Consider this parallel-flow, concentric heat exchanger where here we have hot oil coming
in and it's cooled by water that jackets it. And given the information below, we want to
find how long does the tube have to be in order to perform the necessary cooling. Before
we start, we're going to make a couple of assumptions. So the first assumption we make
is that there is negligible heat loss to the surroundings. And what that means is that
all the heat that's being transferred is transferred between the two fluids. The second assumption
that we make is that that wall in between the two fluids is thin. And what that implies
is that there is no conductive heat resistance between the two fluids. So let's start with
our governing equation. And our equation says that the heat transfer rate is equal to the
overall heat transfer coefficient times the wetted surface area times our delta T log
mean. And we use a delta T log mean where delta T is the difference between the hot
temperature and the cold temperature at some point. However, this delta T varies along
the length of the heat exchanger, so we use a log mean temperature. So now we're going to
rewrite it terms of what we're looking for, or L. And we'll start with this area, which
is going to equal pi times D times L. And let's go back and look at our picture to see which
diameter we're going to use. And since the heat exchange is occurring here, we're going
to use that inside diameter, or 30 millimeters. So now we'll rewrite this such that L is equal
to our heat transfer rate, divided by the overall heat transfer coefficient times pi
times the diameter times our log mean temperature. So let's put together a plan. The first thing
we're going to do is we're going to calculate our heat transfer rate, q dot, using that
q dot equals the mass flow rate of one of the fluids times its heat capacity times the
change in temperature from the start to the end of the heat exchanger. Once we do that,
we're going to calculate our overall heat transfer coefficient using our given convective
heat transfer coefficients. And remember that there's no conduction, so we won't have a
conductive heat transfer coefficient as part of that overall one. Then, we're going to
calculate our delta T log mean from the given temperatures. And finally, once we have all
that information, we can calculate what our L is. Let's start with q. And we're given
the mass flow rate and the heat capacity of the oil, as well as the delta T. So we're
going to use those to calculate q. So our q dot is going to equal 0.15 kilograms per
second, times 2131 joules per kilogram degree Celsius, times 100 degrees minus 60 degrees.
And our q dot therefore is going to equal 12,786 watts. So now we have to calculate
our U. So our U is going to equal one over the convective heat transfer coefficient of
the oil, plus one over the convective heat transfer coefficient of the water, raised
to the negative one. And so when we put in our values for that, our heat transfer coefficients
are given here and here, so this is equal to 38.1 watts per meter squared degrees C.
Now, we need to calculate our delta T log mean. And this is equal to delta T 1 minus
delta T 2, divided by the natural log of delta T 1 divided by delta T 2. So what is this
delta T 1 or T 2? Well the subscripts 1 and 2 correspond to the different ends of the
heat exchanger. So 1 is the difference in temperature at the entrance, and 2 is the
difference in temperature at the exit. Recalling that our delta T is the difference between
the hot and the cold temperature, let's go back to our picture once again, and you'll
note that in a parallel-flow heat exchanger, our delta T 1 is just the difference between the
temperature of the oil coming in and the temperature of the water coming in. So here our delta
T 1 is going to equal 100 degrees C minus 25 degrees C, which is 75 degrees C. We do
the same thing for our delta T2, which is the difference in the temperatures exiting
the heat exchanger, and that's going to equal 10 degrees C. Let's find our delta T log mean.
So our delta T log mean is delta T 1 minus delta T 2, divided by the natural log of delta
T 1 divided by delta T 2. And when we calculate this, we find that this delta T log mean is
32.2 degrees C. And finally we're ready to calculate the length of the heat exchanger.
So that's going to equal the heat transfer rate q dot, and that's going to be divided
by our overall heat transfer coefficient times pi times our diameter (which is 0.03 meters)
times our delta T log mean. And therefore, the length of our heat exchanger is going
to be 110.6 meters.