Heat Conduction in semi-infinite Slab

Transcription

1 1 Module : Diffusive heat and mass transfer Lecture 11: Heat Conduction in semi-infinite Slab with Constant wall Temperature

2 Semi-infinite Solid Semi-infinite solids can be visualized as ver thick walls with one side exposed to some fluid. The other side, since the wall is ver thick remains unaffected b the fluid temperature. The condition applicable can be expressed as T, t T, where T 0 is the initial wall temperature At ( ) 0 Fig Heat transfer in Semi-infinite solids The condition at the exposed side of the wall is called the boundar condition Unstead-state Heat Transfer Consider a heat conduction in -direction in a semi-infinite slab (bounded onl b one face) initiall at a temperature T 0, whose face suddenl at time equal to zero is raised to and maintained at T 1. Assuming constant thermal diffusivit and with no heat generation, a differential equation in one space dimension and time is given b θ θ α, where α is the thermal diffusivit (11.1) t with the following I.C and B.Cs

3 3 At t 0 At 0 At θ 0 θ 1 θ 0 for all for all t for all t > 0 > 0 (11.) An example is heating of a semi-infinite slab, initiall at temperature T 0. The slab surface temperature is raised suddenl to T 1 and maintained at that level for all t>0. For this situation, ( ) T TO k we have considered dimensionless parameter, θ in equation (11.1), and α is ( T1 TO ) ρc p the thermal diffusivit. Solution: One wa to solve this problem is b Combination of Variables Assume θ a t b (11.3) where a and b are constants Differentiating equation (11.3) w.r.t t we get θ t a b t ( b 1) (11.4) Also on differentiating w.r.t we get θ t b a (a -1) ( a-) (11.5) Substituting equation (11.4) and (11.5) into (11.1) we get ( b 1) a b t α ( a-) b t a (a -1) (11.6)

4 4 After some algebraic manipulations, we get α t a ( a 1) b constant Hence the new variable can be an of the form t C α d where C and d are constants Define η (11.7) 4α t 1 1 (Here we choosec and d ) 4 Then θ θ (, t) θ ( η) Differentiating θ (eqn. (11.7)) w.r.t. t we get θ t t 4α t dθ dη (11.8) Also differentiating θ w.r.t. we get θ 1 α t d θ dη (11.10) Substituting eqn. (11.8) and (11.10) into eqn. (11.1) then leads to a second-order ordinar differential equation of the form

5 5 θ dθ + η 0 (11.11) η dη On solving equation (11.11) we get η 0 θ B + A η dη (11.1) e Note For the method of combination of variables to work, the boundar conditions must be expressible in terms of η onl. This method is usuall applied for media of infinite or semiinfinite extend. Boundar Conditions At η 0 θ 1 and At η θ 0 Appling above B.Cs. to equation (11.1) we get η η θ 1 e dη (11.13) π 0 Note

6 6 In mathematics, the error function also called as Gauss error function is a special function (non-elementar) of sigmoid shape which generall occurs in problems related to probabilit, statistics, materials science and partial differential equations. It can be defined as ( ) x t erf x e dt π 0 The complimentar error function denoted b erfc, is defined in terms of the error function: ( ) 1 ( ) erfc x erf x π x e t dt Some useful results of error functions are given below. ( 0) 0, ( ) 1, ( ) ( ) erf erf erf x erf x So utilizing the results given for error function, we can write θ 1 erf 4α t or θ 1 erf( η) (11.14) Fig.(11.3) shows the variation of the surface temperature along the length of the slab with time. Diffusion length

7 7 Diffusion length can be defined as a measure of how far the temperature (or concentration) has propagated in the -direction b diffusion in time t. It can be given as δ 4 α t at this value, the value of η is η 4α t Therefore from equation (11.13) we can write ( ) θ 1 erf This implies that, at a distancel > δ, the field variables has changed b < 0.5%. Solution (11.14) is a good approximation even for finite length slabs if the diffusion length δ << L. The heat flux can be obtained as T ( q) k ( T ) 0 1 To 0 k παt (11.15)

8 8 Fig.11. Error function From the above solution to the problem on semi-infinite slab we can conclude that the diffusion length (or penetration depth) varies as square root of time (t 1/ ) and the wall flux varies as inverse of as square root of time (t -1/ ). Solution b the method of Laplace Transform Definition Let f(z) be an function, then the Laplace transform of f(z) can be given b 0 pt { ( z) } f ( p) e f ( t) f dt (11.16) Some of the standard Laplace results are shown below:

9 9 () f ( ) if f t 1, then after taking Laplace transform, p 1 p In this wa f at () t e f ( p) 1 ( p - a) f () t sin( wt) f ( p) w ( p + w ) Problem Consider a diffusion of component A in the -direction in a semi-infinite (bounded onl b one face) slab initiall at a uniform concentration, C 0, whose face suddenl at time equal to zero is raised to and maintained at C. The governing differential equation is given b C t C D, D is the binar diffusivit (11.17) with following I.C. and B.Cs. C C C 0 O C 0 at at at 0, > 0,, t > 0 t 0 t 0 (11.18) The Laplace transform of equation (11.17) ields

10 10 C D pc C (,0) (11.19) Using I.C. given in equation (11.18), equation (11.19) becomes C D pc (11.0) Appling B.Cs, so C C p C 0 O at as 0 and; Therefore the solution for equation (11.0) is C p O D C e (11.1) p Taking the inverse Laplace transform of equation (11.1) and from the table of Laplace transform C Co 1 erf (11.) 4D t

11 11 Fig.11.3 Temperature variation along the length at different time period