# CHEE 3634 AMDA Process Engineering and Applied Science Questions

Chemical Reaction EngineeringPlug Flow Reactors – Introduction

CHEE 3634 Fall 2020

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Module Overview

PFR Design Equation

PFR Reactors – Course level outcomes

Several course outcomes are associated with this section

Mapping your

problem

Common solution

outcomes

Overview of Section

• Identify key elements in reaction engineering problems and

select appropriate solution methods

• Apply fundamental material balances to derive design

equations for plug flow reaction systems

• Justify and apply appropriate simplifying assumptions

• Apply numerical methods to determine reactor performance

and key design parameters

• Apply Excel (or other tools) to reaction engineering

problems

CHEE 3634 Fall 2020

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Module Overview

PFR Design Equation

Mapping your

problem

Common solution

outcomes

Overview of Section

PFR Design Equations

A reminder: The design equation for a batch reactor can be

written as follows:

𝑑𝑑𝐹𝐹𝑎𝑎

= 𝑟𝑟𝑎𝑎 = 𝑟𝑟𝑎𝑎,1 + 𝑟𝑟𝑎𝑎,2 + ⋯

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

In plug-flow reactor problems, we are often no longer

analyzing things on a time scale… instead, we are solving for

the steady-state “profile” along the length of the reactor

CHEE 3634 Fall 2020

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Module Overview

PFR Design Equation

Mapping your

problem

Common solution

outcomes

Overview of Section

Mapping Your Problem

There are a series of questions you should ask yourself when

considering the analysis of a PFR reactor problem

Many of the questions related to isothermal/non-isothermal

and single/multiple reaction systems don’t change from a

batch-based process… The same applies for feed ratios…

The main additional consideration for a PFR is whether the

fluid is compressible.

CHEE 3634 Fall 2020

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PFR-specific considerations – compressibility

• Is your fluid compressible (i.e. a vapor)?

• If no, the problem is substantially simplified…

volumetric flow rate is treated as a constant.

As you’ll see in this section, we

can estimate pressure drop

from hydrodynamic

• If yes, your volumetric flow rate is a function of

conversion, temperature, and pressure drop

calculations… pipe friction

𝑃𝑃0 𝑇𝑇 𝑁𝑁𝑇𝑇

𝑉𝑉̇ = 𝑉𝑉̇0

𝑃𝑃 𝑇𝑇0 𝑁𝑁𝑇𝑇𝑇

factors for normal flow in a

pipe, or the Ergun equation for

flow through a packed bed.

CHEE 3634 Fall 2020

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Module Overview

PFR Design Equation

Mapping your

problem

Common solution

outcomes

Overview of Section

What are we trying to solve?

The kinetics section looked at interpreting experimental results

to determine the kinetics… Batch reactors looked at a timescale… PFR analysis looks at reactor volume requirements

• Determine volume required to reach a certain conversion

• Determine the optimal conditions for running a reactor to

maximize relative production of one compound over others

(multi-reaction system)

• Determine the physical dimensions a reactor needed to

physically perform the reaction

• Determine reactor heating/cooling requirements

CHEE 3634 Fall 2020

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Reactor volumes or conversions at a certain volume

• Often the most straightforward…

Conversion @ 𝑉𝑉𝑟𝑟𝑟𝑟 , or 𝑉𝑉𝑟𝑟𝑟𝑟 @

conversion are the simplest

problems to solve.

• Start with feed conditions, build your design equation

(and energy balance equation as needed), and solve

from feed conditions until you either reach the desired

conversion (recording the volume) or the desired

volume (recording the conversion)

• For analytical solutions, it’s often easier to solve for a

volume at which you’d reach a certain conversion

(think of the equations you often see… 𝑉𝑉𝑟𝑟𝑟𝑟 is already

isolated on the right hand side of the equations. It is

fairly straightforward to plug in a value for X.

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Optimizing Conditions

• When looking for optimums, you may have to solve a system multiple

times to find the best result.

• How does changes in feed conditions affect the relative reaction rates

for multiple reactions?

• If a secondary reaction consumes your product, how can you minimize

it? Would dilution help? Stopping the reaction suddenly (quenching)?

Or gradual injection of one reactant (multi-port injection system)?

• What is the impact of operating temperature? Heating/cooling? Etc.

• Is your outcome reasonable? Are there other considerations (safety,

materials, etc)? Is your outlet pressure high enough to avoid cavitation or

discharge into your desired exit vessel?

CHEE 3634 Fall 2020

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Sizing a PFR

• PFR sizing is often tied to physical limitations.

For PFR’s, you often have the

volumetric flow rate and outlet

concentration… so total

production capacity is known…

what you’re missing is what is

a physically realistic vessel

that can meet the 𝑉𝑉𝑟𝑟𝑟𝑟 needed.

• Larger pipe diameters have:

• Lower pressure drop, higher likelihood of turbulence

• Lower heat transfer area per unit volume (slower

heat transfer)

• Shorter lengths for equivalent volumes, easier

packing with catalyst

• Increased cost at higher pressures

CHEE 3634 Fall 2020

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Sizing a PFR

For PFR’s, you often have the

volumetric flow rate and outlet

concentration… so total

production capacity is known…

what you’re missing is what is

a physically realistic vessel

that can meet the 𝑉𝑉𝑟𝑟𝑟𝑟 needed.

• Smaller pipe diameters have:

• Higher pressure drop, normally laminar flow (bad for

PFR assumptions)

• High heat transfer areas (good if energy removal is

critical)

• Increased difficulty to pack with catalyst… may rely

on surface coating instead

• Increased potential for plugging if a solids byproduct

is formed

• Longer lengths for equivalent 𝑉𝑉𝑟𝑟𝑟𝑟

CHEE 3634 Fall 2020

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Determining Heating and Cooling

• PFR’s can take many forms…

• A pipe…

• Double-pipe heat exchanger

• Shell and Tube heat exchanger (potentially with the tubes packed with

catalyst)

• Furnace Tube

• A river

• A blood vessel

• A membrane system

CHEE 3634 Fall 2020

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Module Overview

PFR Design Equation

Mapping your

problem

Common solution

outcomes

Overview of Section

Section Overview

The remaining sections will focus on PFR scenarios and

example-based demonstration of how to solve these systems:

• Compressible flowing systems

• Isothermal Single Reaction System

• Compressible and Multi-reaction Systems

• The Energy Balance for Steady State flowing systems

• Non-Isothermal PFR systems

CHEE 3634 Fall 2020

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Chemical Reaction Engineering

Compressible Flowing Systems

CHEE 3634 Fall 2020

1

Module Overview

Compressibility

and C

Predicting

Pressure Drop

Packed Beds

Module Overview

This module explores the basis concepts needed to model

compressible flowing systems. By the end of this module you

should be able to:

• Derive concentration expressions for compressible systems

• Estimate pressure ratios based on flow through a pipe

• Estimate pressure ratios for flow through a packed bed

CHEE 3634 Fall 2020

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Module Overview

Compressibility

and C

Predicting

Pressure Drop

Packed Beds

Compressibility and Concentration

Remembering module 2.6… noting that for a flowing system,

𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

we replace N’s with F’s. i.e. 𝑁𝑁𝐴𝐴𝐴 [𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚] becomes 𝐹𝐹𝐴𝐴𝐴

𝑣𝑣𝑎𝑎

𝑋𝑋𝐹𝐹

𝐹𝐹𝑎𝑎

𝑣𝑣𝑘𝑘 𝑘𝑘𝑘

𝐶𝐶𝑎𝑎 = =

𝑉𝑉̇

̇𝑉𝑉0 𝑃𝑃0 𝑇𝑇 𝐹𝐹𝑇𝑇

𝑃𝑃 𝑇𝑇0 𝐹𝐹𝑇𝑇𝑇

𝐹𝐹𝑎𝑎𝑎 −

𝑣𝑣𝑎𝑎

𝐶𝐶𝑎𝑎𝑎 − 𝑋𝑋𝐶𝐶𝑘𝑘𝑘 𝑃𝑃 𝑇𝑇

𝑣𝑣𝑘𝑘

0

𝐶𝐶𝑎𝑎 =

1 + 𝛿𝛿𝑦𝑦𝑘𝑘𝑘 𝑋𝑋 𝑃𝑃0 𝑇𝑇

CHEE 3634 Fall 2020

𝐿𝐿

𝐹𝐹𝑇𝑇

= 1 + 𝛿𝛿𝑦𝑦𝑘𝑘𝑘 𝑋𝑋

𝐹𝐹𝑇𝑇𝑇

𝐹𝐹𝑎𝑎𝑎

= 𝐶𝐶𝐴𝐴𝐴

̇𝑉𝑉0

𝐹𝐹𝑘𝑘𝑘

= 𝐶𝐶𝑘𝑘0

̇𝑉𝑉0

3

Approaches for Compressible Systems

• When solving for a concentration (for use in your kinetics), you almost

always will be focused on resolving the following ratios

𝑇𝑇

𝑇𝑇0

𝐹𝐹𝑇𝑇

𝐹𝐹𝑇𝑇𝑇

𝑇𝑇

𝑇𝑇

or 0

Determined from the energy balance equation

or 1 + 𝛿𝛿𝑦𝑦𝑘𝑘𝑘 𝑋𝑋

Determined from either adding up all the molar flow

rates of each species, or using the conversion.

𝑃𝑃

𝑃𝑃0

Determined from frictional pressure drop equations.

An Incompressible system for comparison…

• The solution of incompressible PFR systems is almost identical to a batch

system… solved in terms of 𝑉𝑉𝑟𝑟𝑟𝑟 instead of 𝑡𝑡

• For example: Calculate the volume needed to reach 50% conversion for

the liquid-phase reaction A B + C, given that the initial CA0 = 0.0018

mol/L, and that the kinetics second order with respect to A with k = 17.4

L/(mol min). The inlet volumetric flow rate is 10 L/min.

𝑟𝑟𝐴𝐴 = −17.4𝐶𝐶𝐴𝐴2

𝑑𝑑𝐹𝐹𝐴𝐴

= 𝑟𝑟𝐴𝐴

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝐶𝐶𝐴𝐴 = 𝐶𝐶𝐴𝐴𝐴 1 − 𝑋𝑋

𝑑𝑑𝑑𝑑

2

= −17.4𝐶𝐶𝐴𝐴2 = −17.4𝐶𝐶𝐴𝐴0

−𝐹𝐹𝐴𝐴𝐴

1 − 𝑋𝑋 2

𝑑𝑑𝑉𝑉

Batch Reactor Design & Scheduling

𝑚𝑚𝑚𝑚𝑚𝑚

̇

Note that 𝐹𝐹𝐴𝐴𝐴 = 𝐶𝐶𝐴𝐴𝐴 𝑉𝑉0 = 0.0018 10 = 0.018

𝑚𝑚𝑚𝑚𝑚𝑚

𝑑𝑑𝑑𝑑

= −17.4 0.0018 2 1 − 𝑋𝑋 2

−0.018

𝑑𝑑𝑉𝑉

𝑑𝑑𝑑𝑑

= 0.003132 1 − 𝑋𝑋 2

𝑑𝑑𝑉𝑉

𝑑𝑑𝑑𝑑

= 0.003132𝑑𝑑𝑉𝑉

2

1 − 𝑋𝑋

1

1

−

= 0.003132𝑉𝑉𝑟𝑟𝑟𝑟

1 − 0.5

1− 0

2 − 1 = 0.003132𝑉𝑉𝑅𝑅𝑅𝑅

𝑉𝑉𝑟𝑟𝑟𝑟 = 319 𝐿𝐿

An Incompressible system for comparison…

• For the same basic problem, but a compressible system…

• Calculate the volume needed to reach 50% conversion for the gas-phase

reaction A B + C, given the initial CA0 = 10 mol/L and CI0 = 90 mol/L ,

and that the kinetics second order with respect to A with k = 0.174 L/(mol

min). The inlet volumetric flow rate is 10 L/min, the inlet pressure is 𝑃𝑃0 ,

and the system operates isothermally at 300K.

𝑟𝑟𝐴𝐴 = −0.174𝐶𝐶𝐴𝐴2

𝑑𝑑𝐹𝐹𝐴𝐴

= 𝑟𝑟𝐴𝐴

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

2

𝑑𝑑𝑑𝑑

100

1

−

𝑋𝑋

−𝐹𝐹𝐴𝐴𝐴

= −0.174𝐶𝐶𝐴𝐴2 = −0.174

1 + 0.1𝑋𝑋 2

𝑑𝑑𝑉𝑉

𝑃𝑃

𝑃𝑃0

2

𝑣𝑣𝑎𝑎

𝑋𝑋𝐶𝐶

𝑣𝑣𝑘𝑘 𝑘𝑘𝑘 𝑃𝑃 𝑇𝑇0

𝐶𝐶𝑎𝑎 =

1 + 𝛿𝛿𝑦𝑦𝑘𝑘𝑘 𝑋𝑋 𝑃𝑃0 𝑇𝑇

10(1 − 𝑋𝑋) 𝑃𝑃

𝐶𝐶𝑎𝑎 =

1 + 0.1𝑋𝑋 𝑃𝑃0

𝐶𝐶𝑎𝑎𝑎 −

Module Overview

Compressibility

and C

Predicting

Pressure Drop

Packed Beds

Solving for 𝑷𝑷/𝑷𝑷𝟎𝟎

Two examples are provided on how to solve for the pressure

ratio… the same method can be applied to almost any

system, as long as you have reasonable methods of

estimating pressure drop…

In most cases, we have correlations for pressure drop…

𝑃𝑃

𝑃𝑃0

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝑑𝑑

𝑑𝑑

𝑃𝑃

𝑃𝑃0

𝑑𝑑𝑑𝑑

𝑑𝑑

𝑃𝑃

𝑃𝑃0

𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

𝑑𝑑𝑑𝑑 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟 = 𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑𝑑𝑑 =

=

𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

𝑚𝑚3 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

𝜌𝜌𝑠𝑠,𝑏𝑏𝑏𝑏𝑏𝑏

𝜖𝜖𝑠𝑠 3

𝜌𝜌

𝑚𝑚3 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

𝑚𝑚 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑠𝑠,𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑚𝑚3 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

CHEE 3634 Fall 2020

8

Pressure drop in a Pipe

• The simplest example is pressure drop in a pipe using the DarcyWeisbach equation, where you’d have a good correlation available for the

Darcy friction factor, 𝑓𝑓𝑅𝑅𝑅𝑅 .

𝜌𝜌𝑣𝑣 2 Δ𝐿𝐿

Δ𝑃𝑃 = −

𝑓𝑓𝑅𝑅𝑅𝑅

2 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

• Expressing pressure and length in differential format, you get:

𝜌𝜌𝑣𝑣 2 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 = −

𝑓𝑓𝑅𝑅𝑅𝑅

2 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑑𝑑𝑑𝑑

𝜌𝜌𝑣𝑣 2 1

=−

𝑓𝑓𝑅𝑅𝑅𝑅

𝑑𝑑𝑑𝑑

2 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

Pressure drop in a Pipe

• We can express this in many different ways, using the relationship

between velocity and volumetric flow rate, 𝑉𝑉̇ = 𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑣𝑣,

⃑ and the

relationship between Length and volume for a pipe of constant diameter,

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝑑𝑑𝑑𝑑 =

=

2

𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

0.25𝜋𝜋𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑑𝑑𝑑𝑑

𝜌𝜌𝑣𝑣 2 1

𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

=−

𝑓𝑓𝑅𝑅𝑅𝑅

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

2 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑑𝑑𝑑𝑑

𝜌𝜌𝑉𝑉̇ 2

1

𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

=− 2

𝑓𝑓𝑅𝑅𝑅𝑅

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

2𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑑𝑑𝑑𝑑

𝜌𝜌𝑉𝑉̇ 2

1

32𝜌𝜌𝑉𝑉̇ 2

=− 3

𝑓𝑓𝑅𝑅𝑅𝑅 = − 3 7 𝑓𝑓𝑅𝑅𝑅𝑅

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝜋𝜋 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

2𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

Solving for 𝑷𝑷/𝑷𝑷𝟎𝟎

• Since 𝑃𝑃0 is a constant (feed condition), we can divide both sides by 𝑃𝑃0 to

get a differential expression for 𝑃𝑃/𝑃𝑃0

𝑑𝑑

𝑃𝑃

𝜌𝜌𝑣𝑣 2 1

𝑃𝑃0

=−

𝑓𝑓𝑅𝑅𝑅𝑅

𝑑𝑑𝑑𝑑

2𝑃𝑃0 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑃𝑃

𝑑𝑑

32𝜌𝜌𝑉𝑉̇ 2

𝑃𝑃0

=−

𝑓𝑓𝑅𝑅𝑅𝑅

7

3

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝑃𝑃0 𝜋𝜋 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

• Note that you would solve all of these using appropriate units (m3/s, Pa,

m, etc.). If you are working in imperial units, you’d also have a gravity

constant showing up.

For example, what if the flow was laminar?

64𝜇𝜇 0.25𝜋𝜋𝑑𝑑 2

64

64𝜇𝜇

• Bad for a PFR assumption, but what if 𝑓𝑓𝑅𝑅𝑅𝑅 = =

=

̇

𝑅𝑅𝑅𝑅

𝜌𝜌𝑣𝑣𝑑𝑑

𝜌𝜌𝑉𝑉𝑑𝑑

𝑃𝑃

32𝜌𝜌𝑉𝑉̇ 2

32𝜌𝜌𝑉𝑉̇ 2 16𝜇𝜇𝜇𝜇𝜇𝜇

𝑃𝑃0

=−

𝑓𝑓𝑅𝑅𝑅𝑅 = −

7

7

3

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝑃𝑃0 𝜋𝜋 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑃𝑃0 𝜋𝜋 3 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝜌𝜌𝑉𝑉̇

𝑃𝑃

𝑑𝑑

̇

512𝑉𝑉𝜇𝜇

512𝜇𝜇

𝑃𝑃0

̇

=−

=

−

𝑉𝑉

6

6

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝑃𝑃0 𝜋𝜋 2 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑃𝑃0 𝜋𝜋 2 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑑𝑑

• Remembering that we have an expression for 𝑉𝑉̇

𝑃𝑃

512𝜇𝜇

𝑃𝑃0 𝑇𝑇

𝑃𝑃0

=−

𝑉𝑉0̇

1 + 𝛿𝛿𝑦𝑦𝑘𝑘𝑘 𝑋𝑋

6

2

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝑃𝑃 𝑇𝑇0

𝑃𝑃0 𝜋𝜋 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑑𝑑

=

16𝜇𝜇𝜋𝜋𝜋𝜋

𝜌𝜌𝑉𝑉̇

For example, what if it was turbulent?

• What if turbulent flow in a smooth pipe where 𝑓𝑓𝑅𝑅𝑅𝑅 = 0.0012

𝑃𝑃

32𝜌𝜌𝑉𝑉̇ 2

32𝜌𝜌𝑉𝑉̇ 2

𝑃𝑃0

=−

𝑓𝑓𝑅𝑅𝑅𝑅 = −

0.0012

7

7

3

3

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝑃𝑃0 𝜋𝜋 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑃𝑃0 𝜋𝜋 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑃𝑃

𝑑𝑑

0.0384𝜌𝜌 2

𝑃𝑃0

̇

=−

𝑉𝑉

7

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝑃𝑃0 𝜋𝜋 3 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑑𝑑

𝑃𝑃

2

2

0.0384𝜌𝜌

𝑃𝑃

𝑇𝑇

𝑃𝑃0

2

0

2

̇

=−

𝑉𝑉

1

+

𝛿𝛿𝑦𝑦

𝑋𝑋

0

𝑘𝑘𝑘

7

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝑃𝑃

𝑇𝑇0

𝑃𝑃0 𝜋𝜋 3 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑑𝑑

But WAIT! Density changes as well!

• In all of these expressions, one must recognize that the gas density, ρ,

will change along the reactor length

• Recognizing that mass flow rate is constant, ρ0V0= ρV

• And that

𝑃𝑃0 𝑇𝑇

𝑉𝑉̇ = 𝑉𝑉̇0 1 + 𝛿𝛿𝑦𝑦𝑘𝑘𝑘 𝑋𝑋𝑘𝑘

𝑃𝑃 𝑇𝑇0

𝜌𝜌0 𝑉𝑉̇0

𝜌𝜌0

𝑃𝑃 𝑇𝑇0

𝜌𝜌 =

=

1 + 𝛿𝛿𝑦𝑦𝑘𝑘𝑘 𝑋𝑋𝑘𝑘 𝑃𝑃0 𝑇𝑇

𝑉𝑉̇

For example, what if it was turbulent?

• So for the turbulent flow case example… when we account for 𝜌𝜌 changes

𝑃𝑃

2

2

0.0384𝝆𝝆 2 𝑃𝑃0

𝑇𝑇

𝑃𝑃0

2

̇

=−

𝑉𝑉

1

+

𝛿𝛿𝑦𝑦

𝑋𝑋

0

𝑘𝑘𝑘

7

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝑃𝑃

𝑇𝑇0

𝑃𝑃0 𝜋𝜋 3 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑑𝑑

𝑃𝑃

2

2

0.0384

𝑇𝑇

𝜌𝜌0

𝑃𝑃 𝑇𝑇0

𝑃𝑃0

2 𝑃𝑃0

2

̇

=−

𝑉𝑉0

1 + 𝛿𝛿𝑦𝑦𝑘𝑘𝑘 𝑋𝑋

7

3

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝑃𝑃

1 + 𝛿𝛿𝑦𝑦𝑘𝑘𝑘 𝑋𝑋𝑘𝑘 𝑃𝑃0 𝑇𝑇

𝑇𝑇0

𝑃𝑃0 𝜋𝜋 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑃𝑃

𝑑𝑑

0.0384𝜌𝜌0 2 𝑃𝑃0

𝑇𝑇

𝑃𝑃0

̇

=−

𝑉𝑉0

1 + 𝛿𝛿𝑦𝑦𝑘𝑘𝑘 𝑋𝑋

7

3

𝑇𝑇0

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝑃𝑃

𝑃𝑃0 𝜋𝜋 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑑𝑑

Module Overview

Compressibility

and C

Predicting

Pressure Drop

Packed Bed Reactors

Packed Bed reactors are a special case of PFR. They are

commonly found in catalytic processes, and pressure drop

through these systems is determined through the Ergun

equation

Packed Beds

The presence of solids has 2 effects:

1) They increase pressure drop (more wetted surface, and so

the equivalent hydraulic diameter drops)

2) The solids take up reactor volume…. Which needs to be

accounted for.

CHEE 3634 Fall 2020

16

Compressible Flow, Packed Bed

• For a packed bed, we use the Ergun Equation:

1 − 𝜖𝜖𝑔𝑔

𝑑𝑑𝑑𝑑

𝐺𝐺

=−

𝑑𝑑𝑑𝑑

𝝆𝝆𝑔𝑔𝑐𝑐 𝐷𝐷𝑝𝑝

𝜖𝜖𝑔𝑔3

150 1 − 𝜖𝜖𝑔𝑔 𝝁𝝁

+ 1.75𝐺𝐺

𝐷𝐷𝑝𝑝

Where G is the mass flux: 𝐺𝐺 = 𝜌𝜌𝑣𝑣⃑ [density * superficial velocity] [kg / m2 s], or 𝐺𝐺 = 𝑚𝑚/𝐴𝐴

̇ 𝑐𝑐

ρ is the density of the fluid flowing through the packed bed (i.e. gas) [kg/m3]

gc is the gravitational conversion constant (= 1 for SI, = 32.174 ft/s2)

Dp is the diameter of the solid particles in the bed [m]

μ is the dynamic viscosity of the fluid (kg / m s) (May be a function of temperature, or conversion)

𝜖𝜖𝑔𝑔 is the void or gas fraction: 𝜖𝜖𝑔𝑔 = m3 void (or gas) / m3 reactor

Note! The velocity calculated for determining G is what the velocity would be if there were no solids

present in the reactor (i.e. use the full cross-section of the pipe)

Compressible Flow, Packed Bed

• Dividing both sides by P0, and substituting in the expression for the

variable density, the Ergun equation thus reduces to:

𝑑𝑑

𝑃𝑃

1 − 𝜖𝜖𝑔𝑔

𝐺𝐺

𝑃𝑃0

=−

𝑑𝑑𝑑𝑑

𝑃𝑃0 𝜌𝜌0 𝑔𝑔𝑐𝑐 𝐷𝐷𝑝𝑝 𝜖𝜖𝑔𝑔 3

Defining 𝑎𝑎 =

150 1 − 𝜖𝜖𝑔𝑔 𝜇𝜇

+ 1.75𝐺𝐺

𝐷𝐷𝑝𝑝

1−𝜖𝜖𝑔𝑔

𝐺𝐺

𝑃𝑃0 𝜌𝜌0 𝑔𝑔𝑐𝑐 𝐷𝐷𝑝𝑝 𝜖𝜖𝑔𝑔 3

150 1−𝜖𝜖𝑔𝑔 𝜇𝜇

𝐷𝐷𝑝𝑝

𝑃𝑃

𝑑𝑑

𝑇𝑇 𝑃𝑃0

𝑃𝑃0

= −𝑎𝑎

1 + 𝛿𝛿𝑦𝑦𝑘𝑘𝑘 𝑋𝑋𝑘𝑘

𝑑𝑑𝑑𝑑

𝑇𝑇0 𝑃𝑃

𝑇𝑇 𝑃𝑃0

1 + 𝛿𝛿𝑦𝑦𝑘𝑘𝑘 𝑋𝑋𝑘𝑘

𝑇𝑇0 𝑃𝑃

+ 1.75𝐺𝐺

Compressible Flow, Packed Bed

• Special Case1: Near-constant density (𝜌𝜌 = 𝜌𝜌0 ), Isothermal

𝑃𝑃

𝑑𝑑

1 − 𝜖𝜖𝑔𝑔

𝐺𝐺

𝑃𝑃0

=−

𝑑𝑑𝑑𝑑

𝜌𝜌0 𝑃𝑃0 𝑔𝑔𝑐𝑐 𝐷𝐷𝑝𝑝 𝜖𝜖𝑔𝑔 3

𝑃𝑃/𝑃𝑃0

�

1

150 1 − 𝜖𝜖𝑔𝑔 𝜇𝜇

+ 1.75𝐺𝐺

𝐷𝐷𝑝𝑝

𝐿𝐿

𝑃𝑃

𝑑𝑑

= −𝑎𝑎 � 𝑑𝑑𝑑𝑑

𝑃𝑃0

0

𝑃𝑃

= 1 − 𝑎𝑎𝑎𝑎

𝑃𝑃0

Compressible Flow, Packed Bed

• Special Case 2: Variable Density, Isothermal, δyk0 ~ 0

𝑃𝑃

𝑑𝑑

𝑇𝑇 𝑃𝑃0

𝑃𝑃0

𝑃𝑃0

= −𝑎𝑎

1 + 𝛿𝛿𝑦𝑦𝑘𝑘𝑘 𝑋𝑋𝑘𝑘 = −𝑎𝑎

𝑑𝑑𝑑𝑑

𝑃𝑃

𝑇𝑇0 𝑃𝑃

𝑃𝑃/𝑃𝑃0

𝐿𝐿

𝑃𝑃

𝑃𝑃

�

𝑑𝑑

= −𝑎𝑎 � 𝑑𝑑𝑑𝑑

𝑃𝑃0

𝑃𝑃0

1

0

2

𝑃𝑃

− 0.5 = −𝑎𝑎𝑎𝑎

0.5

𝑃𝑃0

𝑃𝑃

= 1 − 2𝑎𝑎𝑎𝑎

𝑃𝑃0

Example: Liquid flowing through a packed tube

• Consider 1 kg/min of water flowing through a 0.1m internal diameter pipe

𝑘𝑘𝑘𝑘

packed with 2mm particles where 𝜌𝜌𝑐𝑐 = 1200 3 . The water enters at 𝑇𝑇0 =

𝑚𝑚

𝑘𝑘𝑘𝑘

298𝐾𝐾 ( 𝜌𝜌𝑤𝑤 = 998 3 , 𝜇𝜇 = 0.001 𝑃𝑃𝑃𝑃 � 𝑠𝑠). Assuming the temperature does

𝑚𝑚

not change significantly, what inlet pressure would be required to have an

outlet pressure of 1 atm after 3 meters of pipe. Note that the solids

holdup in the pipe is ~60% (i.e. 𝜖𝜖𝑆𝑆 = 0.6, 𝜖𝜖𝑔𝑔 = 1 − 𝜖𝜖𝑆𝑆 = 0.4).

• Note that 𝐺𝐺 =

1/60

𝑘𝑘𝑘𝑘

𝑚𝑚̇

=

=

2.122

2

0.25𝜋𝜋𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

𝑚𝑚2 𝑠𝑠

𝐴𝐴𝑐𝑐

CHEE 3634 Fall 2020

21

Example: Liquid flowing through a packed tube

1 − 𝜖𝜖𝑔𝑔 150 1 − 𝜖𝜖𝑔𝑔 𝝁𝝁

𝑑𝑑𝑑𝑑

𝐺𝐺

=−

+ 1.75𝐺𝐺

3

𝑑𝑑𝑑𝑑

𝝆𝝆𝑔𝑔𝑐𝑐 𝐷𝐷𝑝𝑝

𝐷𝐷𝑝𝑝

𝜖𝜖𝑔𝑔

𝑑𝑑𝑑𝑑

2.122

1 − 0.4 150 1 − 0.4 0.001 𝑃𝑃𝑃𝑃 𝑠𝑠

=−

+ 1.75 2.122

3

𝑑𝑑𝑑𝑑

998.2 0.002

0.002 𝑚𝑚

0.4

𝑑𝑑𝑑𝑑

𝑃𝑃𝑃𝑃

= −485.4

𝑑𝑑𝑑𝑑

𝑚𝑚

𝑃𝑃

𝑃𝑃𝑃𝑃 3

� 𝑑𝑑𝐿𝐿

� 𝑑𝑑𝑑𝑑 = −485.4

𝑚𝑚

𝑃𝑃0

0

𝑃𝑃 − 𝑃𝑃0 = −485.4(3)

𝑃𝑃0 = 𝑃𝑃 + 1456.2𝑃𝑃𝑃𝑃 = 101325 + 1456.2 = 102781.3 𝑃𝑃𝑃𝑃

• The inlet pressure required to meet a target outlet pressure of 1 atm at

the specified conditions is 102781.3 Pa (absolute), or 1456.2 Pa (gauge)

CHEE 3634 Fall 2020

22

Compressible Flow, Packed Bed

• While expressing the pressure drop as a function of reactor length is

useful, we may need for other basis:

𝑑𝑑𝑑𝑑 =

Where

𝑑𝑑𝑉𝑉𝑟𝑟𝑟𝑟

𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

=

𝑑𝑑𝑑𝑑

1 − 𝜖𝜖𝑔𝑔 𝜌𝜌𝑐𝑐 𝐴𝐴𝑐𝑐

𝜌𝜌𝑐𝑐 is the density of the catalyst particles [kg solid/m3 particle volume]

Note the shortening of 𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑡𝑡 𝐴𝐴𝑐𝑐

For Example:

1 − 𝜖𝜖𝑔𝑔

𝑑𝑑𝑑𝑑

𝐺𝐺

=−

𝑑𝑑𝑑𝑑

𝜖𝜖𝑔𝑔3

1 − 𝜖𝜖𝑔𝑔 𝜌𝜌𝑐𝑐 𝐴𝐴𝑐𝑐 𝝆𝝆𝑔𝑔𝑐𝑐 𝐷𝐷𝑝𝑝

150 1 − 𝜖𝜖𝑔𝑔 𝜇𝜇

+ 1.75𝐺𝐺

𝐷𝐷𝑝𝑝

Compressible Flow, Packed Bed

• General Case: you are typically solving numerically… note that both

density and viscosity could change depending on the system.

• Reaction kinetics often also result in design equations that cannot be

resolved analytically…

• Note that when you get to reaction kinetics for catalyst systems, often you

will have a rate equation with units mol/(g cat * time). When expressed in

this format, the rate equation is often given as:

𝑑𝑑𝐹𝐹𝐴𝐴

= 𝑟𝑟𝐴𝐴

𝑑𝑑𝑑𝑑