LECTURER: PM DR MAZLAN ABDUL WAHID PM Dr Mazlan Abdul Wahid

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H E A R A N S F E R HEA RANSFER SME 4463 LECURER: PM DR MAZLAN ABDUL WAHID http://www.fkm.utm.

1 H E A R A N S F E R HEA RANSFER SME 4463 LECURER: PM DR MAZLAN ABDUL WAHID C H A P E R 3 Dr Mazlan - SME 4463 H E A R A N S F E R Chapter 5 ranent Conducton PM Dr Mazlan Abdul Wahd Faculty of Mechancal Engneerng Unvert eknolog Malaya C H A P E R 3 Dr Mazlan - SME

2 ranent Conducton: he Lumped Capactance Method Chapter Fve Secton 5.1 through 5.3 ranent Conducton ranent Conducton A heat tranfer proce for whch the temperature vare wth tme, a well a locaton wthn a old. It ntated whenever a ytem experence a change n operatng condton. It can be nduced by change n: urface convecton condton ( h, ), urface radaton condton ( h, ), a urface temperature or heat flux, and/or nternal energy generaton. Soluton echnque he Lumped Capactance Method Exact Soluton he Fnte-Dfference Method r ur

Lumped Capactance Method he Lumped Capactance Method Baed on the aumpton of a patally unform temperature dtrbuton r throughout the tranent proce. Hence,, t t.

3 Lumped Capactance Method he Lumped Capactance Method Baed on the aumpton of a patally unform temperature dtrbuton r throughout the tranent proce. Hence,, t t. Why the aumpton never fully realzed n practce? General Lumped Capactance Analy: Conder a general cae, whch nclude convecton, radaton and/or an appled heat flux at pecfed urface ( A, c, A, r, A, h ), a well a nternal energy generaton Lumped Capactance Method (cont.) Frt Law: de dt t d = ρ = & & + & dt c En Eout Eg Aumng energy outflow due to convecton and radaton and nflow due to an appled heat flux q, d ρ c = q A ha ( ) h A ( ) + E & dt, h, c r, r ur g (5.15) May h and h r be aumed to be contant throughout the tranent proce? How mut uch an equaton be olved? 3

Specal Cae (Neglgble Radaton) Specal Cae (Exact Soluton, ) 0 Neglgble Radaton ( θ θ θ b a), / : a ha, c / ρ c b q A, h + E& g / ρ c he non-homogeneou dfferental equaton tranformed nto a homogeneou

4 Specal Cae (Neglgble Radaton) Specal Cae (Exact Soluton, ) 0 Neglgble Radaton ( θ θ θ b a), / : a ha, c / ρ c b q A, h + E& g / ρ c he non-homogeneou dfferental equaton tranformed nto a homogeneou equaton of the form: dθ = aθ dt Integratng from t = 0 to any t and rearrangng, b / a = exp + 1 exp ( at) ( at) (5.5) o what doe the foregong equaton reduce a teady tate approached? How ele may the teady-tate oluton be obtaned? Specal Cae (Convecton) Neglgble Radaton and Source erm h >> hr, E& g = 0, q = 0 : d ρ c = ha, c ( ) (5.) dt ρ c θ dθ t = dt Note: θ ha θ θ 0, c θ ha, c t = = exp t = exp θ ρ c τt he thermal tme contant defned a 1 τt ha, c ( ρ c) (5.7) (5.6) hermal Retance, R t Lumped hermal Capactance, C t he change n thermal energy torage due to the tranent proce t Et Q = E & t t outdt = ha, c θdt = ( ρ c) θ 1 exp 0 0 τt (5.8) 4

5 Specal Cae (Radaton) Neglgble Convecton and Source erm hr >> h, E& g = 0, q = 0 : Aumng radaton exchange wth large urroundng, d ρ c = ε A σ dt ε A, r ρ c σ t dt = 0 4 4, r ur d 4 4 ur t ρ c + + A ur ur = 1n 1n 4ε 3, rσ ur ur ur tan tan ur ur (5.18) h reult necetate mplct evaluaton of (t). Bot Number he Bot Number and Valdty of he Lumped Capactance Method he Bot Number: he frt of many dmenonle parameter to be condered. Defnton: hl B c k h convecton or radaton coeffcent k thermal conductvty of the o ld L charactertc length of the old ( / A or coordnate c aocated wth maxmum patal temperature dfferenc e) Phycal Interpretaton: B L / ka R = c cond old 1/ ha R conv old/flud Crteron for Applcablty of Lumped Capactance Method: B << 1 See Fg

6 Problem: hermal Energy Storage Problem 5.1: Chargng a thermal energy torage ytem contng of a packed bed of alumnum phere. KNOWN: Dameter, denty, pecfc heat and thermal conductvty of alumnum phere ued n packed bed thermal energy torage ytem. Convecton coeffcent and nlet ga temperature. FIND: me requred for phere at nlet to acqure 90% of maxmum poble thermal energy and the correpondng center temperature. Schematc: Problem: hermal Energy Storage (cont.) ASSUMPIONS: (1) Neglgble heat tranfer to or from a phere by radaton or conducton due to contact wth other phere, () Contant properte. ANALYSIS: o determne whether a lumped capactance analy can be ued, frt compute B = h(r o /3) = 75 W/m K (0.015 m)/150 W/m K = << 1. Hence, the lumped capactance approxmaton may be made, and a unform temperature may be aumed to ext n the phere at any tme. From Eq. 5.8, achevement of 90% of the maxmum poble thermal energy torage correpond to E t = 0.90 = 1 exp ( t / τt ) ρcvθ kg/m m 950 J/kg K τt = ρvc / ha = ρdc / 6h = = W/m K t = τ t ln ( 0.1) = = 984 < From Eq. (5.6), the correpondng temperature at any locaton n the phere ( 984 ) = g, + ( g, ) exp( 6 ht / ρdc) = 300 C 75 C exp 6 75 W/m K 984 / 700 kg/m m 950 J/kg K ( ) ( 984 ) = 7.5 C If the product of the denty and pecfc heat of copper (ρc) Cu 8900 kg/m J/kg K = J/m 3 K, there any advantage to ung copper phere of equvalent dameter n leu of alumnum phere? Doe the tme requred for a phere to reach a precrbed tate of thermal energy torage change wth ncreang dtance from the bed nlet? If o, how and why? < 6

Problem: Furnace Start-up Problem 5.: Heatng of coated furnace wall durng tart-up. KNOWN: hckne and properte of furnace wall. hermal retance of ceramc coatng on urface of wall expoed to furnace gae.

) ASSUMPIONS: (1) Contant properte, () Neglgble coatng thermal capactance, (3) Neglgble radaton. PROPERIES: Carbon teel: ρ = 7850 kg/m 3, c = 430 J/kg K, k = 60 W/m K.

7 Problem: Furnace Start-up Problem 5.: Heatng of coated furnace wall durng tart-up. KNOWN: hckne and properte of furnace wall. hermal retance of ceramc coatng on urface of wall expoed to furnace gae. Intal wall temperature. FIND: (a) me requred for urface of wall to reach a precrbed temperature, (b) Correpondng value of coatng urface temperature. Schematc: Problem: Furnace Start-up (cont.) ASSUMPIONS: (1) Contant properte, () Neglgble coatng thermal capactance, (3) Neglgble radaton. PROPERIES: Carbon teel: ρ = 7850 kg/m 3, c = 430 J/kg K, k = 60 W/m K. ANALYSIS: Heat tranfer to the wall determned by the total retance to heat tranfer from the ga to the urface of the teel, and not mply by the convecton retance. Hence, wth U = ( Rtot ) = + R f = + 10 m K/W = 0 W/m K. h 5 W/m K UL 0 W/m K 0.01 m B = = = << 0.1 k 60 W/m K and the lumped capactance method can be ued. (a) From Eq. (5.6) and (5.7), = exp / t = exp / t t = exp / ( t τ ) ( t R C ) ( Ut ρlc) 3 ρlc ln 7850 kg/m 0.01 m 430 J/kg K t ln = = U 0 W/m K t = 3886 = 1.08 h. < 7

8 Problem: Furnace Start-up (cont.) (b) Performng an energy balance at the outer urface (,o), ( ) = ( ) / h,o,o, R f h / 5 W/m K 1300 K 100 K/10 - m +, R f + K/W,o = = h + 1/ W/m K ( R f ),o = 10 K. < How doe the coatng affect the thermal tme contant? ranent Conducton: Spatal Effect and the Role of Analytcal Soluton Chapter 5 Secton 5.4 through 5.8 8

9 Plane Wall Soluton to the Heat Equaton for a Plane Wall wth Symmetrcal Convecton Condton If the lumped capactance approxmaton cannot be made, conderaton mut be gven to patal, a well a temporal, varaton n temperature durng the tranent proce. For a plane wall wth ymmetrcal convecton condton and contant properte, the heat equaton and ntal/boundary condton are: 1 = x α t x,0 x x= 0 (5.9) = (5.30) = 0 k = h ( L, t) x x= L Extence of eght ndependent varable: (,,,,,, α, ) x t L k h (5.31) (5.3) = (5.33) How may the functonal dependence be mplfed? Plane Wall (cont.) Non-dmenonalzaton of Heat Equaton and Intal/Boundary Condton: θ Dmenonle temperature dfference: θ * = θ x Dmenonle pace coordnate: x * L αt Dmenonle tme: t * Fo L Fo the Fourer Number hl he Bot Number: B k Exact Soluton: θ * = f x *, Fo, B old co * = C n exp n Fo nx * n= 1 θ ζ ζ C = 4nζ n ζ tanζ B ζ + n = ( ζ ) n n n n n (5.4a) (5.4b,c) See Appendx B.3 for frt four root (egenvalue ζ,..., ζ ) of Eq. (5.4c)

10 Plane Wall (cont.) he One-erm Approxmaton ( Vald for Fo > 0.) : Varaton of mdplane temperature (x * = 0) wth tme ( Fo) : ( ) * o θ o C 1 exp( ζ 1 ) Fo able 5.1 C and ζ a a functon of B 1 1 Varaton of temperature wth locaton (x * ) and tme Fo : θ = θ co ζ ( x ) * o * 1 * Change n thermal energy torage wth tme: t (5.44) (5.43b) E = Q (5.46a) nζ 1 Q = Q 1 * o θ o (5.49) ζ 1 Q = ρcv (5.47) o Can the foregong reult be ued for a plane wall that well nulated on one de and convectvely heated or cooled on the other? Can the foregong reult be ued f an othermal condton ntantaneouly mpoed on both urface of a plane wall or on one urface of a wall whoe other urface well nulated? Heler Chart Graphcal Repreentaton of the One-erm Approxmaton he Heler Chart, Secton 5 S.1 Mdplane emperature: 10

11 Heler Chart (cont.) emperature Dtrbuton: Change n hermal Energy Storage: Radal Sytem Radal Sytem Long Rod or Sphere Heated or Cooled by Convecton. B = hr / k o Fo = αt / r o One-erm Approxmaton: Long Rod: Eq. (5.5) and (5.54) Sphere: Eq. (5.53) and (5.55) C, ζ able Graphcal Repreentaton: Long Rod: Fg. 5 S.4 5 S.6 Sphere: Fg. 5 S.7 5 S.9 11

12 Sem-Infnte Sold he Sem-Infnte Sold A old that ntally of unform temperature and aumed to extend to nfnty from a urface at whch thermal condton are altered. Specal Cae: Cae 1: Change n Surface emperature ( ) ( 0, ) (,0) t = x = (, ) x t x = erf αt (5.60) ( ) k q = παt (5.61) Sem-Infnte Sold (cont.) Cae : Contant Heat Flux ( q = q ) 1 α π q o t / x ( x, t) = exp k 4αt q ox x erfc k αt (5.6) Cae 3: Surface Convecton, h k = h ( 0, t) x x= 0 (, ) x t x = erfc αt hx h αt x h αt exp + erfc k k + αt k (5.63) o 1

Multdmenonal Effect Multdmenonal Effect Soluton for multdmenonal tranent conducton can often be expreed a a product of related one-dmenonal oluton for a plane wall, P(x,t), an nfnte cylnder, C(r,t),

13 Multdmenonal Effect Multdmenonal Effect Soluton for multdmenonal tranent conducton can often be expreed a a product of related one-dmenonal oluton for a plane wall, P(x,t), an nfnte cylnder, C(r,t), and/or a em-nfnte old, S(x,t). See Equaton (5 S.1) to (5 S.3) and Fg. 5 S.11. Conder uperpoton of oluton for two-dmenonal conducton n a hort cylnder: (,, ) r x t (, ) C ( r, t) = P x t (, ) x t r,t = Plane Wall Infnte Cylnder Object wth Contant or q Object wth Contant Surface emperature or Surface Heat Fluxe ranent repone of a varety of object to a tep change n urface temperature or heat flux can be unfed by defnng the dmenonle conducton heat rate: q q L k c * = ( ) (5.67) where L c a charactertc length that depend on the geometry of the object. Conder the varaton of q* wth tme, or Fo, for Interor heat tranfer: Heat tranfer nde object uch a plane wall, cylnder, or phere, Exteror heat tranfer: Heat tranfer n an nfnte medum urroundng an embedded object. 13

Object wth Contant or q (cont.) When q* plotted veru Fo n Fgure 5.10, we ee that: All object behave the ame a a em-nfnte old for hort tme. q* approache a teady tate for exteror object.

14 Object wth Contant or q (cont.) When q* plotted veru Fo n Fgure 5.10, we ee that: All object behave the ame a a em-nfnte old for hort tme. q* approache a teady tate for exteror object. q* doe not reach a teady tate for nteror object, but decreae contnually wth tme (Fo). Contant Contant q Why do all object behave the ame a a em-nfnte old for hort tme? Object wth Contant or q (cont.) Approxmate Soluton for Object wth Contant or q Eay-to-ue approxmate oluton for q*(fo) are preented n able 5. for all the cae preented n Fgure A an example of the ue of able 5., conder: Infnte cylnder ntally at ha contant heat flux mpoed at t urface. Fnd t urface temperature a a functon of tme. Look n able 5.b for contant urface heat flux, Interor Cae, Infnte cylnder. Length cale L c = r o, the cylnder radu. Exact oluton for q*(fo) a complcated nfnte ere. Approxmate oluton gven by: 1 π π 1 q* = for Fo < 0. or q* = Fo for Fo 0. Fo It then a mple matter to fnd from the defnton, 1 q q L k c * = ( ) 14

15 Problem: hermal Energy Storage Problem 5.80: Chargng a thermal energy torage ytem contng of a packed bed of Pyrex phere. KNOWN: Dameter, denty, pecfc heat and thermal conductvty of Pyrex phere n packed bed thermal energy torage ytem. Convecton coeffcent and nlet ga temperature. FIND: me requred for phere to acqure 90% of maxmum poble thermal energy and the correpondng center and urface temperature. SCHEMAIC: Problem: hermal Energy Storage (cont.) ASSUMPIONS: (1) One-dmenonal radal conducton n phere, () Neglgble heat tranfer to or from a phere by radaton or conducton due to contact wth adjonng phere, (3) Contant properte. ANALYSIS: Wth B h(r o /3)/k = 75 W/m K (0.015m)/1.4 W/m K = 0.67, the lumped capactance method napproprate and the approxmate (one-term) oluton for one-dmenonal tranent conducton n a phere ued to obtan the dered reult. o obtan the requred tme, the pecfed chargng requrement ( Q / Q o = 0.9) mut frt be ued to obtan the dmenonle center temperature, θ o *. From Eq. (5.55), 3 ζ1 Q θo = 1 3 n( ζ1) ζ1co( ζ1) Qo Wth B hr o /k =.01, ζ 1.03 and C from able 5.1. Hence, 3 ( ) θ o = = = 15

16 Problem: hermal Energy Storage (cont.) From Eq. (5.53c), the correpondng tme r o θ t ln o = αζ C α = k / ρc = 1.4 W/m K / 5kg/m 835J/kg K = m /, ( m) ln ( / 1.48) m / (.03) t = = 1,00 < From the defnton of θ o *, the center temperature o = g, g, = 300 C 4.7 C = 57.3 C he urface temperature at the tme of nteret may be obtaned from Eq. (5.53b) wth r = 1, ( ζ1) ζ θo n = g, + ( g, ) = 300 C 75 C = 80.9 C 1.03 I ue of the one-term approxmaton approprate? < < Problem: hermal Repone Frewall Problem: 5.93: Ue of radaton heat tranfer from hgh ntenty lamp ( q 4 = 10 W/m ) for a precrbed duraton (t=30 mn) to ae ablty of frewall to meet afety tandard correpondng to maxmum allowable temperature at the heated (front) and unheated (back) urface. KNOWN: hckne, ntal temperature and thermophycal properte of concrete frewall. Incdent radant flux and duraton of radant heatng. Maxmum allowable urface temperature at the end of heatng. FIND: If maxmum allowable temperature are exceeded. SCHEMAIC: 16

17 Problem: hermal Repone of Frewall (cont.) ASSUMPIONS: (1) One-dmenonal conducton n wall, () Valdty of emnfnte medum approxmaton, (3) Neglgble convecton and radatve exchange wth the urroundng at the rradated urface, (4) Neglgble heat tranfer from the back urface, (5) Contant properte. ANALYSIS: he thermal repone of the wall decrbed by Eq. (5.6) ( α π ) 1/ x qo t / q (, ) exp o x x x t = + erfc k 4αt k αt 7 where, α = k / ρc p = m / and for ( α π ) 1/ t = 30 mn = 1800, q t / / k = 84.5 K. Hence, at x = 0, o ( 0,30 mn) = 5 C C = C < 35 C At x = 0.5 m, ( x / 4α t ) = 1.54; q / 1, 786 K, and / 1/ o x k = x α t = Hence, m, 30 mn = 5 C C C ~ 0 5 C Problem: hermal Repone of Frewall (cont.) Both requrement are met. < I the aumpton of a em-nfnte old for a plane wall of fnte thckne approprate under the foregong condton? COMMENS: he foregong analy may or may not be conervatve, nce heat tranfer at the rradated urface due to convecton and net radaton exchange wth the envronment ha been neglected. If the emvty of the urface and the temperature of the urroundng are aumed to be ε = 1 and ur = 98K, radaton exchange at = C would be 4 4 q rad = εσ ur = 6080 W/m K, whch gnfcant (~ 60% of the precrbed radaton). However, under actual condton, the wall would lkely be expoed to combuton gae and adjonng wall at elevated temperature. 17

18 Problem: Mcrowave Heatng Problem: 5.101: Mcrowave heatng of a phercal pece of frozen ground beef ung mcrowave-aborbng packagng materal. KNOWN: Ma and ntal temperature of frozen ground beef. Rate of mcrowave power aborbed n packagng materal. FIND: me for beef adjacent to packagng to reach 0 C. SCHEMAIC: Beef, 1kg = -0 C Packagng materal, q & ASSUMPIONS: (1) Beef ha properte of ce, () Radaton and convecton to envronment are neglected, (3) Contant properte, (4) Packagng materal ha neglgble heat capacty. Problem: Mcrowave Heatng (cont.) PROPERIES: able A.3, Ice ( 73 K): ρ = 90 kg/m 3, c = 040 J/kg K, k = 1.88 W/m K. ANALYSIS: Neglectng radaton and convecton loe, all the power aborbed n the packagng materal conduct nto the beef. he urface heat flux q& 0.5P q = = 4π A R he radu of the phere can be found from knowledge of the ma and denty: 4 m = rv = ρ π r3 o 3 1/3 1/3 3 m 3 1 kg ro = = = m 4π ρ 4π 3 90 kg/m hu 0.5(1000W) q = = 9780 W/m 4π ( m) q 9780 W/m m * r q o = = = 16.6 k ( ) o o 1.88 W/m K ( 0 C -(-0 C) ) he beef can be een a the nteror of a phere wth a contant heat flux at t urface, thu the relatonhp n able 5.b, Interor Cae, phere, can be ued. We begn by calculatng q* for =0 C. 18

19 Problem: Mcrowave Heatng (cont.) We proceed to olve for Fo. Aumng that Fo < 0., we have 1 π π q* - Fo 4 - π Fo = π ( q* + ) = Snce th le than 0., our aumpton wa correct. Fnally we can olve for the tme: t = Fo r / = o α Fo ro ρc / k = (0.006 ( m) 90 kg/m J/kg K)/(1.88 W/m K) = 10.6 < COMMENS: At the mnmum urface temperature of -0 C, wth = 30 C and h = 15 W/m K from Problem 5.33, the convecton heat flux 750 W/m, whch le than 8% of the mcrowave heat flux. he radaton heat flux would lkely be le, dependng on the temperature of the oven wall. 19

Transient Conduction: Spatial Effects and the Role of Analytical Solutions

Transient Conduction: Spatial Effects and the Role of Analytical Solutions Transent Cnductn: Spatal Effects and the Rle f Analytcal Slutns Slutn t the Heat Equatn fr a Plane Wall wth Symmetrcal Cnvectn Cndtns If the lumped capactance apprxmatn can nt be made, cnsderatn must be