Lecture 2: Design of Reactors for First Order Reactions

Aims of lecture:

To introduce the three main types of chemical reactor
To outline the mathematics of reactor sizing for reactors with first order reactions

Learning Outcomes:

Students who have followed this lecture will be able to:

Estimate the sizes required for batch and continuous reactors controlled by 1st order kinetics.

Designing Reactors

We will restrict ourselves to 1st order reactions to illustrate the principles of reactor design. The same concepts are relevent to more comples mechanisms, but the mathematics is harder!

We will consider the 1st order isomerisation reaction:

A -> B

except where stated otherwise.

As well as having a very simple kinetic equation, for this system we can always say that:

amount of B produced = amount of A consumed

regardless of whether we are using mass or molar units.

Also:

rate of B production = rate of A consumption

in kmol/hr, kg/s or whatever units we wish.

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Types of Process

There is a fundamental difference between two ways of operating any chemical process.

Batch processes are like bench chemistry on a large scale. A reactor is filled with reactant(s) and this/these change over time into product(s). Such a process is easy to visualise, although its mathematics are harder to handle than the continuous process.

Most chemical processes are continuous : reactants are constantly fed to a reactor and products are continuously withdrawn. Such a process is said to operate at steady state since, unlike a batch process, its operating conditions, e.g. the amount of reactant in a reactor, do not change with time. The mathematics of continuous processes is generally simpler than that of batch processes.

Large scale processes, with production rates greater than, say 20,000te/yr, are invariably continuous. Small scale processes, say less than 2000te/yr are usually batch.

There are practical advantages and disadvantages to both approaches.

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Types of Reactor

The reactor in a batch process is normally a tank which is filled with reactants which are allowed to react, often for several hours. The contents are then withdrawn and generally processed to separate or purify the final product. Since batch separations are difficult, the aim of a batch reactor is normally to get a high conversion of reactants into product.

There are two distinct types of reactor in continuous processes. One is called a Continuous Stirred Tank Reactor (CSTR) which is again a tank but to which reactants are continuously fed and from which product is constantly withdrawn. although it looks very like a batch reactor it behaves quite differently.

The other is effectively a pipe through which material constantly flows. This is called a Plug Flow Reactor (PFR). this has quite different characteristics from the CSTR.

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The Batch Reactor

Consider a tank which at an instant in time contains m_a kmols of reactant A.

The reaction has rate constant k in e.g. hr ^-1

The instantaneous rate of reaction in e.g. kmol/hr is given by the kinetic equation:

R_a = - k m_a

This gives the rate of A production R_a , which is negative since A is used up.
As A is used up m_a decreases and so the reaction rate falls.
So R_a is a function of time , R_a(t). We can write formally that at time t=0 :

R_a(0) = - k m_a(0)

More generally at sone time t after the start of the reaction:

R_a(t) = - k m_a(t)

In fact is the rate of change of the amount of A in the reactor which we can write as:

dm_a/dt

Hence the design equation for a 1st order batch reactor:

dm_a/dt = R_a(t) = - k m_a(t)

With initial condition m_a = m_a(0) at t=0.

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Solution

The above is an ordinary differential equation which has a simple analytical solution obtained, e.g. by separating the variables.

Since this form of equation occurs frequently in chemical engineering, it is worth remembering the solution.

For

dy/dt = - k y(t)

With initial condition y = y(0) at t=0:

y(t) = y(0) e^-kt

Hence for:

dm_a/dt = - k m_a(t)

m_a(t) = m_a(0) exp(-kt)

Or:

m_a(t)/m_a(0) = exp(-kt)

This is the fraction of the original A remaining at time which we can define as:

Z_a = amount of A left / amount of A supplied
= exp(-kt)

We define the conversion of A as:

X_a = amount of A consumed / amount of A supplied
= (m_a(0)-m_a(t)) / m_a(0) = 1 - m_a(t)/m_a(0) = 1 - Z_a

Hence:

X_a = 1 - exp (-kt)

m_a(t)/m_a(0) = 1 - Xa

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Behaviour of Batch Reactors

We can see that the initial rate is highest, and it gradually falls off as A is consumed. In fact we can NEVER use up all the A except when exp(-kt)=0 which only occurs at t equal to infinity, which is not operationally possible.

Let if we call the batch time or residence time in the reactor t_r we can tabulate the product (k tr) for all 1st order reactions and so work out the batch time required for a given conversion for any known k.

(k tr)  exp(-ktr)  100 (1 - exp(-ktr)) (%)
0.5      0.607             39.3
1        0.368             63.2
2        0.135             86.5
2.303    0.100             90
3        0.0495            95
4        0.183             98.2
5        0.0067            99.3

To check that you understand this table, try this question.

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Size of a Batch Reactor

If we know how long a batch takes to reach a given conversion, then we can work out how large the reactor must be for a given production or consumption over a period of time.

A process operates 24hr/day with a required production of 48 te/day. The batch time for 90% conversion is 2 hours. How much material will the reactor have to contain for each batch?

Calculation:

We can get 24/2 = 12 batches per day.

We therefore need to produce  48/12 = 4  tonne per batch.

At 90%  (0.9 fractional) conversion we will need 
to fill the reactor with:

4/0.9 = 4.4 tonne/batch.

From the density of the material we can determine reactor volume
required.

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The Continuous stirred tank reactor (CSTR)

A batch reactor is typically a tank with a stirrer. What happens if we take such a tank and arrange to feed reactant(s) at a constant rate and to withdraw a product stream so that the amount of material in the tank stays constant?

Once such a system has settled down to its steady state there will be no change in the composition of the material in the tank either. If the mixer operates effectively then the composition everywhere in the tank will be the same.

Also note that the composition of the material leaving the tank must be exactly the same as that of the material in the tank.

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Rate of production of a CSTR

In general the tank will contain both reactants A and products B. let the molar quamtities of these be m_a and m_b.

The extensive rate of reaction will be equal to:

km_a

and will be equal to the rate of consumption of A or minus the rate of A production:

i.e. it will be constant and depend only on the amount of A in the reactor.

The rate of production of B will be minus the rate of A production:

Hence the production rate in a CSTR depends on the amount of reactant m_a it contains, and not directly on the size of the reactor.

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Effect of Reactor Volume

The reactor volume however is important because it determines the composition of the product mixture.

What is not A is B!

A:B in reactor A:B in product

A high concentration of B is desirable to reduce the cost of downstream separations.

The required size of a reactor can be calculated using:

the required production rate
the required product concentration

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CSTR Sizing Example

The rate constant for a 1st order reaction is 5hr^-1. A production rate of 100 kmol/hr is required. How much reactant must the reactor contain?

Solution:


-Ra = production rate = 100 kmol/hr = k ma = 5 ma
So  ma = 20 kmol

The product stream must contain 60% (molar) of product. What is the total molar holdup in the reactor?


B:A   60:40 in both reactor and product.

So B in reactor = 20 x 60 / 40 = 30 kmol

Total holdup is thus  (20+30) = 50 kmols.

The reactor volume can be determined from knowledge of molecular weight and density for a liquid system, or temperature and pressure for an ideal gas system.

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Reactor Material Balances: 1

Do not attempt to memorise the following equations! All can be written down when required from an understanding of what is going on.

Assumptions:

Reaction is A -> B
Feed is pure A

See diagram

Let -R_a kmol/s be rate at which A is converted by reaction to B
(Remember that R_a is negative!)
Let R_b be rate at which B is produced
Let p_b be rate at which B leaves reactor

R_b = p_b = -R_a = km_a

Let feed rate of pure A to reactor be f_a kmol/s and
Let unreacted A leave reactor at p_a kmol/s

f_a = p_a + R_b = p_a + p_b

Fraction of feed A leaving reactor unconverted:

Z_a = p_a / f_a

Fraction of A converted (`per pass conversion'):

X_a = rate of A conversion / rate of A supply = -R_a/f_a = km_a / f_a
X_a = 1 - (fraction of A not converted) = (1 - Z_a)

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Example - Continued

In the above example the feed to the reactor is pure A.
To produce 100 kmol/hr of B we had calculated that the reactor should contain 20kmol A and 30 kmol of B.
At what rate does unreacted A leave the reactor?

NB: Composition in reactor and of product stream are equal:

m_a / m_b = p_a / p_b

So here: pa/100 = 20/30
So pa = 66.7 kmol/hr

What is the A feed rate to the reactor?

fa = pa + pb = 100 + 66.7 = 166.7 kmol/hr

What is the reactor per pass conversion?


By definition, e.g. :

Xa = pb / fa = 100/166.7 = 0.6
 
Conversion is 0.6 or 60%

Note that if reactor feed is pure A then:

B mol fraction in reactor & product = reactor fractional conversion

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The Plug Flow Reactor (PFR)

Question: Does a molecule of A in a reactor care whether it sits stationary in a tank for a time t_r or whether it drifts downstream in a pipe for the same length of time?

Answer: no it doesn't.

A plug flow reactor (PFR) is another type of continuous reactor which, while being physically totally different, behaves mathematically just like a batch reactor.

The key quantity is the residence time t_r,

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Residence Time Calculation

Let total molar flowrate through reactor be F kmol/hr
Let total molar holdup in reactor be M kmol, then:

M / F = t_r = residence time, hr

As before, a reactor volume may be calculated from the relevant properties of the species at the reactor conditions.

An analogous relationship holds for any consistent flow and holdup units.

For a volumetric flow rate of F_v m3/hr and a reactor volume V m3:

t_r = V / F_v

For a mass flowrate of G kg/s and a mass holdup of W kg:

t_r = W / G

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