TEACHER: Many heat exchangers use saturated steam
as a means to heat a colder fluid.
This allows us to solve design
and performance problems very easily
using the NTU effectiveness method.
Let's look at an example
of how to do this.
So, we're gonna start with a well insulated heat exchanger.
We have saturated steam.
What we're gonna do
is we're going to heat up
one kilogram per second of water
which has a heat capacity
that's equal to 4200 joules per kilogram Kelvin,
and we wanna heat it from 30 degrees C,
to 90 degrees C.
And we have an overall heat transfer coefficient
that's been found to be 32 watts per meter squared Kelvin.
And what we wanna know is what is the heat transfer area.
So our NTU by definition,
is equal to U times
this area divided by C-min,
where C-min is the minimum heat capacity rate
and we can calculate that by our m,
mass flow rate of a fluid times it's heat capacity.
So to find the area,
this is just going to be equal
to our NTU times our C-min,
divided by that overall heat transfer coefficient.
So our NTU is a function of the effectiveness
as well as the heat capacity ratio.
So our C sub r is equal to C-min divided by C-max.
So we're gonna need to find both
this effectiveness and this ratio.
So we're gonna start with the effectiveness
and this is kind of like an efficiency,
so this is equal to the heat transfer rate
over the maximum heat transfer rate we can have.
And our heat transfer rate
is equal to the mass flow rate
times the heat capacity
times the change in temperature
of whatever fluid you're looking at.
If you note we have all
the necessary values for the cold stream.
So this is gonna be equal
to one kilogram per second
times 4200 joules per kilogram Kelvin
times our Delta T, which is 60 degrees C.
And when we do those calculations
we find that this is equal
to 252000 watts
or 252 kilowatts.
So next we have to calculate
our q-max which is equal
to the minimum heat capacity rate
times the inlet temperature of the hot stream
minus the inlet temperature of the cold stream.
So let's assume the steam
comes in at atmospheric pressure.
So the steam coming in
the temperature is equal to a 100 degrees C.
Now to find our q-max
we need C-min.
So we calculate our C-min
and that's equal to 420 watts per Kelvin.
So we have enough information now
to calculate our q-max.
So our q-max is just gonna be equal
to this 4200 watts per Kelvin
times that a 100 degrees C minus 30 degrees C
and our q-max is gonna be equal to 294 kilowatts.
And so now if we divide
our q over here by our q-max,
we're gonna end up
with an effectiveness of 0.86.
So now we're going to need our heat capacity ratio
which is equal again to C-min
divided by C-max.
As it turns out for all heat exchangers
where C sub r is equal to zero,
on other words with a phase change
the expression for our NTU
is always minus the natural log of one
minus the effectiveness.
So if we perform the rest of the calculations,
we find that our NTU
is going to be equal to 1.97
and therefore our area
is going to be equal to that NTU
times our C-min
so that's 4200 watts per K,
divided now by that overall heat transfer coefficient
32 watts per meter squared K
and our area is going to be equal to 258 meters squared.
So, just remember,
whenever you have a saturated steam
that's condensing
in order to heat up in colder fluid,
our C sub r is always equal to zero
and regardless of what
the heated type of heat exchanger you have,
our NTU is gonna be equal to minus
the natural log of one minus the effectiveness.