3.4 More on transient problems
3.4.5 Transient problems in radial systems
t=0.01 t=0.2
t=1
Figure 3.6: temperature profile: constant heat flux BC
3.4.5 Transient problems in radial systems The solid cylinder
The basic analytical methods introduced above for the 1–D (in space) slab apply directly to prob-lems in cylindrical and spherical coordinates; only the basis functions for the eigenfunctions will change. To illustrate, consider the following problem: A long solid circular cylinder is initially at temperature T1. At t = 0 the surface temperature is instantaneously brought to T2. Find the temperature distribution within the cylinder as a function of time and radial position.
It is assumed that the cylinder is sufficiently long, relative to the diameter, so that there is no temperature variation in the z (axial) direction. In addition, there is no φ dependence in the problem. The problem is therefore 2–D in r and t. The nondimensional variables are defined in the usual manner:
T → T − T2
T1− T2, r → r
R, t → tα
R2 (3.97)
where R is the cylinder radius. The dimensionless problem appears as
∂T
∂t = 1 r
∂
∂rr∂T
∂r (3.98)
T (r = 0, t) is finite (3.99)
T (r = 1, t) = 0 (3.100)
T (r, t = 0) = 1 (3.101)
The BCs and the DE are homogeneous, and SOV can proceed in the usual manner. Let T (r, t) = u(r) · v(t), and replace this into the DE and separate:
v′ v = 1
ru(ru′)′= constant = −λ2 (3.102) Choose −λ2 as the separation constant because this will give the correct time–decaying solution.
The solution for v is the same as before:
v = e−λ2t (3.103)
For the u variable, the characteristic DE is
(ru′)′+ rλ2u = 0 (3.104)
or
r2u′′+ ru′+ r2λ2u = 0 (3.105) From the previous chapter, this is recognized as Bessel’s equation and has the solution
u = AJ0(λr) + BY0(λr) (3.106)
The function Y0 is singular at the origin, so set B = 0 to keep the centerline temperature finite.
Application of the BC at r = 1 gives
J0(λ) = 0
This condition provides the eigencondition to the problem. A plot of J0(x) in the previous chapter shows that the function oscillates about 0 much in the same way as the triginometric functions.
The eigenvalues λn correspond to the first, second, etc. roots to the above equation, i.e.,
J0(λn) = 0, n = 1, 2, 3, . . . (3.107) An explicit formula for the eigenvalues cannot be obtained for this particular eigenfunction. Rather, one must resort to an appropriate rootfinding method to obtain the λn’s – much in the same way as for convection–type eigenconditions. A table of the first few roots of the above equation is included in the Appendix.
Continuing with the SOV procedure, the general solution to the problem is given as a series expansion of the eigenfunctions (which are J0(λnr)) times the time–dependent part of the solution:
T (r, t) = X∞ n=1
AnJ0(λnr)e−λ2nt (3.108) At t = 0 the condition is
1 = X∞ n=1
AnJ0(λnr) (3.109)
One could anticipate at this point that J0(λn) is orthogonal and proceed to obtain the integral for-mula for the Ans – yet the precise nature of the orthogonality relation (in particular, the weighting function) are unknown. Equation (3.104) is in the Sturm–Liouville form (Eq. (3.42)) with w(r) = r as the weighting function. Each side of the equation is therefore multiplied by rJ0(λmr) and inte-grated from 0 to 1. To make the notation compact, denote J0(λnr) as φn(r), i.e., the eigenfunction.
Again, this function satisfies the DE
(rφ′n)′+ rλ2nφn= 0 (3.110)
and the BCs
φ′n(0) = 0, φn(1) = 0 (3.111)
The orthogonality relation is then Z 1
Recognize again how the BC’s on φnwere used to eliminate the boundary terms in the above. The integral is which provides the desired orthogonality proof.
With this information in hand, the expansion coefficients are obtained from Eq. (3.109) as An=
Using the integral formulas from the previous notes (and the appendix) results in Z 1
In the second integral the eigencondition was used to eliminate J0(λn). Also, J1(0) = 0 by the properties of Bessel functions. The final formula for An is
An= 2
λnJ1(λn) (3.116)
and the complete solution for the temperature is
T (r, t) = 2 X∞ n=1
J0(λnr)
λnJ1(λn)e−λ2nt (3.117)
Annular cylindrical regions
A more complicated problem is now examined. Consider a cylindrical pipe, as illustrated in Fig.3.7.
Initially the pipe is at a uniform temperature of T1. At time t = 0 a uniform heat flux of q0′′ is applied to the inner surface. The heat is removed from the outer surface by convection, which is characterized by a heat transfer coefficient h and an ambient temperature T∞.
On a dimensional basis, the problem is
∂T
∂t = k r
∂
∂rr∂T
∂r
−k ∂T
∂r
Ri
= q0′′
−k ∂T
∂r
Ro
= h(T (r = Ro, t) − T∞) T (r, t = 0) = T1
Following the usual procedure, the problem is recast in nondimensional form. The characteristic temperature is obviously T∞, yet we two choices exist for the characteristic temperature difference, namely ∆TC = q0′′L/k and T1− T∞. The choice is arbitrary – for lack of a better reason choose the first. The variable definitions become
T → (T − T∞)k
q0′′Ro , r → r
Ro, t → tα
R2o (3.118)
Three dimensionless parameters fall out of the problem, which are
a = Ri
Ro, Bi = hRo
k , T1 → (T1− T∞)k
qo′′Ro (3.119)
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Figure 3.7: annular pipe
The dimensionless problem is now
∂T
The problem has a homogeneous DE, yet the inner BC and the IC are inhomogeneous. Conse-quently, it cannot be attacked as–is with SOV. Rather, partial solutions are needed to split the problem into sub–problems which, individually, admit solutions amenable to our analtical tech-niques.
This particular system will eventually attain a steady state, and the solution is therefore for-mulated in terms of a steady–state part and a transient part, i.e.,
T (x, t) = w(r, t) + s(r) (3.124)
The steady–state part has a simple, 1–D conduction solution, details of which need not be repeated here. The problem is
(rs′)′= 0 (3.125)
s′(a) = −1 (3.126)
s′(1) = −Bi s(1) (3.127)
which has the solution
s = a 1
Bi− ln(r)
(3.128)
The superposition T = w + s is now substituted into Eqs. (3.120–3.123);
and the DE and BCs for s are used to cancel terms. The problem statement for w becomes
∂w
The problem for w now has homogeneous BCs, and can be solved directly with the SOV method.
The time–dependent part to w will be identical to those in previous solutions (exponential decay). The spacial dependence will be of the form,
u(r) = AJ0(λr) + BY0(λr)
The Y0 Bessel function must now be retained, because the origin (r = 0) is not included in the domain. The condition at r = a is
u′(a) = 0
which gives (after employing formulas for the derivatives of Bessel functions)
0 = −λ [AJ1(λa) + BY1(λa)] (3.133)
A general solution for u, which satisfies the DE and the inner BC, is therefore
u = A [J0(λr)Y1(λa) − J1(λa)Y0(λr)] (3.134) The condition at r = 1 is
u′(1) = −Bi u(1) or, after using Eq. (3.134) and cancelling constant terms:
λn[J1(λn)Y1(λna) − J1(λna)Y1(λn)] = Bi [J0(λn)Y1(λna) − J1(λna)Y0(λn)] (3.135)
This provides the eigencondition for the problem, in which λnrepresents the nthroot to Eq. (3.135).
Do not be overly concerned about the apparent complexity of this equation. In general, numerical rootfinding techniques are required to solve the most simple of eigenconditions for radial problems – such as that encountered in Eq. (3.107) from the previous example. Providing that an appropriate rootfinding ‘black box’ is available in your numerical tools (which is the case in Mathematica), the only extra overhead involved in finding the roots to Eq. (3.135) is coding the equation into the package.
To condense some of the notation, let φn(r) denote the eigenfunction of this problem, i.e., φn(r) = J0(λr)Y1(λa) − J1(λa)Y0(λr) (3.136) The general solution is then
T (r, t) = X∞ n=1
Anφn(r)e−λ2nt (3.137)
and the initial condition is
T1− s(r) = X∞ n=1
Anφn(r) (3.138)
The eigenfunctions φn are orthogonal on the interval (a, 1) (not (0, 1) as have been all previous problems), with a weighting function r. The expansion coefficients are then
An=
Obviously, the difficult part in wrapping up this example is evaluation of the integrals. The eigenfunction φn represents a linear combination of ordinary Bessel functions of order zero (see Eq. (3.136)), so let
φn(r) = J0(λnr)Y1(λna) − J1(λna)Y0(λnr) ≡ C0(λnr) (3.139) where C0 denotes the linear combination of J0 and Y0. The integral formulas for combinations of Bessel functions, given in the previous notes, can now be applied. In particular,
Z 1 Alternatively, the formulas in the appendix could have been used in the above. As has been done in practically every previous example, the BC at r = a and the eigencondition were used to simplify the result. The other integral appearing in the formula for An is
Z 1 a
[T1− s(r)] φnr dr
This can be split into two parts. The first half is
and the second half is Z 1
Pay attention to the fact that the explicit forms of s (Eq. (3.128)) and φn (Eq. (3.136)) are never used in evaluation of the integral: all that is needed are the DEs (which provides the rules for integration by parts) and BCs (which are used to evaluate the boundary terms) that s and φn satisfy. This property has been exploited in practically every example of the chapter. The final formula for the Anis
An= 2[T1Biφn(1) − aφn(a)]
[(λ2n+ Bi2)φ2n(1) − a2λ2nφ2n(a)] (3.143) and the dimensionless temperature in the cylinder is
T = a 1
Calculation results are presented in Fig. 3.8 for a system with a = 1/2, Bi = 5, and T1
= 1. Perhaps the most noteworthy aspect of the results is the initial increase in temperature at the inner wall (r = a = 0.5), followed by a decrease in temperature to a steady–state value that is less than the starting value of unity. Such behavior is easily explained from a physics perspective: the initial increase in temperature at the inner surface results from the instantaneous application of the heat flux at t = 0. On the other hand, the steady–state inner surface temperature (= a(1/Bi − ln a) = 0.446) is less than the initial temperature of unity for the conditions used here.
0.5 0.6 0.7 0.8 0.9 1
x
0.2 0.4 0.6 0.8 1
T
t=0.001
0.01
0.1
1
Figure 3.8: solution of Eq. (3.144), Bi = 5, a = 0.5, T1= 1
Consequently, the inner surface temperature will attain a maximum value at a finite time period into the process, and will then relax to a minimum as t → ∞.
It is not easy, however, to anticipate this behavior solely from inspection of the analytical solution, Eq. (3.144). As is the case will all solutions in this chapter that have a time–independent steady–state limit, the time dependence in the solution appears in the series terms as decaying exponentials. Since the magnitude of each term in the series must therefore monotonically decease in time, one might expect that the series, as a whole, would display the same monotonic behavior – which would imply that the maxima and minima in the series would occur at t = 0 and/or t → ∞.
This behavior does not occur because of the delicate balancing of terms in the series; the terms have different signs (some are +, others -) and they decay at different rates. With this property, it is entirely possible for the series to have a local maximum/mininum in time.