1D heat conduction problems

Transcription

1 Chapter 1D heat conduction problems.1 1D heat conduction equation When we consider one-dimensional heat conduction problems of a homogeneous isotropic solid, the Fourier equation simplifies to the form: cρ T t = λ ) T + Q (.1) x ) If there is no heat generation, as is usually the case, such equation reduces to: T t = k ) T x ) (.) where k =.. Furthermore, if the temperature distribution does not depend on time: /0 ) T x ) = 0 (.3) The stationary case of heat conduction in a one-dimension domain, like the one represented in figure.1, is particular simple to be solved. Figure.1: the temperature within a solid media with prescribed Dirichelet boundary conditions In fact, the general solution of the equation is in this case: T x = Ax + B (.4) with A and B coefficients to be determined by imposing the boundary conditions.

2 Prescribed temperatures at both left and right surfaces: T 456 = T 7 T 458 = T 9 (.5) we have: T 7 = B and T 9 = Al + T 7 or In conclusion: ; < =; > 8 = A (.6) T x = T 9 T 7 l x + T 7 (.7) Prescribed temperature at left surface ad assigned heat flux at the right surface. Figure.: the temperature within a solid media with prescribed temperature at left surface ad assigned heat flux at the right surface q AB + q C q DEF = 0 steady state q AB = q 458, q C = q 9, q DEF = 0 q q 9 = 0 T 456 = T 7 q q 9 = 0 λ dt dx q 9 = 0 λ dt dx 458 = q 9 λdt = q 9 dx λ[t] 6 4 = q 9 [x] 6 4 λ T x T 0 = q 9 x 0

3 λ T x T 7 = q 9 x T x = q 9 λ x + T 7 Prescribed temperature at right side and assigned heat flux at left side Figure.3: the temperature within a solid media with prescribed temperature at right side and assigned heat flux at left side q AB + q C q DEF = 0 steady state q AB = 0, q C = q 7, q DEF = q 456 q 7 q 456 = 0 T 458 = T 9 q 456 = q 7 λ dt dx 456 = q 7 λ dt dx 456 = q 7 λdt = q 7 dx λ[t] 4 8 = q 7 [x] 4 8 λ T l T x λ T 9 T x = q 7 l x = q 7 l x T x = q 7 λ l x + T 9

4 Convective and prescribed heat flux at both surfaces Figure.4: the temperature distribution with convective and prescribed heat fluxes Governing equation: d ) T dx ) = 0 (.8) Boundary conditions: T q S4F + q 7 q 456 = 0 h 7 T V> T + q 7 + λ dt = 0 on x = 0 (.9) dx TT q q 9 q S4F = 0 λ dt dx + q 9 h 9 T T V< on x = l (.10) where T V> and T V< are the temperatures of the surrounding media, q 7 and q 9 are surface heat generations per unit area per unit time, h 7 and h 9 are the heat transfer coefficients and subscripts a and b denote boundaries at x = 0 and x = l, respectively. A general solution of eq..8 is: T x = Ax + B (.11) By substitution of the solution in eq..9 and.10 we obtain: λa + q 7 = h 7 Ax + B T a λa + q 9 = h 9 Ax + B T b λa + q 7 = h 7 B T a per x = 0 λa + q 9 = h 9 Al + B T b per x = l The coefficients A and B are: A = h 7 h 9 [ T b T a + q 9 q 7 ] h 9 h 7 λ h 7 + h 9 + h 7 h 9 l

5 B = T a + q λh 9 T b T a + ( q 9 q 7) 7 h + 9 h 7 h 7 λ h 7 + h 9 + h 7 h 9 l If no surface heat generation q 7 = q 9 = 0 h 9 (h 7 x + λ) T x = T V> + (T V< T V> ) λ h 7 + h 9 + h 7 h 9 l (.1) It is no possible to find a stationary solution by imposing an heat flux condition on both sides. In fact if q a and q b are not equal, the temperature response is not stationary since in presence of a variation of heat content, the temperature of the body should vary with time. If q a = q b the slope of the line is determined and the difference of temperature only can be obtained by no assumption on the temperature value. The case is analog to the mechanical response of an elastic spring free in the space at the end of which two forces with opposite sign are applied rigid body motion

6 . 1D heat conduction: transient Let us now consider a transient problem in which the temperature at x=0 is equal to T a, the temperature at x=l is equal to zero and the initial condition is set as T=T i (x). The governing equations read as follows T t = λ ) T cρ x ), λ k = cρ (.13) T 456 = T 7 T 458 = 0 T F56 = T A x The solution is found by separation of variables, as we assume that the temperature can be expressed by the product of a function of position only f(x) and a function of time g(t) ( ) = f ( x) g( t) T x, t, by substituting in the governing equation we obtain dg t f( x) ( ), dt dx (.14) () d f ( x) = kg t (.15) that can also be written as 1 dg( t) 1 d f ( x) =, kg() t dt f ( x) dx (.16) since the first member does not depend upon x, the second one does not depend upon t and the two members are equal, they can be set equal to a constant. By setting the constant equal to -s we obtain two separate equations dg() t dt And + ks g t = () 0, (.17) d f x ( ) dx + s f( x) = 0. (.18) The general solution of the first equation can be easily obtained by searching solution of the kind g t = e bf and by finding the characteristic equation α + ks = 0, (.19) that leads to the general solution

7 g t = c c for s ) = 0, g t = c ) e =ecf F for s ) 0 (.0) The general solution of the second equation can be sought in the same way or set directly as f x = c i x + c j for s ) = 0 f x = c k sin sx + c n cos sx for s ) 0 (.1) In conclusion the general solution of the original equation, that should be valid for arbitrary values of s can be written as T x, t = e =ecff A sin sx + B cos sx + Cx + D (.) with A= c c, B= c c, C = cc, D= cc (.3) We now impose the satisfaction of boundary conditions by substituting then in the latter expression: T 0, t = Be =ecff + D = T 7, T l, t = e =ecff A sin sl + B cos sl + Cl + D = 0 (.4) In order to satisfy the first equation for every value of t since e =ecf F is never equal to zero it is necessary that B= 0, D= T a. For the second equation we have now (.5) ks t Ae sin sl + Cl + T a = 0 (.6) that can only be satisfied for sin sl = 0 and C = ; > 8 (.7) The values of k for which sin sl = 0 are s n nπ = n= 1,, 3... l they are the eigenvalues of our problem. (.8) The temperature can then be expanded in an infinite series of form x n T( x, t) = T 1 ks t a + Ae n sin snx, l n= 1 (.9) when A n are unknown coefficients still to be determined. We now impose the initial condition T x, 0 = T A (x) by substitution in the preceding equation we obtain

8 T A x T 7 1 x l V = A B sin s B x B5c (.30) In order to determine the coefficients A n corresponding to initial condition we can profit of the properties of the sinusoidal function. In fact by multiplying both sides of the equation by sin s u x and integrating it from 0 to l (that is by finding the value of the scalar product between the terms at the first and the second member of the equation and the generic sinusoidal function sin s u x ) we obtain: sin s B x sin s u xdx = 0 sin ) s B x dx = l for m n for m = n (.31) And, for the coefficients A n l x An = T( ) 1 sin. 0 i x Ta snxdx l l (.3) In conclusion the expression of the temperature can be written as x l x ksnt T( x, t) = Ta 1 + { T( ) 1 sin }sin. 0 i x Ta snxdx snxe l l n= 1 l (.33) From the form of the solution it is clear that the first term represents the stationary response that will eventually reached after a transient. The second term is on the contrary a time-dependent one, with the tendency of decaying to zero as time increases. In the case that different boundary conditions are imposed, say on both sides x=0,l imposed heat flux by convection, as expressed below Figure.5: The temperature profile within a solid media which separates two semi-infinite fluid media. T V> and T V< correspond, respectively, to the temperature of the external media for x<0 and for x>l.

9 T q S4F = h 7 T V> T TT q S4F = h 9 T T V< (.34) with T q S4F + q 7 q = 0, q 7 = 0 h 7 T V> T + λ dt = 0 on x = 0 (.35) dx TT q + q 9 q S4F = 0, q 9 = 0 λ dt dx h 9 T T V< = 0 on x = l (.36) It is possible to use the same expression of the solution obtained before: T x, t = e =ecff A sin sx + B cos sx + Cx + D (.37) For obtaining the solution in this case similar steps can be followed to the ones used for the previous set of boundary conditions. The solution in this case will result as follows T x, t = T V> + T V< T V> h 9 h 7 x + λ λ h 7 + h 9 + h 7 h 9 l + + V B5c λ ) s ) ) B + h 9 h 7 sin s B x + λs B cos s B x l λ ) s ) B + h) 7 λ ) s ) B + h ) 9 + k h 7 + h 9 λ ) s ) B + h 7 h 9 e =ecff.. 8 h 9 h 7 x + λ T A x T V> + T V< T V> λ h 7 + h 9 + h 7 h 9 l h 7 sin s B x 6 + λs B cos s B x dx, (.38) with s n being any positive root of the transcendental equation ( + h ) λs ha b tan sl =. λ s h h a b (.39) The stationary part of solution T x, t = T V> + T V< T V> h 9 h 7 x + λ λ h 7 + h 9 + h 7 h 9 l (.40) can be further examined. In fact, the boundary condition in x=0 reads q 7 = λ dt dx = h 7 T T V> (.41) if the conductivity of the material is low or the convention thermal coefficient is high, the temperature on the wall reaches the temperature of the fluid T V>. In fact, with low λ and high h a we have. { > 1 and

10 λ T h 7 x = (T T V > ) (.4) with the first term of the equation equal to zero and T = T V>. This results can also obtained from the above expression of the stationary part of the solution in presence of convective conditions on both sides, that can be written as T x = T V> + T V< T V> h 9 h 7 x + λ λ h 7 + h 9 + h 7 h 9 l h 7 h 9 x + λ h = T V> + T V< T 7 V> λ h 7 + h 9 + h 7 h 9 l (.43) = T V> + T V< T V> x + λ h 7 and, considering the same assumption. T x = T V> + T V< T V> 4 λ 1 h h 7 + l { > 1 and. { < 1 we finally have 8. (.44) That is the temperature of the media at both sides of the solid can be directly assumed as the wall temperature. It is also interesting to note that, also with entering fluxes at both sides q a and q b, the temperature of the body T in the stationary response cannot have temperatures higher than the highest of the media by which the body is surrounded in perfect agreement with the physics of heat transfer. It is also obvious that, also in the presence of an entering flux, the body cannot increase indefinitely its temperature that cannot increase higher than the one from which the flux is generated. Figure.6: The temperature within a wall with prescribed Dirichlet and Neumann boundary conditions The 1D thermal behaviour just described can be applied to the case of an indefinite plate of thickness l with the side at isothermal condition where either the temperature, also variable with

11 the time t, or the heat flux can be applied. The same results are also valid for a one-dimensional solid (like a bar or a cable) with uniform section, provided that the lateral surface of the bar is isolated, that is no heat flux is present through it.

12 .3 Honeycomb panel Figure.7: Structure of a honeycomb sandwich panel: assembled view (A), and exploded view (with the two face sheets B, and the honeycomb core C). Ribbons run along the x direction, and are glued side by side in counter-phase along the y direction as detailed. Honeycomb panels (Figure.7) are structural elements with great stiffness-to-mass ratio, widely used in aerospace vehicles. Heat transfer through honeycomb panels is non-isotropic and difficult to predict. If the effect of the cover faces is taken aside, and convection and radiation within the honeycomb cells can be neglected in comparison with conduction along the ribbons (what is the actual case in aluminium honeycombs), heat transfer across each of the dimensions is: T 4 Q 4 = λf 4 A 4 with F l 4 = 3 δ 4 s Q ƒ = λf ƒ A ƒ T ƒ l ƒ with F ƒ = δ s (.45) T Q = λf A with F l = 8 δ 3 s where F is the factor modifying solid body conduction (the effective conductive area divided by the plate cross-section area), which is proportional to ribbon thickness, δ, divided by cell size, s (distance between opposite sides in the hexagonal cell, not hexagon side, a, in Figure.7; s = 3a), and depends on the direction considered: x is along the ribbons (which are glued side by side), y is perpendicular to the sides, and z is perpendicular to the panel. For instance, for the rectangular unit cell pointed out in Figure.7, of cross-section area 3as, the solid area is 8aδ, and the quotient is F z =(8/3)δ/s.

13 Example Evaluate the mean core-panel values of density and thermal conductivity through-thethickness for a core made of aluminium foil with ρ=700 kg/m 3, λ=150 W/(m K), thickness δ=30 μm and s=3 mm cell pattern. Solution. F z = (8/3)(δ/s) = (8/3)(0.03/3) = 0.07 ρ honeycomb = ρ F z = =73 kg/ m 3 λ z honeycomb = λ F z = = 4 W/(m K).