This is part 2 of the material and energy balances final review. In this screencast
we will be going through the energy balance portion of the problem. In the previous part
1 we went through a combustion problem and we completed the material balance and now
we are going to calculate how much energy that reaction produce in our reactor. We have
ethane and air coming into a combustion chamber. The ethane is at 25 degrees Celsius and the
air is heated to 100 degrees Celsius prior to entering the reactor. The products leave
the chamber at 300 degrees Celsius. This includes carbon monoxide since we are told in the original
problem that the combustion was incomplete. To determine the energy that comes out of
this reactor from this reaction we are going to use the heat of formation method. In doing
this since we have multiple species and at different conditions the best way to start
any energy balance is to set up a table. In this table we will have a column that represent
our species and each row will be a different species in our material balance. Then we keep
track of the flow rate and specific enthalpy associated with that species and that stream.
For a single inlet and a single outlet we should get 4 columns where we represent our
moles that are coming in and our moles that are coming out of our reactor. Now this was
already calculated in part 1 so see that to determine these values. We want to determine
the specific enthalpy of each species at the conditions stated. We know that for moles
in we have ethane at 25 degrees Celsius but we have oxygen at 100 degrees Celcius and
nitrogen at 100 degrees Celsius. For our outlet we are at 300 degrees Celsius. This is where
the table will come in handy in terms of keeping track of each enthalpy associated with each
species. So how are we going to do this? The first step once we have our table is choice
appropriate reference so that we could look up or calculate the specific enthalpies of
each species. Since we are using the heat of formation method I am going to say that
our references are going to be the standard elements. So carbon solid, oxygen gas, hydrogen
gas and nitrogen gas at 25 degrees Celsius and 1 atmosphere. Now you will see why I do
this in a minute, but basically since we are using the heat of formation method we want
the reference state to be the individual elements at 25 degrees Celsius mainly because that's
a nice reference temperature to use and it's easy to look up those values in multiple different
places. To determine the specific enthalpy of a species using the heat of formation method
we are taking the heat of formation at the reference state, 25 degrees Celsius, and adding
any sensible heat change from our reference state of 25 degrees Celsius to our temperature
that that species is at. Now to account for this sensible heat difference it's the heat
capacity which is a function of temperature integrated between those two temperatures.
Now we have our heat of formation and our sensible heat change which we could calculate
or look up in tables. So let's work through this table and start filling in some of these
values. Let's start with ethane. We have ethane entering at 25 degrees Celsius and exiting
at 300 degree Celsius. So the enthalpy of ethane that is entering is just going to be
the heat of formation of ethane at 25 degrees Celsius since we are not evaluating it at
a different temperature the sensible heat difference is zero. We look this up in a table.
You can that the heat of formation of ethane is -84.68 kilo Joules per mole. This means
in our table we can plug in the specific enthalpy is -84.68, however for the outlet we need
to evaluate this integral. We look up the coefficients associated with the heat capacity
as a function of temperature. Again these could be looked up in the references I mentioned
earlier. So to evaluate this integral it is important to note whether it's in Celsius
or Kelvin. The values I looked up were for Kelvin. So we are evaluating from 298 Kelvin
to 573 Kelvin. So the heat capacity as a function of temperature has the following form. "A"
plus "BT" plus "CT^2" plus "DT^3" and we evaluate that integral. So now you just have to look
up the values for those coefficients. So now let me plug in these coefficients for ethane.
Note that the coefficients are usually to some power so make sure to include that when
your doing your calculations. If I evaluate this integral we get the following sensible
heat difference of roughly 19.27 kilo Joules per mole. That is this part of the equation
so it is not the specific enthalpy of the stream. We have to include the heat of formation
at our reference state of 25 degrees Celsius. So when we take that -84.68 and add our sensible
heat difference we can get the enthalpy at 300 degrees Celsius. So I will plug that into
our table. I get -65.42. For all of the other species we can follow the exact same format
however there are some certain shortcuts. I am going to lump oxygen and nitrogen together
so that you can see what we are going to do here. Now the table in the back of the Felder
and Rousseau textbooks list these common combustion species at various temperatures using a reference
of 25 degrees Celsius. This table, table B8 in the appendices, gives you the specific
enthalpy of the selected gas using a reference of 25 degrees Celsius and 1 atmosphere. So
what we could do is look up oxygen at 100 degrees for what's entering and we see that
it is 2.24 kilo Joules per mole. Now this is this part of the equations. Since the oxygen
is already in it elemental state and that was our references used for our method above
the heat of formation of oxygen is 0. So that part cancels out and you can see that if we
were at 25 degrees Celsius we would have an enthalpy of 0 which is our reference state.
So this 2.24 is just our enthalpy associated with the temperature difference. So we look
up the same for nitrogen and now we have our enthalpies associated with those two species
at 100 degrees Celsius. So for oxygen we have 2.24. For Nitrogen we have 2.19. So as you
can see now we have everything we need for our inlet stream. We have to finish this for
outlet. We could do the exact same thing for oxygen and nitrogen at 300 degrees Celsius.
Let's take a look at that table. Now at 300 we have 8.47 and 8.12. Now for water, carbon
dioxide, and carbon monoxide these aren't elemental species that are our reference state
so the heat of formation does need to be included in our calculation. So we could look those
up in a table and add what we see in this other table to get our final specific enthalpy
at those temperatures. Let me show you. Let's take water at 300 degrees Celsius. It has
a specific enthalpy of 9.57 when using a reference temperature of 25. Again this is this part
of the equation, the sensible heat change, we add the heat of formation. So this looks
like -241.84 plus 9.57. Which gives us a specific enthalpy of water at 300 degrees Celsius of
-232.27 kilo Joules per mole. We could do this calculation for carbon dioxide and carbon
monoxide and finish our enthalpy table above. So now we have our enthalpy table with our
molar flow rates. All that is left to do is calculate our overall energy balance cause
recall the problem asked how much energy is absorbed or generated from this reactor. Our
overall energy balance is going to have the following form where we have the heat absorbed
or generated by the reaction is going to be equal to the enthalpy difference between the
outlet and inlet streams. What this means is that we can set this equal to the sum of
each component times the specific enthalpy. This is going to be for the outlet and we
are going to subtract that same for for the inlet. Just to be clear let's set this up
for ethane. If we look at the outlet we have 2.5 moles times the specific enthalpy associated
with that stream. Now we are going to subtract out the flow in the inlet stream times the
specific enthalpy associated with ethane in that stream and now we continue this for each
species. When you do this you get this nice drawn out equation with all of your values
plugged in and I get a "Q" equal to -8,388 or roughly -8,400 kilo Joules. Notice that
we do have a heat that is produced by the reaction. It is coming out of the reactor
which is something we would expect for the combustion of ethane. But again this does
depend on the inlet and outlet temperatures of our streams so it is always good to double
check those values. So after looking at a reaction like this think about how we could
produce more heat. in this problem we used air and we had to heat up our nitrogen and
it comes out at a higher temperature so it absorbs some of the heat produced by the reaction.
Now if we used pure oxygen we would not have this enthalpy lost to this nitrogen so that's
one way we might able to get maybe even a more efficient combustion as well as more
energy produced by that combustion. I hope this helps show you how to set up an energy
balance table and the overall energy balance. After this visit the material and energy balances
final review part 3.