Models for measurement of thermophysical parameters

Transcription

1 Models for measurement of thermophysical parameters P. Dieška Department of Physics, STU FEI Ilkovičova 3, Bratislava, Slovakia October 16,

2 Model as a image of real experiment Heat equation: Initial and boundary conditions: cρ T t = λ T (1) adequate to the experimental setup (2) Solution: T (t, r) T EXP (t, r) c... Specific heat capacity ρ... Density λ... Thermal conductivity 1 45

3 Specification of the problem The parameters are temperature independent Linearisation of the problem The sample have certain symmetry Possibility of spatial dimension reduction The heat source is external Time dependent boundary conditions Used methods Laplace transform for time dependence Specific integral transform for spatial dependence Superposition method 2 46

4 Solved models - 1D 1. Ideal model. Instantaneous planar heat source placed between two semiinfinit samples. The thermal contact of heat source and samples is ideal. Heat equation: Initial condition: 1 k T t = 2 T x 2 (3) T (0, x) = 0 (4) Boundary conditions: λ T = q1(t) (5) x x=0 3 47

5 T (t, ) = 0 (6) " # e Solution T (t, x) = T u2 0 πu Φ (u) u = x 2 kt T 0 = qx λ T... temperature t... time x... Cartesian coordinate q... heat flow density at source 1(t)... Heaviside unit step function λ... thermal conductivity k... thermal diffusivity C... heat capacity per unit area of source α s... heat trasfer coefficient for sample - heat source interface Φ (u) is the complementary error function. 4 48

6 2. Slabs - symmetric disposition Boundary conditions: λ T = α s (T s T ) (7) x x=0 x=0 q1(t) = C T s t + α s(t s T ) (8) x=0 Solution T (t, L) = 0 (9) T (t, x) = T 0 n (1 x L ) + 2a P ν e kt L 2 ν2 ν sin(ν L x ) (a bν2 ) cos(ν L x o ) ν 2 [b 2 ν 4 (2ab b 1)ν 2 +a(a+1)] 5 49

7 T 0 = ql λ a = λl Ck b = λ Lα s T s... temperature of heat source L... thickness of sample C... heat capacity per unit area of source α s... heat transfer coefficient of source - sample interface ν is a root of equation (a bν 2 ) cos ν ν sin ν =

8 Solution 3. Slabs - asymmetric disposition Boundary conditions: λ T + λ T = 2q1(t) x x x=0 + x=0 (10) T x=0 + T x=0 = 0 (11) T x= L1 = T x=l2 = 0 (12) T (t, x) = 2q kt λ P j= ( 1) jh e u2 i j π u j Φ (u j ) u j = x + 2Lj (L 1 L 2 )[1 ( 1) j ] 2 kt 7 51

9 L 1, L 2... thicknesses of samples L = 1 2 (L 1 + L 2 ) 2q... sum of heat flow densities at source 4. Sandwich disposition Boundary conditions: T (t, ) = 0 (13) T (t, ) = 0 (14) λ T x x=0 + + λ 0 T x x=0 = 2q1(t) (15) 8 52

10 Solution λ T x T 13 (t, x) = T 0 γ u 0 T x=0 + T x=0 = 0 (16) T + λ 0 = 0 (17) x x=h x=h + T x=h T x=h + = 0 (18) P j=0 δ jh e u2 i j π u j Φ (u j ) T 0 = qh λ u j = (n )h kt, γ = λ 0 λ q k k 0! 2, δ = 1 λ 0 λ q k k λ 0 λ q k k 0! 2 h... thickness of the sample λ 0... reference thermal conductivity k 0... reference thermal diffusivity 9 53

11 Solved models - 2D 5. Semi-infinite cylinder Heat equation: 1 k T t = 2 T x r r r T r (19) Initial condition: T (0, x) = 0 (20) 10 54

12 Boundary conditions: λ T = αt (21) r r=r r=r λ T = q1(t) (22) x x=0 T = 0 (23) x= Solution T (t, x, r) = T 0 R x P ξ β ξ(ξ 2 +β 2 ) J 0 (ξ r R ) J 0 (ξ) F (u, v) F (u, v) = e 2uv Φ (u v) e 2uv Φ (u + v) T 0 = qx λ β = Rα λ u = x 2 kt v = ξ kt R 11 55

13 x... axial space coordinate r... radial space coordinate R... radius of the sample q... heat flow density at source α... heat trasfer coefficient for sample - ambient interface ξ is the root of the equation βj 0 (ξ) ξj 1 (ξ) =

14 Solved models - 3D 5. Finite cuboid Heat equation: Initial condition: 1 k T t = 2 T x T y T z 2 (24) T (0, x, y, z) = 0 (25) 13 57

15 Boundary conditions: λ T z T = 0 (26) x x=0 λ T = αt (27) x x=a x=a T = 0 (28) y y=0 λ T = αt (29) y y=a y=a T = T = 0 (30) z= L1 z=l2 T z=0 = T (31) + z=0 + z=0 + + λ T z z=0 = 2q 1(t) (32) 14 58

16 Solution: T (t, x, y, z) = T 0 w 2 P P n=1 m=1 bnbm vnm ϕ n x a ϕ y P h i m a F (u 1j, v nm ) F (u 2j, v nm ) j= ϕ n (s) = s 2β β + sin 2 cos(µ n s) µ n F (u, v) = e 2uv Φ (u v) e 2uv Φ (u + v) q T 0 = qa λ, β = aα λ, w = kt a, v nm = w u 1j = z... axial space coordinate L 1... length of left sample L 2... length of right sample z + 4Lj 2 kt µ 2 n + µ2 m, b n = ϕ n (0) sin(µ n) µ n u 2j = z + 4Lj + 2L 1 2 kt 15 59

17 L = (L 1 + L 2 )/2 x, y... transversal space coordinates 2a... transversal size of the sample q... heat flow density at source α... heat trasfer coefficient for sample - ambient interface Φ (u) is the complementary error function µ n are the roots of equation β cos µ µ sin µ =

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