Fall 2002 10 34 Numerical Methods Applied to Chemical Engineering Homework 5 Boundary Value Problems Assigned Friday 10 18 02 Due Friday 10 25 02 Problem 1 Green s function calculation of a 1 D temperature profile Consider the heat transfer problem outlined in the figure below We have a slab of material of thickness L On one side of the slab is a fluid at a temperature Tf 1 On the other is a fluid at a temperature Tf 2 These fixed temperatures set the boundary conditions on the temperature field within the catalyst at x 0 and x L respectively The slab may be considered to be infinitely long and tall so that the only variation of temperature that we wish to consider occurs in the x direction Within the slab are located several regions in which electrical resistance is used to dissipate energy as heat Within these heat generating regions the rate of heat produced per unit volume is equal to a uniform specified value S in SI units of J s m3 We wish to calculate the temperature profile within the slab and the rate of heat transfer per unit area to the fluids on either side To do so we will apply Green s function analysis The derivation is outlined below You are to perform the calculation and plot the resulting temperature field Then compute the heat flux per unit area to the fluids on each side of the slab FIGURE 1 Geometry of 1 D heat transfer problem First we derive the governing differential equation for the temperature field Let the rate of local heat generation per unit volume be a specified function s x that we write as October 18 2002 1 s x S H x EQ 1 where H x is equal to 1 if the point x is within a heat generating region and is 0 everywhere else The boundary value problem for the steady state temperature field is then 2 dT dx T 0 T f 1 2 s x EQ 2 T L T f 2 is the thermal conductivity To solve this boundary value problem we will use the technique of Green s functions outlined below To do so we first must convert the boundary conditions by defining the excess temperature from the expression x T x T f 1 T f 2 T f 1 x L EQ 3 In the absence of heat generation this excess temperature will be uniformly equal to zero We substitute this form of the temperature field into the heat equation to obtain the following boundary value problem for the excess temperature 2 d 2 s x f x dx 0 0 EQ 4 L 0 This modified boundary value problem is of a form that we can solve using the Green s function approach outlined below Your assignment is the following 1 A Using the Green s function method outlined above compute the temperature profile for the following set of parameter values T f 1 L 1 1 T f 2 2 x 1 0 2 t 1 0 1 x 2 0 7 t 2 0 1 S 10 EQ 5 1 1 B From this solution compute the heat flux across each side of the slab into the surrounding fluids October 18 2002 2 A Primer on the use of Green s functions We wish to solve a boundary value problem of the following form where known data function f x is some 2 du 2 dx u 0 0 f x EQ 6 u L 0 For this boundary value problem we define the Green s function the related boundary value problem g x as that satisfying 2 dg 2 dx g 0 0 x EQ 7 g L 0 The Dirac delta function x is a function that is zero everywhere but at that at that point blows up to infinity such that x and x d 1 EQ 8 We can represent the Dirac delta function by taking the following limit 2 1 x x lim exp 2 0 2 2 EQ 9 The figure below shows how the Dirac delta function is approached as we shrink the standard deviation of the normal distribution to zero As we reduce the distribution becomes more sharply peaked but the area under the integral remains equal to one October 18 2002 3 FIGURE 2 Dirac delta function as limit of Gaussian normal distribution as standard deviation approaches zero We will never have to evaluate the Dirac delta function but will only use the property that a function f x defined on the domain 0 x L may be written as L f x f x d EQ 10 0 To see that this equation is valid we note the Dirac delta function is zero except at x We therefore can choose a very small but non zero number and write this integral as x f x x f x d 0 x f x f 0 d L f x d x x x L f x d x 0 x f x f x d f 0 d x EQ 11 x f x d f x x x d x f x f x October 18 2002 4 It is this property of the Dirac delta function that upon integration it extracts the value of a function at a specific point that makes it useful in the solution of boundary value problems involving linear differential equations We propose that the Green s function g x defined above can be used to express the solution to the boundary value problem as L u x f g x d EQ 12 0 To see that this is true we substitute this trial form of the solution into the differential equation 2 du dx d dx 2 2 L f g x d f x d 0 f 0 f EQ 13 L 2 d g x d 2 dx L 0 f x L L 0 2 f x d 0 2 d g x x d 0 2 dx We therefore find that we can satisfy the differential equation for all possible functions f x as long as the Green s function satisfies the differential equation 2 dg dx 2 x EQ 14 In particular we want to find solutions to the differential equation that satisfy the boundary conditions L u 0 0 L f g 0 d u L 0 0 f g L d EQ 15 0 This implies that the Green s function must satisfy the boundary conditions g 0 0 g L 0 October 18 2002 EQ 16 5 as specified in the original definition of g x The whole idea behind the use of Green s functions is that once we find the functional form of g x we can solve the problem for any given function f x merely by evaluating a single definite integral L u x f g x d EQ 17 0 There are two approaches that we may use to compute the Green s function for this problem The first is a rather straight forward method of …
View Full Document
Unlocking...