We want to find the overall heat transfer coefficient for a thin-walled - and the reason
we say it's thin-walled is that we can neglect conduction through the wall - counter-flow,
concentric heat exchanger. It's going to heat pressurized water using hot gases. The water
comes in at a mass flow rate of 1 kilogram per second. It's heat capacity is 4197 joules
per kilogram K. It comes in at 40 degrees C and exits at 140 degrees C. The gas mass
flow rate is 1.9 kilograms per second. It's heat capacity is equal to 1000 joules per
kilogram kelvin. It comes in at a temperature of 350 degrees C and exits at a temperature
of 100 degrees C. What we're looking for is that U, or overall heat transfer coefficient.
As I mentioned before, our NTU is defined as U times A, divided by C min. Therefore,
we can find our U by taking the NTU, multiplying it by our C min, and dividing it all by the
area. And in this particular problem, our area is 20 meters squared. The next thing
we have to do is figure out what the NTU is. And if we look it up on a table, we find for
a concentric, counter-flow heat exchanger that the NTU is one divided by this ratio
C sub r minus one, multiplied by the natural log of the effectiveness minus one divided
by the effectiveness times C sub r minus one. The next thing we have to do is define this
effectiveness, which is our heat transfer rate over the maximum heat transfer rate, which
is equal to (and we'll look at the hot fluid) the heat capacity rate of the gases times
the temperature coming in minus the temperature coming out, all divided by C min T h in minus
T c in. Our plan is as follows. First, we need to find all of these heat capacity rates,
determine which one is minimum, and find the ratio. Next, we're going to solve for effectiveness.
This will then allow us to solve for our NTU. And finally, we can solve for our overall
heat transfer coefficient. Let's go ahead and do this. First, we have these heat capacity
rates. The cold heat capacity rate is its mass flow rate times its heat capacity. So
C sub c is 4197 watts per K. Our hot heat capacity rate is the mass flow rate of the
gas times the heat capacity of the gas, and that's going to equal 1900 watts per K. So
this is going to be our C min. So finally, we find our ratio of C min over C max. So
our C sub r is 0.45. Next we have to find the effectiveness, and note I could've used
for our q dot either one of the fluids, the hot or the cold, but I chose the hot fluid
so that we could cancel these two out. And so we're left with an effectiveness of 0.8.
Now, we have to use the relationship between the NTU and the effectiveness for our particular
flow arrangement. And when we do the calculations, we find that our NTU is 2.1. And finally,
we can use this number to solve for our overall heat transfer coefficient. And when we calculate
this, we find that our overall heat transfer coefficient is 201 watts per meter squared
K.