In this screencast, we're going to use the NTU-Effectiveness method for a parallel-flow
heat exchanger to find the overall heat transfer coefficient, or U. In this concentric heat
exchanger, we have water that comes in at 40 degrees C and is heated up to 140 degrees
C by hot gases that come in at 350 degrees C. How are we going to solve for the overall
heat transfer coefficient? By definition, the Number of Transfer Units, or NTU, is equal
to U times A, divided by C sub min. And our C, which is known as the heat transfer rate,
can either be the mass flow rate of the hot fluid times its heat capacity or the mass
flow rate of the cold fluid times its heat capacity. And you calculate those heat capacity
rates and C min is the one that is the smaller of the two. Our U is going to equal NTU times
this C min, divided by the area. Next we have to find our NTU. This NTU can be found as
a function of effectiveness and tables for appropriate flow relations. And for parallel-flow,
this is going to equal minus the natural log of 1 minus this effectiveness multiplied by
1 plus C sub r, and this is divided by 1 plus C sub r. This is the ratio of the minimum
heat capacity rate divided by the maximum heat capacity rate. Next we have to find our
effectiveness, and the effectiveness is the heat transfer rate divided by the maximum
heat transfer rate. And this is equal to the heat capacity rate of the hot fluid times
the difference in temperature of the fluid coming in and out, divided by our minimum
heat capacity rate times the temperature of the hot fluid coming in minus the temperature
of the cold fluid coming in. Now that we have that effectiveness, we can calculate our NTU.
So what's our plan? First thing that we're going to do is calculate all of our heat capacity
rates. So that means our hot heat capacity rate, our cold heat capacity rate, from that
we can figure out what our minimum heat capacity rate is and then find the ratio C min over
C max. Once we know that, we can solve for our effectiveness. This allows us to use the
relationship between the two to solve for NTU. And finally, once we have NTU, we can
solve for our overall heat transfer coefficient. Let's get started. C sub c, or our cold fluid,
which is the water, this is going to equal 4197 watts per kelvin. We do the same for
the hot fluid, the gas, and this is going to equal 1900 watts per kelvin. So this is
going to be equal to our C min. Finally, we have to find the ratio of these heat capacity
rates, and that's equal to 0.45. The next thing we do is calculate our effectiveness,
and when we calculate this, we find that we have an effectiveness of 0.8. Now we use that
effectiveness to find our NTU. However, when we try to calculate our NTU, what we find
out is that this does not exist. So physically, using a parallel-flow, concentric heat exchanger,
you cannot take water that comes in at 40 degrees C and have the gas cool down from
350 degrees to 100 degrees C. So instead, let's change some of the numbers, and change
one in particular. Let's say that the gas, instead of exiting at 100 degrees C, we're
going to have it exit at 150 degrees C. The first thing is that that's going to change
our effectiveness. And we're left with an effectiveness of 0.645. Now when we put this
back in to our equation for our NTU, we end up with an NTU of 1.9. And we end up with
an overall heat transfer coefficient of 181 watts per meter squared K.