In this screencast we are going to look at an example of an isothermal mixing process,
and use the enthalpy concentration diagram for H2SO4. We have 2 lb of pure water, and
100 degF, and that is mixed isothermally with 1lb of 60 wt % of H2SO4 solution that is also
at 100 degF. We want to find the heat that is released upon mixing. So when we say that
it is being mixed isothermally we know that the inputs are 100 degF. Isothermally means
constant temperature. So the outlet is also at 100 degF. Mixing of H2SO4 with water is
an exothermic processes. So if we are keeping the temperature constant that means heat is
released. So we are trying to find how much heat is released for mixing this particular
processes. So we will need an energy balance, but before we get there we will need to find
the mass and concentration for the mixture. So for that we can use mass balances. We know
that mass is conserved and if there is no accumulation in the mixing unit we know that
the mass balances will have a form of in is equal to out. We can use a total mass balance
to figure out the total mass of the mixture. The mass going into the mixing unit is 2 lb
plus 1 lb. So Mout must be equal to 3 lb. Now we can use an H2SO4 balance or a water
balance to find the concentration. I will use H2SO4. The mass fraction of the H2SO4
stream entering the mixing unit is 0.6 we multiply that by the total mass, which is
1 lb. That has to be equal to the mass fraction of the mixture, which has a composition variable
of x times the total mass. We just calculated, which is 3 lb. So in this case the mass fraction
of H2SO4 of the mixture is equal to 0.2. Lets use these values to set up an energy balance
that is going to allow us to calculate the heat that is released upon mixing. We will
draw a flow chart to account for the variables that we just calculated. If we want to calculate
the heat that is released we need an energy balance. In this case our energy balance is
our Q is equal to delta H, which is our change in enthalpy. So we can re-write the change
in enthalpy as the outlet enthalpy minus the inlet. In each case the total enthalpy is
related to the amount of material that we have in each stream. So we can rewrite our
energy balance in terms of the masses for each stream as well as their specific enthalpy.
Were specific enthalpy is on a per unit mass basis. I have assigned a specific enthalpy
variable for each stream, which we can use to rewrite the energy balances. So we have
Q is equal to on the outlet side 3 lb times the specific enthalpy H3, minus 2 lb minus
H1, and 1 lb times H2. So the total inlet enthalpy is the sum of the material flow rates
times the specific enthalpy of each stream on the inlet side. To find values for the
specific enthalpy of the stream we can look at a concentration diagram. Since we know
both the concentration of each stream as well as the temperature. The diagram that I have
shown shows the specific enthalpy of units of BTU/lb of solution as a function of the
mass fraction of H2SO4. So the specific enthalpy is shown as a function of concentration at
a variety of temperature ranging from 32 degF to 200 degF. So we have 3 specific enthalpies
that we need to fine for the 3 streams that are in our process. The first stream is pure
water. So the mass fraction of H2SO4 is equal to 0, and we know the temperature of 100.
So it is somewhere on the 100 degF isotherm. If we trace that back up to x is equal to
0. We can get the specific enthalpy. So it is slightly less then 70 BTU/lb. So I will
use the value of 60, and our 2nd stream mass fraction of H2SO4 was 0.6, and tjhe stream
was also at 100 degF. So we trace the 100 degF isotherm direction to x is equal to 0.6
we can get specific enthalpy. So at x is equal to 0.6 it is slightly above negative 110 BTU/lb.
So I will use a value of -108. For our mixture on the outlet side we know that the mixture
has a mass fraction of H2SO4 of 0.2, which we calculated from the mass balance, because
it is an isothermal mixing process we also know that the temperature is being constant
at 100 degF. So if we follow the 100 degF isotherm to x is equal to 0.2, 20% H2SO4 we
can find the specific enthalpy, and that is represented by this point in the enthalpy
diagram. So we note that the specific enthalpy is between 0 or negative 10. I am going to
use a value of -6 BTU/lb. With the specific enthalpy of each stream known from the diagram.
We can go back to our energy balance to solve for the heat that is lost. Alright going back
to our energy balance we know that the heat that is released is equal to the change in
total energy, and the total enthalpy of each stream is the product of the mass times the
specific enthalpy, which we just determine from the enthalpy diagram. So plugging those
values into this energy balance we can find that the heat that is released is equal to
the negative 46 BTU. So here the negative indicates an exothermic process. So 46 BTU
of heat is being released.