In transient conduction it is often important
to calculate the total energy transfer during
a certain time.
So in order to do so we need to start with
the simplified overall energy balance such
that E(in)-E(out) is equal to the change in
the stored energy.
Depending on whether energy is being put in
or taken out either E(in) or E(out) drops
out.
So here we're going to assume that energy is transferred
out of the system, which means that E(in)
is going to be equal to 0.
So our Q is going to be equal to our -E(out),
and the way to look at that then is that Q
equals minus the energy at some time minus
the energy at t=0.
So we can rewrite this as Q is equal to minus
rho, density, times the heat capacity times
our T, whatever T we're looking at, some x
and some t, and I am calculating this currently
for a plane wall with convection, minus T(i)
dV, where this is integrated over the volume
of the wall.
So we now normalize this Q by introducing
Q(0), which is that rho times the c times the
volume times T(i)-T(infinity).
And so this Q(0) is considered the maximum
amount of energy transfer that could occur.
One way to look at it is that T(i)-T(infinity)
is the largest possible change in temperature
that we could get.
So now let's write the whole thing out.
So Q divided by Q(0) is equal to the integral
of -T(x,t)-T(i) divided by now T(i)-T(infinity)
dV over the volume, and we're going to assume
that that volume is constant, and we can rewrite
this as, pull out that one over V, the integral
over the total volume , (1-theta*)dV.
And by definition this theta* is equal to
T-T(infinity) divided by T(i)-T(infinity),
and this can also be rewritten as this theta*(0),
which is the center line temperature times
the cosine of our eigenvalue, times x*, where
x* is the normalized length into the wall.
So if we integrate this we're left with that
Q over Q(0) is equal to 1 minus, now we have
the sign of our eigenvalue, divided by the
eigenvalue itself, times theta*(0).
So our theta*, and again this is considered
the mid-plain temperature, is equal to c(1)
exponential minus this eigenvalue squared
times the Fourier number.
So the total energy transfer for a long cylinder
sphere is done the exact same way.
We do use the same normalization term Q(0),
but remember that volume is going to change.
So for a cylinder we write it as theta* is
equal now to this theta(0)* times the Bessal
function of the first kind of this eigenvalue
time r*, where r* is that dimensionless radius.And
then our sphere, theta*, is equal to theta(0)*
one over our eigenvalue times our *, and then
our sign of our eigenvalue times our r*.
Again our theta(0) is as shown above.
So the integration produce the following expressions
for these geometries.
So the first one is our cylinder, Q over Q(0)
is equal to 1 minus 2 times this theta(0)*
divided by the eigenvalue times the Bessel
function again of the first kind of the eigenvalue.
And a sphere we get the ratio of our Q divided
by Q(0) is equal to a slightly more complex
1-3 times this theta(0)* divided by, and now
we have a cubed term here, of the eigenvalue,
and this whole thing is multiplied by the
sign of the eigenvalue one minus the eigenvalue
times the cosine again of that eigenvalue.
So as a reminder the eigenvalues are obtained
through a one term approximation of an infinite
series solution, and these Bessel functions
are again Bessel functions of the first kind
that can be calculated or looked up in the
appropriate table.
So to find the total energy transfer during
a certain time you need to calculate the normalized
energy transfer, and then multiply it by Q(0).
The way to look at it is that your original
solution shows you the fraction of the maximum
energy transfer, and that's what would occur
as time approaches infinity.
So in future screencasts we'll show you how
to use these equations and find the total
energy transfer for different geometries.
