INSTRUCTOR: Here we're gonna explain the terms
and energy balance for a plug flow reactor.
I'm not gonna derive the equations,
but just explain them.
We'll do this and then we'll look at
for case if we have reaction a plus b goes to c
and the rate of reaction is first order n a.
So tubular reactor, so this figure is supposed
to represent a cross section of tubular reactor
where we use notion to distance equal zero
and then z equals length of reactor
or v equals zero in a cumulative volume
so v is the total value of the reactor here.
And the way we're going to do the balance is look
at a slice, differential slice.
So thickness d z
and so we have so flow rate of a, b, and c
at temperature t entering that slice
and then we have reaction in that slice
and that's where we'll look at the energy pounds.
Also let's label a couple of other things.
D is diameter of the tube
and t c is the temperature surrounding the tube.
For example it might be in a heat exchanger
and the temperature t c might be constant.
So a couple of things.
These flow rates are moles per time.
For example moles per hour.
So if I now write the energy balance.
And I'm gonna sum the molar flow rates
times molar heat capacity and then d t is the change,
the differential change in temperature
in this differential volume we're looking at.
And temperature change is because we have a reaction,
heat of reactions where this is
for example kilojoules per mole of reactant a.
So per mole of a that reacts and therefore I write rate
of reaction of a.
And then I want the volume of this small disk,
differential disk.
So pie d squared over four is cross sectional area,
times d z.
And then heat is removed by heat transfer,
so it is a heat transfer coefficient.
Pie, diameter, d z t in the reactor, minus t c.
So note this is the heat transfer area.
This is the heat transfer coefficient.
And this cross sectional area,
let's call that a x.
So we're gonna divide through,
divide the equation through by a x times d z.
So let me pause and do that division.
I also made the substitution of rate of reaction.
Notice that these cancel and one of the ds cancel.
Now we have an energy balance.
If I write it for the reaction a plus b goes to c.
Let me pause and do that.
So here's the equation for energy balance.
And we're going to solve this simultaneously
with mass balance.
This is when we would write in general mass balance
for each component.
So for example for the mass balance for a would be,
rate of reaction of a.
Also note that this could be written
as d t, d volume
and likewise this could be the change in flow rate of a
with cumulative volume down to reactor.
Now to solve these equations we need initial conditions.
And initial conditions mean
that the reactor inlets.
So that z equals zero,
we'd have temperature come into reactor t zero.
We'd have the flow rate of each of the components
as the inlet flow rate.
So flow rate of a, a zero.
Flow rate b, b zero
and in this case flow rate c would be zero.
So then with initial conditions
we can solve these equations numerically.
And we typically wanna do it numerically
because this term
is gonna have an exponential dependence on temperature.
And therefore that's the only practical way
to solve these equations.