JANET: An energy balance on a single-phase system
means that the latent heat,
or the heat that's caused by a phase change,
is not included, only the sensible heat.
And sensible heat is created by a temperature change,
and can be calculated using a heat capacity.
So, if we're looking for a specific enthalpy,
this is gonna be equal to the integral
from the initial temperature to the final temperature,
our heat capacity, which is a function of temperature, dT.
Where do we find these heat capacities?
You look them up in the back of the book,
or, you can find them online.
A primary example of a single-phase system,
is a heat exchanger.
So you have one fluid which is hot, that exchanges heat,
sensible heat, with a another fluid
raising that colder fluid's temperature.
The heat exchanger is generally adiabatic,
so all the heat lost by one fluid, is gained by the other.
And if you wanna write it as an energy balance,
it's that Q of the system,
which is the heat exchanger, is equal to zero.
Again, what that means is that enthalpy lost by one fluid,
is gained by the other.
So, let's look at an example.
So, here we have an adiabatic shell and tube heat exchanger,
and it uses water at 90 degrees.
So we have hot water that comes in at 90 degrees C,
and it heats methanol that comes in at 20 degrees C,
and heats it to 50 degrees C.
And the flow rate of the hot water,
so our n dot of the water,
is going to be equal to 0.5 moles per second.
And the flow rate of the methanol,
is gonna be equal to 0.3 moles per second.
And what we want to know
is, what is the final temperature of the water?
So, we're given the heat capacities.
So the heat capacity of the water,
is equal to 75.3 joules per mole kelvin,
and the heat capacity of the methanol,
is given as 75.86 joules per mole kelvin.
So, we'll go back to the energy balance.
So we know because it's adiabatic, that Q equals zero.
And so, that's gonna be equal to the H of the water,
plus the H of the methanol.
So, that equals, so we have our n of the water,
and our integral from our T final two our 90 degrees,
because that's the initial temperature.
Our C sub P of the water dT,
plus the mole or flow rate of the methanol.
And now here, the initial temperature is 20 degrees C,
and the final temperature is 50 degrees C,
and this is C sub P of the methanol dT.
And again, that's gonna be equal to zero.
So, if you note our heat capacities are one term,
so we can take everything out of the integral,
and call it n C sub P delta T.
So let's do that.
We have that zero is equal to, and here's the water,
0.5 moles per second, times our heat capacity,
75.3 joules per mole kelvin, times our delta T,
which is T final, minus 90 degrees C.
And we add to that, our flow rate,
0.3 moles per second of the methanol,
times 75.86 joules per mole kelvin.
And now, our delta T here, is 50 degrees minus 20 degrees C.
And so if we do the calculation,
we find that the final temperature of the water,
is equal to 72 degrees Celsius.
Now, an easier way perhaps for you to see it
is that, the enthalpy of one,
is equal to the negative enthalpy of the other.
So we can rewrite this as 0.5 moles per second.
Again, this is the water, times 75.3 joules per mole Kelvin,
and that's the heat capacity,
times, and now we're switching this,
90 degrees C minus T final,
and that's gonna be equal to 0.3,
again this is the methanol,
moles per second, times 75.86 joules per mole kelvin.
And now this is 50 degrees minus 20 degrees C.
And actually you can calculate this,
which is equal to 683 joules per second or Watts.
So one thing you have to be careful of
when you're doing a problem like this,
if you note, I only had one term for the heat capacity,
and that is reasonable because we have liquids,
but once you have gases,
you have to have the entire expression,
for the heat capacity, which usually runs to four terms.