Thermal characterization of an insulating material through a three- - PowerPoint PPT Presentation

Laboratoire d’Energétique et de Mécanique Théorique et Appliquée METTI 5 Mesures en Thermique et Techniques Inverses Roscoff, June, 13-18 2011 Tutorial T9 – 1h30 Thermal characterization of an insulating material through a three- layers transient method (experimental) Vincent FELIX Yves JANNOT Alain DEGIOVANNI

Plan of the presentation I. Introduction II. The classical approach 1. Example : the flash method 2. Example : the hot plate method III. The three-layers device method 1. Proposed approach 2. Principle 3. Modeling 4. Sensitivity analysis 5. Numerical simulations IV. Experimental part 1. The device 2. Materials 3. Measurements and data processing V. Conclusion 2

I. Introduction � Classical approach in thermal characterization • Model under the form of a h,Rc,mc s … heat flow / temperature relationship T(x,y,z,t) ( ) ( ) , , , , , , , , ... = Φ λ ρ T x y z t f c h Rc mc Φ S λ , ρ c,a… • The estimation is performed on the whole unknown parameters with two risks: � The accuracy is decreasing with the number of parameters (correlations, local minimums) � Some parameters are not easily quantifiable: hypothesis may introduce bias in the model • What if the material is very insulating, has a low density or a high porosity ? � A qualitative answer through 2 examples : the flash method and the hot plate method • How does the proposed three-layers device method avoid these drawbacks ? • The technical specificity • The original modeling approach : a temperature / temperature relationship 3

II. Classical approach 1. Example: the flash method • Model under the form: ( ) ( ) , , 1 , = Φ T t f a h h h 1 2 T(t) r • Assuming the hypothesis that: Φ x = = = h h h cst h 2 1 2 a • The following parameters are estimated: Φ , a and h . Φ Φ Φ • In case of an insulating material, � T heated surface >> T and thus h 1 ≠ h 2 � T heated surface has to be very high to have a good signal T � The more insulating the material is, the more the hypothesis is contestable. � If h 1 and h 2 are estimated separately, h 1 , Φ and a get correlated between them. • In case of a low density material � The measurement of the surface temperature is difficult 4 � The heat flow may not be absorbed only at the surface but inside the sample (volume absorbtion)

II. Classical approach 2. Example: the hot plate method • Model under the form: ( ) ( ) , , , = Φ T t f E Rc mc mc s s T(t) • Assuming the following hypothesis: r • Uniform contact resistance Φ x • Homogeneous heating element • Thin heating element (capacity) E • The estimated parameters are: E, Rc and Rc mc S . • In case of a low density material � The thermal capacity of the heating element is no more negligible compared to the capacity of the sample � The effusivity may be correlated with the thermal capacity of the heating element • In case of an insulating material � The transfer into the heating element (parallel to the contact surface) is not negligible � To limit bounds effects, large sample are needed 5

III. The three-layers method 1. Proposed approach: temperature / temperature relationship � Input / Output representation where h 2 ,h 3 ,… the system (sample) is characterized by its transfer function H(p): T 1 (x,y,z,t) T 2 (x,y,z,t) [ ] ( ) L T t ( ) = 2 H p Φ [ ] ( ) L T t 1 λ , ρ c,… h 1 ,Rc,mc s … ( ) ( ) , , , , , ( ) = λ ρ ⊗ T t F t c a h h T t • Time dependent model under the form of a convolution product: 2 2 3 1 ( ) ( ) , , , , , , , , , = φ λ ρ T t F t c a h h h R c m c Instead of: 2 1 2 3 S � As a consequence: the model is independent of all the parameters before the input T 1 : Φ ,h 1 ,mc S ,Rc… 6

III. The three-layers method 2. Principle • Flash method = surface temperature measurement impossible for low density materials � Technical solution: flash method on a three-layers system Brass, λ b , ρ c b � Brass discs: thin, high thermal T 2 (t) conductivity so their temperature h 1 r T 1 and T 2 are uniform Φ (t) x h 2 λ , ρ c T 1 (t) • The problem of the correlation between h 1 , Φ and λ remains if a standard model is considered • Trying to describe the heat flow is difficult: is it a perfect pulse ? A rectangular function ? � Mathematical solution: to measure T 1 (t) and to use it as known data in a convolution model [ ] ( ) ( ) − 1 , , , , ( ) = λ ρ ⊗ T t L H p c h h T t 7 2 2 3 1

III. The three-layers method 3. Modeling Hypothesis • 3D heat transfers with cylindrical symmetry: T=T(r,x,t) • Contact resistances between the brass discs and the sample are neglected (insulating sample) • Isotropic and macroscopically homogeneous sample • Optically thick sample, no radiative transfers at the considered temperatures (ambient) • Uniform brass discs temperature • The convective and radiative exchange with the ambient medium are considered in a constant h • The system is at thermal equilibrium at initial time 8

III. The three-layers method 3. Modeling Energy conservation equation for a single layer system 1 ∂ ( , , ) ∂ 2 ( , , ) 1 ∂ ( , , ) ∂ 2 ( , , ) ∂ 2 θ ( , , ) 1 ∂ θ ( , , ) ∂ 2 θ ( , , ) T r x t T r x t T r x t T r x t p r x p r x p r x p ( , , ) = + + θ r x p = + + 2 2 ∂ ∂ ∂ ∂ ∂ 2 ∂ ∂ 2 a t r r r x a r r r x Laplace transform ( , , ) ( , , ) ∂ θ r x p ∂ T r x t ( ) ( ) − λ = − θ ( , 0 , ) + Φ ( ) − λ = − ( , 0 , ) − + ϕ h r p p h T r t T t 1 1 ∞ ∂ x ∂ x x = 0 = 0 x ∂ ( , , ) ∂ θ ( , , ) T r x t r x p [ ] ( ) ( , , ) ( , ) ( , ) ( , , ) − λ = − θ = − − λ = θ h T r e t T x p L T x t T h r e p 2 ∞ ∞ 2 ∂ ∂ x x x = e x = e ∂ θ ( , , ) ∂ ( , , ) r x p T r x t 0 = 0 − λ = ∂ ∂ r r 0 r = 0 r = T(r,x,t) R h 1 ∂ θ ( , , ) r x p ∂ ( , , ) T r x t ( ) − λ = θ ( , , ) r h R x p − λ = ( , , ) − h T R x t T 3 ∂ r 3 ∞ ∂ r e r = R h 2 r = R x Φ (t) ( , , 0 ) = T T r x ∞ λ , ρ c h 3 9

III. The three-layers method 3. Modeling Solution of the single layer system θ ( , , ) = ( , ) ( , ) r x p X x p R r p Integral transform + separataion of the variables:  1  2 ( , ) 1 ( , )  ∂ ∂ R r p R r p   2 + = − α    ( , ) ∂ 2 ∂ R r p r r r    A system of two differential equations to solve:  1 2 ( , ) ∂ X x p  − γ 2 = 0  ( , ) ∂ 2 X x p x  Solutions for x=0 and x=e as a function of r : = ∞ = ∞ n n ( ) [ ( ) ( ) ] ( ) ∑ ∑ θ ( , 0 , ) = α β + β β θ ( , , ) = α β r p F J r H sh ch r e p F J r 0 2 0 n n n n n n n n 1 1 n = n = Solutions for x=0 and x=e averaged on r : 2 2 = ∞ = ∞ n n [ ] ( ∑ ( ) ∑ ( ) ( ) ) θ ( 0 , ) = β ω θ ( , ) = β β + β ω p F J e p F ch H sh J 1 2 1 moy n n n moy n n n n n ω ω n = 0 n = 0 n n 10

III. The three-layers method 3. Modeling Including the brass discs Brass layers heat balance Single layer limit conditions ( ) ∂ θ , x p 2 ∂ θ ( , , )  e h  r x p   − λ isomoy = −  ρ + + 1 3  θ ( ) + Φ ( ) − λ = − θ ( , 0 , ) + Φ ( ) e c p h p p h r p p   1 1 1 1 1 ∂ x R ∂ x     0 x = ( ) ∂ θ , ∂ θ ( , , ) x p 2 r x p  e h    ( ) − λ = θ ( , , ) − λ isomoy =  ρ + + 2 3  θ h r e p e c p h p   2 2 2 2 2 ∂ ∂ x x R     x = e � Considering the average value of the temperature on the variable r, the brass layers heat balance is similar to the limit conditions at x=0 and x=e of the single layer system (the sample) � The brass discs are taken into account in the single layer solution by introducing the modified coefficients: 2 2  e h   e h  * * h = e ρ c p + h + 1 3 h = e ρ c p + h + 2 3     1 1 1 1 2 2 2 2  R   R  11