1-Dimensional Steady Conduction

Transcription

1 1-Dimnsional Stady Conduction On of th simplst transport procsss ncountrd in many nginring applications is 1- dimnsional conduction. On-dimnsional conduction will b usd to discuss som of fundamntals associatd with th numrical mthod introducd in this cours. All th dvlopmnt discussd in this chaptr is qually prtinnt to mor complx transport procsss prsntd in latr chaptrs. On-dimnsional conduction quation may b obtaind from th gnral form of transport quation as discussd. With φ, Γk/cv, and V0, w gt an nrgy quation ( ρ ) t x ( k c x ) + y ( k c y ) + z ( k c v v v z ) + S (1) For incomprssibl substanc, ρ constant, CvCpC, and dcd. hus, Eq. (1) can b writtn as ρc t (k x x ) + y (k y ) + z (k z ) + S () Not that w hav not mad any assumption on th spcific hat, C. hat is C can b a function of spac and tmpratur. Assuming th tmpratur variation is in x-coordinat alon, Eq. () rducs to 1- dimnsional transint conduction quation, ρ c t (k x x ) + S (3) For 1-dimnsional stady conduction, this furthr rducs to d d (k dx dx ) + S 0 (4) 1

2 Eq. (4) is a simpl transport quation which dscribs stady stat nrgy balanc whn th nrgy is transportd by diffusion (conduction) alon in 1-dimnsional spac. Finit Volum Equation Finit diffrnc approximation to Eq. (4) can b obtaind by a numbr of diffrnt approachs. W will considr a control volum mthod [1]. W will us notations and symbols commonly adoptd for finit volum mthod (s Fig.1). Rwriting Eq. (4), w hav - d dx (J ) + S 0 x () whr Jx -kd/dx is th conduction flux in th x-dirction. W ar not making any assumption on th conductivity and th sourc, so that thy may hav diffrnt valus at diffrnt control volums and may dpnd on tmpratur as wll. Figur 1 1-dimnsional uniform control volums Intgrating Eq. () ovr a control volum containing, w hav 1 1 x + S) dxdydz (1 x 1)[ - dj x + Sdx] (6) 0 0 w w w (- d J dx Carrying out th intgration for th first trm in th RHS of Eq. (6), w gt w - d J - [ ( J ) - ( J ) ] x x x w - ( - k d d x ) + ( - k d d x ) w (7) k ( d d x ) - k ( d d x ) w w

3 Diffusion flux at th intrfacs "" and "w" can b approximatd by E dx ) k - ( δ x ) k ( d (8a) and W w w w dx ) k - ( δ x ) w k ( d (8b) Eqns. (8a) and (8b) ar scond-ordr accurat whn uniform control volums ar usd. If control volums ar non-uniform, ths xprssions must b modifid to rflct th diffrnt grid sizs. Howvr, w will still us Eqns.(8a) and (8b) to valuat diffusion flux vn for non-uniform control volums for th sak of simplicity in rsulting finit volum quations. Accuracy will not b affctd significantly as long as th grid sizs adjacnt to ach othr ar not xtrmly diffrnt. hysical maning of this approximation is to assum a picwis linar tmpratur variation btwn th tmpraturs in two adjacnt control volums as shown in Fig.. Intgration of th scond trm in th RHS of Eq. (6) yilds Sdx S x (9) w whr S rprsnts th avrag valu of S in th givn control volum. hysically, this assumption implis a stpwis variation of sourc as shown in Fig.. utting ths approximations into Eq. (6), w obtain k (1 x 1) ( δ x ) ( - ) - E k w (1 x 1) ( δ x ) w ( - ) + S(1 x 1) x 0 E Rarranging this, w hav a ae E + aw W + b (10) whr 3

4 a E k (1x1) ( δ x ) a w W k (1x1) ( δ x ) w (11a) (11b) a a E + a W (11c) and b S(1x1) x (11d) Figur icwis linar tmpratur btwn adjacnt control volums (a) and stp-wis uniform sourc in a control volum (b) Eq. (10) is th finit volum quation that dscribs nrgy consrvation for 1-dimnsional stady stat conduction. mpratur of th control volum undr considration is influncd by tmpratur in th nighboring control volums, W at th wst and E at th ast. Cofficints ae, and aw rprsnt conductanc btwn th control volum undr considration and th adjacnt control volums. Conductanc is proportional to th intrfacial conductivitis and th cross-sctional ara prpndicular to th x-coordinat (1x1) and is invrsly proportional to th distancs btwn two tmpratur nods. All cofficints apparing in Eq. (10) ar positiv. hus if w incras th tmpraturs in th nighboring volums, tmpratur will also 4

5 incras, rflcting ralistic physical situations. If cofficints ar not positiv, howvr, incrasing nighboring tmpratur will dcras. Having positiv cofficints in finit volum quations is ssntial for a succssful numrical simulation. In convction, ths cofficints ar not always positiv and rquir som spcial tratmnts to mak thm positiv (discussd in a latr chaptr). In Eq. (10), b rprsnts gnration or dstruction of nrgy in th control volum undr considration. Finit Volum Equation in Indx Notations Computr programming rquirs that th finit volum quations b writtn in indx notations. Following practics in dfining th control volums and associatd indx notations ar adoptd. Each control volums ar dfind by idntifying ach control volum surfacs by assigning thir locations in x, i.., x1, x,...,xi-1, xi, xi+1,...xn, xn+1, xn+, and xn+3 as shown in Fig. 3. mpratur and thrmal conductivity ar dfind xplicitly at th cntr of ach control volum. hr ar N+ control volums. wo control volums at th nd of calculation domain ar fictitious control volums with zro thicknss. hy ar introducd to handl th boundary conditions. whr In trms of ths notations, finit volum quation can b writtn as, i N+1, ai i bi i+1 + ci i-1 + d i (1) k i+1/ (1x1) δ 0.( x + x ) b i k (1x1) ( x ) w c i k (1x1) ( x ) w i i+1 k i-1/ (1x1) δ 0.( x + x ) i-1 i (1a) (1b) and i i i a b + c (1c) i i i d S x (1x1) (1d)

6 Figur 3 Indx notations Writing Eq. (1) for vry control volums, i N+1, w hav N simultanous quations: 6

7 a b + c + d 3 1 a b + c + d a b + c + d i-1 i-1 i-1 i i-1 i- i-1 a b + c + d i i i i+1 i i-1 i a b + c + d... i+1 i+1 i+1 i+ i+1 i i+1 a b + c + d N N N N+1 N N-1 N a N+1 N+1 b N+1 N+ + cn+1 N + d N+1 (13) whr tmpraturs 1 and N+ ar givn by th boundary conditions. Unknown tmpraturs,, 3,...,N, and N+1 ar found by solving Eq.(13) simultanously. Sinc Eq. (13) is in a tri-diagonal matrix form, ri-diagonal-matrix-algorithm (DMA) can b usd. his will b dscribd latr. Boundary Conditions Spcification of boundary tmpraturs, 1 and N+ dpnds upon th physical boundary conditions at th nds of calculation domain. Lt us considr various boundary conditions at th lft boundary. Cas 1: Boundary tmpratur, B, is givn. his is th simplst boundary condition. hn 1B. Cas : Hat flux at th boundary, q, is givn. Boundary tmpratur (1) is not known, '' in howvr. W can find a rlation btwn th givn hat flux and th tmpratur insid th calculation domain by considring an nrgy balanc. Considr th fictitious control volum (Fig. 4) and taking nrgy balanc w gt t ( ρ ca x1 1) q in"a- q out"a (14) 7

8 Figur 4 A known hat flux boundary at x0 h first trm in Eq. (14) is zro sinc thr is no mass in this volum. Now valuating th conduction flux through th right surfac (Jx)w and quating with th givn hat flux, q, w gt Solving for th unknown boundary tmpratur, 1, w hav '' - 1 q in -k (1) 0.( x1 + x) 1 x '' 1 + q in (16) k '' in If '' q in 0, i.., an insulatd boundary, Eq. (16) rducs to 1 (17) Cas 3: Convctiv hat transfr cofficint, hf and fluid tmpratur, f ar givn. his is similar to Cas xcpt th convction flux is givn by " q conv h f (f - 1) (18) 8

9 Not that th convction flux is positiv in th x-dirction if f is gratr than 1. By taking an nrgy balanc at th fictitious control volum at x0, w obtain - 1 hf (f - 1) -k 0.( x1 + x ) (19) Solving for 1, w hav hf f + k x 1 (0) hf + k x If hf>>k, Eq. (0) rducs to 1f, and if hf << k, Eq.(0) rducs to Eq.(17), as xpctd. Similar approach can b usd at th right boundary. For th thr boundary conditions considrd so far, it is possibl to solv xplicitly for th boundary tmpratur 1. Suppos w hav a radiativ boundary at x0. hn th nrgy balanc givs σε (f - 1) - k (1) 0.( x1 + x) W cannot solv for 1 xplicitly from this quation. A mor gnral tratmnt of such nonlinar boundary conditions will b discussd in dtail latr in this chaptr. W hav formulatd finit volum quation for on dimnsional conduction and considrd how to incorporat som boundary conditions. Lt us xamin fw xampl problms to rinforc our undrstanding of th proposd numrical mthod. Exampl 1: Considr a stady conduction in 1-dimnsional bar with known tmpratur at x0 (1 0 K) and xl (6 16 K). h lngth of th bar is 8.0 m, conductivity is constant at 1. W/K.m, and thr is a uniform hat sourc that gnrats 3.0 W/m 3. W would lik to find th tmpratur at th cntr of 4 qual control volums as shown. 9

10 Solution: Evaluating th cofficints at i, 3, 4 and, w hav b (1.)(1x1) 0.( + ) 1. ;c (1.)(1x1) 0.(0 + ) 1.;a b3 ;c3 ;a3 b 4 b 1. ;c 1.;c 4 1. ;a 1. ;a Sourc trms ar, for all i's; d i Si x i (1x1) (3.0 W/ m )( mx1m ) 6 W With ths cofficints and sourc trms, Eq. (13) bcoms, aftr multiplid by, (a.1) (a.) (a.3) (a.4) (a.4) bcom Substituting th givn boundary tmpraturs, 10 and 616, Eqns.(a.1) 10

11 (b.1) (b.) (b.3) (b.4) hs quations can b solvd by forward and backward substitution mthod. o do that lts rwrit ths quations as Forward substitution bgins by introducing into (c.), Rarranging this (c.1) (c.) (c.3) (c.4) ( ) (d) Substituting this into (c.3), w gt Finally substituting 4 into (c.4) and solving for, w hav. Backward substitution dtrmins 46, 3 and 10. Exact solution of this problm is With th givn conditions, Eq. (f) bcoms () 7 7 (x) + ( - ) x L + S (L- x) x k 0 L 0 (f) 11

12 (x) 10 x- x (g) Numrical solution is on dgr largr than th xact valus at all control volums. his is du to small numbr of control volum usd in th solution. o chck th solution furthr, lt s calculat th nrgy balanc. Enrgy gnratd in th bar is q gn 3 (3 W/ m )(8 mx1m ) 4 W 1 x / q 0 q x 0 - k - (1 x 1) (-1.) 10-0 / (1) -1 W 6 x / q L q x L - k - (1 x 1) / (1) 9 W hus q out - q 0 + q L 1 W+ 9 W 4 W W s th nrgy consrvation is satisfid. Exampl : An aluminum bar is xposd to a convctiv boundary at xl and a constant tmpratur at x0. Calculat stady-stat tmpratur at four locations as shown. Assum k14 W/m.K, 0373 K, f98 K and hf10 W/m.K. 1

13 Solution: Evaluating th cofficints for ach i, w hav b (14)(1) 0.( ) 8; c (14)(1) 0.(0 + 0.) 6; a 84 b 3 8; c 3 8; a 3 6 b 4 8; c 4 8; a 4 6 b (14)(1) 0.(0. + 0) 6; c 8; a 84 hr is no sourc, thus di0, for all i's. With ths cofficints, simultanous quations ar (a.1-a.4) Boundary tmpratur 1 is 0373 K. 6 is dtrmind by nrgy balanc at th right boundary. hus, with N4, 13

14 6 hf f + k x + k hf x (b) Substituting 0 into (a.1), Substituting (b) into (a.4) and rarranging, w hav (a.1') (a.4') Solving (a.1'), (a.), (a.3) and (a.4') simultanously, w obtain th numrical solutions. Exact solution of th givn problm is (x) -.06 x abl blow shows a comparison of ths rsults. nod numrical xact Enrgy balanc can b usd to chck th accuracy of numrical solution. Enrgy ntring through th lft boundary by conduction is q0" -k (-14) W/ m x 0.(0.) Enrgy laving through th right boundary du to convction is 14

15 q " hf (6 - f) L (10)( ) W/ m About 0.8 % rror is obsrvd in nrgy consrvation. Exampl 3 Solution: Considr th sam physical and boundary conditions as givn in Exampl. But allow uniform nrgy gnration in th aluminum bar at a rat of 100 W/m 3. Only chang w hav to mak is to add a sourc trm bi(100 W/m 3 )(0. m)(1m )0 W. h simultanous quations ar: Solving ths w gt, ; ; ; Enrgy balanc is q (14) W 0.(0.) q L q conv "(A) (10)( )(1.0) W and q gn 3 & q(volum) (100 W/ m )( mx1m ) 00 W 1

16 hus, th rror in nrgy balanc is Error q gn + q q in out - q out (.8 %) o improv th accuracy w nd to hav mor control volums. hr xampl problms shown abov illustrat th basic mthodology of th finit volum numrical mthod. W find that thr ar basically no diffrncs in formulating th numrical solutions for all thr problms. Only diffrncs appar in th valuation of sourc trms and th boundary tmpraturs. hus a gnral solution approach for any combinations of boundary conditions and sourc trms can b formulatd. his flxibility of numrical mthod is an advantag ovr th xact analytical mthod. Sourc rm Linarization W hav considrd sourc trms that rmain constant throughout th domain. hr ar many typs of sourc trms that ar not uniform but dpnd on dpndnt variabls. Lt us rvisit Exampl considrd in a prvious sction. If w rmov th insulation around th aluminum bar and subjct its surfac to th sam convctiv boundary condition imposd at xl, nrgy balanc quation bcoms [], ρ c t x (k x ) + h f ( f - ) A () Figur A rctangular fin whr is th wttd primtr of th cross sction and A is th cross sction ara. hus w s that sourc trm for this cas is 16

17 S h f A - h f f A S C + S (3) whr S c is th constant part and S p is th slop of tmpratur dpndncy. his is a linar quation in. hus sourc trm is alrady in a linar form. In many cass howvr, w hav to linariz th sourc trm for succssful numrical calculations. With a linar form of sourc trm, stady 1-D conduction quation can b writtn as - dj X dx + S C + S 0 (4) Intgrating th sourc trm in Eq. (4) ovr a control volum surrounding "p" as w did bfor, w obtain 1 1 C C 0 0 w (S + S ) dxdydz S x + S x h rsulting finit volum quation is whr a ae E + aw W + b () a E k (1x1) ( δ x ) a W kw (1x1) ( δ x ) w (6.a-d) a a E + a W - S x(1x1) and b S h diffrnc btwn Eq. (10) and Eq. () is du to th linarizd sourc trm. If S p 0, thy ar th sam quation. In linarizing th sourc trm, it is absolutly ncssary to hav S p 0. Othrwis a can bcom ngativ, and th rsulting numrical solution can bcom nonphysical [1]. Exampl 4 C x(1x1) 17

18 Rxamin Exampl with a chang such that nrgy loss through th surfac du to convction is includd. Us th sam physical and boundary conditions. Assum 1 by 1 cross sctional ara of th bar. Solution: Linarizd sourc trm is hus, for all i's, S h f A - h f f A S + S C f S C i ( h f i A ) (10 W/ m K)(4 m) 1m (98 K) 1190 W/ m 3 and F S - ( h i A 3 ) i - 40 W/ m K 3 i C x (A i) (1190 W/ m )(0. m)(1m ) 960 W d S i i With ths changs, simultanous quations bcom (s xampl ) Solving ths, w hav 3 -S x A - (-40 W/ m K)(0. m)(1m ) 0 W/ K i i i aking an nrgy balanc, w hav q (84 + 0) (6 + 0) (6 + 0) ( ) x L 1 373; ; ; 30.49; q x (1x1)(-14)( 0.(0.) ) W (1x1)(10 W/ m K)( ) K 38.1 W 18

19 q wall A Si hf (i - f) (0 m )( ) W W s about 0.1 % rrors in th nrgy balanc. Furthr Considrations on Sourc rm Linarization h sourc trms ar, in gnral, nonlinar function of dpndnt variabls. Suppos w hav a sourc trm as S 4-3 (7) his is a nonlinar quation sinc cofficint (- ) dpnds on tmpratur itslf. Svral mthods can b usd to linariz this xprssion. Mthod 1: Lt S c 4 and S p 0. p * is th known tmpratur from th prvious calculation. Mthod : Lt S c 4 and S p -. Mthod 3: Us th aylor sris xpansion of S() about p *. S( ) S( ) + ( ds * ) ( - * * )+... d * (0-1 ) * ( - )+... * *3 *3 * *3 * p hus S c and S p --1. Graphical intrprtation of thr mthods is shown in Fig. 6. Mthods 1 and ar undrstimating th dpndncy of S on whil mthod 3 givs an accurat dpndncy. Not that p and p * ar vry clos. 19

20 Figur 6 Graphical rprsntation of linarizd sourc trm Sourc trm linarization can b usd to spcify th dsird valus of dpndnt variabl at any control volums. Suppos w lik to spcify a tmpratur at an intrior control volum, say p dsird. o accomplish this lt S c 10 0 x dsird and S p With this choic of sourc trm, w hav E E W W a + a + b a E E W W C a + a + S x(1x1) a E + a W - S x(1x1) C S x(1x1) - S x(1x1) 0 dsird (-10 ) dsird his is a vry usful mthod to spcify a valu of dpndnt variabl insid a calculation domain. Sourc trm linarization can also b usd to stabiliz numrical calculation whn th physics dictats that th dpndnt variabl should rmain always positiv, such as dnsity, absolut tmpratur, and nrgy. Suppos w want to calculat th absolut tmpratur,, for a problm in which sourc trm is givn by S S const and S const is a vry larg ngativ numbr. Undr crtain conditions, p can go ngativ bcaus of ngativ larg sourc trm. o prvnt such occurrnc, lt th sourc trm b linarizd by S 0 + S const * 0

21 hn const S C 0 ; S S * 0 hus p can approach zro but nvr bcoms ngativ. Intrfacial Conductivity Up to this point of our discussions, it is implicitly assumd that th conductivity at th intrfacs btwn two control volums is known and is qual to th conductivity of matrial in an adjacnt control volum. his is prfctly accptabl if conductivity is constant and uniform throughout th matrial as has bn th cas for all xampl problms. If conductivity is not uniform du to diffrnt matrials or du to a dpndncy on tmpratur, intrfacial conductivity should b carfully valuatd to nsur consrvational principl. Considr two control volums with diffrnt conductivity as shown in th Fig. 7. It is vry tmpting to valuat th conductivity at th intrfac "" by linar intrpolation. hat is k k + (k E - k ) ( x ) - ( δ x ) δ (8) Figur 7 Intrfac conductivity at ast intrfac If two control volums ar th sam siz, Eq.(8) rducs to k 0. (k + k E ), which is th arithmtic man valu of conductivity of two adjacnt control volums. his approach is simpl yt it can lad to nonphysical solutions. As an xampl, lt k E 0. (insulator), thn conduction flux at th intrfac should b zro. But, k 0.k and th hat flux at th intrfac is 1

22 E q - k - ( δ x ) E -0.k - ( δ x ) 0 which is not physically corrct. his will introduc an nrgy sink or sourc through th intrfac "". o driv a physically corrct intrfacial conductivity, w considr an nrgy balanc at th intrfac. Dfin a fictitious control volum of zro thicknss at th intrfac (Fig. 8). Enrgy ntring from th lft surfac is, using th first-ordr accurat diffrnc schm, q - - k - ( δ x ) - (9a) Figur 8 Enrgy balanc at th intrfac control volum And th nrgy laving through th right surfac is q E + - k - E ( δ x ) + (9b) whr is th tmpratur at th intrfac. Rarranging Eqs. (9a) and (9b), w hav (q ) ( δ x ) - k - - (30a) and

23 (q ) ( δ x ) + ke + - E (30b) Adding Eqs.(30a) and (30b) and noting that q - q + q, w hav q ( δ x ) k - E - + ( δ x ) ke + k k E ( δ x ) - k ( δ x ) + k ( δ x ) ( δ x ) E - + E. - (31a) Conduction flux through th intrfac "" is xprssd as E q - k - ( δ x ) (31b) Comparing Eqs.(31a) and (31b) w conclud that k k ke ( δ x ) k ( δ x ) + k ( δ x ) E - + (3) If two control volums ar of th qual siz, Eq.(3) bcoms E k k k E k + k (33) which is th harmonic man valu of adjacnt conductivitis. If k E 0, thn k 0 as xpctd. mpratur at th intrfac "" can b calculatd by using Eq.(30a) and Eq.(30b), k ( δ x ) - k ( δ x ) - ke + ( δ x ) ke + ( δ x ) + + E (34) All diffusion cofficints at th intrfac "" will b valuatd by this mthod. hus 3

24 Γ ( δ x ) ( δ x ) / Γ + ( δ x ) / Γ - + E Γ Γ ΓE ( δ x ) ( δ x ) + Γ ( δ x ) E - + (3) his formulation of diffusion cofficint at th intrfac allows th prsnt numrical mthod work for conduction in composit matrials, fluid flow in complx gomtris, conjugat hat transfr and hat transfr with phas changs and othr complx transport procsss. Intrfacial conductivity including a contact surfac rsistanc, R t, can b obtaind by th sam mthod (Fig. 9). Hat flux from th lft, through th contact surfac and to th right surfac is, rspctivly; q - k - ( δ x ) and q - 1 R ( - ) R q + t + - E - k - E ( δ x ) + + Figur 9 Contact Rsistanc 4

25 Eliminating - and + from ths quations with q - q + q R q, w gt k k E ( δ x ) k ke ( δ x ) + k ( x ) + + k - δ E k R t (36) mpraturs at th contact surfac ar (lft as an xrcis), - k E E E t ( x ) + k + R k k δ - ( δ x ) + ( δ x ) - ( δ x ) + k + k + R k k ( δ x ) ( δ x ) δ δ - E E t + ( x ) - ( x ) + (37a) and + k E E E t ( x ) + k + R k k δ ( δ x ) ( δ x ) ( δ x ) k + k + R k k ( δ x ) δ δ δ E E t ( x ) ( x ) ( x ) E (37b) Exampl Considr a fin problm as discussd in Exampl 4 with two changs. Conductivity of th bar is 14 W/m.K for th first half of th bar and 4 W/m.K for th rmaining half. Enrgy is bing gnratd uniformly at 100 W/m 3. All othr conditions rmain sam as in Exampl 4; lngth of th bar is m, cross sctional ara is 1.0 m, bas tmpratur is 373 K and all surfacs ar xposd to convctiv hat transfr with h f 10 W/m.K and f 98 K.

26 Solution: Diffrntial quation for this problm is givn by d d (k dx dx ) + h f A ( - ) + &q 0 f hrfor, th sourc trm is S &q+ h f A - h f f A Rcall that a E k (1x1) ( δ x ) whr th intrfac conductivity at '' is givn by Eq.(36), i.., k k E k ( δ x ) ( δ x ) k + ( δ x ) k - E + Substituting k, a E bcoms a E ke k (1x1) ( δ x ) k + ( δ x ) k - E + hus 6

27 b i a E ki+1ki (1x1) 1 1 xi ki-1 + x i+1 k i ki ki+1(1x1) xi ki+1 + xi+1k i and ci a W ki-1 ki (1x1) xi-1 ki + xi k i-1 Evaluating th cofficints for ach i,..,, w hav b k k 3 (1 x 1) x k 3 + x 3 k ()(14)(14 0.(14) + )(1 x 1) 0.(14) 8 c k 1 j (1 x 1) x 1 k + x k 1 ()(14)(1 x 1) (14) 6 b ; c 3 8 b 4 48; c b 96; c 48 From th sourc trm, w hav f Sp i x i (1x1) - h x i (1x1) -0 A hus th cofficints a i b i + c i - Sp i x i (1x1) ar a b + c - S x (1x1) (-0) a ; a ; a Constant parts of th sourc trms ar, for all i's, f i C x i (1x1) (&q+ h d S i A f) x i (1x1) ( )(0.)(1x1)

28 hus th simultanous quations ar Boundary conditions ar, K and 6 h f h f f + + x x k k Substituting th boundary conditions and aftr rarranging, w hav Solving for th unknown tmpraturs, w gt 344.1; ; ; 3 4 Boundary tmpratur 6 is * mpratur at th intrfac btwn two diffrnt matrials is givn by Eq.(34), 8

29 k3 0. x3 k3 0. x 3 3 k x k x K his tmpratur is lowr than th linarly intrpolatd tmpratur ( K). Fig. E shows th tmpratur distribution. Figur E mpratur distribution 9

30 o chck th nrgy balanc, w calculat 1 q x 0 k (1x1) W 0. x 3 q gn q& ( volum) ( 100 W / m )( m x 1m ) 00 W q q x L surfac h f ( i 6 h f f )(1x1) 68.6 W As ( i i f ) 17 W hus, q in + q gn W, and q out W. h rror is about 0.1 %. Nonlinarity Finit volum quation, Eq.(13) can b solvd for th unknown tmpraturs if all cofficints and sourc trms ar xplicitly known valus. For th linar problms, cofficints and sourc trms ar known. hrfor, simultanous quations ar to b solvd only onc for th unknown tmpraturs. For th nonlinar problms, howvr, cofficints and sourc trms cannot b valuatd xactly bcaus thy dpnd on th tmpraturs that ar unknown. Suppos th thrmal conductivity is a function of tmpratur, k k(). Actually, thrmal conductivity of most matrials dpnds strongly on th tmpratur. o valuat th cofficints, w nd to know th valu of thrmal conductivity, which, in turn, rquirs th known tmpraturs. o initiat th calculation, w hav to assum tmpratur, * and valuat kk( * ). With th gussd cofficints, simultanous quations can b solvd for th unknown tmpratur,. If nwly calculatd tmpratur,, is clos nough to th gussd tmpratur *, w hav solution. If not, lt * and rpat th calculation until * - Error abs( ) < ε (38) whr ε is an rror tolranc. Nonlinarity can also aris from th nonlinar sourc trms. Rturning to Exampl 4 w not that sourc trm contains a convctiv cofficint, h f. Convctiv cofficint dpnds, in gnral, on th film tmpratur that is th avrag of th unknown surfac tmpratur () and th fluid tmpratur ( f ). 30

31 Nonlinar boundary conditions also mak th problm nonlinar. Rturning to Exampl, and considr th convctiv boundary at xl. Rcall that th boundary tmpratur 6 was givn by (Eq.(b) in Exampl ), which contains an unknown tmpratur. his is an xampl of nonlinar boundary conditions. In Exampl, 6 was substitutd into th finit volum quation for th fifth (i) control volum and manually rarrangd to liminat th nonlinarity through an algbraic manipulation; which dos not contain any unknown trms , Without this algbraic stp, howvr, finit volum quation for th fifth control volum would contain nonlinar trm, 6 *, which is * * whr * dnots valus at th prvious itration. Substituting this into Eq.(b.4) in Exampl, w hav * ( * ) which contains nonlinar sourc trms du to *. Sinc sourc trm contains th unknown tmpraturs, simultanous quations must b solvd itrativly until solution convrgs within a prscribd tolranc. hat is * - abs ε Nonlinar boundary conditions, such as radiativ boundary condition, rquir such itrations sinc th boundary tmpratur oftn cannot b xplicitly xprssd in trms of th intrior tmpraturs. 31

32 Nonlinarity is quit common in most transport procsss involving fluid flow. Numrical solution of nonlinar problms do not ntail additional nw concpts othr than that simultanous quations must b solvd many tims until convrgd solutions ar obtaind. his is a task computr can handl rathr rapidly. ri-diagonal-matrix-algorithm (DMA) By introducing th boundary conditions at both nds of calculation domain, N- simultanous quations can b put into a tri-diagonal matrix form. ri-diagonal form of matrix quations can b asily solvd by th forward and backward substitution mthod as has bn utilizd in our xampl problms. homas algorithm is a mthod basd on th sam principl and is usful to solv a larg tri-diagonal matrix [1]. Finit volum quation in indx notation can b rarrangd into th following form; -ci i-1 + ai i - bi i+1 di (39) for i,3,...i-1,i,i+1,...,n,n+1. Implmnting th boundary conditions at x0 and xl, and xpanding Eq.(39), w gt a tri-diagonal matrix: a -b d -c3 a3 -b3 3 d3 -c4 a4 -b4 4 d ci ai -bi i di (40) cn an -bn N dn -cn+1 an+1 N+1 d whr d ' and d N+1 ' ar d ' d + c 1 (41a) and ' d N + 1 N+1 N+1 N+ d + b (41b) du to th boundary conditions at x0 and xl. Not that c b N+1 0 aftr implmnting th boundary conditions. 3

33 In th forward substitution, w sk a rlation i i i+1 + Q i (4) right aftr w obtain i-1 i-1 i + Q i-1 (43) Substituting Eq.(43) into Eq.(39) for i-1, w hav Solving for i, w gt -c i (i-1i + Q i-1 ) + a i i - b i i+1 d i i bi i i i-1 i+1 a i - ci + d + c Q i-1 i i i-1 a - c (44) Now comparing Eq.(4) and Eq.(44), w obtain two rcursiv rlations i bi a - c i i i-1 (4a) and Q i d + c Q a - c i i i-1 i i i-1 (4b) Forward substitution is to valuat i and Q i for i,3,...,n+1; b i b ; Q d ' + c Q a - c 1 a a - c b3 i 3 ; Q d 3 + c 3Q 3 3 a 3 - c3 a 3 - c3... bi d i + ci Qi-1 i i i ; Q i ai - ci i-1 ai - ci i-1... ' 1 d a 1 33

34 i N i N+1 N N+1 bn a N - cn N-1 bn+1 a N+1 - CN+1 N 0 ; Q d + c Q N a - c dn +1 + cn+1qn ; QN+1 a N+1 - cn+1 N N N N-1 N N N-1 With th known i and Q i, Eq.(4) yilds 3 + Q Q 3... i i i+1 + Q i... N N N+1 + Q N N+1 N+1N+ + QN+10+QN+1 Sinc Q N+1 is known, N+1 is found from th last quation. hn backward substitution yilds N, N-1,..., 3, and. DMA is a simpl mthod to us and rquirs small computr mmory spac that is proportional to N. It can b usd for ADI (Altrnating-Dirction-Implicit) mthod commonly adoptd for th multi-dimnsional problms. Quasi 1-Dimnsional Conduction In 1-dimnsional conduction along th x-dirction, th ara prpndicular to th hat flow dirction rmains constant, i.., A(x)constant. W hav assumd (1x1) cross sctional ara in all of our discussions so far. hr ar many situations in which ara changs along th hat flow dirction. If tmpratur variation prpndicular to th x-dirction is vry small compard to th variation in th x-dirction, tmpratur variation prpndicular to th x- dirction may b nglctd. In this cas, hat transfr bcoms a quasi-on-dimnsional in x- dirction and th stady stat hat conduction quation bcoms 1 d d (Ak A dx dx ) + S 0 (46) By intgrating Eq.(46), it can b asily shown that 34

35 a a E E + a W W + b (47) whr a E k A ( δ x ) a W k w A ( δ x ) w w (48a) (48b) a a E + aw - S A x (48c) and b SC A x (48d) his is a gnralization of 1-dimnsional finit volum rprsntation of hat conduction quation. h ara A(x) can b any arbitrary valus. If A(x)x, 1-D cylindrical coordinat and if A(x)x, 1-D sphrical coordinat systm rsult. Figur 10 Quasi-on-dimnsional conduction 3

36 Stady 1-Dimnsional Conduction rogram W ar now in a position to writ a computr program that can solv th gnral 1- dimnsional stady conduction problms. Conductivity of th matrial may vary throughout th matrial and may also dpnd on th tmpratur as wll. h cross-sctional ara can b arbitrary. Varity of sourc trms can b includd. Dtails of a computr program do vary significantly from on to anothr, rflcting th styls of individuals who writ th programs vn though th programs ar basd on th sam numrical mthod and ar dsignd to solv th sam physical problm. hr is absolutly no possibility that two programs can b idntical in vry dtail. A MALAB program basd on th mthodology discussd in prvious sctions is cratd. his program is namd as std1da.m, which solvs 1-dimnsional stady conduction problms. h flow chart shown in Fig. 11 shows th structur of th program. Dtail of th program is illustratd by using an application to conduction through a circular fin. Figur 111 Flow chart for stady 1-D conduction 36

37 Dscription of th problm Govrning diffrntial quation for 1-D hat conduction through a varying cross sctional ara is 1 A d d (Ak dx whr A(x) is th cross sctional ara, is th wttd primtr of th cross-sction and q& is th sourc trm. For th prsnt xampl problm, cross sctional ara is of circular shap with. cm-dia. h fin is mad of coppr. h lngth is 1. m and th conductivity is assumd to b constant at 401 W/m.K. h bas tmpratur is 473 K and th fin surfacs including th right sid boundary ar xposd to air at 98 K with a constant convctiv cofficint of 10 W/m K, and q. 0. h purpos of this xampl is to highlight th ingrdints of th finit volum mthod discussd so far and compar th rsults with known xact solution. It is always a good ida to compar numrical solutions with availabl data ithr xact or xprimntal. Exact solution of this problm is givn by [] dx ) + h. f (f - ) + q 0 (49) A θ θ b f coshm(l- x) + ( h mk )sinhm(l- x) f coshml+ ( h mk )sinhml whr h f (x) m ; θ b B - f ; and θ ka(x) - f Structur of th program Numbr of control volums usd in this xampl is 10 and thy ar uniform in sizs. Cross sctional ara is also constant, ac(x)4.91x10-4 m. Aftr dfining th gomtry of th problm, initial tmpratur is assumd at 373 K. his initial tmpratur can b arbitrary valu. If w assumd initial tmpratur clos to th stady solution, numbr of itrations will dcras. mpratur, tp * is th prvious itration lvl tmpratur savd to chck th convrgnc of th solution. IER is th itration countr. Maximum numbr of itration is st to maxitr00. hrmal conductivity of th fin is nxt givn in th program. It is 37

38 assumd to b constant in this xampl. Boundary conditions at both nds of th fin ar nxt prscribd. At x0, 1 b 473 K. At xl, a convctiv boundary condition is usd and tmpratur N+ is found by nrgy balanc which contains unknown tmpratur, N+1. A bttr tratmnt of nonlinar boundary condition is discussd in th nxt sction. Conductanc btwn th control volums ar nxt valuatd by using Eqns.(11a,b), whr intrfacial conductivitis ar valuatd by Eq.(33). For th prsnt problm, of cours, intrfacial conductivity rmains constant sinc th matrial is homognous and conductivity is assumd to b tmpratur indpndnt. Constant part (Sc) and tmpratur dpndnt part (Sp) of sourc trms ar h f Sc A q and Sp h f f x A and th sourc trm b is thn valuatd. h cofficints for th simultanous quations ar thn calculatd. Simultanous quations ar put into a tri-diagonal-matrix form by incorporating th boundary conditions. DMA mthod is usd to solv ths st of simultanous quations. Aftr solution is obtaind, convrgnc is tstd by * abs( i - i ) ε i for i1,...,n+. If convrgnc tst is mt, solution is obtaind (iflag0). Othrwis (iflag1), th gussd tmpraturs ar rplacd by th just calculatd tmpraturs and th solution stps ar rpatd until solution is convrgd or itration count (itr) xcds a prst maximum itration numbr (maxitr). h lattr option is ndd to scap from an ndlss doloop cratd by programming mistaks. At th nd of calculation, numbr of itrations rquird to rach th solution and th tmpraturs at ach control volums ar printd. %std1da.m. %stady, 1-dimnsional conduction with varying cross sction ara %,nonuniform condutivity and sourcs. Finit volum formulation %using matlab program. (By Dr. S. Han, Sp 3, 008) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %spcify th numbr of control volums n10;%numbr of control volums maxitr100;%maximum itration numbr np1n+1; npn+; np3n+3; %dfin calculation domain tl1.; %total lngth of th bar dlxtl/n; dxons(1,np); dxdlx*dx; %rplac fictitious boundary volum siz dx(1)1.0-10; 38

39 dx(np)1.0-10; %assign x-coordinat x(1)0; for m1:np x(m+1)x(m)+dx(m); nd %dfin cross-sctional ara dia0.0; for i1:np3 ac(i)pi/4*dia^; %constant cross sction nd %prscrib intitial tmpraturs for all control volums for i1:np t(i)373; tp(i)t(i); nd %itration for convrgnc itr0; iflag1; %itration loop for th convrgnc whil iflag1 % nd is at th nd of program ********* %prscrib thrmal conductivity for i1:np tk(i)401; nd %prscrib boundary tmpratur t(1)473;%at th lft boundary givn tmpratur hf10;% right boundary givn flux tf98; t(np)t(np1)-hf*(t(np)-tf)/(tk(np1)/... (.*dx(np1))); %initializ th cofficints for tdma matrix %valuat th diffusion conductanc and sourc trms for i:np1 % diffusion conductanc ktk(i)*tk(i+1)*(dx(i)+dx(i+1))/... (dx(i)*tk(i+1)+dx(i+1)*tk(i)); d.0*k*ac(i+1)/(dx(i)+dx(i+1)); ad; kwtk(i)*tk(i-1)*(dx(i)+dx(i-1))/... (dx(i-1)*tk(i)+dx(i)*tk(i-1)); dw.0*kw*ac(i)/(dx(i-1)+dx(i)); awdw; %sourc trm valuation % sp-1600; % sc4.768; prpi*dia; sp-hf*pr/ac(i); schf*pr/ac(i)*tf; apa+aw-sp*dx(i)*0.*(ac(i+1)+ac(i)); bsc*dx(i)*0.*(ac(i)+ac(i+1)); %stting cofficints for tdma matrix ta(i)ap; tb(i)a; tc(i)aw; 39

40 td(i)b; %modify using boundary conditions if i td(i)td(i)+aw*t(1); lsif inp1 td(i)td(i)+a*t(np); nd nd %solv th simultanous quations by using tdma nqn; nqp1nq+1; nqm1nq-1; %forward substitution bta()tb()/ta(); alpa()td()/ta(); for i3:nqp1 bta(i)tb(i)/(ta(i)-tc(i)*bta(i-1)); alpa(i)(td(i)+tc(i)*alpa(i-1))/... (ta(i)-tc(i)*bta(i-1)); nd %backward substitution dum(nqp1)alpa(nqp1); for j1:nqm1 inqp1-j; dum(i)bta(i)*dum(i+1)+alpa(i); nd %nd of tdma %updat th tmpratur for i:np1 t(i)dum(i); nd %chck th convrgnc for i1:np rrot(i)abs(t(i)-tp(i))/t(i); nd rror1.0-6; if (max(rrot)>rror) itritr+1; tpt; iflag1; ls iflag0; nd if itr>maxitr brak nd nd % this nd gos with th whil iflag1 at th top******** %compar with xact solution to mak sur program works corrctly %locat th midpoint of control volums for i1:np xc(i)0.*(x(i)+x(i+1)); nd % msqrt(hf*/(tk(1)*1.-)); thtabt(1)-tf; 40

41 for i1:np thta(i)(cosh(m*(tl-xc(i)))+hf/(m*tk(1))*sinh(m*(tl-xc(i))))/... (cosh(m*tl)+hf/(m*tk(1))*sinh(m*tl)); nd thtathtab*thta; txactthta+tf; txacttxact' %print th rsults fprintf('itration numbr is %i \n',itr) disp('stady stat tmpraturs ar') fprintf('%9.3f \n',t') %plot th rsult xc.vs.t plot(xc,t,'-',xc,txact,'--o') grid on titl('stady stat tmp'),xlabl('x(m)'),ylabl('(k)') lgnd('numrical','xact') stady stat tmp numrical xact (K) x(m) Figur 1. Comparison of numrical and xact solution A stady stat solution is obtaind aftr 76 itrations. Numrical solution and xact solution ar in good agrmnt as shown in fig. 1. A larg numbr of itrations was rquird 41

42 du to a nonlinar tratmnt of th convctiv boundary condition at xl. A mor gnral tratmnt of linarization of nonlinar boundary conditions is discussd in th nxt sction. Linarization of Boundary Conditions Boundary tmpraturs, 1 and N+ ar not xplicitly known whn th boundary conditions ar givn in trms of a hat flux, q ''. By taking an nrgy balanc at x0, it was prviously shown that 1 x 1 + q" (0) k whr q" is a boundary flux at x0. Not that q" is assumd to b positiv whn it is dirctd along th positiv x-dirction. Linarizd flux q" can b xprssd by q" q c + qp 1 (1) whr q c is a constant and q p is th slop of linarizd flux quation. Not that w do not rquir q bing lss than zro in th boundary flux linarization. Substituting Eq.(1) into Eq. (0) and rarranging, w hav x k 1 x qc k x q q k () Substituting Eq.() into th first quation in th simultanous quations (13) and aftr rarranging, w gt a + (3) ' ' b 3 d whr a' and d' ar th modifid cofficints and sourc givn by 4

43 1 a ' a - c (4a) 1 1- x q p k and 1 x qc k d ' d + c (4b) 1 1- x q k Likwis whn th boundary condition at xl is givn in trms of a hat flux, boundary tmpratur is givn by N+ N+1-1 [ x N + 1 k N+1 q"] () h hat flux q" is xprssd as q" q C + q N+ Substituting this in Eq.() and solving for N+, w hav N x k q 1 xn - k 1 x 1+ k + 1 C N+1 N+1 N + 1 N + 1 q N+1 N+1 q (6) Rplacing N+ in th last quation in th simultanous quations (13) and aftr rarrangmnt, w hav ' ' N + 1 N+1 c N + 1 N + d N + 1 a (7) 43

44 whr 1 a' N + 1 a N+1 - bn+1 (8a) 1 1+ xn+1 q k N+1 and 1 xn+1 qc ' kn+1 dn + 1 dn+1 - bn+1 (8b) 1 1+ xn+1 q k N+1 h fin problm discussd arlir is solvd again with th linarizd convctiv boundary at xl. h convctiv hat flux boundary at xl is q conv" h f (N+ - f) q C + q N+ (9) hus Not that in writing th hat flux at xl, hat flux dirctd along th positiv x- dirction is assumd to b positiv. Hnc Eq.(9) is usd. Modifications ar mad on std1da.m program to rflct th linarizd convctiv boundary at xl. his modifid program is calld std1d.m. In this program, w introduc boundary typs 1 (known tmpratur), (known hat flux) and 3 (priodic). Numrical solution for th fin problm with boundary typ 1 at x0 (bx01) and boundary typ (bx1) was obtaind with only itrations which is a dramatic dcras from 76 rquird without boundary linarization. h copy of th program can b found from CAE lab wbsit, riodic Boundary Condition q C - hf f and q h f (60) 44

45 W hav considrd boundary conditions givn by in trms of a known tmpratur (typ 1 boundary) or a known hat flux (typ boundary), which may b nonlinar. hr is on additional boundary condition, which is priodic (typ 3 boundary), frquntly occurs in transport procsss. As an xampl, considr a closd loop of aluminum wir with uniform cross sctional ara. Enrgy is gnratd non-uniformly in th wir and th surfac is xposd to a convctiv nvironmnt as shown in th figur. o dtrmin th stady stat tmpratur in th wir, on has to dfin physical boundary and apply appropriat boundary conditions. hr ar no obvious physical boundaris sinc physical domain is an ndlss loop. If thr is symmtry, symmtry boundary conditions can b applid. Howvr, it is not clar whthr thr is symmtry in th prsnt problm. If w masur tmpratur of th wir starting from any arbitrary point in th wir, tmpratur distribution along th wir may chang as w mov along th wir until w mak a full circl. Aftr on circl, tmpratur chang will rpat th sam distribution ovr and ovr. his typ of boundary condition is calld "priodic" boundary condition (Fig.13). A lazy way of handling th priodic boundary condition is to lt 1 N+1 * and N+ * (61) Whr "*" mans th tmpraturs at th prvious itration valus and thrfor act as nonlinar sourcs, rsulting in many itrations for a convrgnc. Figur 13 Closd loops with non-uniform nrgy gnration o acclrat th convrgnc, nonlinar sourc trms must b liminatd from th quations. his is accomplishd by using a spcializd MDA which xplicitly taks advantag of th priodic boundary condition. his mthod is calld th Circular-ri-diagonal-Matrix- Algorithm (CDMA). A dtail dscription of th CDMA is prsntd in th Appndix [1]. An xampl is usd to illustrat th procdur of using th CDMA for a problm with a 4

46 priodic boundary condition. h modifid program std1d.m is dsignd also to handl priodic boundary condition as discussd blow. roblm dscription o apprciat th CDMA, lt us considr a coppr wir of. cm-dia forming a loop. h lngth of th loop is 1. m. Enrgy is gnratd uniformly at 10 W/m 3 in th half of th loop and uniformly at x10 W/m 3 in th rmaining half of th loop. h conductivity of th coppr is 401 W/m.K and th convctiv conditions ar hf10 W/m.K and f98 K. W want to calculat th stady-stat tmpratur distribution in th coppr loop. Figur 14 Coppr wir loops with nrgy gnration Bcaus of th priodic boundary condition, fictitious boundary control volums ar no longr ndd. Instad, from th gomtry, w hav x1 x N+ 1 and x N+ x (6) h CDMA solvs for th tmpraturs,...,and N+1. It is obvious from th gomtry that 1 N+1 and N+ (63) 46

47 Not th changs mad on implmnting th boundary conditions for CDMA routin and th CDMA routin, which is addd nxt to DMA routin. hs changs ar highlightd in boxs. A nw variabl IRIOD is introducd. IRIOD1 mans a priodic boundary and IRIOD0 mans a non-priodic boundary condition. h prsnt problm calls for IRIOD1. Numrical Rsults Numrical rsults ar obtaind aftr itrations. If w usd a lazy man's approach calling DMA instad of CDMA, convrgd solution would tak 47 itrations. (his is lft as an xrcis.) Numrical rsults show a priodic variation of tmpratur distribution along th loop as xpctd. o chck th accuracy of th rsults, w can chck th nrgy balanc. his can b don convnintly by adding a fw lins of statmnts at th nd of th program. (his is lft as an xrcis). IER I I E 3.987E E E E E E+0 E 4.011E E E E E E+0 Rfrncs 1. atankar, S. V., Numrical Hat ransfr and Fluid Flow, Hmisphr, Incropra, F.., and DWitt, D.., Fundamntals of Hat ransfr, 3rd Ed., John Wily &. Sons, atankar, S. V., Liu, C. H. and Sparrow, E. M.,"Fully Dvlopd Flow and Hat ransfr in Ducts Having Stramwis-riodic Variations of Cross-Sctional Ara," ASME J. of Hat ransfr, Vol. 99, p , roblms 47

48 1. Starting from a 3-dimnsional hat conduction quation writtn in th sphrical coordinat systm, show that 1-d stady stat conduction in th radial dirction is givn by th following quation. Driv finit volum quation by intgrating th abov quation. Explain th physical maning of ach trm apparing in th finit volum quation.. Starting from a 3-dimnsional conduction quation writtn in th cylindrical coordinat systm show that 1-d stady stat conduction in th thta dirction ( θ ) is givn by Driv finit volum quation by intgrating th abov quation. Explain th physical manings of ach trm in th finit volum quation. 3. Convctiv and radiativ boundary condition at x0 is givn by 1 r 1 r d dr d (r k dr ) + S 0 d d (k 1 d θ r d θ ) + S 0 q"hf(f - 1) + C(f ), whr C is a constant. Dscrib how you can implmnt this boundary condition into a finit volum quation. 4. Considr 1-dimnsional stady conduction as shown in th figur. whr q00 W/m, L8 m, L0, S4 W/m 3, k W/m.K. Calculat tmpraturs at two points as shown and chck th nrgy balanc.. Considr a 1-dimnsional stady conduction through a bar as shown in th figur. 48

49 whr L8 m, 00, L16 K, k W/m.K, S4 W/m 3. Calculat th tmpraturs at two points and chck th nrgy balanc. 6. Angular dirction momntum quation for a circular Coutt flow has a sourc trm (s problm 1 in chaptr ), v S -µ r Discuss possibl linarization schms for this sourc trm. θ ρ vr v - r θ 7. Driv intrfacial conductivity including contact surfac rsistanc and tmpratur as shown by Eqs.(36) and (37a,b). 8. Dtrmin th tmpratur at 4 qually spacd points in a taprd cylindrical fin. h bas is at C and th rmaining surfacs ar xposd to convctiv nvironmnt with hf10 W/m.K and f 0 C. Assum constant conductivity of 400 W/m.K. 9. Considr 1-D conduction in a rod that is bnt into a circular shap to form an ndlss loop. It thus has no xposd nds whr boundary conditions can b prscribd. Indd all grid points ar intrnal grid points. If a cut is mad in th loop, cyclic 49

50 boundary condition can b assumd at th cut. hat is 1N+1 and N+. How would you handl such cyclic boundary conditions? A spcial typ of DMA suitabl for cyclic boundary conditions is discussd in Rf.[3] that is calld circular DMA. (S Appndix) rojct 1. A prob of ovrall lngth L0 cm and diamtr D1. cm is insrtd through a duct wall such that a portion of its lngth, rfrrd to as th immrsion lngth Li is in contact with a gas stram at 000 K that is to b masurd. h rmaining lngth of th prob is xposd to air at 98 K. h convctiv hat transfr cofficint in th duct is 1000 W/m.K and in th air is 10 W/m.K. h conductivity of th prob matrial is 00 and its missivity is 0.1. [Rf. ] Immrsion rror is dfind by th tmpratur diffrnc btwn th gas tmpratur in th duct and th tmpratur at th tip of th prob, rr tip gas, which dpnds on th immrsion lngth Li. (a) Driv appropriat conduction quation for this problm. W/m.K (b) By using th numrical mthod, calculat th tmpratur distribution in th prob whn Li/L 1.0. Assum 1 uniform control volums. lot th rsults and chck th nrgy balanc. 0

51 (c) Estimat rr as a function of Li/L and plot th rsults. Us Li/L 0., 0., 0.7 and W want to valuat triangular fin fficincis by numrically solving stady hat conduction quation. riangular fin gomtry is shown in Fig. a. Fin fficincy is dfind by ηf qf/qmax, whr qf is th actual hat transfr from th fin and qmax is th maximum hat transfr whn th ntir fin surfac tmpratur is assumd to b at th bas tmpratur. [Rf. ] In ordr to vrify our approach to this problm, w will first chck th accuracy of numrical solution by solving a similar problm with a known solution. Considr a truncatd rctangular fin as shown in Fig. B. Fin surfacs is insulatd and tmpraturs at th bas and tips ar maintaind at B and L1, rspctivly. (a) By intgrating th quasi-on-dimnsional stady conduction quation without sourc trms, show that hat transfr rat along th x-dirction is givn by B L1 q x - Dtk( - ) Lln 1 - L 1 L and th tmpratur distribution is givn by 1

52 (b) Solv th abov truncatd fin problm numrically by using 10 uniform control volums. Assum following valus: t0.0 m; L0. m; D1.0 m; k400 W/m.K; B373 K; L198 K; L1L/ Compar th numrical rsults with th xact solution in trms of tmpratur distribution. Aftr chcking th fasibility of numrical approach, w now rturn to th fin problm. ak th insulation off th fin surfac and considr ntir fin (not th truncatd fin). h fin surfac is now xposd to a convctiv cooling with known hf and f. (c) Show that th conduction quation is givn by whr A is th cross-sctional ara and is th wttd primtr. (d) Calculat th stady stat tmpraturs by using 10 uniform control volums with th following conditions. Chck nrgy balanc to s whthr your solution is rasonabl. 0.0 m; L0. m; D1.0 m; k400 W/m.K; B373 K; f98 K; hf10 W/m.K (x) + q L Dtk ln(1 - x x B );0 x L1 L 1 d d (Ak A dx dx ) + h f A ( - ) 0 () Calculat th fin fficincy and compar with th rsults shown in Fig. C. (f) By changing th conductivity whil maintaining all othr valus, you can calculat th fin fficincy as a function of Lc 3/ (hf/kap) 1/ as shown in Fig. C. Obtain an fficincy curv and compar with th rsults in Fig.c. f

53 Figur (a) riangular fin, (b) runcatd triangular fin, (c) Fin fficincy 3

54 4. Considr a circular fin with rctangular profil as shown in th diagram. (a) Show that th stady conduction along th radial dirction in th fin is dscribd by th following quation 1 d d h Aκ + ( f ) 0 A dr dr A whr A is th cross sction ara and is th wttd primtr of th cross sction, A πrt ; 4πr (b) Lt r1 mm, r0 mm, t0. mm, κ0 W/m.K, f300 K, and h8. W/m.K. Calculat th tmpratur along th fin using th numrical mthod discussd in this chaptr. Us 10 uniform control volums. h bas tmpratur is 380 K and th tip of th fin is insulatd. (c) Calculat th hat transfr rat fro, th bas of th fin and valuat th fin fficincy η f q fin q max (d) Rpat th calculations with svral cass and obtain a fin fficincy curv. Compar with th known rsults. () Add th radiation ffcts on th hat transfr and modify th govrning quation in part (a). Lt th missivity varis 0.1 ~ 0.9. Compar with th rsults without radiation for part (b). 4

55 . Considr a pin fin with circular cross sction as shown in th figur blow. h fin is mad of pur iron. h total volum of th fin is fixd at.0-7 m 3. h thrmal conductivity of th matrial can b approximatd by κ ( ) W / mk, whr is in Klvin. Hf r f bas X0 h x h bas tmpratur is 800 K and th convctiv conditions ar hf10 W/m K and f98 K. W would lik to solv this problm by modifying std1db.m. h radius of th fin bas (r) changs as w chang th hight (h) sinc th total volum of th fin is fixd. 1 h volum of th pin fin is givn by V π r h. 3 a. Optimiz th pin hight (h) that givs th maximum conduction hat transfr rat from th bas by plotting q x 0 as a function of h, varying h from 0.0 m to 0.1 m. b. Fin fficincy can b dfind by η q / q x0 max, whr q max A sid h f ( 1 f ). lot η vs h. Is th maximum fin fficincy cas also th maximum hat transfr rat cas?