In this screencast I'm going to solve the mass and energy balance for an adiabatic batch
reactor. It's a liquid-phase reaction. We're given the initial temperature as 100 degrees
C and of course constant volume. We start out with just A in the reactor and then we
have things like heat of reaction, heat capacity, reaction rate in terms of concentration and
the rate constant. Of course the rate constant has an exponential temperature dependence.
The two questions we want to answer are what is the reactor temperature when 60% of A is reacted,
and then how long does it take to reach a conversion of 60%?
So first I'm going to write the mass
balance for a batch reactor and I'll do the balance on A. So if we have the change in the total
number of moles of A, N_A is total number of moles, with respect to time is equal to the
rate of reaction of A times the volume. Now we can simplify this because it's liquid phase
and constant volume. So the number of moles of A is the concentration of A times the volume.
The volume is constant so I can rewrite this in terms of concentration. I'm going to substitute
in the rate expression here. So we see that the volume cancels out. We have the mass balance.
We can likewise write the energy balance for this system. The energy balance is the number
of moles of A, heat capacity of A plus the number of moles of B, heat capacity of B, because
the number of moles of A and B are changing with time. Heat capacities are constant.
This is equal to the rate of reaction of A times the volume, so this is number of moles per
time, times the heat of reaction per mole. Now I'm going to simplify this because the
heat capacities of A and B are the same. So I can write this as the heat capacity of A.
The right side I have the rate constant, the concentration, the volume and the heat of
reaction. N_A plus N_B is just equal to the number of moles of A initially. The reactor
did not contain any B so I can write this as N_A0, the heat capacity of A and then this
change in temperature with time. The right side hasn't changed. Now the number of moles
of A is the volume of A times the initial concentration of A which means that the volume
is going to drop out of this equation also. The system doesn't depend on volume so we
cancel out the volume. So I can solve these two equation simultaneously and numerically.
I want to solve them numerically because k, which appears in both equations, is an exponential
function of temperature. However, first if I just want to determine the adiabatic temperatures,
I can combine the two equations. What I can do is substitute into these equations and
I can substitute for this term here. So I can substitute d C_A with respect to time
and I'll get a linear relationship between concentration change and temperature change.
Now I can basically cancel the dT and integrate and when I do that C_A0 and heat capacity
are constant so T - T_0 equals heat of reaction, C_A at any given temperature, C_A0. I know
everything in this equation for 60% conversion. So 60% conversion means that C_A is 40% unreacted.
So I can make this substitution and I can solve for T. So I've made the substitution.
I've kept C_A0 in here because you can see this also cancels. I can solve for the temperature
at 60% conversion. This temperature is 478 Kelvin, or the temperature is 205 degrees C. So
the adiabatic temperature of course is independent of the type of reactor that we're looking at.
If we want to get the time however we're now going to have to go back and solve these
differential equations simultaneously and the best way to do that is to solve numerically.
It's an initial value problem. At time equals 0 we know the concentration. We know the temperature.
The program will then calculate the temperature, the concentration, conversion at any later
time. Here are the equations that will substitute into a program that solves ordinary differential
equations. These are in the format for POLYMATH. We have the mass balance, the energy balance,
our initial time and we just guessed the final time and then ran it a few times to get a
value close to 60% that we could then read off the exact value. The rate constant, it's
important to remember that the temperature here is the absolute temperature so therefore
the initial temperature, 100 degrees C, is 373 K. Then we calculate conversion from the
concentrations. Because it's a constant volume reaction we can do this, and what we find is that
the time is 3,904 seconds or about 65 minutes to reach a conversion of 60%, which corresponds then
to a concentration of 2 mol/L and the temperature of 478 as we also determined. This is a plot
of conversion on the y-axis. So this is conversion verses time, showing that we get to 60% conversion
here and it shows how conversion changes as a function of time. So this is output from
the POLYMATH program when we solve these ordinary differential equations.