Lecture 9, Thermal Notes, 3.054


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1 Lecture 9, Thermal Notes, Thermal Properties of Foams Closed cell foams widely used for thermal insulation Only materials with lower conductivity are aerogels (tend to be brittle and weak) and vacuum insulation panels Low thermal conductivity of foam arises from: low volume fraction of solid high volume fraction of gas with low λ small cell size suppresses convection and radiation (through repeated absorption and reflection) Applications: buildings, refrigerated vehicles, LNE tankers Foams also have good thermal shock resistance since coefficient of thermal expansion of foam equals to that of the solid; plus the modulus is much lower (ɛ = α T σ = Eα T = σ f ) used as heat shields Ceramic foams used as firebrick ceramic has high T foam  low λ  low heat loss low heat capacity  lowers energy to heat furnace to temperature good thermal shock resistance 1
2 Thermal conductivity, λ Steady state conduction (T constant with time) Fourier Law: q = λ T 1D dt q = λ dx Nonsteady heat conduction (T varies with time t) q = hect flux [J/(m 2 /s)] λ = thermal conductivity [W/mK] T = temperature gradient T = i + j T x y + k T z T τ = a 2 T x 2 a = thermal diffusivity = λ ρ = density C p = specific heat  heat required to raise the temperature of unit mass by 1 K ρ C p 2 ρ C p = volumetric heat capacity [J/m 3 K] [m /s] Values for λ, a Table 7.1 2
3 Data for thermal conductivity and thermal diffusivity Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, Table courtesy of Lorna Gibson and Cambridge University Press. 3
4 Thermal diffusivity, a Materials with a high value of a rapidly adjust their temperature to that of surroundings, because they conduct hear rapidly in comparison to their volumetric heat capacity; do not require much energy to reach thermal equilibrium e.g. Cu a = m 2 /s nylon a = m 2 /s wood a = m 2 /s Thermal conductivity of a foam, λ. λ contributions from conduction through solid, λ s conduction through gas, λ g convection within cells, λ c radiation through cell walls and across voids, λ r λ = λ s + λ g + λ c + λ r Conduction through solid: λ s = ηλ s (ρ /ρ s ) η = efficiency factor 2/3 Conduction through gas: λ g = λ g (1 ρ /ρ s ) 4
5 For example, 2.5% dense closedcell polystyrene foam: λ = W/mK; λ s = 0.15 W/mK; λ g = W/mK (air) λ s + λ g =2/3 (0.15)(0.025) + (0.025)(0.975) = =0.027 W/mK Most of conductivity comes from conduction through gas Foams for isolation blown with low λ g gases Problem with aging low λ g gases diffuse out of foam over time, air diffuses in; λ g Convection within the cell Gas rises and falls due to density changes with temperature Density changes buoyancy forces Also have viscous forces from drag of gas as it moves past cell wall Convection is important when Rayleigh number > 1000 ρ = density of gas T c = temp. diff. across the ρgβ Tc l 3 R a = g = grav. acceleration cell µa β = volume expansion l = cell size for a gas = 1/T (isobaric) µ = dynamic viscosity of gas a = thermal diffusion 5
6 Convection For R a = 1000 air p = p atm T = room temp β = 1/T = 1/300 ( K 1 ). T c = 1 K µ air = Pa s ρ 3 air = 1.2 kg/m aair = m 2 /s l = 20 mm 5 Convection important if cell size > 20 mm Most foams: cell size < 1 mm convection negligible Radiation Hect flux passing by radiation, q 0 r, from surface at temperature T 1, to one at a lower temperature T 0, with a vacuum between them, is: q 0 4 r = β 1 σ(t 1 T 4 0 ) Stefan s law σ = Stefan s constant = W/m 2 K 4 β 1 = constant (< 1) describing emissivity of the surfaces (emitted radiant flux per unit area of sample relative to black body radiator at same temperature and conditions; black body absorbs all energy; black body emissivity =1) 6
7 Radiation If put foam between two surfaces, heat flux is reduced, since radiation is absorbed by the solid and reflected by cell walls Attenuation q r = qr 0 exp ( K t ) Beer s law K = extinction coefficient for foam t = thickness of foam For optically thin walls and struts (t < 10µm) (transparent to radiation) K = (ρ /ρ s ) K s Heat flux by radiation then: dt q r = λ r dx 4 4 dt q r = β 1 σ(t 1 T 0 ) exp [ (ρ /ρ s )K s t ] = λ r dx Obtain λ r using some approximations 7
8 Approximations: dt T dx 1 T 0 = T t t ( T1 4 T0 4 4 T T 3 T1 T ) 0 T = 2 q r = β 1 σ4 T T 3 exp [ (ρ /ρ s )K s t ] = λ r r = 4β 1 σt 3 t exp [ (ρ /ρ s )K s t ] as ρ /ρ s λ r λ T dx Thermal conductivity Relative contributions of λ s, λ g, λ r shown in Fig. 7.1 largest contribution λ g λ plotted against relative density Fig. 7.2 minimum between ρ /ρ s of 0.03 and 0.07 at which point λ only slightly larger than λ s at low ρ /ρ s, λ increases  increasing transparency to radiation (also, walls may rupture) tradeoff: as ρ /ρ s goes down, λ s goes down, but λ r goes up 8
9 Thermal Conductivity Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, Figure courtesy of Lorna Gibson and Cambridge University Press. 9
10 Cond. Vs. Relative Density Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, Figure courtesy of Lorna Gibson and Cambridge University Press. 10
11 Cond. vs. Cell Size Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, Figure courtesy of Lorna Gibson and Cambridge University Press. 11
12 λ plotted against cell size Fig. 7.3 λ increases with cell size Radiation reflected less often Note: aerogels Pore size < 100nm Mean free path of air at ambient pressure = 68 nm average distance molecules move before collision with another molecule Aerogels pore size < mean free path of air reduced conduction through gas Specific hear C p Specific heat energy required to raise temperature of unit mass by unit temperature C p = C ps [J/kg K] Thermal expansion coefficient α = α s (consider foam as framework) (but if closedcell foam cooled dramatically gas can freeze, collapsing the cells; or if heated gas expands, increasing the internal pressure and strains) 12
13 Thermal shock resistance If material subjected to sudden change in surface temperature  induces thermal stresses at surface, plus cracking and spalling Consider material at T 1 dropped intp water at T 2 (T 1 > T 2 ) Surface temperature drops to T 2, contracting surface layers Thermal strain ɛ T = α T If surface bonded to underlying block of material  constrained to original dimensions σ = E α T 1 in the surface Cracking/spalling when σ = σ f 1 ν T = σ f = critical T to just cause cracking Eα For foam: (open cells) σfs(ρ /ρs) /2 (1 ν ) 0.2 σ fs (1 ν) T c = = = E s (ρ /ρ s ) 2 α s (ρ /ρ s ) 1/2 E s α s 0.2 T /ρ s ) 1/2 (ρ cs As foam density goes down, T c goes up firebrick  porous ceramic 13
14 Case study: optimization of foam density for thermal insulation There is an optimal foam density for a given thermal insulation problem Already saw λ has a minimum as a f(ρ /ρ s ) Typically, have a constraint on the foam thickness, t, t =constant 2 λ = (ρ /ρ s )λ s + (1 ρ /ρ s )λ g + 4β 1 σt 3 t exp[ K s (ρ /ρ s )t ] 3 What is optimum ρ /ρ s for a given t? dλ d(ρ /ρ s ) = 0 (ρ /ρ s ) opt = 1 [ K s t ln 4Ks β 1 σ T 3 t 2] 2 3 λ s λ g As given thickness t increases, (ρ /ρ s ) opt decreases As T increases, (ρ /ρ s ) opt increases e.g. coffee cup t = 3mm (ρ /ρ s ) opt = 0.08 refrigerator t = 50mm (ρ /ρ s ) opt = 0.02 (see PP slide Table 7.3 for data used in calculations) 14
15 Case Study: Optimization of Relative Density Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, Figures courtesy of Lorna Gibson and Cambridge University Press. 15
16 Case Study: Optimum Relative Density Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, Table courtesy of Lorna Gibson and Cambridge University Press. 16
17 Case study: insulation for refrigerators Insulation reduces energy cost, but has a cost itself Total cost is the cost of insulation plus the cost of energy lost by hear transfer through walls Objective function: minimize total cost given: x=thickness of insulation C M =cost of insulation/mass T =temp. diff. across insulation C E =cost of energy / joule t l =design life of refrigerator C T =total cost/area T C T = x ρ C M + λ t l C E (heat flux q = λ T x x Define: M 1 = 1 M 1 ρ 2 = λ C T x = 1 M 1 + C M [ ] T 1 t l C E x 2 M 2 J m 2 s ) 17
18 The terms are equal when: [ ] T M 2 = t l C E } x 2 {{ } coupling constant M 1 Family of parallel straight lines of constant value T t x 2 l C E Fig T = 20 x = 10mm C E = 0.01/µJ Two lines for t 2 = 10 years and t l = 1 month (note error in book t l = 10 years line should be moved over) Also plotted a set of curved contours  plots of C T /x: As move up and to the right of plot, the value of C T /x decreases For t l = 10 years phenolic foam ρ = Mg/m 3 For t l = 1 month EPS ρ = 0.02 Mg/m 3 PP ρ = 0.02 Mg/m 3 18
19 Case Study: Insulation for Refrigerators Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, Figures courtesy of Lorna Gibson and Cambridge University Press. 19
20 MIT OpenCourseWare / 3.36 Cellular Solids: Structure, Properties and Applications Spring 2014 For information about citing these materials or our Terms of Use, visit:
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