October 18, 2002 1Fall 2002. 10.34. Numerical Methods Applied to Chemical EngineeringHomework # 5. Boundary Value ProblemsAssigned Friday 10/18/02. Due Friday 10/25/02Problem 1. Green’s function calculation of a 1-D temperature profileConsider the heat transfer problem outlined in the figure below. We have a slab of material of thickness . On one side of the slab is a fluid at a temperature . On the other is a fluid at a temperature . These fixed temperatures set the boundary conditions on the temperature field within the catalyst at and respectively. The slab may be con-sidered to be infinitely long and tall, so that the only variation of temperature that we wish to consider occurs in the 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 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 problemFirst, we derive the governing differential equation for the temperature field. Let the rate of local heat generation per unit volume be a specified function that we write asLTf1,Tf2,x 0=xL=xSsx()October 18, 2002 2(EQ 1)where is equal to 1 if the point is within a heat generating region and is 0 every-where else. The boundary value problem for the steady state temperature field is then(EQ 2) is the thermal conductivity. To solve this boundary value problem, we will use the tech-nique of Green’s functions outlined below. To do so, we first must convert the boundary conditions by defining the “excess” temperature from the expression(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,(EQ 4)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 following1.A. Using the Green’s function method outlined above, compute the temperature profile for the following set of parameter values.(EQ 5)1.B. From this solution, compute the heat flux across each side of the slab into the sur-rounding fluids.sx()SHx()×=Hx()xλx22ddT– sx()=T0() Tf 1,= TL() Tf2,=λθTx() Tf 1,Tf 2,Tf 1,–()xL---θ x()++=x22dd θfx()sx()λ----------–==θ 0() 0= θ L()0=L 1=Tf 1,1= Tf 2,2=x10.2= t10.1=x20.7= t20.1=S 10= λ 1=October 18, 2002 3A Primer on the use of Green’s functionsWe wish to solve a boundary value problem of the following form, where is some known “data” function.(EQ 6)For this boundary value problem, we define the Green’s function as that satisfying the related boundary value problem,(EQ 7)The Dirac delta function is a “function” that is zero everywhere but at , and that at that point blows up to infinity such that(EQ 8)We can represent the Dirac delta function by taking the following limit,(EQ 9)The figure below shows how the Dirac delta function is approached as we shrink the stan-dard 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.fx()x22ddufx()=u0() 0= uL()0=gxξ,()x22ddgδ x ξ–()=g0 ξ,()0= gLξ,()0=δ x ξ–()xξ=δ x ξ–()ξd∞–∞∫1=δ x ξ–()1σ 2π--------------x ξ–()22σ2------------------–expσ 0→lim=σOctober 18, 2002 4FIGURE 2. Dirac delta function as limit of Gaussian normal distribution as standard deviation approaches zeroWe will never have to evaluate the Dirac delta function, but will only use the property that a function defined on the domain may be written as(EQ 10)To see that this equation is valid, we note the Dirac delta function is zero except at . We therefore can choose a very small, but non-zero, number and write this integral as(EQ 11)fx()0 xL≤≤fx() f ξ()δx ξ–()ξd0L∫=xξ=εfx() f ξ()δx ξ–()ξd0xε–()∫f ξ()δx ξ–()ξdx ε–()xε+()∫f ξ()δx ξ–()ξdx ε+()L∫++=fx() f ξ()0[]ξd0x ε–()∫f ξ()δx ξ–()ξdx ε–()x ε+()∫f ξ()0[]ξdx ε+()L∫++=fx() f ξ()δx ξ–()ξdx ε–()x ε+()∫fx() δx ξ–()ξdx ε–()x ε+()∫≈=fx() fx()=October 18, 2002 5It 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 defined above can be used to express the solu-tion to the boundary value problem as(EQ 12)To see that this is true, we substitute this trial form of the solution into the differential equation,(EQ 13)We therefore find that we can satisfy the differential equation for all possible functions as long as the Green’s function satisfies the differential equation(EQ 14)In particular, we want to find solutions to the differential equation that satisfy the bound-ary conditions,(EQ 15)This implies that the Green’s function must satisfy the boundary conditions,(EQ 16)gxξ,()ux() f ξ()gxξ,()ξd0L∫=x22ddufx()=x22ddf ξ()gxξ,()ξd0L∫f ξ()δx ξ–()ξd0L∫=f ξ()x22ddgxξ,()ξd0L∫f ξ()δx ξ–()ξd0L∫=f ξ()x22ddgxξ,()δx ξ–()– ξd0L∫0=fx()x22ddgδ x ξ–()=u0() 0 f ξ()g 0 ξ,()ξd0L∫== uL() 0 f ξ()gLξ,()ξd0L∫==g 0ξ,()0= gLξ,()0=October 18, 2002 6as specified in the original definition of .The whole idea behind the use of Green’s functions is that once we find the functional form of , we can solve the problem for any given function merely by evaluating a single definite integral,(EQ 17)There are two approaches that we may use to compute the Green’s function for this prob-lem. The first is a rather straight-forward method of writing the Green’s function as a lin-ear combination of the eigenfunctions of the second derivative.(EQ 18)We substitute this expression into the differential equation and use the fact that is an