4.3 Superposition
4.3.1 Superposition example #1
2
dx (4.35)
= 2k(T∞− T2) X∞ n=1
Ansin(λn)
sinh(2λn) +Bi
λn cosh(2λn)
(4.36) The terms in the series still approach 1/n as n → ∞ – yet they will not all have the same sign and the series will converge for Bi < ∞. Presented in Fig. 4.5 is the dimensionless heat transfer rate q′, defined by
q′ = q′
2k(T∞− T2)Bi = q′ hL(T∞− T2),
vs. the Biot number Bi. This particular scaling for q′ is chosen so that in the limit of Bi → 0, the dimensionless q′ → 3. This limit has a relatively simple physical explanation which you should be able to figure out. The quantity q′ goes to zero for Bi → ∞ – because q′ ∼ q′/Bi – yet the actual heat transfer would also go to infinity in this limit.
4.3 Superposition
It is not uncommon to have conduction problems that have inhomogeneous boundary conditions in both directions or that have an inhomogeneous DE (through heat generation). Separation of variables, however, can only work if a separated ODE and the associated BCs conform to the Sturm–Liouville system, which constrains the method to conduction problems with a homogeneous DE and homogeneous BCs in all but one direction. The way around this dilemma is to apply the technique of superposition – which is a generalized form of the partial solutions technique that was learned in Ch. 3. The superposition method exploits the fact that the heat conduction equation is linear. Because of this, any sum of solutions to the DE is also a solution. If solutions to the DE can be found which, when summed together, satisfy all the BCs to the original problem, then the original problem has been solved.
Devising a superimposed solution involves some creativity and imagination. In general, the procedure is to sequentially replace the inhomogeneous conditions with homogeneous conditions of the same type, and then solve the resulting problems with appropriate techniques. If N inhomo-geneous conditions appear in the original problem, then the complete solution will need at most N partial solutions that are pasted together. Often this number can be reduced by inspection or appropriate combinations.
4.3.1 Superposition example #1
The application of the method is best seen through application. Consider a square region of width
= height = L. The surface at y = 0 is maintained at T1, and convection occurs at the y = L
surface, characterized by h and T∞. The surface at x = 0 is adiabatic, and a uniform heat flux of q0′′is applied at x = L. Finally, a uniform heat generation, of strength q0′′′ occurs within the region.
The dimensional problem is
∂2T Define the dimensionless temperature by
T = T − T∞
T1− T∞
(4.37) and the dimensionless problem becomes
∂2T
in which the dimensionless parameters are q′′= q0′′L
k(T1− T∞), q′′′= q′′′0 L2
k(T1− T∞), Bi = hL k The dimensionless domain is shown in Fig.4.6
The problem has an inhomogeneous DE and inhomogeneous BCs at x = 1 and y = 0. SOV cannot be applied to the total problem at hand, but the problem can be split into several sub–
problems which, individually, can be solved using SOV or more simple methods.
∇2T = −q′′′
∂T
∂x = 0 ∂T
∂x = q′′
T = 1
∂y
6 -y
x 1
1
Figure 4.6: a complicated problem
For this particular case, it will now be demonstrated the problem can be split into the two superimposed problems illustrated in Fig.4.7. Problem A consists of a square region with adiabatic conditions at x = 0 and 1, uniform temperature of unity at y = 0, convection at y = 1 and uniform heat generation of q′′′. Problem B has an adiabatic condition at x = 0, uniform flux of q′′ at x = 1, zero temperature at y = 0, convection at y = 1, and no heat generation.
Denote as TA and TB the solutions to these individual problems, which satisfy
∂2TA
∂x2 +∂2TA
∂y2 + q′′′= 0 (4.43)
∂TA
∂x
0
= 0 (4.44)
∂TA
∂x
1
= 0 (4.45)
TA(0, y) = 1 (4.46)
− ∂TA
∂y
1
= Bi TA(0, y) (4.47)
∇2TA= −q′′′
Figure 4.7: the two superimposed problems
and You should be able to convince yourself, by substitution of the above equations into the original problem in Eqs. (4.38–4.42), that the superposition T = TA+ TB solves the original problem.
Specifically, TAtakes care of the inhomogeneous bondary condition at y = 0 and the heat generation term, whereas TB takes care of the inhomogeneous boundary condition at x = 1.
It is important to recognize that in formulating the problem for A, the heat flux condition at x = 1 was replaced with the corresponding homogeneous condition – which is an adiabatic condition.
Likewise, in formulating the problem for B the inhomogenous, unit–temperature condition at y = 0 was replaced with the corresponding homogeneous condition of TB= 0. This procedure is critical:
always replace an inhomogeneous BC in a superimposed solution with a homogenous on of the same type.
It turns out that the problem for A has a relatively simple solution because the A configuration presents a one dimensional problem. That is, the adiabatic conditions at x = 0 and 1 eliminates any heat flow in the x direction, for which the temperature will depend only y. The corresponding 1–D solution for TA is
TA= 1 − q′′′y2
2 +q′′′(2 + Bi) − 2Bi
2 + 2Bi y (4.54)
The B problem must be solved by separation of variables. The y direction has the homogeneous boundary conditions, and the separation constant is chosen to give eigenfunctions in the y direction.
Since TB(x, 0) = 0, the eigenfunction will be of the form
φn(y) = sin(λny) (4.55)
The convection condition at y = 1 provides the eigencondition
λncos(λn) + Bi sin(λn) = 0, n = 1, 2, . . . (4.56) The special case of λ = 0 is a solution to the above eigencondition, and this case must therefore be examined further. The zeroth eigenfunction would be (from solution of the characteristic ODE in the homogeneous direction)
φ0(y) = A + By
The constant A = 0 from the BC at y = 0. At y = 1 the convection condition has B = −Bi B
which can be satisfied for arbitrary Bi only if B = 0. Consequently, the zeroth eigenfunction is zero, and can be dismissed from the solution. Recognize that λ1 will now denote the first non–zero root to Eq. (4.56).
In the x direction the zero–gradient condition at x = 0 will eliminate the sinh term in the general solution of Eq. (4.18). The solution for TB will then appear as
TB= X∞ n=1
Ancosh(λnx) sin(λny)
At x = 1 the specified heat flux BC gives q′′=
X∞ n=1
λnAnsinh(λn) sin(λny) (4.57)
Figure 4.8: superimposed solution: q′′′ = Bi = 0, q′′= 1 (left), q′′′ = 10, q′′= −10, Bi = 1 (right)
from which the expansion coefficients are obtained as
An=
q′′
Z 1 0
sin(λny) dy λnsinh(λn)
Z 1 0
sin2(λny) dy
= 2q′′[1 − cos(λn)]
λnsinh(λn)[λn− cos(λn) sin(λn)] (4.58) The complete solution to the problem is now given by T = TA+ TB, or
T = 1 −q′′′y2
2 +q′′′(2 + Bi) − Bi
2 + 2Bi y + 2q′′
X∞ n=1
[1 − cos(λn)] cosh(λnx) sin(λny)
λnsinh(λn)[λn− cos(λn) sin(λn)] (4.59) in which the eigenvalues λn are obtained from the roots of Eq. (4.56).
The solution to this problem is plotted in Fig. 4.8 for Bi = q′′′ = 0, q′′ = 1 (left plot) and Bi = 1, q′′= −10, q′′′ = 10 (right). Setting Bi = 0 results in an adiabatic top surface, and this is observed in the perpendicular intersection of the isotherms with the upper surface. Recognize that the first set of parameters results in TA= 1 throughout the domain; this result therefore provides a check on the correctness of the SOV solution for TB. It is difficult to gauge from a contour plot whether or not a constant, non–zero flux condition is satisfied at a boundary (as occurs along x = 1); to examine the results in a different perspective a plot of T vs. x, using q′′ = Bi = 0 and q′′= 1, is shown in Fig.4.9. The sloping lines represent the temperature profile at a constant values of y; increasing height of the lines correspond to increasing y. This view shows that the flux condition is met at x = 1, in that the lines uniformly intersect the surface with a slope of unity.
The one point where this condition would fail would be at y = 0 and x = 1 – which is a result of the contradiction implied in maintaining the bottom surface at T = 1 while providing a uniform
0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.81 1
1.2 1.4 1.6 1.8
0
Figure 4.9: T vs. x for q′′′ = Bi = 0, q′′= 1
flux from the side. This is solely a mathematical artifact and, for reasons discussed above, would not occur physically; it would be impossible to maintain the uniform surface temperature and/or the constant flux in the vicinity of this point.