Consider the element shown in Figure 2.1. The statement of energy conservation applied to this element in a time period t is that:

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1 . Conduction. e General Conduction Equation Conduction occurs in a stationary medium wic is most liely to be a solid, but conduction can also occur in s. Heat is transferred by conduction due to motion of free electrons in metals or atoms in non-metals. Conduction is quantified by Fourier s law: te eat flu, q, is proportional to te temperature gradient in te direction of te outward normal. e.g. in te -direction: d q. d q W / m. e constant of proportionality, is te termal conductivity and over an area A, te rate of eat flow in te -direction, is d A W.3 Conduction may be treated as eiter steady state, were te temperature at a point is constant wit time, or as time dependent or transient were temperature varies wit time. e general, time dependent and multi-dimensional, governing equation for conduction can be derived from an energy balance on an element of dimensions, y, z. Consider te element sown in Figure.. e statement of energy conservation applied to tis element in a time period t is tat: eat flow in + internal eat generation = eat flow out + rate of increase in internal energy or y z g yy zz mc t.4 y yy z zz g mc t 0.5

2 As noted above, te eat flow is related to temperature gradient troug Fourier s Law, so: Figure - Heat Balance for conduction in an infinitismal element

3 d z y d A dy d z dy d A y.6 d y dz d A z Using a aylor series epansion: !!.7 For small values of it is a good approimation to ignore terms containing and iger order terms, So: z y.8 A similar treatment can be applied to te oter terms. For time dependent conduction in tree dimensions,y,z, wit internal eat generation z y m W q g g / / 3 : t C q z z y y g.9 For constant termal conductivity and no internal eat generation Fourier s Equation: t C z y.0 Were / C is nown as, te termal diffusivity m/s. For steady state conduction wit constant termal conductivity and no internal eat generation 0 z y. 3

4 Similar governing equations eist for oter co-ordinate systems. For eample, for D cylindrical coordinate system r, z,. In tis system tere is an etra term involving /r wic accounts for te variation in area wit r. z r r r 0. For one-dimensional steady state conduction in say te -direction d 0.3 It is possible to derive analytical solutions to te D and in some cases 3D conduction equations. However, since tis is beyond te scope of tis tet te interested reader is referred to te classic tet by Carlslaw and Jaeger 980 A meaningful solution to one of te above conduction equations is not possible witout information about wat appens at te boundaries wic usually coincide wit a solid- or solid-solid interface. is information is nown as te boundary conditions and in conduction wor tere are tree main types:. were temperature is specified, for eample te temperature of te surface of a turbine disc, tis is nown as a boundary condition of te st ind;. were te eat flu is specified, for eample te eat flu from a power transistor to its eat sin, tis is nown as a boundary condition of te nd ind; 3. were te eat transfer coefficient is specified, for eample te eat transfer coefficient acting on a eat ecanger fin, tis is nown as a boundary condition of te 3rd ind... Dimensionless Groups for Conduction ere are two principal dimensionless groups used in conduction. ese are: e Biot number, Bi L / and e Fourier number, Fo t It is customary to tae te caracteristic lengt scale L as te ratio of te volume to eposed surface area of te solid. / L. e Biot number can be tougt of as te ratio of te termal resistance due to conduction L/ to te termal resistance due to convection /. So for Bi <<, temperature gradients witin te solid are negligible and for Bi > tey are not. e Fourier number can be tougt of as a time constant for conduction. For Fo, time dependent effects are significant and for Fo tey are not. 4

5 .. One-Dimensional Steady State Conduction in Plane Walls In general, conduction is multi-dimensional. However, we can usually simplify te problem to two or even one-dimensional conduction. For one-dimensional steady state conduction in say te -direction: d 0.4 From integrating twice: C C were te constants C and C are determined from te boundary conditions. For eample if te temperature is specified st Kind on one boundary = at = 0 and tere is convection into d / a surrounding 3rd Kind at te oter boundary at = L ten:.5 wic is an equation for a straigt line. 5

6 o analyse -D conduction problems for a plane wall write down equations for te eat flu q. For eample, te eat flows troug a boiler wall wit convection on te outside and convection on te inside: q inside inside / L q outside outside q Rearrange, and add to eliminate and wall temperatures outside inside otside inside q.6 Note te similarity between te above equation wit I = V / R eat flu is te analogue of electrical current, temperature is of voltage and te denominator is te overall termal resistance, comprising individual resistance terms from convection and conduction. In building services it is common to quote a U value for double glazing and building eat loss calculations. is is called te overall eat transfer coefficient and is te inverse of te overall termal resistance. inside outside U.7..3 e Composite Plane Wall e etension of te above to a composite wall Region of widt L, termal conductivity, Region of widt L and termal conductivity.... etc. is fairly straigtforward. outside inside otside inside L L L q

7 Eample. e walls of te ouses in a new estate are to be constructed using a cavity wall design. is comprises an inner layer of bric = 0.5 W/m K and 0 mm tic, an air gap and an outer layer of bric = 0.3 W/m K and 0 mm tic. At te design condition te inside room temperature is 0ºC, te outside air temperature is -0ºC; te eat transfer coefficient on te inside is 0 W/m K, tat on te outside 40 W/m K, and tat in te air gap 6 W/m K. Wat is te eat flu troug te wall? Figure -: Conduction troug a plane wall Note te arrow sowing te eat flu wic is constant troug te wall. is is a useful concept, because we can simply write down te equations for tis eat flu. Convection from inside air to te surface of te inner layer of bric q in in Conduction troug te inner layer of bric q in / Lin Convection from te surface of te inner layer of bric to te air gap 7

8 q gap gap Convection from air gap to te surface of te outer layer of bric q gap 3 gap Conduction troug te outer layer of bric q out / Lout 3 4 Convection from te surface of te outer layer of bric to te outside air q out 4 out e above provides si equations wit si unnowns te five temperatures,, 3, 4 and gap and te eat flu q. ey can be solved simply by rearranging wit te temperatures on te left and side. q / in in 8

9 / q in / L in q / gap gap 3 q / gap gap 4 q / out / L out 3 4 q / out out en by adding, te unnown temperatures are eliminated and te eat flu can be found directly q L in in in gap in out gap L out out out q W / m It is instructive to write out te separate terms in te denominator as it can be seen tat te greatest termal resistance is provided by te outer layer of bric and te least termal resistance by convection on te outside surface of te wall. Once te eat flu is nown it is a simple matter to use tis to find eac of te surface temperatures. For eample, 4 q out out / 7.3/ C 4 ermal Contact Resistance In practice wen two solid surfaces meet ten tere is not perfect termal contact between tem. is can be accounted for using an appropriate value of termal contact resistance wic can be obtained eiter from eperimental results or publised, tabulated data. 9

10 . One-Dimensional Steady-State Conduction in Radial Geometries: Pipes, pressure vessels and annular fins are engineering eamples of radial systems. e governing equation for steady-state one-dimensional conduction in a radial system is d dr r d dr 0.9 From integrating twice: C ln C, and te constants are determined from te boundary conditions, e.g. if = at r = r and = at r = r, ten: ln r / r ln r / r.0 Similarly since te eat flow A d / dr, ten for a lengt L in te aial or z direction te eat flow can be found from differentiating Equation.0. LK ln r / r. o analyse -D radial conduction problems: Write down equations for te eat flow not te flu, q, as in plane systems, since in a radial system te area is not constant, so q is not constant. For eample, te eat flow troug a pipe wall wit convection on te outside and convection on te inside: r Linside inside L / ln r / r r Loutside outsied Rearrange, and add to eliminate and wall temperatures 0

11 r inside L inside outside ln r / r r outside. e etension to a composite pipe wall Region of termal conductivity, Region of termal conductivity.... etc. is fairly straigtforward. Eample. e Figure below sows a cross section troug an insulated eating pipe wic is made from steel = 45 W / m K wit an inner radius of 50 mm and an outer radius of 55 mm. e pipe is coated wit 00 mm ticness of insulation aving a termal conductivity of = 0.06 W / m K. Air at i = 60 C flows troug te pipe and te convective eat transfer coefficient from te air to te inside of te pipe as a value of i = 35 W / m K. e outside surface of te pipe is surrounded by air wic is at 5 C and te convective eat transfer coefficient on tis surface as a value of o = 0 W / m K. Calculate te eat loss troug 50 m of tis pipe.

12 Solution Figure -3: Conduction troug a radial wall Unlie te plane wall, te eat flu is not constant because te area varies wit radius. So we write down separate equations for te eat flow,. Convection from inside air to inside of steel pipe r L in in Conduction troug steel pipe L steel / ln r / r Conduction troug te insulation L insulation 3 / ln r3 / r Convection from outside surface of insulation to te surrounding air r L 3 3 out out

13 Following te practice establised for te plane wall, rewrite in terms of temperatures on te left and side and ten add to eliminate te unnown values of temperature, giving r in ln r L / r ln r steel i o 3 / r r3 insulation out ln0.55 / 0.50 ln0.55 / Watts Again, te termal resistance of te insulation is seen to be greater tan eiter te steel or te two resistances due to convection. Critical Insulation Radius Adding more insulation to a pipe does not always guarantee a reduction in te eat loss. Adding more insulation also increases te surface area from wic eat escapes. If te area increases more tan te termal resistance ten te eat loss is increased rater tan decreased. e so-called critical insulation radius is te largest radius at wic adding more insulation will create an increase in te eat loss r / crit ins et Eample.3 Find te critical insulation radius for te previous question. Solution: r / crit ins et r crit 0.06 /0 r crit 6mm So for r3 > 6 mm, adding more insulation, as intended, will reduce te eat loss. 3

14 .3 Fins and Etended Surfaces Figure -4: Eamples of fins a motorcycel engine, b eat sin Fins and etended surfaces are used to increase te surface area and terefore enance te surface eat transfer. Eamples are seen on: motorcycle engines, electric motor casings, gearbo casings, electronic eat sins, transformer casings and eat ecangers. Etended surfaces may also be an unintentional product of design. Loo for eample at a typical bloc of oliday apartments in a si resort, eac wit a concrete balcony protruding from eternal te wall. is acts as a fin and draws eat from te inside of eac apartment to te outside. e fin model may also be used as a first approimation to analyse eat transfer by conduction from say compressor and turbine blades. 4

15 .3. General Fin Equation e general equation for steady-state eat transfer from an etended surface is derived by considering te eat flows troug an elemental cross-section of lengt, surface area As and cross-sectional area Ac. Convection occurs at te surface into a were te eat transfer coefficient is and te temperature. Figure -5 Fin Equation: eat balance on an element Writing down a eat balance in words: eat flow into te element = eat flow out of te element + eat transfer to te surroundings by convection. And in terms of te symbols in Figure.5 As.4 From Fourier s Law. A c d.5 and from a aylor s series, using Equation.5 d A c d.6 and so combining Equations.4 and.6 d A c d As 0.7 5

16 e term on te left is identical to te result for a plane wall. e difference ere is tat te area is not constant wit. So, using te product rule to multiply out te first term on te left andside, gives: d A c dac d A c das 0.8 e simplest geometry to consider is a plane fin were te cross-sectional area, Ac and surface m p / A area As are bot uniform. Putting and letting c, were P is te perimeter of te cross-section d m 0.9 e general solution to tis is = Cem + Ce-m, were te constants C and C, depend on te boundary conditions. It is useful to loo at te following four different pysical configurations: N.B. sin, cos and tan are te so-called yperbolic sine, cosine and tangent functions defined by: e cos e e, sin e and sin tan cos.30 Convection from te fin tip L tip base cos m L { cos ml { tip tip / m sin m L } / m sin ml}.3 0 were base. P A / c base sin ml { cos ml { tip tip / m cos ml} / m sin ml}.3 Adiabatic ip tip = 0 in Equation.3 base cos m L cos ml.33 6

17 tan / ml A P base c ip at a specified temperature = L sin sin sin ml L m m base L base.35 sin cos / ml ml A P base L base c.36 Infinite Fin at m base e.37 / base c A P.38 7

18 .3. Fin Performance e performance of a fin is caracterised by te fin effectiveness and te fin efficiency Fin effectiveness, fin fin fin eat transfer rate / eat transfer rate tat would occur in te absence of te fin fin / Ac base.39 wic for an infinite fin becomes, given by fin P / A c Fin efficiency, / fin.40 fin actual eat transfer troug te fin / eat transfer tat would occur if te entire fin were at te base temperature. fin / A base s.4 wic for an infinite fin becomes, wit given by Equation.38 fin PA A / / c s.4 Eample.3 e design of a single pin fin wic is to be used in an array of identical pin fins on an electronics eat sin is sown in Figure.6. e fin is made from cast aluminium, = 80 W / m K, te diameter is 3 mm and te lengt 5 mm. ere is a eat transfer coefficient of 30 W / m K between te surface of te fin and surrounding air wic is at 5 C.. Use te epression for a fin wit an adiabatic tip to calculate te eat flow troug a single pin fin wen te base as a temperature of 55 C.. Calculate also te efficiency and te effectiveness of tis fin design. 3. How long would tis fin ave to be to be considered infinite? 8

19 Figure -6 Pin Fin design Solution For a fin wit an adiabatic tip P A / c base tan ml For te above geometry P d m A c d 6 / m 0 / ml P / Ac tan ml 0. L / Watts / Fin efficiency fin / As base A s dl m 9

20 fin 0.5 / fin % Fin effectiveness fin fin A / c base 0.5 / fin 9.6 For an infinite fin, = L =. However, te fin could be considered infinite if te temperature at te tip approaces tat of te. If we, for argument sae, limit te temperature difference between fin tip and to 5% of te temperature difference between fin base and, ten: L b

21 Using equation.33 for te temperature distribution and substituting = L, noting tat cos 0 =, implies tat / cos ml = So, ml = 3.7, wic requires tat L > 47 mm. Simple ime Dependent Conduction e -D time-dependent conduction equation is given by Equation.0 wit no variation in te y or z directions: t.43 A full analytical solution to te -D conduction equation is relatively comple and requires finding te roots of a transcendental equation and summing an infinite series te series converges rapidly so usually it is adequate to consider alf a dozen terms. ere are two alternative possible ways in wic a transient conduction analysis may be simplified, depending on te value of te Biot number Bi = L /..3.3 Small Biot Number Bi << : Lumped Mass Approimation A small value of Bi implies eiter tat te convective resistance /, is large, or te conductive resistance L/ is small. Eiter of tese conditions tends to reduce te temperature gradient in te material. For eample tere will be very little difference between te two surface temperatures of a eated copper plate 400 W/m K of say 5 or 0 mm ticness. Wereas for Perspe 0. W/m K, tere could be a significant difference. e copper tus beaves as a lumped mass. Hence for te purpose of analysis we may treat it as a body wit a omogenous temperature. A simple eat balance on a material of mass, m, density, specific eat C, ecanging eat by convection from an area A to surrounds at, gives m C d dt A s s.44 define: s / s, initial initial and : A / m C, f in forced convection wen s. is gives te simple solution: ln - t or ep- t.45 In free convection wen te eat transfer coefficient depends on te surface to temperature difference, say s n, ten te solution becomes: n n initial t.46

22 .3.4 Large Biot Number Bi >>: Semi - Infinite Approimation Wen Bi is large Bi >> tere are, as eplained above, large temperature variations witin te material. For sort time periods from te beginning of te transient or to be more precise for Fo <<, te boundary away from te surface is unaffected by wat is appening at te surface and matematically may be treated as if it is at infinity. ere are tree so called semi-infinite solutions: N.B. erfc = - erf; erf = error function Given by te series: erf ! 5 3! 7.47 Constant surface eat flu,t -initial q t q ep erfc 4t 4t / 4 /.48 Constant surface temperature, t s erf initial s 4t /.49 Constant surface eat transfer coefficient, t s erfc initial s 4t t ep erfc 4t / t / /.50 As well as being useful in determining te temperature of a body at time, tese low Biot number and large Biot number metods can also be used in te inverse mode. is is te reverse of te above and maes use of te temperature time istory to determine te eat transfer coefficient. Eample.4 A titanium alloy blade from an aial compressor for wic = 5 W / m K, = 4500 g / m3 and C = 50 J/g K, is initially at 40ºC. Altoug te blade ticness from pressure to suction side varies along te blade, te effective lengt scale for conduction may be taen as 3mm. Wen eposed to a ot gas stream at 350ºC, te blade eperiences a eat transfer coefficient of 50 W / m K. Use te lumped mass approimation to estimate te blade temperature after 50 s.

23 Firstly, cec tat Bi << Bi = L / = / 5 = So te lumped mass metod can be used. Solution ep t initial ep A m C t However, te mass m, and surface area A, are not nown. It is easy to reprase te above relationsip, since mass = density volume and volume = area ticness, were tis ticness is te conduction lengt scale, L. So initial ep t C L From wic 3

24 C L initial ep t ep C.4 Summary is capter as introduced te mecanism of eat transfer nown as conduction. In te contet of engineering applications, tis is more liely to be representative of te beaviour in solids tan s. Conduction penomena may be treated as eiter time-dependent or steady state. It is relatively easy to derive and apply simple analytical solutions for one-dimensional steadystate conduction in bot Cartesian plates and walls and cylindrical pipes and pressure vessels coordinates. wo-dimensional steady-state solutions are muc more comple to derive and apply, so tey are considered beyond te scope of tis introductory level tet. Fins and etended surfaces are an important engineering application of a one-dimensional conduction analysis. e design engineer will be concerned wit calculating te eart flow troug te fin, te fin efficiency and effectiveness. A number of relatively simple relations were presented for fins were te surface and cross sectional areas are constant. ime-dependent conduction as been simplified to te two etreme cases of Bi << and Bi >>. For te former, te lumped mass metod may be used and in te latter te semi-infinite metod. It is wort noting tat in bot cases tese metods are used in practical applications in te inverse mode to measure eat transfer coefficients from a now temperature-time istory. In many cases, te boundary conditions to a conduction analysis are provided in terms of te convective eat transfer coefficient. In tis capter a value as usually been ascribed to tis, witout eplaining ow and from were it was obtained. is will be te topic of te net capter. 4