one of the easiest ways to solve a
problem that has more than one mode of
heat transfer or different conduction or
convection coefficients is the use of
thermal circuits what is an example of
this well let's look at a plain wall
here it has a thermal conductivity of K
then we have out here some surrounding
temperature will call T infinity 1 and
then a surface temperature TS 1 a
surface temperature TS 2 and finally a
surrounding temperature of T infinity to
the surrounding temperature on the left
has a heat transfer coefficient of H 1
and out here is H 2 and H 1 does not
equal H 2 in this particular problem how
do we solve for Q in this situation so
what we're going to do is use an analogy
between heat and electrical charge both
of them have a driving force divided by
resistances for example current has
change in voltage over an electrical
resistance here our heat rate is a
change in temperature over a thermal
resistance how do we find our
resistances let's start with our
conductive resistance so we go back to
that equation for Q conductive
resistance equals delta T over Q we just
rearrange this which is delta T over Q
conductive which is K a delta T divided
by L so
our conductive resistance is l / k times
a where L is the length of the wall and
a is the area right here that Q goes
through now let's look at our convective
resistance here we use delta T over Q or
our convective resistance which is delta
T and now convection is H our heat
transfer coefficient times this area as
shown before times delta T so our
convective resistance is 1 over H times
a for trying to find our Q again its
delta T divided by our total resistances
so if we go back here and look at our
picture the first resistance is the
convective resistance when we're looking
for our total we start with 1 over H 1 a
now we go to TS 1 and the next thing we
have is our conductive resistance
because we have conduction through the
wall we had convection that caused the Q
between the temperatures of T infinity 1
and TS 1 now we're going to look at the
difference between TS 1 and t s 2 and
its resistance we have a conductive
resistance now we have this third
resistance between T s 2 and T infinity
2 and this is a convective resistance so
it is 1 over H 2 a because we have a
different heat transfer coefficient
and so now when we find Q our delta T is
from T infinity 1 minus T infinity 2
because those are the two temperatures
that we're going from over the sum of
the resistances the beauty of using
thermal circuits is that we can find the
temperature difference between any two
points the reason for this is that Q of
X is a constant how do we know that Q
sub X is a constant in other words it's
not a function of X well let's go back
to our temperature profile which is T of
x equals a constant times X plus a
second constant we determine this from a
previous screencast which use the heat
diffusion coefficient and simplified it
for a plane wall remember that Q is
minus K a DT DX and when we take the
derivative of T with respect to X what
we end up with is c1 which is a constant
which is not a function of X so let's
see how we can use something like that
if we go back to our picture we want to
find out something between T infinity 1
and T s 2 then we know that Q equals T
infinity 1 minus T s 2 over the total
resistances between the two of those
temperatures and the resistance is our
first our convective resistance and then
our conductive resistance so once you
determine Q then you can figure out
another temperature just using the idea
of thermal circuits