PASAR A TRATAMIENTO DE SEGUNDA LÍNEA
C. Obstrucción del flujo aéreo resistente a corticoides (descenso progresivo del VEF1 ≥ 10%
Two dimensional conduction problems in spherical coordinates typically occur for the independent variable pairs (r, z) and (r, φ). An example of the first type was encountered in the previous section. The latter case will occur whenever the surface of the cylinder is subjected to an uneven heating/cooling situation.
R1 R2
h1, T∞,1
h2, T∞,2 q′′0
φ
Figure 4.15: pipe configuration
To pose an example, consider an annular pipe, having inner and outer radii of R1 and R2, that is carrying a fluid at temperature T∞,1. Convection occurs between the inner surface and the fluid with a heat transfer coefficient h1, and likewise the outer surface is exposed to a convection environment characterized by h2 and T∞,2. In addition , the outside of the pipe is exposed to a collimated source of thermal radiation, which has a flux of q0′′. The problem is schematically illustrated in Fig. 4.15.
Taking into account the angle between surface normal and the incident radiation, the normal component of the absorbed heat flux at the exterior pipe surface is
q′′r =
αq0′′cos(φ), −π
2 ≤ φ ≤ π 2
0, π
2 < φ < 3π 2
where α is the surface absorptivity of the pipe. The objective of the problem is to predict the temperature distribution in the pipe and the heat flux at the inner wall.
Assuming that the temperature varies only in the r and φ directions, the dimensional problem
is
The BC on φ simply states a ‘continuation’ principle – in that the temperature at φ has to be the same as the temperature at φ + 2π (i.e., once around the pipe).
Let the nondimensional radial position and temperature be defined r = r
R2, T = (T − T∞,1)k αq0′′R2 The dimensionless problem becomes
1
The dimensionless problem has a homogeneous DE and homogeneous BCs in the φ direction, and an inhomogeneous convection BC occurs at r = 1. The 1/a in Eq. (4.107) comes from the fact that we are defining Bi1 with respect to R1 instead of R2.
Separation of variables can be applied directly to this problem. Let T = u(r)v(φ), which leads to
r(ru′)′± λ2u = 0 v′′− (±λ2)v = 0
The homogeneous direction of the problem is φ and eigenfunctions are required in the φ direction.
We then choose −λ2 in the above which will give a solution for v in terms of the trigonometric functions. Specifically,
v = A cos(λφ) + B sin(λφ)
Because of the symmetry of the problem, the temperature field must have T (r, φ) = T (r,−φ) – i.e, the temperature is even in φ. The constant B can therefore be set to zero. From Eq. (4.109) the eigencondition of the problem is, simply,
λn= n, n = 0, 1, 2, . . . This will give the desired periodic behavior in φ.
The ODE for u becomes
r2u′′+ ru′− n2u = 0
Two cases need to be examined – depending on whether or not n is zero. If n 6= 0 the DE takes the form of an equidimensional equation, in which each term has the same net dimension in r.
The solution to such equations is typically obtained by setting u = Crβ, where β is a constant.
Replacing this into the DE, the solution to the above becomes u = Anrn+ Bnr−n, n = 1, 2, . . . For the case of n = 0 the DE for u will appear as
(ru′)′ = 0 Integrating twice over r gives
u = A0+ B0ln(r), n = 0
which is recognized simply as the 1–D radial temperature profile in cylindrical coordinates.
At this point the general solution to the problem appears as T = A0+ B0ln(r) +
X∞ n=1
Anrn+ Bnr−n
cos(nφ) (4.111)
The BC at r = a is now used to eliminate the B coefficients. Since this BC is homogeneous it can be applied to each individual term in the series. For n > 0 the result is
n Anan−1− Bna−n−1
= Bi1
a Anan+ Bna−n which gives
Bn= Ana2nn − Bi1
n + Bi1
and for n = 0: To make the notation more compact, define the function gn(r) as
gn(r) = rn+ a2nn − Bi1
n + Bi1r−n (4.112)
With this convention, the general solution can now be written as T = A0
Application of the remaining inhomogeneous BC at r = 1 results in
A0 Bi1 Multiply Eq. (4.114) by cos(mφ) and integrate over φ from 0 to π. The orthogonality relation is
Z π
Three distinct cases are obtained from Eq. (4.114), depending on whether m = 0, 1, or > 1. For m = 0 we find that
The m = 1 case gives The integral has the value π/4, so the formula for A1 is
A1 = 1 The value of the integral is now
Z π/2 0
cos(φ) cos(nφ) dφ = cos(nπ/2) 1 − n2 and the corresponding formula is
An= 2 cos(nπ/2)
π (1 − n2) (gn′(1) + Bi2gn(1)), n > 1 (4.118) Observe that the An in the above equation will be zero for odd n. Taking this into consideration, the complete solution for the temperature is
T (r, φ) = 1 + πBi2T∞,2 Bi1 = 1000, Bi2 = 5 (right plot). The external ambient temperature is T∞,2 = 0 for both plots.
The left plot (for which external convection is absent) shows how the exterior surface becomes adiabatic for φ > π/2. The second plot illustrates a case that would be typical of a liquid flow in the pipe (high Bi) and free convection to air on the outside.
The total rate of heat transfer into the fluid would be (per unit length) q′ = 2R2k
Figure 4.16: isotherms in the pipe: Bi1= 5, Bi2= 0 (left); Bi1 = 1000, Bi2= 5 (right)
By replacing the solution for T into the above and integrating, it turns out that the only term in the series that contributes anything to the total heat transfer is the one associated with ln(r). The final result is
q′ = 2αq0′′R2 1 + πBi2T∞,2 Bi1
[Bi1+ Bi2(1 − Bi1ln(a))] (4.120) What may be remarkable to you is the fact that the net heat transfer does not depend, in any way, on the SOV–derived expansion coefficients for the temperature distribution. Indeed, the above formula could have been derived by integration of the governing DE (Eq. (4.106)) over φ and solution of the subsequent 1–D boundary value problem. This is left as an exercise.
By setting Bi2= 0 (i.e., no exterior convection), the result shows (as it must) that q′= 2R2αq0′′, i.e., the net heat transfer to the fluid is equal to the incident flux times the absorptivity times the projected area of the pipe. This makes sense since all of the absorbed radiant energy has to go to the fluid – none can escape back to the environment. Conversely, by setting Bi1 = 0 (insulated inner surface) the result q′= 0 is obtained – because none of the absorbed heat can reach the fluid.