ERT 208

REACTION ENGINEERING

MISMISURAYA MEOR AHMAD School of bioprocess engineering

Unimap

CONVERSION & REACTOR SIZING (part 1)

Course Outcome No.3:

•Ability to calculate conversion and sizing for reactors and explain steady-state isothermal reactor.

objectives DEFINE conversion and REWRITE all the balance equation in term of conversion EVALUATE the size of CSTR & PFR based on conversion, DETERMINE the reactor volume necessary to achieved a specified conversion, COMPARE CSTRs & PFRs and overall conversion for reactor arranged in series. IDENTIFY the best arrangement of reactors in series.

How to define conversion

1) Let’s choose one of the reactants as

the basis of calculation

2) Then relate the other species involved in the reaction to this basis.

CONVERSION

The uppercase letters (A, B, C & D) represent chemical species

The lowercase letter (a, b, c & d) represent stoichiometric coefficients

Consider the general reaction:

Taking species A as the basis of calculation.

a "per mole of A" basis limiting reactant.

HOW? Devide the general reaction expression

through by the stoichiometric coefficient of basic calculation of species A, a.

How can we quantify how far a reaction proceeds to the

right?

How many moles of C are formed for every mole A consumed?

The conversion XA is the number of moles of A that have reacted per mole of A fed to the system:

Question:

Answer:

• For irreversible reactions, the maximum

conversion is 1.0 (complete conversion) [Xmax = 1.0]

• For reversible reactions, the maximum conversion is the equilibrium conversion, Xequilibrium [Xmax = Xequilibrium]

DESIGN EQUATIONS OF REACTORS

BASED ON CONVERSION, X

Batch Reactor

• In most batch reactors, the longer a reactant stays in the reactor, the more the reactant is converted to product until either equilibrium is reached or the reactant is exhausted.

Design Equation

• Consequently, in batch systems the conversion

X is a function of the time the reactants spend

in the reactor.

If NAO is the number of moles of A initially in the reactor, then the total number of moles of A that have reacted after time t is:

REWRITE the mole balances in terms of conversion.

The number of moles of A that remain in the reactor after a time t, NA, can be expressed in terms of NAO and X:

The number of moles of A in the reactor after a conversion X has been achieved is:

When no spatial variations in reaction rate exist, the mole balance on species A for a batch system is given by:

Multiply both sides with -1 because A is disappearing:

The rate of disappearance of A, -rA, in this reaction can be given by the rate law, -rA = kCACB

• For batch reactor, we are interested with time = how long the reactants are left in the reactors to achieve a conversion, X.

• Our task now is to get the

length of time, t

1. Write the mole balance equation in terms of conversion, X.

Length of Time, t in Batch Reactor Steps:

2. Then, differentiate with time,

[bear in mind, that NAO is the initial number of moles of A present and it is a constant with time]

4. So that, the design equation for batch reactor in the differential form is:

3. Combine with the mole balance equation on batch reactor :

5. If constant-volume batch reactor, V=Vo

6. To determine the time,t to achieve certain

conversion, rearrange the equation:

7. That equation integrated with the limits

At t= 0, X= 0 (the reaction begins at times equal zero where there is no conversion initially)

And at t= t, we achieve X conversion in batch reactor:

• For continuous flow systems, conversion increases with increasing reactor volume. [Meaning that, the bigger the reactor, the more time it will take the reactants to completely flow through the reactor]

• Therefore, the more time to react.

• Hence, the conversion X is a function of reactor volume V.

Flow Reactors Design Equation

• If FAO is the molar flow rate of species A fed to the system, operating at a steady state, the molar flow rate at which A is reacting within the entire system will be FAOX.

• The molar feed rate of A to the system minus the molar rate at which A is consumed within the system equals the molar flow rate at which A leaves the system.

• Rearranging gives:

• The entering molar flow rate of species A, FAO (mol/s), is just the product of the entering concentration, CAO (mol/dm3) and the entering volumetric rate, :

• If liquid systems, CAO is commonly given in terms of

molarity CAO = 2 mol/dm3

• If gas systems, CAO can be calculated from the

entering temperature and pressure using the ideal gas law or some other gas law.

• Ideal gas:

• So that, the entering molar flow rate in gas system will be:

Example 2.1

We will use this value to size & evaluate a no. of reactor schemes

Flow Reactors

We will discussed 3 types:

1)CSTR

2) Tubular Flow Reactor (PFR)

3) Packed Bed Reactor (PBR)

The CSTR mole balance as applied to species A is:

1) CSTR Design Equation

Concept: Is modeled as being well mixed so that, there are no spatial variation in the reactor.

Consider this reaction

Now substitute FA in terms of FAO and X:

So that:

• Simplifying:

CSTR volume to achieve a specified

conversion

Why

Because CSTR is perfectly mixed, the exit

composition from the reactor is identical to the composition inside the reactor that’s why the rate of reaction evaluate at the exit conditions.

Concept: Model the tubular reactor as having the

fluid flowing in plug Flow no radial gradients in conc., temp. or reaction rate.

2) Tubular Flow Reactor (PFR) Design Equation

As the reactants enter & flow axially down the reactor, they are consumed & conversion increase along the length of the reactor.

To develop the PFR design equation first multiply both sides of the mole balance in PFR by -1.

Mole Balance in PFR

• So, that balance for species A:

• For flow system, FA in term of the entering molar flow rate FAO & the conversion.

Differentiated

• Substitute into previous equation, to get:

• Separate the variable & Integrate with limits

V=0 when X=0, to obtain the PFR volume necessary to achieved specified conversion, X

Concept: PBR is tubular reactor filled with catalyst

particles. The derivation of the differential & integral forms of the design equations for PBR same as PFR.

2) Packed-Bed Reactor (PBR) Design Equation

If absence of pressure drop, we can integrate with limit X = 0 at W = 0

The catalyst weight, W necessary to achieved a conversion, X when total pressure remains constant.

Differential form used when analyzing reactors that have pressure drop along the length of the reactor.

The Design Equations (based on conversion) for:

1. CSTR

2. PFR

Remember:

Question

1) Derive the design equations for CSTR.

2) Derive the design equations for PFR.

Applications of

Design Equations for Continuous-Flow Reaction

Design equation to size CSTRs and PFRs (reactor volume) from knowledge of the rate of reaction, -rA as a function of conversion, X.

The rate of disappearance of A, -rA is almost always a function of the concentrations of the various species present.

When only one reaction occur, each of the conc. can be expressed as a function of the conversion, X (Chapter 3) -rA can be expressed as a function of X.

Example simple function dependence (1st order dependence):

k: specific reaction rate & is a function only of temp. & CAO is the entering conc.

Noted that, the reactor volume (CSTR & PFR) in a function of the reciprocal –rA

-rA = kCA = kCAO(1 - X)

For 1st order dependence, a plot of the the reciprocal rate of reaction (1/ –rA ) as a function of conversion, X:

1/-rA = 1/kCAO (1/1 - X)

Steps to apply design equation (Reactor Sizing)

From experiment (Temp. 500 K (440 °F), total pressure=830 kPa (8.2 atm) & pure A used as reactant), we can determine the rate of chemical reaction as a function of the conversion of reactant A (given in Table below).

X 0.0 0.1 0.2 0.4 0.6 0.7 0.8

-rA 0.45 0.37 0.30 0.195 0.113 0.079 0.05

Consider the isothermal gas-phase isomerization:

Note: If we know –rA as a function of X we can size any

isothermal reaction system.

X 0.0 0.1 0.2 0.4 0.6 0.7 0.8

-rA 0.45 0.37 0.30 0.195 0.113 0.079 0.05

(1/-rA) 2.22 2.70 3.33 5.13 8.85 12.7 20

• Plot 1/-rA vs conversion, X

• In CSTR & PFR design equation, the reactor volume, V varies with the reciprocal of -rA , 1 /-rA. So that, convert –rA to 1/-rA

This figure can used to size flow reactors for different entering molar flow rates

For isothermal reaction: • Rate is greatest or highest at the start of the

reaction when the concentration of reactant is greatest (neglect conversion, X=0). Therefore, 1/-rA is small.

• Near the end of the reaction, the reactants are mostly consumed, concentration of A is small (conversion is large). But, 1/-rA is large.

For all irreversible reaction >zero order: • Complete conversion where all the limiting

reactant is used up, X = 1, the reciprocal rate approaches infinity as does the reactor volume:

We see that an infinity reactor volume is necessary to

obtain the reach complete conversion, X = 1.0

For reversible reaction : • The maximum conversion is the equilibrium

conversion, Xe. At equilibrium, the reaction rate is zero:

We see that an infinity reactor volume is necessary to

obtain the exact equilibrium conversion, X = Xe

• Use FAO = 0.4 mol/s, to add another row. So that:

X 0.0 0.1 0.2 0.4 0.6 0.7 0.8

-rA 0.45 0.37 0.30 0.195 0.113 0.079 0.05

(1/-rA) 2.22 2.70 3.33 5.13 8.85 12.7 20

FAO/-rA 0.89 1.08 1.33 2.05 3.54 5.06 8.0

From this equation, rearrange

• Plotting FAO/-rA vs X using data above:

Levenspiel Plot

Let’s try this:

X 0.0 0.1 0.2 0.4 0.6 0.7 0.8

-rA 0.45 0.37 0.30 0.195 0.113 0.079 0.05

(1/-rA) 2.22 2.70 3.33 5.13 8.85 12.7 20

FAO/-rA 0.89 1.08 1.33 2.05 3.54 5.06 8.0

Sizing a CSTR

using previous tables

Species A enters a CSTR at a molar flow rate of 0.4 mol/s. Calculate the volume necessary to achieve 80% conversion in a CSTR.

Have 2 methods to solve

Example 2.2

Answer 1st method

From table

So that

2nd method: Shade the area for Levenspiel Plot

Levenspiel CSTR Plot

Comment:

The CSTR volume necessary to achieved 80 % conversion is 6.4 m3 when operated at 500 K, 830 kPa & with an entering molar flow rate of A 0.4 mol/s. This volume corresponds to a rector about 1.5 m in diameter & 3.6 m high. It’s a large CSTR but this is a gas –phase reaction & normally not used for gas-phase reaction. CSTR are used primary for liquid-phase reaction.

The reaction is to be carried out in a PFR. The entering molar flow rate is 0.4 mol/s. Determine the PFR reactor volume necessary to achieve 80% conversion.

# Sketch the profiles of –rA and X down length of the reactor

Sizing a PFR

Example 2.3

Answer 1st method

2nd method: Shade the area for Levenspiel Plot

# Sketch the profiles of –rA and X down length of the reactor

• Let’s compare the volumes of CSTR and

PFR based on previous example.

• Let’s learn which reactor require smaller volume to achieve 80% conversion.

• Is it CSTR or PFR????

Comparing CSTR and PFR Sizing

• The data: Use FAO = 0.4 mol/s

X 0.0 0.1 0.2 0.4 0.6 0.7 0.8

-rA 0.45 0.37 0.30 0.195 0.113 0.079 0.05

(1/-rA) 2.22 2.70 3.33 5.13 8.85 12.7 20

FAO/-rA 0.89 1.08 1.33 2.05 3.54 5.06 8.0

•Plot FAO/-rA vs X

• The CSTR volume is 6.4 dm3 while PFR volume is 2.165 dm3.

• If we combine the 2 figures of CSTR and PFR

CSTR PFR

The reason the isothermal CSTR volume is usually greater than the PFR is:

The crosshatched area above the curve is the difference in CSTR and PFR reactor volumes.

The CSTR is always operating at the lowest reaction rate. (i.e. for CSTR: -rA= 0.05 mol/m3)

The PFR starts at a high rate at the entrance & gradually decrease to the exit rate, thereby requiring less volume (the volume is inversely proportional to the rate)

Q &A

THANK YOU