless heat rate to solve some 2D 88 conduction problems Understand the basics of the finite difference technique and how the heat equation can be put into a finite difference form Derive nodal equations for finite difference problems Identify methods to ensure finite difference solutions are accurate

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