SPEAKER: Since the purpose of a heat exchanger
is to use a hot fluid to heat up a colder fluid,
you can't use the same convective
or h heat transfer coefficient
to find your heat transfer rate.
Then you add conductive resistances
from the pipe wall fouling factors et cetera
and that makes it apparent
that you need an overall heat transfer coefficient,
which will call you to describe your system
and this overall heat transfer coefficient
is going to have units of watts per meter squared Kelvin.
So not surprisingly,
you know need to use the resistances
to find this coefficient
and in fact, our U times our area
is equal to the sum of all the thermal resistances
raised to the minus one, in other words the reciprocal.
So note that this is written in terms of UA not U.
So the question is, what area are we talking about?
Well, as it turns out,
the U will change depending on what area
the heat transfer takes place on.
However, UA will always stay the same.
So when you talk about an overall
heat transfer coefficient U
you have to state what area does this apply to?
So let's look at an example.
So we have a pipe and this pipe is insulated,
you have a fluid A in here
and it has its convective heat transfer.
Then you have a wall here that has an inside
and an outside diameter and a thermal conductivity
and finally, this wall separates fluid B
that has its convective heat transfer coefficient.
Our hA is not equal to hB and what we want to do
is to find our UA
so as we put before, UA is equal
to the inverse of the sum of the thermal resistances.
So let's take a look at what resistances we have.
So the first is the convective resistance
with the outside fluid
so that's equal to one times hA times the area
that the heat transfer occurs on
and that's gonna be equal to our pi
times our outside diameter times our length
and let's draw this out a little bit better.
So this we're gonna call the wall in between the two fluids
and here is the outside diameter
and then there is the inside diameter.
So that's our resistance for our convective heat transfer
in fluid A.
Now we need our conductive heat transfer resistance
that occurs in the wall
and so that's equal to the natural log of the ratio
of the two diameters divided by two pi K
thermal conductivity of the wall times the length
and this L that we're talking about
is the length of this pipe
and now finally
we have our convective heat transfer resistance
for the inner fluid, so that's h times B times pi times
now our inside diameter times L
and so to find our UA we add all these up,
we raise it to the minus one or we take the reciprocal
and that's gonna be equal to UA.
So again, we wanna find a U and you can find it
in terms of the inside diameter
or in terms of the outside diameter.
So let's look at it and we write it here
in terms of this outside diameter
and so what this means is that U outside
times A outside is equal to UA
which is equal to what we had here.
So basically what we're going to do
is divide this expression that we put together
by that outside area and from that we get our U outside.
So one of the easiest ways to do this
is to say that our UA is equal to
we'll call this resistance one, that's the first one
plus resistance two plus resistance three to the minus one.
So the easiest way to write this
is that one over U outside is just equal to now our area
pi outside diameter times L
and we multiply this by R one plus R two plus R three
and so now let's write that out
when you can see what cancels.
So separate this R one times pi diameter L
plus R two and again, here's this outside area
plus R three and this outside area
and now if you go back and take a look
at what these resistances are here
you can notice that a number of these values
will cancel out and what we're left with
is one divided by Rh sub A plus Ln
and again we have this ratio the diameters times
and everything else here will cancel out plus
and now we have one over hB the ratio of these diameters.
And so if we want to actually find
our outside heat transfer coefficient,
let me just take this entire expression solve for it
and take the reciprocal of it.
One of the things that can make this an easier problem
is you can assume that the wall between the two fluids
is thin therefore that conductive resistance is negligible.
So, now we can actually find U
so this is just U outside
because the areas are all the same.
So our U is going to be equal to one over hA
plus one over hB to the minus one.
By the way, this generally is a valid assumptions
that you want as much heat transfer
as possible between the two fluids.