In this screencast we are going to look at an example of adiabatic mixing and the problem
that we will look at requires the use of an enthalpy concentration diagram to solve.
In this problem we have 2 lb pure water at 100 degF, and it is mixed adabatically with
1 lb of 60 wt% of H2SO4 solution also at 100 degF. We want to find the outlet temperature.
So when we say the process is adiabatic mixing, it implies that the mixing unit is well insulated
and that there is no heat exchange with those surroundings. So Q is equal to 0, that means
that any heat that is adsorbed or released upon mixing is transferred to the final mixture.
Now we know exactly what is going into the mixing unit. We do not know anything about
the outlet. So we do not know the total mass. We do not know the concentration leaving the
mixing unit. We can use mass balances to solve. We know that mass is conserved, and if we
assumed there is no accumulation in the mixing unit the mass balances will have a form of
in equals out. We can quickly find out the total mass leaving the mixing unit by a total
mass balance. The mass leaving the mixing unit is simply the sum of what goes in, which
is 2 plus 1. So Mout is equal to 3 lb. To find the composition we can use a balance
on either H2SO4 or water. So I will use a balance on H2SO4. So on the inlet side the
mass fraction of the H2SO4 solution going into the mixing unit is 0.6. We multiply that
by the total mass of that stream, which is 1 lb. That must be equal to the mass fraction
leaving the mixture, which is x times the total mass, which we just calculated. So solving
for x here. X is equal to 0.2. We want to find the outlet temperature of the mixture
we are going to need an energy balance. So scrolling down here I have re-drawn the process
flow chart with the variables we just calculated. Next we need an energy balance. So in our
energy balance in this case. This Q is equal to delta H. For an adiabatic process Q is
equal to 0. So if our change in enthalpy is equal to 0. That implies that the inlet enthalpy
has to be equal to the outlet enthalpy. The total enthalpy of each stream is dependent
on how much material is associated with that stream. So we can rewrite the total enthalpy
in terms of the specific enthalpy and the mass associated with each stream. So here
I have labeled the specific enthalpy of each stream in the chart. Our energy balance changes
as follows. In the inlet side the enthalpy is the sum of the enthalpies of each stream.
Which can be written as 2 lb times the specific enthalpies of stream 1 H1, plus 1 lb times
the specific enthalpy off the second stream and that has to be equal to 3 lb times H3
on the outlet. On the inlet side we know that both of these streams are at 100 degF. So
we should be able to use an enthalpy concentration diagram to find the specific enthalpy of these
solutions at a specific temperature and concentration. We do not know what the temperature is on
the outlet side, but if we use the energy balance we should be able to calculate each
stream. So we will go to the enthalpy diagram to find values for H1 and H2. I am showing
an example of the enthalpy concentration diagram for H2SO4. So the diagram shows the specific
enthalpy of the solution of units of BTU/lb of solution, and that is a function of mass
fraction of H2SO4. The specific enthalpy of the solution is shown for several isotherms.
So these curves each represent different temperatures. In our case the inlet side both of the streams
that are mixed are at a temperature of 100 degF. So we can use these diagrams to find
the specific enthalpy of pure water and our H2SO4 going into the mixture. Our first stream
was pure water. So the mass fraction of H2SO4 then is equal to 0, and we are at 100 degF
in the inlet side. So all the way on the left the mass fraction of H2SO4 is 0, and the isotherm
is 100 degF. So if we trace this up to x is equal to 0 we can get a reading for the specific
enthalpy at that temperature. We can see that it is pretty close to 70 so I will use a value
of 68, and again the units there are specific enthalpy in this case ate BTU/lb of solution.
Our second solution with a 60% H2SO4 also at 100 degF. So we can follow the 100 degF
isotherm to x is equal to 0.6 to find our specific enthalpy, which is about this point
here. So it is less then 100. So I will use a value of -108. Going back to the energy
balance that we set up. We now have values for both H1 and H2, which we can use to solve
for the specific enthalpy of our mixture. According to the values that we found in the
enthalpy diagram we can calculate H3, which I find is 9.3 BTU/lb. So we know the specific
enthalpy of the mixture. We also know the concentration from the mass balance. We found
that the concentration is 0.2. We can use the enthalpy concentration diagram to find
at what temperature does the H2SO4 solution of 20 mass fraction equal 9.3 BTU/lb. So lets
go back to the enthalpy concentration diagram to find out temperature. We know the specific
enthalpy of the mixture is 9.3 BTU/lb. So that is right bellow this horizontal line
on the enthalpy diagram. We also know the concentration of the mixture is 20%. So if
we bring that up we can find the isotherm that this mixture has the specific enthalpy
of 9.3 BTU/lb. It is in between 100 and 120 in this region. If we zoom in a little bit
here we see that it is barly underneath the 120 degF isotherm. So our outlet temperature
is approximately 118 degF. In this case the mixing of pure water with H2SO4 is an exothermic
process. So energy is released when mixing these 2 streams, which has the affect of driving
the temperature up from 100 degF to 118 degF.