# 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
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.
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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.
(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?
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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
• Increased potential for plugging if a solids byproduct
is formed
• Longer lengths for equivalent 𝑉𝑉𝑟𝑟𝑟𝑟
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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
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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
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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

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