Consider heat flow in the x direction through a material slab of arbitrary cross section,
as shown in the figure below. The edges of the slab (somewhere in y and z) are
assumed to be perfectly insulated. (The slab might be a wall or ceiling in a house, for
example.) The left edge of the slab is at x = 0 and the right edge is at x = L, the
thickness of the slab. The temperature of the ambient medium (e.g., air) to the left of
the slab is Th and the temperature of the ambient medium to the right of it is Ta.
Th
Ta
The temperature distribution u(x,t), a function of position x and time t, inside the slab,
assuming one-dimensional heat flow, satisfies a partial differential equation:
,
(1)
where k = κ/ρC, and κ, ρ, and C are the material’s thermal conductivity, mass density,
and specific heat, respectively. (k is called the diffusivity.) Equation 1, together with
boundary conditions at x = 0 and L, along with specification of the temperature at some
fixed time (such as t = 0), form a boundary value problem.
The boundary conditions are taken to be convective:
(2a)
and
(2b)
where the convection coefficient h is taken to be the same on both sides of the wall to
keep things simple. The initial condition is
u(x,0) = f(x),
(2c)
where f(x) is some as yet unspecified function of position within the slab.
Basic physics and experience indicate that the temperature function u(x,t) is, or can be
thought of as, a sum of two functions, a steady state function of position only, V(x), and
a transient function, W(x,t), a function of position and time that vanishes as t
approaches infinity. (No, the time never becomes infinite but it becomes large enough
that W becomes indistinguishable from zero. V(x) describes the situation that exists
when the slab’s temperature has ceased changing with time. It has ceased heating up
or cooling down from the state described by f(x) (equation 2c). In the steady state, the
slab is in thermodynamic equilibrium with its surroundings.)
In the steady state, u(x,t) = V(x), and V(x) satisfies the ordinary differential equation
(3)
The solution of equation 3 is
V(x) = C1x + C0, 0 ≤x ≤ L,
the equation of a line with slope C1 and intercept C0. Equations 2a and 2b translate
directly to V(x) and lead to
(4)
(5a)
and
.
(5b)
(Do not confuse all the “h” subscripts with the convection coefficient “h.”)
Before moving on to the transient solution, there’s at least one interesting thing that can
be done with the steady state solution. The rate at which heat is transferred through the
wall per unit area in the steady state is -κdV/dx = -κC1. But if the slab is taken to be a
sheet of housing insulation, the rate of heat transferred per unit area is, for T h > Ta,
where R is the insulation’s thermal resistance, the so called R-value generally quoted
when selling or advertising insulation. Hence, an equation for R in terms of more
fundamental thermal parameters is
.
(6)
With h generally being significantly larger than unity, unless the thickness is
unreasonably small, the L/κ term dominates, thus explaining the often made statement
that doubling the thickness doubles the R-value. Equation 6 also shows that the
convection coefficient is generally not a large driver in determining R. That’s a good
thing because h on the outside of a wall where the wind is strong increases significantly
over nonwindy atmospheres (thus invalidating this analysis’s assumption of h being the
same on both sides of the wall). Hence, good insulation makes wind somewhat
irrelevant. (However, this is in the context of both sides of the insulation being directly
exposed to air. Things may be somewhat different in real applications, where the
insulation is between slabs of other materials, with somewhat higher thermal
conductivities.) See http://www.stumpff.me/rvalue.doc for plots showing the relationship
of R to h, κ, and L.
As was just discussed, the steady state solution of equation (1) is good for evaluating
the performance of housing insulation, as one example. If an analysis is being
performed that’s treating the wall as just one of many outside surfaces of a house to
evaluate what the house’s furnace is competing with while it maintains the inside air
temperature (Th here) at the thermostat’s set point, the steady state solution could be
useful for that. But if you’re interested in what your furnace must deal with while it’s
getting the house up to the thermostat’s set point from an initial cold state (or an initial
hot state in the case of air conditioning), or you’re interested in the fine thermal details
involved in returning the house to the set point after the air drops down to the
temperature that triggers the thermostat, that involves knowing what the slab is doing
thermally as a function of time.
The transient solution, W(x,t), is a little more involved. With the second spatial derivative
of V(x) and its time derivative both being zero, equation 1 also applies to W(x,t). It’s just
a matter of translating the boundary and initial conditions to W(x,t). With
W(x,t) = u(x,t) – V(x),
The boundary conditions on the transient function become
(7a)
and
.
(7b)
The initial condition on u(x,t) translates to
.
(7c)
Note that both equations 7a and 7b involve a linear combination of functions of W(x,t)
that equal zero. Getting to that so called condition of homogeneous boundary
conditions was the mathematical reason for separating the actual temperature function
u(x,t) into separate steady state and transient functions. But this is not a textbook on
boundary value problems. (There are any number of such texts that you are free to
consult. I used “Boundary Value Problems,” by David L. Powers, as a model for this
derivation, but this particular example was not worked out there. You aren’t reading
something coped from a textbook.)
The standard technique of solving equation 1 for W with homogeneous boundary
conditions is to use the product method. It is assumed that W(x,y) is a product of a
function of x alone and a function of t alone:
,
and the new goal becomes to find the functions Ɵ and T.
Taking u(x,t) in equation 1 to be the above proposed product solution for W(x,t)
produces
,
where the standard simplifying practice of using a prime to signify differentiation with
respect to the independent variable is used. (Note that it’s “k,” the diffusivity, in the
above equation, not κ, the conductivity.) Dividing the above equation through by ƟT
yields an equallity between ratios that can only exist if each ratio equals the same
constant, independent of x or t:
.
(8a)
The above mentioned constant is signified here by -α2. The notation is used to force the
constant to be negative. Choice of a positive constant leads to a nonphysical solution.
(A mathematician would say that -α2 is an eigenvalue of the system. I’ll be generally
more relaxed and won’t make a big distinction as to whether it’s α or -α2 that’s the
eigenvalue.) Equation 8a represents two ordinary differential equations:
Ɵ’’ + α2Ɵ = 0
and
(8b)
T’ + kα2T = 0.
(8c)
The solution to equation 8c is almost trivial:
.
(9a)
(There’s usually an undetermined coefficient out in front of the exponential. That
coefficient is suppressed here and will be absorbed in the coefficients in the solution for
Ɵ(x).) The negative exponent in equation (9a) is why the transient portion of u(x,t)
becomes negligible for large t.
The solution for equation 8b is
Ɵ(x) = b1cosαx + b2sinαx.
(9b)
With the temporal function of equation 9a dividing out, the boundary conditions
represented by equations 7a and 7b apply to Ɵ(x). Equation 7a yields
hb1 = ακb2.
With the requirement that b1 and b2 cannot both be zero, the above relationship can be
used to represent b2 in terms of b1 and then equation 7b and equation 9b combine to
yield
.
(10)
This is the system characteristic equation defining the eigenvalues α. There are an
infinite number of eigenvalues, both positive and negative (but not zero) that satisfy
equation 10. All of the positive ones are associated with an independent solution of
equation 1 for W(x,t). (The negative values do not contain any additional information
and can be disregarded.) The full solution for W is a summation of all those independent
functions. But I’m getting ahead of myself. Equation 10 is a transcendental equation; it
cannot be solved via standard algebraic or trigonometric techniques. There are various
numerical techniques, however, that work as long as specific values for the parameters
other than α are available. (I use Regula Falsi.)
For now, just imagine that you’ve obtained the infinite set of positive solutions to
equation 10. Arrange them in sequential order and signify each solution with a
subscript n = 1, 2, 3, …, ∞. There is a different undetermined coefficient b1 for each αn;
b1 becomes bn. (The relationship between b1 and b2 avoids turning b2 into a subscripted
variable. Further, the sequence of coefficients bn might be called an eigenvector if one
wanted to be mathematical about things.) The transient solution is the sum of all ƟT
products:
,
(11)
where
.
(12)
(And again, watch out for word processors not making a big distinction between “k”
(equation 11—diffusivity) and “κ” (equation 12—conductivity).
gn(x) is the key to determining the coefficients bn. It is an orthogonal function over the
interval 0 to L as long as αn obeys equation 10. Orthogonallity means that
So, applying the initial condition (equation 7c) on W(x,t), multiplying through by gm(x),
and then integrating over x, the coefficients end up being given by
,
where
(13a)
.
(13b)
Because I’m a really all around great guy, I did that integral for you:
,
and I also worked out the known part of bn:
.
(Note the αn multiplying the integral that you’ll have to divide through by at some point in
whatever you might be using this for, and then at some point you’ll need to divide
through by Dn.) The unknown part of bn will become known upon specification of f(x),
the initial temperature distribution, u(x,0).)
Just add V(x) (equation 4) to W(x,t) (equation 11) and you have u(x,t), the spatial and
temporal temperature distribution inside the slab. Should you actually go to compute
numbers, you don’t need to worry about the impossibility of evaluating the infinite sum in
equation 11. You can truncate the series. The coefficients decrease in magnitude as n
increases and the exponential time function contributes to terms for large n becoming
small (except for very small times). (But watch out for |bn| showing a temporary
maximum close to but after n = 1 before tapering down to insignificant values. I don’t
know if that can happen but I don’t know that it can’t.) I’ve been using 200 terms and
things appear to work okay except for small times (t) or large thicknesses (L). Note that,
as L increases, αn decreases. That increases the time constant for each term in the
series and makes offending small times uninteresting.)
The rate at which heat flows through the slab per unit area is
.
(14)
It is a simple matter to integrate dq/dt over t from 0 to some time t’ to find the total heat
transferred per unit area in time t’. Only the leading term involving C1 and the
exponential factor within the summation are involved in the integration process. If the
left side of the wall is important, use x = 0. If the right side is the important one, use x =
L. Either way, positive dq/dt means that heat flow is to the right and negative dq/dt
means that |dq/dt| heat per unit time and unit area is flowing to the left.
Note that dq/dt is independent of x in the steady state. The exact same amount of heat
that flows into one side of the slab in that state flows out the other side. (That
equalization is what establishes the stabilized temperature distribution that exists in the
steady state.) However, before that state is reached, less heat flows out of the opposite
side of the slab than what flowed into it. That’s because, during the transient phase,
heat is being stored in the material. That’s why the temperature of the slab increases
from its value at t = 0. (Or if the slab somehow starts out hotter than Th or Ta, heat
already stored conducts away and that causes the slab to cool.) Thus, the transient
portion of u(x,t) represents heat stored in the slab. (A more fundamental way of seeing
that is to work through the derivation of equation 1 from conservation of energy
principles, where you will see that the time derivative on the right arises because of heat
storage in the first place.)
It is left as an exercise for the reader to use equation 14 to derive a non-steady state
formula for the R-Value.
For another exercise for the reader, prove the following assertions about what happens
when the convection coefficient is different on the two sides of the slab, say, h1 on the
left and h2 on the right. It turns out that equation 12 for gn(x) stays exactly the same if
“h” is replaced with “h1,” and thus, any formula you’ve worked out for bn remains
unchanged except for h →h1.1 (Also, make that change in the worked out expression
for Dn, the expression for the integral over gn(x)V(x), and in equation 14 for dq/dt.) The
characteristic equation for the eigenvalues (equation 10) becomes
.
(15)
(The fact that the function gn(x) without change remains orthogonal even though the
characteristic equation changed is somewhat interesting.)
The steady state slope and intercept formula become
(16a)
1
But note the new formula for the steady state coefficients, C0 and C1.
and
.
The term in equation 6 for R is replaced by
(16b)
.
(Note that all formula involving two convection coefficients reduce to the single-h
formula when h1 = h2 = h.)
And that’s the solution to the one dimensional convective heat transfer problem.