Study Unit 1.1: Mole balances
Fogler (Chapter 1)
SU 1: Fogler, Ch. 1-4, more or less introductory (setting the scene, learning the jargon)
1.1 Ch 1: Mole balances
1.2 Ch 2: Conversion and Reactor Sizing
1.3 Ch 3: Rate Laws
1.4 Ch 4: Stoichiometry
SU 2: Fogler, Ch. 5-7, reactor design and analysis; the chemical engineer's bread and
butter (still introductory …..)
2.1 Ch 5: Isothermal Reactor Design: Conversion
2.2 Ch 6: Isothermal Reactor Design: Moles and Molar Flow rates
2.3 Ch 7: Collection and Analysis of Rate Data
SU 3 + P: Fogler, Ch. 8 and 11, and Project, somewhat more to practice … (still introductory …..)
3.1 Ch 8: Multiple Reactions
3.2 Ch 11:Nonisothermal Reactor Design - The Steady State Energy Balance and
Adiabatic PFR Applications
1.1. Study Unit 1.1
1.1.1 Further background to reactors (SA context)
1.1.2 Reaction rate
1.1.3 The general mole balance
Reactor classifications:
1.1.4 The Batch Reactor (BR)
1.1.5 The Continuous Stirred Tank Reactor (CSTR)
1.1.6 The Plug Flow Reactor (PFR) and the Packed
Bed Reactor (PBR)
1.1.7 Reactors in practice (self study)
1.1.1 Further background to reactors (SA context)
The Aim of a chemical process, is often to
produce a value increased product
In South Africa, the largest chemical engineering activity is the
conversion of coal to fuels and chemicals (Secunda)
Coal (~R 200/ton)
Petrol (~R 10 000/ton)
About 40 million ton of coal is converted each year at Sasol Secunda
Multiple reactive and
separation processes
C + H2O = CO + H2
Coal
H2 O
O2
n CO + (2n+1) H2 =
CnH2n+2 + n H2O
Chemicals
Coal
CO Fischer Tropsch
Gasification H2 Synthesis
Product
Upgrading
Fuel
Picture taken from http://www.google.co.za/imgres?imgurl=http://www.netl.doe.gov/technologies/coalpower/gasification/gasifipedia/images/
Coal
Coal
CO
Gasification H2
H2 O
O2
Entrained flow reactor
(dp ~ 50 Pm)
The conversion of coal into syngas
is carried out in different form of
reactors, all over the world.
The choice of reactor is often related
to the type of coal used.
Fluidised bed reactor
(dp ~ 5 mm)
Fixed bed reactor
(dp ~ 50 mm)
Simplified Process Flow Diagram (PFD) of the
HydroDeAlkylation (HDA) of toluene:
Heart of the process
1.1.2 Reaction rate
Simplified Process Flow Diagram (PFD) of the
HydroDeAlkylation (HDA) of toluene:
Heart of the process
C7H8 + H2 Æ C6H6 + CH4:
One of the important factors in reactor design (and sizing) is the
rate of reaction (how fast a reaction proceeds)
The rate of reaction (- rA) is the amount of matter that reacts per unit
time per unit volume; and can be expressed on various ways:
Mass based:
kg of Toluene that has reacted per hour per Liter [kg/hr L]
pounds Toluene that has reacted per minute per gallon [lb/min gal]
Mol based (preferred)
kmol Toluene that has reacted per hour per cubic feet [kmol/hr ft3]
mol Toluene that has reacted per second per m3 [mol/s m3]
(Standard SI unit)
Read also § 1.1
Exercise:
Consider that 100 kton/yr of toluene is converted in the HDA
process, and that on average the process operates 8000 h/yr.
a) Estimate the reactor volume if the average reaction rate of the
conversion of toluene equals 1.2 mol/m3 s.
b) Suggest a shape and dimensions of the reactor
2 C6H6 Æ C12H10 + H2:
The production of Diphenyl (C12H10) from Benzene (C6H6) can
schematically be given as:
2AÆB+C
(A = Benzene, B = Diphenyl, C = Hydrogen)
Convention: if a species disappears, the rate is negative, and if the
species is formed, the rate is positive
rA = - 10 mol/m3 s, or - rA = 10 mol/m3 s
means 10 mol of Benzene reacts per m3 per second
Formation and disappearance rates are linked via stoichiometry
rA = - 10 mol/m3 s Æ rB = 5 mol/m3 s, rC = 5 mol/m3 s
Read also § 1.1
rj: rate of reaction
Definition of rj: "the rate of formation of species j per unit volume"
Convention: rj is negative for species that react (disappear) and positive for
species that are produced (formed)
The rate equation (i.e., rate law) for rj is an algebraic equation that is solely
a function of the properties of the reacting materials and reaction
conditions (e.g., species concentration, temperature, pressure, or type of
catalyst, if any) at a point in the system.
The rate equation is independent of the type of reactor (e.g., batch or
continuous flow) in which the reaction is carried out (intrinsic kinetics).
For the reaction A Æ B, the rate of reaction may be: (examples)
(1st order reaction, (in A)) െ‫ݎ‬஺ = ݇‫ܥ‬஺
(2nd order reaction, (in A))
െ‫ݎ‬஺ = ݇‫ܥ‬஺ଶ
Where k is normally a function
of temperature (SU 1.3)
More complex forms: e.g.:
െ‫ݎ‬஺ =
݇ଵ ‫ܥ‬஺
1 + ݇ଶ ‫ܥ‬஺
Read also § 1.1, p7
1.1.3 The general mole balance
Example: A Æ B,
A
FA =100 mol/s
െ‫ݎ‬஺ = ݇‫ܥ‬஺ ,
k = f (T)
rA = f (CA, T)
High CA Æ
high rA
mixture of
low CA Æ
low rA
A/B
FA =25 mol/s
FB = 75 mol/s
Picasso reactor (Fig. 1-1)
In the reactor, the concentration of A (and temperature) is not everywhere the
same, so the rate of reaction also varies on the local position in the reactor
So the overall conversion of A ((FA,0 – FA)/FA0) is an integration of the local
reaction rates in the reactor
So the overall conversion of A, is dependent on both
reaction rate equation (rA) AND reactor configuration
In a nutshell, this is where reactor theory and design is about
Flash-back, slide from CEMI121 on mass balances (courtesy of Prof M le Roux)
Same approach only generated (+/-) (Fogler) includes the sum
of generated (+) and used (+) (as in Felder)
§1.2: The general Mole Balance Equation
Chemical Engineering is for a large part:
Accounting for momentum, mass and energy (via balances)
For the fundaments in reactor theory, we make use of mole balances
We account for the amounts of moles of each species in a (continuous) system
(species can flow in and flow out, accumulation and generation in the system)
Species can be generated)
Gj (+ formed, - reacted)
Species flow in
(Fj0) (always +)
System
Species can accumulate
dNj/dt (+/-)
Species flow out
(Fj) (always +)
The accumulation is the net result of the –inflow, -outflow and generation:
݀ܰ௝
‫ܨ‬௝଴ െ ‫ܨ‬௝ + ‫ܩ‬௝ =
݀‫ݐ‬
[mol/s]
݀ܰ௝
‫ܨ‬௝଴ െ ‫ܨ‬௝ + ‫ܩ‬௝ =
݀‫ݐ‬
The generation term (Gj) is the product of the system volume (V) and rate
of formation (rj):
Gj [mol/s] = rj [mol/m3s] · V [m3]
The reaction rate term (rj) is only seldom constant in the reactor, but
often a function of the local position in the reactor
A
FA =100 mol/s
rA = f (CA, T)
High CA Æ
high rA
low CA Æ
low rA
Picasso reactor (Fig. 1-1)
mixture of
A/B
FA =25 mol/s
FB = 75 mol/s
The reactor can be segmented in a number of M small volumes, with a
volume of 'Vi
for each cell:
'Gj,i = rj,i·'Vi
Reactor
And Gj (generation in total reactor) can be given by:
ெ
‫ܩ‬௝ = ෍ ο‫ܩ‬௝௜
௜ୀଵ
ெ
= ෍ ‫ݎ‬௝௜ οܸ௜
௜ୀଵ
If 'V Æ0
௏
‫ܩ‬௝ = න ‫ݎ‬௝ ܸ݀
General Mole balance (Eq. 1.3):
݀ܰ௝
‫ܨ‬௝଴ െ ‫ܨ‬௝ + ‫ܩ‬௝ =
݀‫ݐ‬
௏
With generation term (for a reactor):
‫ܩ‬௝ = න ‫ݎ‬௝ ܸ݀
௏
Gives the generic equation (1.4):
݀ܰ௝
‫ܨ‬௝଴ െ ‫ܨ‬௝ + න ‫ݎ‬௝ ܸ݀ =
݀‫ݐ‬
Equation 1.4 forms the basis of the analysis of various reactors,
and is universally applicable.
So the overall conversion of A, is dependent on both
reaction rate equation (rA) AND reactor configuration
The reaction rate term (rj) is only seldom constant in the reactor, but
often a function of the local position in the reactor
To introduce reactor theory 4 model reactors (ideal reactors) are going to be discussed:
In general, we can classify reactions into 2 categories:
Homogeneous (or single-phase):
reactants and products are in the same phase (often G or L)
Heterogeneous (or multi phase):
reactants and products are not in the same phase (e.g. G/S, G/L, L/S, or even
G/L/S), which can further be divided into catalytic / non-catalytic systems
Reactor classifications:
1.1.4 The Batch Reactor (BR)
In Section 1.1.4 – 1.1.6, we are going to discuss 5 types of ideal
reactors, which are classified as batch and continuous reactors,
where (for the further classification of continuous reactors)
homogeneous and heterogeneous reactors are discussed
Operation mode:
Homogeneous
Batch
Heterogeneous
The Batch Reactor (BR 1.1.4)
Continuous:
Perfect mixed
CSTR
FB (1.1.5)
Plug Flow (non mixed)
PFR
PBR (1.1.6)
In general, we can classify reactors into 2 categories:
Batch (Afr: enkellading) reactors:
1. reactants are loaded/charged into a reactor,
2. the reaction takes place inside the reactor,
3. the mixture of reactants and products is charged from the reactor.
Continuous flow (Afr: Kontinue Vloei) reactors:
1. Reactants flow into the reactor and a mixture of reactants and products flow
out of the reactor continuously
In general, batch reactors are used for small scale operation,
while continuous reactors are used for large scale operation
(Some reactors are operated semi-continuously)
Batch reactors:
Batch Reactor
mainly used for small scale operation
suitable for slow reactions
mainly used for liquid-phase reactions
charge-in/clean-up times can be large
Often stirred/agitated (and therefore often modelled as
ideally mixed (discussed in the next section)
Generic mole balance
௏
݀ܰ௝
‫ܨ‬௝଴ െ ‫ܨ‬௝ + න ‫ݎ‬௝ ܸ݀ =
݀‫ݐ‬
During operation: No in- and outflow: Fj0 = Fj = 0:
௏
݀ܰ௝
= න ‫ݎ‬௝ ܸ݀
݀‫ݐ‬
Go through and study Professional Reference Shell (PRS) on batch reactors
on web page that is associated to the book, for additional information
Reactor classifications:
1.1.5 The Continuous Stirred
Tank Reactor (CSTR)
and Fluidized Bed Reactor (FBR)
https://commons.wikimedia.org/wiki/File:Agitated_vessel.svg
Perfect mixing
Mixing / Agitation is extremely important
in reactors.
For initial reactor analysis, we often assume PERFECT (or IDEAL) mixing,
which means that at any time, the reaction rate (concentrations of the species)
in the reactor is independent of the location in the reactor
(in other words: the reaction rate is only a function of time, not of space)
௏
For a batch reactor:
݀ܰ௝
= න ‫ݎ‬௝ ܸ݀
݀‫ݐ‬
Perfect
mixing
݀ܰ௝
= ‫ݎ‬௝ ܸ
݀‫ݐ‬
Continuous flow reactors:
1) Continuous stirred tank reactor (CSTR)
steady state operation; used in series
good mixing leads to uniform concentration
and temperature
mainly used for liquid phase reaction
Generic mole balance
suitable for viscous liquids
௏
݀ܰ௝
‫ܨ‬௝଴ െ ‫ܨ‬௝ + න ‫ݎ‬௝ ܸ݀ =
݀‫ݐ‬
If steady state:
௏
‫ܨ‬௝଴ െ ‫ܨ‬௝ + න ‫ݎ‬௝ ܸ݀ = 0
If perfectly mixed:
‫ܨ‬௝଴ െ ‫ܨ‬௝ + ‫ݎ‬௝ ܸ = 0
‫ܨ‬௝଴ െ ‫ܨ‬௝
ܸ=
െ‫ݎ‬௝
Go through and study Professional Reference Shell (PRS) on CSTR's
An important consequence of perfect mixing is that:
concentration in the reactor = outflow concentration
FA0
ܸ=
‫ܨ‬஺଴ െ ‫ܨ‬஺
െ‫ݎ‬஺
FA
Where rA is based on concentrations
in reactor (= outflow concentration)
݉‫݆ ݂݋ ݏ݈݁݋‬
݉‫݆ ݂݋ ݏ݈݁݋‬
‫݁݉ݑ݈݋ݒ‬
= ࡯࢐ [
] ȉ ࢜[
]
ࡲ࢐
‫݁݉݅ݐ‬
‫݁݉ݑ݈݋ݒ‬
‫݁݉݅ݐ‬
‫ݒ‬଴ ‫ܥ‬௝଴ െ ‫ܥݒ‬௝
‫ܨ‬௝଴ െ ‫ܨ‬௝
ܸ=
ܸ=
െ‫ݎ‬௝
െ‫ݎ‬௝
v is termed
volumetric flow rate
The Fluidized Bed Reactor (FBR)
The fluidized bed (FB) is a heterogeneous reactor, typically housing a gas and
a solid (solid can be a reacting component (e.g. coal) or a non reacting
component (a catalyst))
An FB is such operated that the content of the reactor is well mixed and as
such that for modelling purposes, often ideal mixing can be assumed.
The description of this reactor therefore shows great similarities to the CSTR,
See Professional Reference Shelf
Reactor classifications:
1.1.6 The Plug Flow Reactor (PFR)
and the Packed Bed Reactor (PBR)
2) Tubular reactor
suitable for fast reactions
often used for gas phase reactions
temperature control is difficult
there are no moving parts
Fj
Fj0
Picture from: http://www.stamixco-usa.com/products/plug-flow-reactors/default.html
Plug Flow
Fj
Fj0
Although the shape of a cylindrical pipe is simple, the flow pattern
can be complex (CEMI311); there are 2 extremes:
Laminar flow (low Re),
parabolic velocity profile
Plug flow (high Re, turbulent (idealised),
flat velocity profile; uniform velocity
profile
Plug Flow Reactor (PFR)
Fj0
Small part of the PFR with volume 'V and generation 'Gj (where rj = constant):
Fj0
Fj
'V
Fj
V V+'V
௏
݀ܰ௝
‫ܨ‬௝଴ െ ‫ܨ‬௝ + න ‫ݎ‬௝ ܸ݀ =
݀‫ݐ‬
Steady
State
‫ܨ‬௝ ቚ െ ‫ܨ‬௝ ቚ
௏
௏ାο௏
+ ‫ݎ‬௝ οܸ = 0
Fj0
Fj
‫ܨ‬௝ ቚ െ ‫ܨ‬௝ ቚ
+ ‫ݎ‬௝ οܸ = 0
lim 'V Æ 0
݀‫ܨ‬௝
= ‫ݎ‬௝
ܸ݀
௏
௏ାο௏
'V
Fj
Fj
‫ܨ‬௝ ห
௏ାο௏
െ ‫ܨ‬௝ ቚ
οܸ
݀‫ܨ‬௝
ܸ݀ =
‫ݎ‬௝
ܸ=න
ிೕ ݀‫ܨ‬
௝
ிೕబ
‫ݎ‬௝
௏
= ‫ݎ‬௝
Packed Bed Reactor (PBR)
3) Packed Bed Reactor = (Tubular reactor packed
with a catalyst which is fixed)
suitable for fast catalytic reactions
often used for catalytic gas phase reactions
temperature control is difficult
there are no moving parts
In stead of reactor volume, the mass (or area) of catalyst is the determining factor in
the reaction rate, and the reaction rate for catalytic reactions -rA' (or -rA'') is defined
as:
࢓࢕࢒ ࡭ ࢘ࢋࢇࢉ࢚ࢋࢊ
࢓࢕࢒ ࡭ ࢘ࢋࢇࢉ࢚ࢋࢊ
െ࢘ᇱ࡭ =
࢙ ࢑ࢍࢉࢇ࢚ࢇ࢒࢙࢚࢟
െ࢘ᇱᇱ
࡭ =
࢙ ࢓૛ࢉࢇ࢚ࢇ࢒࢙࢚࢟
The derivation for the design equation of a PBR (plug flow assumed) goes analogue
with that of a PFR:
ிೕ
ிೕ
݀‫ܨ‬௝
ܸ=න
ிೕబ ‫ݎ‬௝
PFR:
PBR:
݀‫ܨ‬௝
ܹ=න
ᇱ
‫ݎ‬
ிೕబ ௝
Plug flow
Plug flow
1.1.7 Reactors in practice (self study)
Self Study
Section 1.5 is a collection of some other industrial reactors.
Study this part in connection with the Professional Reference Shelf
(PRS), provided on the website
This information will make you better prepared for
understanding and designing reactors
http://www.umich.edu/~elements/5e/index.html
http://www.umich.edu/~elements/5e/index.html
Please note that this website is for the 5th edition:
Read the Preface (p xvii – xxxii) critically
http://www.umich.edu/~elements/5e/index.html
http://www.umich.edu/~elements/5e/index.html
Professional reference shell:
Each chapter is supported by
additional material, also to
visualize and conceptualise
the topic of reactor theory
(Study these sections for a
better understanding)
http://encyclopedia.che.engin.umich.edu/Pages/Reactors/menu.html
It is highly recommended to make
use of this additional material
Study Unit 1.2: Conversion and reactor sizing
Fogler (Chapter 2)
SU 1: Fogler, Ch. 1-4, more or less introductory (setting the scene, learning the jargon)
1.1 Ch 1: Mole balances
1.2 Ch 2: Conversion and Reactor Sizing
1.3 Ch 3: Rate Laws
1.4 Ch 4: Stoichiometry
SU 2: Fogler, Ch. 5-7, reactor design and analysis; the chemical engineer's bread and
butter (still introductory …..)
2.1 Ch 5: Isothermal Reactor Design: Conversion
2.2 Ch 6: Isothermal Reactor Design: Moles and Molar Flow rates
2.3 Ch 7: Collection and Analysis of Rate Data
SU 3 + P: Fogler, Ch. 8 and 11, and Project, somewhat more to practice … (still introductory …..)
3.1 Ch 8: Multiple Reactions
3.2 Ch 11:Nonisothermal Reactor Design - The Steady State Energy Balance and
Adiabatic PFR Applications
1.2. Study Unit 1.2
1.2.1 Conversion
1.2.2 The Levenspiel plot
1.2.3 Reactors in series
1.2.4 Reactors in parallel
1.2.5 The Liquid Hourly Space Velocity (LHSV) and
Gas Hourly Space Velocity (GHSV)
1.2.1 Conversion
Conversion (X) is an important parameter
in reactor design, and is defined as:
݉‫݀݁ݐܿܽ݁ݎ ܣ ݂݋ ݏ݈݁݋‬
ܺ஺ =
݉‫݂݀݁ ܣ ݂݋ ݏ݈݁݋‬
Also sometimes termed fractional conversion, it indicates that part of a
species that is converted in a reactor
Batch:
[mol]
[mol/s]
ܰ஺଴ െ ܰ஺
ܺ஺ =
ܰ஺଴
‫ܨ‬஺଴ െ ‫ܨ‬஺
ܺ஺ =
‫ܨ‬஺଴
Continuous:
Consider that 100 kton/yr of toluene is converted in the HDA process,
and that on average the process operates 8000 h/yr.
a) Estimate the reactor volume if the average reaction rate of the
conversion of toluene equals 1.2 mol/m3 s (31 m3, SU 1.1)
b) Suggest a shape and dimensions of the reactor (shape: mostly cylindrical;
dimensions: e.g. D = 2.4 and L = 7.1, if aspect ratio = 3; more options possible here)
c) If toluene is fed to the reactor with 80 mol/s, what is the (single
pass) conversion of toluene (XT)?
1.2.2 The Levenspiel plot
Plug flow
Plug flow
The conversion can be "plugged in" the mole balance equations:
Batch:
ܰ஺଴ െ ܰ஺
ܺ஺ =
ܰ஺଴
Continuous:
‫ܨ‬஺଴ െ ‫ܨ‬஺
ܺ஺ =
‫ܨ‬஺଴
If we inspect the design equations of the CSTR and the PFR a bit closer
CSTR:
‫ܨ‬஺଴ ܺ
ܸ=
െ‫ݎ‬஺ ௢௨௧
௑
PFR:
ܸ = ‫ܨ‬஺଴ න
଴
݀ܺ
െ‫ݎ‬஺
‫ܨ‬஺଴
ܺ
െ‫ݎ‬஺ ௢௨௧
ܸ=
௑
ܸ=න
଴
‫ܨ‬஺଴
݀ܺ
െ‫ݎ‬஺
The ratio (FA0/-rA) is important in sizing reactors, and the
reactor volume is inversely proportional to (1/-rA)
Reaction rate data can only be obtained from experiments, and the following
example is on the gas-phase isomerisation reaction A Æ B, carried out at 500
K and 830 kPa, with pure A as initial charge to the reactor
The following experimental data are obtained:
3
(mol/m3 s)s)
XA r-rA A(mol/m
0.45
0
0.1
0.37
0.2
0.30
0.4
0.195
0.6
0.113
0.7
0.079
0.8
0.05
Can you explain the reaction rate
decreasing with conversion?
This is typical for all reactions with an
order > 0 (more on this in SU 1.3)
25
(1/-rA) as a function of
conversion
1/-rA (m3 s/mol)
(mol/m33s)s) - 1/rA (m3 s/mol)
XA rA-rA(mol/m
0
0.45
2.22
0.1
0.37
2.70
0.2
0.30
3.33
0.4
0.195
5.13
0.6
0.113
8.85
0.7
0.079
12.7
0.8
0.05
20.0
20
15
10
5
0
0
0.2
0.4
0.6
XA
0.8
1
If FA0 = 0.4 mol/s
9
8
7
6
FA0/-rA (m3 )
3
33
XA rA-r(mol/m
- FA0/rA (m3)
A (mol/m s)s) - 1/rA (m s/mol)
0
0.45
2.22
0.89
0.1
0.37
2.70
1.08
0.2
0.30
3.33
1.33
0.4
0.195
5.13
2.05
0.6
0.113
8.85
3.54
0.7
0.079
12.7
5.06
0.8
0.05
20.0
8.00
5
4
3
Irreversible reaction:
(A Æ B): V Æ f as X Æ1
Reversible reaction (A l B):
V Æ f as X ÆXeq
2
1
0
0
0.2
0.4
0.6
XA
0.8
1
CSTR:
ܸ=
௑
‫ܨ‬஺଴
ܺ
െ‫ݎ‬஺ ௢௨௧
PFR: ܸ = න
଴
‫ܨ‬஺଴
݀ܺ
െ‫ݎ‬஺
9
What volume do I need to obtain
an 80% conversion (XA = 0.8)?
8
7
VCSTR
= (0.8)*8 = 6.4 m3
FA0/-rA (m3 )
6
5
4
3
VPFR ~ 2.15 m3
2
1
0
0
0.2
0.4
0.6
XA
Levenspiel plot
0.8
1
1.2.3 Reactors in series
Reactors in series
Often not only 1 reactor is used, but reactors are used in series or parallel
We start by analysing CSTR's in series
Let's consider two CSTR's in series (again for the irreversible reaction A Æ B,
in which the reaction rate was studied in the previous section), and find again
solutions (for V) for 80% conversion.
CSTR 1
Design equation for CSTR:
CSTR 2
ܸ=
‫ܨ‬஺଴
ܺ
െ‫ݎ‬஺ ௢௨௧
Design equation for CSTR:
FA0
FA1
CSTR 1
ܸଵ =
ܸ=
‫ܨ‬஺଴
ܺଵ
െ‫ݎ‬஺ଵ ௢௨௧
‫ܨ‬஺଴
ܺ
െ‫ݎ‬஺ ௢௨௧
FA2
CSTR 2
ܸଶ =
‫ܨ‬஺ଵ
ܺଶ
െ‫ݎ‬஺ଶ ௢௨௧
ܸଶ =
‫ܨ‬஺଴
െ‫ݎ‬஺ଶ ௢௨௧
ܺଶ െ ܺଵ
We can e.g. design the reactors in such way that the same amount of A is
converted in each CSTR.
FA0
FA2
FA1
CSTR 1
X1 = 0.4
CSTR 2
X2 = 0.8
9
8
7
VCSTR1 = (0.4)*2.05 = 0.82 m3
Vtotal
= 4.02 m3
Compare to the volume of
6.4 m3 if 1 CSTR is used and
comment.
FA0/-rA (m3 )
VCSTR2 = (0.4)*8 = 3.2 m3
6
5
4
3
2
1
0
0
0.2
0.4
0.6
XA
0.8
1
FA0
FA2
FA1
CSTR 1
CSTR 2
X1
X2 = 0.8
7.00
Total
6.00
CSTR 2
CSTR 1
V (m3)
5.00
4.00
3.00
2.00
1.00
0.00
0
0.2
0.4
X1 (-)
0.6
0.8
1
VCSTR1, VCSTR2 and VTOTAL as a function of conversion in the CSTR 1 (X1)
Minimum reactor volume at VCSTR1 = 1.71 m3 (X1 = 0.57), VCSTR2 = 1.96 m3;
Vtotal = 3.67 m3 (compared to V = 6.4 m3 if 1 CSTR is used)
The total volume of the CSTRs depends on how many CSTRs are used in series
If equally sized CSTRs are used:
N = 1:
N = 2:
N = 3:
FA0
X =0.8
Vtotal = VCSTR = = 6.4 m3
VCSTR = 1.83 m3
Vtotal = 3.66 m3
X =0.8
FA0
X1 =0.57
X =0.8
FA0
X1 =0.45
X2 =0.67
Exercise: Confirm yourself (by calculation),
that for N=3, 3.09 m3 is needed
VCSTR = 1.03 m3
Vtotal = 3.09 m3
N = 5: VCSTR = 0.53 m3;Vtotal = 2.65 m3
N = 10: VCSTR = 0.24 m3;Vtotal = 2.40 m3
7
6
Vtotal (m3)
5
CSTRs in series
4
3
2
PFR
1
0
1
2
3
4
5
6
7
Number of CSTR's in series
8
9
10
11
Total CSTR volume as a function of amount of CSTRs in series in comparison
to a PFR (for chemical system as detailed in Example 2.2)
A PFR can be modelled as infinite CSTRs in series, which can be illustrated
using a Levenspiel plot
9
8
7
FA0/-rA (m3 )
6
5
4
3
2
1
0
0
0.2
0.4
0.6
XA
0.8
1
௑
଴
‫ܨ‬஺଴
݀ܺ
െ‫ݎ‬஺
0.6
0.8
PFR: ܸ = න
9
What do you think of carrying out
this reaction in a series of PFR's,
in terms of conversion?
8
7
FA0/-rA (m3 )
6
5
4
3
2
1
0
0
0.2
0.4
XA
1
Exercise
Consider an exothermic reaction, carried out in an adiabatic reactor:
AÆB
-rA = kA(T) ˜ CA
'h = -
What happens with the reaction rate if I carry out an exothermic reaction in
an adiabatic reactor?
In terms of concentration (w.r.t. conversion)?
-rA p if X n
In terms of temperature (w.r.t. conversion)?
T n if X n so -rA n
So in this case, the dependence of the reaction rate with conversion is more
complex, and it may well be possible that the reaction rate increases with
conversion (so -1/rA decreases with conversion)
Exercise - continued
If for a certain reaction, the Levenspiel
plot has this shape, and 1 reactor will
be used, what reactor would you
choose
b) If a high conversion is required
c) If you can choose different reactors,
what series of reactors would you
choose for optimal (minimal) reactor
volume, if a conversion of 60% is
required?
2
FA0/-rA (m3 )
a) If a low conversion is required
3
2
1
1
0
0
0.2
0.4
XA
0.6
1.2.4 Reactors in parallel
ܸ=
‫ܨ‬஺଴
ܺ
െ‫ݎ‬஺ ௢௨௧
Reactors are sometimes also staged in a parallel configuration
FA0,R1
R1
FA0
FA
If e.g. FA0, R1 = 0.5 FA0
then: FA0,R2 = 0.5 FA0
FA0,R2
R2
FA0 = FA0,R1 + FA0,R2
It is then easy to see that VR1 = VR2 =
0.5V0 (if V0 is the volume of a single
CSTR (with a feed of A of FA0))
Why should one use a parallel configuration, in the chemical process industry?
Configurations with combinations of reactors is also possible
FA0
FA1
CSTR 1
PFR 1
FA3
FA2
CSTR 2
An infinite amount of series/parallel configurations are possible
Sometimes a good configuration of reactors in series can optimise conversion
Practically, it is not always economic to design a reactor / reactor
configuration to a specific size
Suppliers only fabricate standard sizes
Sometimes old (already used) reactors are available
For mechanical and safety reasons, reactors can often not be too large
In Tutorial: Exercise Problem 2-3 for training in the use of
series/parallel configuration with given reactor volume!
1.2.5 The Liquid Hourly Space Velocity (LHSV)
and Gas Hourly Space Velocity (GHSV)
Some further definitions
The space time, W, is defined as the ratio of the volume of the reactor
(V) and the volumetric flow rate entering the reactor (v0)
s
ܸ
߬‫ؠ‬
‫ݒ‬଴
[m3]
[m3/s]
Volumetric flow rates can change during reaction (specifically gas
phase reactions); therefore W is an indication for the residence time of
species in the reactor
The Liquid Hourly Space Velocity (LHSV) and Gas Hourly Space Velocity
(GHSV), are the reciprocal of the space time, but the volumetric flow rate
is often not necessarily measured at the point of entrance of the reactor
‫ݒ‬଴ ]௟௜௤௨௜ௗ
‫= ܸܵܪܮ‬
ܸ
‫ݒ‬଴ ]௚௔௦
‫= ܸܵܪܩ‬
ܸ
Where the v0 in the GHSV is often expressed at STP (Nm3/hr)
As the definition says, the LHSV and GHSV are normally expressed on a
per hour basis
Have fast gas phase reactions a high or low GHSV?
Study Unit 1.3: Rate laws
(Fogler, Ch 3)
SU 1: Fogler, Ch. 1-4, more or less introductory (setting the scene, learning the jargon)
1.1 Ch 1: Mole balances
1.2 Ch 2: Conversion and Reactor Sizing
1.3 Ch 3: Rate Laws
1.4 Ch 4: Stoichiometry
SU 2: Fogler, Ch. 5-7, reactor design and analysis; the chemical engineer's bread and
butter (still introductory …..)
2.1 Ch 5: Isothermal Reactor Design: Conversion
2.2 Ch 6: Isothermal Reactor Design: Moles and Molar Flow rates
2.3 Ch 7: Collection and Analysis of Rate Data
SU 3 + P: Fogler, Ch. 8 and 11, and Project, somewhat more to practice … (still introductory …..)
3.1 Ch 8: Multiple Reactions
3.2 Ch 11:Nonisothermal Reactor Design - The Steady State Energy Balance and
Adiabatic PFR Applications
1.3. Study Unit 1.3
1.3.1 Introduction
1.3.2 The rate equation
1.3.3 The rate equation for equilibrium reactions
1.3.4 The Arrhenius equation
A1
Information on Assessment 1
1.3.1 Introduction
Classification of reactions:
In general, reactions can be classified in homogeneous (single phase) and
heterogeneous (multi phase) reactions:
Homogeneous reactions; reactants and products form 1 single phase (normally
either liquid or gas).
e.g. the Water-Gas-Shift reaction at elevated temperatures is a homogeneous gas
phase reaction
Heterogeneous reactions; reactions and products are in different phases (e.g.
G/S, G/L, G/L/S).
e.g. Coal combustion is a heterogeneous gas solid (G/S) reaction, since both solids
and gases are present in the reaction system
Catalytic reactions are also termed heterogeneous reactions, although the
catalyst does not react.
e.g. Catalytic production of ammonia is a heterogeneous catalysed gas reaction,
since both solids and gases are present in the reaction system
Definition of reaction rate (SU 1.1):
The rate of reaction (- rA) is the amount of matter that reacts per unit time per unit
volume; and can be expressed on various ways:
Example hydrodealkylation of toluene (C7H8) to form benzene (C6H6)
Mass based:
C7H8 + H2 Æ C6H6 + CH4:
kg of Toluene that has reacted per hour per Liter [kg/hr L]
pounds Toluene that has reacted per minute per gallon [lb/min gal]
Mole based (preferred)
kmol Toluene that has reacted per hour per cubic feet [kmol/hr ft3]
mol Toluene that has reacted per second per m3 [mol/s m3]
(Standard SI unit)
The production of Diphenyl (C12H10) from Benzene (C6H6)
2 C6H6 Æ C12H10 + H2
can schematically be given as:
2AÆ B+C
(A = Benzene, B = Diphenyl, C = Hydrogen)
Convention: if a species disappears, the rate is negative, and if the species is
formed, the rate is positive
rA = - 10 mol/m3 s, or –rA = 10 mol/m3 s
means 10 mol of Benzene reacts per m3 per second
Formation and disappearance rates are linked via stoichiometry
rA = - 10 mol/m3 s Æ rB = 5 mol/m3 s, rC = 5 mol/m3 s
In general for a arbitrary reaction:
ܽ‫ ܣ‬+ ܾ‫ ܤ‬՞ ܿ‫ ܥ‬+ ݀‫ܦ‬
‫ݎ‬஺
‫ݎ‬஻
‫ݎ‬஼ ‫ݎ‬஽
=
= =
െܽ െܾ
ܿ
݀
Reaction rate for catalytic reactions:
The rate of reaction (- r’A) for catalytic reactions is often the amount of matter that
reacts per unit time per unit mass of catalyst:
[mol/s kgcat]
or amount of matter that reacts per unit time (r”A) per unit surface area of catalyst:
[mol/s m2cat]
1.3.2 The rate equation
The rate of reaction is a function of various process parameters, of which temperature
and concentration (partial pressure) are the most important.
െ‫ݎ‬஺ = ݂ ܶ, ‫ܥ‬஺ , ‫ܥ‬஻ … … . .
The effect of temperature is often incorporated in the reaction rate constant (k), which
is a function of temperature only
The effect of reactant (and product) concentration is incorporated separately:
െ‫ݎ‬஺ = ݇஺ (ܶ)݂ ‫ܥ‬஺ , ‫ܥ‬஻ … … . .
The rate dependence on concentrations is fundamentally related to the reaction
mechanism and can be considerate complex (see next slide on mechanism of watergas-shift reaction over a Cu-catalyst)
Even for the Cu-catalysed Water
Gas Shift reaction, the reaction
mechanism is complex
Fishtik I and Ravindra Datta, Surface Science, 512, 2002, 229–254
Callaghan, Fishtik, Datta, Carpenter, Chmielewski and Lugo, Surface Science, 541, 2003, 21-30
Power rate law
Normally 1, or a few of the intermediate reactions are rate determining, and from this
(these) reaction(s) a rate law can be determined (more in CEMI415 on this).
In practice, the concentration dependence is determined experimentally, and is often
represented in a power rate law, according to:
ఉ
െ‫ݎ‬஺ = ݇஺ ‫ܥ‬஺ఈ ‫ܥ‬஻
In which D and E represent the order in A and B respectively
The overall order (n) is the sum of the orders of the reactants, here (D + E)
In general for a arbitrary reaction:
ܽ‫ ܣ‬+ ܾ‫ ܤ‬՞ ܿ‫ ܥ‬+ ݀‫ܦ‬
Examples of the reaction rate (and associated dimensions for k):
െ‫ݎ‬஺ = ݇஺ ‫ܥ‬஺
First order in A
[k] = [s-1]
െ‫ݎ‬஺ = ݇஺ ‫ܥ‬஺ ‫ܥ‬஻
[k] = [m3/ mol s]
െ‫ݎ‬஺ = ݇஺ ‫ܥ‬஺ଶ
Second order in A
[k] = [m3/ mol s]
First order in A and B
ଵ/ଶ
െ‫ݎ‬஺ = ݇஺ ‫ܥ‬஺
Fractional orders are
possible
[k] = [mol1/2/ m1.5 s]
Elementary rate law
If the reaction rate equation includes orders that are the same as the stoichiometry
constants, it follows an elementary rate law.
e.g. if for the oxidation of NO to NO2 the rate law is second order in NO and first
order in O2 by, it follows an elementary rate law
2 ܱܰ + ܱଶ ՜ ʹܱܰଶ
ଶ
െ‫ݎ‬ேை = ݇ேை ‫ܥ‬ேை
CO2
In general, reactions do not follow elementary rate laws
General remark on rate laws
Reaction rates can almost take any mathematical form and therefore also can be
more complex functions, of temperature and concentration.
e.g. the formation of HBr from H2 and Br (H2 + Br2 Æ 2 HBr) has the
following rate equation (under certain conditions):
ଵ/ଶ
݇ଵ ‫ܥ‬ுమ ‫ܥ‬஻௥మ
‫ݎ‬ு஻௥ =
‫ܥ‬
݇ଶ + ு஻௥
‫ܥ‬஻௥మ
(In which k1 and k2 are a f(T))
Rate equations and the associated parameters (e.g. reaction rate constants,
activation energy) have to be obtained from experiments (and are not so readily
available as for thermodynamic constants)
1.3.3 The rate equation for
equilibrium reactions
Consider the Water Gas Shift reaction:
CO H 2 O œ CO 2 H 2
1.0
3.5
0.9
3.0
0.8
2.5
0.7
2.0
0.6
1.5
0.5
1.0
0.4
0.5
0.3
0.0
If elementary reaction rates apply, reaction
rate can be written(in terms of partial
pressure):
0
200
400
-0.5
600
800
T (oC)
1000
0.2
1200
0.1
0.0
-1.0
Kc and XCO as a f(T)
(feed ratio CO/H2 = 1), P = 1 bar
Forward reaction rate:
‫ݎ‬՜ = ݇՜ ‫݌‬஼ை ‫݌‬ுమ ை
Backward reaction rate:
‫ݎ‬՚ = ݇՚ ‫݌‬஼ைమ ‫݌‬ுమ
‫݇ = ݎ‬՜ ‫݌‬஼ை ‫݌‬ுమ ை െ ݇՚ ‫݌‬஼ைమ ‫݌‬ுమ
݇՜ ‫݌‬஼ை ‫݌‬ுమ ை = ݇՚ ‫݌‬஼ைమ ‫݌‬ுమ
݇՜ ‫݌‬஼ைమ ‫݌‬ுమ
=
݇՚ ‫݌‬஼ை ‫݌‬ுమ ை
At equilibrium, r = 0
The equilibrium coefficient for this reaction:
So that:
݇՜
݇՚ =
‫ܭ‬௘௤
and:
Net reaction rate:
‫݌‬஼ைమ ‫݌‬ுమ
‫ܭ‬௘௤ =
‫݌‬஼ை ‫݌‬ுమ ை
‫݇ = ݎ‬՜
‫݌‬஼ைమ ‫݌‬ுమ
‫݌‬஼ை ‫݌‬ுమ ை െ
‫ܭ‬௘௤
Conversion of CO (-)
ln Keq (-)
Rate laws including equilibrium
4.0
Fundamentally (as you all know from CEMI313) the equilibrium constant is
dimensionless.
However, in practice the equilibrium constant is often used with a dimension,
and one should carefully watch how the equilibrium constant is defined.
See e.g. P3-16:
Calculate the equilibrium conversion and concentrations for each of the
following reactions:
(a) The liquid-phase reaction A + B l C,
with CA0 = CB0 = 2 mol/dm3 and KC = 10 dm3/mol
Seemingly the equilibrium constant is defined as:
(with [Ci] in mol/dm3)
‫ܥ‬஼
‫ܭ‬஼ =
‫ܥ‬஺ ‫ܥ‬஻
1.3.4 The Arrhenius equation
െ‫ݎ‬஺ = ݇஺ (ܶ)݂ ‫ܥ‬஺ , ‫ܥ‬஻ … … . .
The temperature dependency of reaction rate constants (k) is nearly always in the
form of an Arrhenius equation (also used for different phenomena (see P3-4)):
݇஺ ܶ
ିா
= ‫ ݁ܣ‬ோ்
A = pre-exponential factor [same dimension as k]
E = activation energy
[J/mol]
R = gas constant
[8.314 J/mol K]
T = absolute temperature [K]
Arrhenius in linear form suggests a relation between ln (kA) and (1/T)
Compare also to the temperature dependency of the equilibrium constant, if the
బ
reaction enthalpy ('H0Rx = constant):
ିοுೃೣ
‫ܭ = ܭ‬଴ ݁
ோ்
reaction rate collision theory
Reaction kinetics has its fundamentals in the reaction rate collision theory,
which can be illustrated by the reaction between Ethylene (C2H4) and Hydrogen
Chloride (HCl) to form chloroethane (C2H5Cl)
‫ܥ‬ଶ ‫ܪ‬ସ + ‫ ݈ܥܪ‬՜ ‫ܥ‬ଶ ‫ܪ‬ହ ‫݈ܥ‬
The reaction will only proceed:
if the molecules collide (related to order of reaction)
െ‫ݎ‬஼మ ுర = ݇ ‫ܥ‬஼మ ுర ‫ܥ‬ு஼௟
if the collision is effective and causes a reaction (related to T)
See for more detail: http://www.chemguide.co.uk/physical/basicrates/introduction.html
The effectiveness of a collision depends on orientation of molecules (see
previous slide) and the amount of energy the molecules have.
The amount of energy per molecule is strongly related with the temperature, and
must be high enough to overcome the energy barrier (Activation energy) for
reaction
• Atoms and/or molecules of matter are in constant motion
- Gas phase: Atoms and molecules move chaotically
- Velocities vary in magnitude and direction
- Average distribution of velocities are constant
• Kinetic theory of gases
Maxwell-Boltzmann Distribution
- T is independent of
• the nature of the gas
- Two gases at same T
• have equal ekmolecular
• Temperature scale
ekmolecular
3 / 2 kT
- Absolute T-scale
- T = 0 when ekmolecular = 0
1 uur2
T | mV
2
ekmolecular
݇஺ ܶ = ‫ܣ‬
ିா
݁ ோ்
Only that part of the collisions that
have sufficient energy to overcome
the activation energy
E
E
R
120 kJ/mol
120000 J/mol
8.314 J/mol K
T ( oC)
25
400
600
800
1000
1500
T (K)
288.15
663.15
863.15 1063.15 1263.15 1763.15
exp(-E/RT) 1.76243E-22 3.53E-10 5.47E-08 1.27E-06 1.09E-05 0.000278
Strong function of T
Arrhenius plot
݇஺ ܶ = ‫ܣ‬
ln ݇஺
ିா
݁ ோ்
‫ܧ‬
= ݈݊ ‫ ܣ‬െ
ܴܶ
ln ݇஺
1 1
( )
ܶ ‫ܭ‬
‫ܧ‬
‫ = ݁݌݋݈ݏ‬െ
ܴ
Exercise:
The irreversible liquid phase reaction A Æ B is
studied and the following reaction rate constants T (oC)
(kA) were found as a function of temperature.
30
a) Determine the activation energy and preexponential factor for this reaction in the
temperature interval of 30 – 90oC.
b) If the activation energy was 100 kJ mol-1 (with a
reaction rate constant of 0.45 at 30oC), draw the
relation of kA with T in a figure and compare to the
relationship with the activation energy found in a)
40
50
60
70
80
90
-1
kA (s )
0.45
0.62
1.0
1.8
2.9
3.6
5.1
Study Unit 1.4: Stoichiometry
(Fogler, Ch 4)
SU 1: Fogler, Ch. 1-4, more or less introductory (setting the scene, learning the jargon)
1.1 Ch 1: Mole balances
1.2 Ch 2: Conversion and Reactor Sizing
1.3 Ch 3: Rate Laws
1.4 Ch 4: Stoichiometry
SU 2: Fogler, Ch. 5-7, reactor design and analysis; the chemical engineer's bread and
butter (still introductory …..)
2.1 Ch 5: Isothermal Reactor Design: Conversion
2.2 Ch 6: Isothermal Reactor Design: Moles and Molar Flow rates
2.3 Ch 7: Collection and Analysis of Rate Data
SU 3 + P: Fogler, Ch. 8 and 11, and Project, somewhat more to practice … (still introductory …..)
3.1 Ch 8: Multiple Reactions
3.2 Ch 11:Nonisothermal Reactor Design - The Steady State Energy Balance and
Adiabatic PFR Applications
1.4. Study Unit 1.4
1.4.1 Introduction
1.4.2 Batch systems
1.4.3 Flow (continuous) systems
1.4.1 Introduction
Introduction
9
8
Chapter 2: Levenspiel plot:
7
relation between conversion (X) and reaction rate (r)
FA0/-rA (m3 )
6
5
4
Chapter 3: rate laws:
3
relation between reaction rate (r) and concentration (Ci)
1
2
0
0
0.2
0.4
0.6
0.8
1
XA
െ‫ݎ‬஺ = ݇஺(ܶ)݂ ‫ܥ‬஺, ‫ܥ‬஻ … … . .
Levenspiel plot
Normally reaction rate is not measured directly (but indirectly via the concentration
and flow rates),
So relations between conversion (X) and concentration (Ci) are of practical value
Stoichiometric tables
For this the following arbitrary reversible reaction is chosen:
ܽ‫ ܣ‬+ ܾ‫ ܤ‬՞ ܿ‫ ܥ‬+ ݀‫ܦ‬
Reaction rate of each component is related to stoichiometry
‫ݎ‬஺
‫ݎ‬஻
‫ݎ‬஼ ‫ݎ‬஽
=
= =
െܽ െܾ
ܿ
݀
The reaction can be re-written in (normalised per 1 mole of A)
ܾ
ܿ
݀
‫ܣ‬+ ‫ ܤ‬՞ ‫ܥ‬+ ‫ܦ‬
ܽ
ܽ
ܽ
1
2
e.g. ܰଶ + 3 ‫ܪ‬ଶ ՞ ʹܰ‫ܪ‬ଷ can be written as ‫ܪ‬ଶ + ܰଶ ՞ ܰ‫ܪ‬ଷ
3
3
1.4.2 Batch systems
ܾ
ܿ
݀
‫ܣ‬+ ‫ ܤ‬՞ ‫ܥ‬+ ‫ܦ‬
ܽ
ܽ
ܽ
Batch systems
The conversion (in a batch system) is defined as number of moles of a certain
component that has reacted over the moles that were initially fed:
݉‫݀݁ݐܿܽ݁ݎ ݁ݒ݄ܽ ݐ݄ܽݐ ܣ ݂݋ ݏ݈݁݋‬
ܰ஺଴ െ ܰ஺
ܰ஺
=
ܺ=
=1െ
݉‫ݐ݊݁ݏ݁ݎ݌ ݕ݈݈ܽ݅ݐ݅݊݅ ݁ݎ݁ݓ ݐ݄ܽݐ ܣ ݂݋ ݏ݈݁݋‬
ܰ஺଴
ܰ஺଴
So that NA can be expressed in terms of conversion according to:
ܰ஺ = ܰ஺଴ 1 െ ܺ
The moles of A that have generated can be expressed as:
െ(ܰ஺଴ ܺ)
The moles of B that have generated can be expressed as:
ܾ
െ (ܰ஺଴ ܺ)
ܽ
The moles of C that have generated can be expressed as:
The moles of D that have generated can be expressed as:
ܿ
(ܰ ܺ)
ܽ ஺଴
݀
(ܰ ܺ)
ܽ ஺଴
Batch systems
Initially
present
ܾ
ܿ
݀
‫ܣ‬+ ‫ ܤ‬՞ ‫ܥ‬+ ‫ܦ‬
ܽ
ܽ
ܽ
So for any reaction in a batch reactor, the
amount of moles for each species can be
obtained, for the example (with an inert):
Gene
rated
ܰ஺ = ܰ஺଴ െ ܰ஺଴ ܺ
ܾ
ܰ஻ = ܰ஻଴ െ ܰ஺଴ ܺ
ܽ
ܿ
ܰ஼ = ܰ஼଴ + ܰ஺଴ ܺ
ܽ
݀
ܰ஽ = ܰ஽଴ + ܰ஺଴ ܺ
ܽ
+
ܰூ = ܰூ଴
்ܰ = ்ܰ଴ +
ௗ
௖
௕
+ െ െ1
௔
௔
௔
ܰ஺଴ X
G = change in total number of moles
per mole of A reacted
Batch systems
ܾ
ܿ
݀
‫ܣ‬+ ‫ ܤ‬՞ ‫ܥ‬+ ‫ܦ‬
ܽ
ܽ
ܽ
݀ ܿ ܾ
+ െ െ 1 ܰ஺଴ ܺ
்ܰ = ்ܰ଴ +
ܽ ܽ ܽ
்ܰ = ்ܰ଴ + ߜܰ஺଴ ܺ
What is G for the following reaction?
G = change in total number of moles
per mole of A reacted
ܰଶ + 3‫ܪ‬ଶ = ʹܰ‫ܪ‬ଷ
G = - 2 (when based on N2)
1
2
‫ܪ‬ଶ + ܰଶ = ܰ‫ܪ‬ଷ
3
3
G = - 2/3 (when based on H2)
Choose the limiting reactant as the basis of calculation; See example 4-2 (3-3)*
*In the following of the course the values in brackets relate to the 4th edition of Fogler
CSTR:
PFR:
9
Reaction rate is
normally expressed in
concentrations (Ch 3)
8
7
6
VCSTR = (0.8)*8 = 6.4 m3
FA0/-rA (m3 )
For design purposes,
we need a relation
between conversion
and reaction rate
What volume do I need to obtain
an 80% conversion (XA = 0.8)?
5
4
3
VPFR ~ 2.15 m3
From Ch 2
2
1
0
0
0.2
0.4
0.6
XA
Levenspiel plot
ܰ஺
For a batch system:
[mol/m3]
‫ܥ‬஺ =
ܸ
And a new dimensionless parameter, 4, is introduced, which is
defined as:
ܰ௜଴
‫ܥ‬௜଴
‫ݕ‬௜଴
4௜ =
=
=
ܰ஺଴ ‫ܥ‬஺଴ ‫ݕ‬஺଴
0.8
1
ܾ
ܿ
݀
‫ܣ‬+ ‫ ܤ‬՞ ‫ܥ‬+ ‫ܦ‬
ܽ
ܽ
ܽ
ܰ஺
‫ܥ‬஺ =
ܸ
ܰ௜଴
‫ܥ‬௜଴
‫ݕ‬௜଴
4௜ =
=
=
ܰ஺଴ ‫ܥ‬஺଴ ‫ݕ‬஺଴
With these the following concentration relations can be derived:
ܰ஺ ܰ஺଴ (1 െ ܺ)
‫ܥ‬஺ =
=
ܸ
ܸ
ܾ
ܰ஺଴ (4஻ െ ܺ)
ܽ
‫ܥ‬஻ =
ܸ
ܿ
ܰ஺଴ (4஼ + ܺ)
ܽ
‫ܥ‬஼ =
ܸ
݀
ܰ஺଴ (4஽ + ܺ)
ܽ
‫ܥ‬஽ =
ܸ
The question that arises now, is
how V is related to conversion?
For batch reactors, gas-phase reactions are normally carried out in a
constant-volume reactor, so V = V0 = constant
Liquid reactions are practically mostly carried out at negligible volume
differences (constant density), although there are exceptions.
For constant-volume
batch reaction systems:
ܰ஺ ܰ஺଴ (1 െ ܺ)
‫ܥ‬஺ =
=
ܸ0
ܸ0
ܾ
ܰ஺଴ (4஻ െ ܺ)
ܽ
‫ܥ‬஻ =
ܸ0
ܿ
ܰ஺଴ (4஼ + ܺ)
ܽ
‫ܥ‬஼ =
ܸ0
݀
ܰ஺଴ (4஽ + ܺ)
ܽ
‫ܥ‬஽ =
ܸ0
1.4.3 Flow (continuous) systems
ܾ
ܿ
݀
‫ܣ‬+ ‫ ܤ‬՞ ‫ܥ‬+ ‫ܦ‬
ܽ
ܽ
ܽ
Flow systems
Mole-balances:
Batch-system:
ܰ஺ = ܰ஺଴ 1 െ ܺ
Flow system:
ܾ
ܰ஻ = ܰ஺଴ 4஻ െ ܺ
ܽ
ܿ
ܰ஼ = ܰ஺଴ 4஼ + ܺ
ܽ
ܾ
‫ܨ‬஻ = ‫ܨ‬஺଴ 4஻ െ ܺ
ܽ
ܿ
‫ܨ‬஼ = ‫ܨ‬஺଴ 4஼ + ܺ
ܽ
݀
4஽ + ܺ
ܽ
݀
4஽ + ܺ
ܽ
ܰ஽ = ܰ஺଴
ܰூ = ܰ஺଴ 4ூ
‫ܨ‬஺ = ‫ܨ‬஺଴ 1 െ ܺ
‫ܨ‬஽ = ‫ܨ‬஺଴
‫ܨ‬ூ = ‫ܨ‬஺଴ 4ூ
ܾ
ܿ
݀
‫ܣ‬+ ‫ ܤ‬՞ ‫ܥ‬+ ‫ܦ‬
ܽ
ܽ
ܽ
Flow systems
Batch-system:
ܰ஺
‫ܥ‬஺ =
ܸ
ܰ஺ ܰ஺଴ (1 െ ܺ)
=
ܸ
ܸ
ܾ
ܰ஺଴ (4஻ െ ܺ)
ܽ
‫ܥ‬஻ =
ܸ
ܿ
ܰ஺଴ (4஼ + ܺ)
ܽ
‫ܥ‬஼ =
ܸ
݀
ܰ஺଴ (4஽ + ܺ)
ܽ
‫ܥ‬஽ =
ܸ
ܰ஺଴ 4ூ
‫ܥ‬ூ =
ܸ
‫ܥ‬஺ =
Concentration:
Flow system:
mol/s
‫ܨ‬
mol/m3 ‫ܥ‬஺ = ஺
‫ ݒ‬m3/s
‫ܨ‬஺ ‫ܨ‬஺଴ (1 െ ܺ)
=
‫ݒ‬
‫ݒ‬
ܾ
‫ܨ‬஺଴ (4஻ െ ܺ)
ܽ
‫ܥ‬஻ =
‫ݒ‬
ܿ
‫ܨ‬஺଴ (4஼ + ܺ)
ܽ
‫ܥ‬஼ =
‫ݒ‬
݀
‫ܨ‬஺଴ (4஽ + ܺ)
ܽ
The question that
‫ܥ‬஽ =
‫ݒ‬
arises now, is how v is
‫ܨ‬஺଴ 4ூ
related to conversion?
‫ܥ‬ூ =
‫ݒ‬
‫ܥ‬஺ =
Flow systems
Liquid phase reactions:
Volume changes can normally be neglected (constant density) so
that v = vo
‫ܨ‬஺ ‫ܨ‬஺଴ (1 െ ܺ)
=
‫ݒ‬0
‫ݒ‬0
ܾ
‫ܨ‬஺଴ (4஻ െ ܺ)
ܽ
‫ܥ‬஻ =
‫ݒ‬0
ܿ
‫ܨ‬஺଴ (4஼ + ܺ)
ܽ
‫ܥ‬஼ =
‫ݒ‬0
݀
‫ܨ‬஺଴ (4஽ + ܺ)
ܽ
‫ܥ‬஽ =
‫ݒ‬0
‫ܨ‬஺଴ 4ூ
‫ܥ‬ூ =
‫ݒ‬0
‫ܥ‬஺ =
So rate law can be easily written
in terms of concentration (and
seizing can be done)
Gas-phase reactions (that have a change in the total number of moles)
If the volume of a batch reactor or the volumetric flow in a flow
reactor does not change with conversion, the concentrations can be
relative easily expressed in terms of conversion.
For gas-phase reactions, the flow rate is a function of the reaction
(are more or less moles formed due to the reaction), temperature
and pressure
First let's consider (bit hypothetical) a batch
reactor with variable volume, and let's see how
volume is related to conversion, temperature
and pressure
V = V (X, P, T)
Gas phase reaction in batch reactor with variable volume
General equation of state (always valid):
ܸܲ = ்ܼܰ ܴܶ
General equation of state, at t = 0:
ܲ଴
ܸ = ܸ଴
ܲ
Combined:
Previously found:
Rewritten:
ܲ଴ ܸ଴ = ܼ଴ ்ܰ଴ ܴܶ଴
ܶ
ܶ଴
்ܰ = ்ܰ଴ +
ܼ
ܼ଴
்ܰ
்ܰ଴
ௗ
௖
௕
+ െ െ1
௔
௔
௔
்ܰ
்ܰ଴
ܰ஺଴
=
+ߜ
ܺ = 1 + ߜ‫ݕ‬஺଴ ܺ
்ܰ଴ ்ܰ଴
்ܰ଴
ܰ‫்ܰ = ܺͲܣ‬଴ +G ܰ‫ܺͲܣ‬
்ܰ
= 1 + Hܺ
்ܰ଴
Where H is defined as: H = ya0G
ܲ଴
ܸ = ܸ଴
ܲ
ܶ
ܶ଴
ܼ
ܼ଴
்ܰ
= 1 + Hܺ
்ܰ଴
்ܰ
்ܰ଴
ܲ଴
ܸ = ܸ଴
ܲ
ܶ
ܶ଴
ܼ
ܼ଴
1 + Hܺ
And if Z does not significantly change with reaction conditions (Z ~ Z0)
ܲ଴
ܸ = ܸ଴
ܲ
ܶ
ܶ଴
1 + Hܺ
Now all batch systems can be scaled
Similar for flow systems:
‫ݒ = ݒ‬଴
ܲ଴
ܲ
ܶ
ܶ଴
1 + Hܺ
Flow systems
‫ݒ = ݒ‬଴
ܲ଴
ܲ
ܶ
ܶ଴
1 + Hܺ
Recall that for any species j:
‫ܨ‬௝ = ‫ܨ‬஺଴ 4௝ + Q௝ ܺ
‫ܨ‬௝
‫ܥ‬௝ =
‫ݒ‬
‫ݒ = ݒ‬଴
ܲ଴
ܲ
ܶ
ܶ଴
1 + Hܺ
where Qj is the stoichiometric coefficient, in the normalised reaction equation
ܽ‫ ܣ‬+ ܾ‫ ܤ‬՞ ܿ‫ ܥ‬+ ݀‫ܦ‬
ܾ
ܿ
݀
‫ܣ‬+ ‫ ܤ‬՞ ‫ܥ‬+ ‫ܦ‬
ܽ
ܽ
ܽ
QA = -1; QB = -b/a: Qc = c/a; Qd = d/a
So that the concentration of any species can be written as:
‫ܥ‬஺଴ 4௝ + Q௝ ܺ
‫ܥ‬௝ =
1 + Hܺ
ܲ
ܲ଴
ܶ଴
ܶ
Which than can be used for reactor sizing
Modification for equilibrium reactions, see Example 4-5 (3.6)
Study Unit 2.1:
Isothermal reactor design: Conversion
Fogler (Chapter 5)
SU 1: Fogler, Ch. 1-4, more or less introductory (setting the scene, learning the jargon)
1.1 Ch 1: Mole balances
1.2 Ch 2: Conversion and Reactor Sizing
1.3 Ch 3: Rate Laws
1.4 Ch 4: Stoichiometry
SU 2: Fogler, Ch. 5-7, reactor design and analysis; the chemical engineer's bread and
butter (still introductory …..)
2.1 Ch 5: Isothermal Reactor Design: Conversion
2.2 Ch 6: Isothermal Reactor Design: Moles and Molar Flow rates
2.3 Ch 7: Collection and Analysis of Rate Data
SU 3 + P: Fogler, Ch. 8 and 11, and Project, somewhat more to practice … (still introductory …..)
3.1 Ch 8: Multiple Reactions
3.2 Ch 11:Nonisothermal Reactor Design - The Steady State Energy Balance and
Adiabatic PFR Applications
2.1. Study Unit 2.1
2.1.1 Introduction
2.1.2 Isothermal CSTR design
2.1.3 Isothermal PFR design
2.1.4 Pressure drop
2.1.1 Introduction
In carrying out reactions (industrially) the conversion of a reaction is
dependent on many parameters, of which the 5 most important are:
•
•
•
•
•
Temperature
Pressure
Concentration of species in reactor
Presence of catalyst
Type or reactor
In the following section, we will inspect reactor design in the absence
of temperature variations (Isothermal reactor design)
Levenspiel Plot
General algorithm
for isothermal
reactor design
Example:
First order gas
phase reaction in a
plug flow reactor
with no significant
temperature effect
The idea
Chemistry
A few grams
(batch)
Pilot plant operation
A few kilograms per hour
(continuous)
Large scale chemical operation
A few hundred kilotons per year
(continuous)
Chemical Engineering
Stage of development
Reactor design for model reactors
Initially we assume ideal reactors (perfect mixed)
• Batch:
– constant volume, well-mixed (Self study §5.2
(4.2))
• CSTR:
– constant volumetric flow rate, well mixed
• PFR:
– constant volumetric flow rate, perfect plug flow
2.1.2 Isothermal CSTR design
Damköhler number
For order estimation, the Damköhler number (Da) is a handy
parameter, which is defined as the ratio of conversion rate at X = 0
[mole/s] over the feed rate of a certain species
Da
rA0 V
FA0
rate of reaction at entrance
entering flow rate of A
For a 1st order irreversible reaction:
For a 2nd order irreversible reaction:
Rule of thumb:
if Da < 0.1, X ~ 0.1;
if Da > 10, X ~ 0.9
" A reaction rate"
" A convection rate"
Da
rA0 V
FA0
kC A0 V
v 0 C A0
IJN
Da
rA0 V
FA0
kC 2A0 V
v 0 C A0
IJN& A0
Design of CSTR's (in series)
Design equation for CSTR:
ܸ=
‫ܨ‬஺଴
ܺ
െ‫ݎ‬஺ ௢௨௧
Step 1) Mole balance
Since FA0 = v0CA0
ܸ=
‫ݒ‬0‫ܥ‬஺଴
ܺ
െ‫ݎ‬஺ ௢௨௧
Space time was defined as W = V/v0, resulting in:
ܸ
‫ܥ‬஺଴ ܺ
߬=
=
‫ݒ‬଴
െ‫ݎ‬஺
ܸ
‫ܥ‬஺଴ ܺ
߬=
=
‫ݒ‬଴
െ‫ݎ‬஺
If we consider a first-order irreversible reaction:
െ‫ݎ‬஺ = ݇‫ܥ‬஺
Step 2) Rate law
If we consider a liquid phase reaction (no significant volume change):
‫ܥ‬஺ = ‫ܥ‬஺଴ 1 െ ܺ
Step 3) Stoichiometry
These 3 equations can be combined to:
1
ܺ
߬=
݇ 1െܺ
In terms of conversion:
Step 4) Combine
݇߬
ܺ=
1 + ݇߬
‫ܽܦ‬
ܺ=
1 + ‫ܽܦ‬
‫ܥ‬஺଴
‫ܥ‬஺ =
1 + ݇߬
‫ܥ‬஺଴
‫ܥ‬஺ =
1 + ݇߬
CSTR's in series
CA0
CA1
CSTR 1
(-rA1, V1)
With general design ‫ܥ = ܥ‬஺଴
஺ଵ
1 + ݇ଵ ߬ଵ
equation for CSTR 1:
CA2
CSTR 2
(-rA2, V2)
With general design
equation for CSTR 2:
‫ܥ‬஺ଶ =
‫ܥ‬஺ଵ
1 + ݇ଶ ߬ ଶ
‫ܥ‬஺଴
‫ܥ‬஺ଶ =
1 + ݇ଵ ߬ଵ 1 + ݇ଶ ߬ଶ
‫ܥ‬஺଴
‫ܥ‬஺ଶ =
1 + ݇ଵ ߬ଵ 1 + ݇ଶ ߬ଶ
CA0
CA1
CA2
CSTR 1
(-rA1, V)
CSTR 2
(-rA2, V)
If CSTR's are of equal size (V1=V2=V, hence W1=W2=W), and operate at the same
temperature (k1=k2=k):
‫ܥ‬
‫ܥ‬஺ଶ =
For n identical CSTR's in series:
In terms of conversion:
஺଴
1 + ݇߬ ଶ
‫ܥ‬஺଴
‫ܥ‬஺଴
‫ܥ‬஺௡ =
=
௡
1 + ݇߬
1 + ‫ ܽܦ‬௡
1
1
ܺ =1െ
=1െ
௡
1 + ݇߬
1 + ‫ ܽܦ‬௡
CSTR's in parallel
FA01
FA0
CA0
ܸ1 =
1
‫ܨ‬
2 ஺଴
ܺ
െ‫ݎ‬஺ ௢௨௧
ܸ2 =
1
‫ܨ‬
2 ஺଴
ܺ
െ‫ݎ‬஺ ௢௨௧
CA1
CA0
FA02
ܸ݅ =
‫ܨ‬஺଴௜
ܺ
െ‫ݎ‬஺௜ ௢௨௧
CA2
If CSTR's are of equal size (V1=V2=V, hence W1=W2=W), and operate at the same
temperature (rA1=rA2=rA), and the feed-stream is equally divided between the 2 CSTR's:
The conversion is identical to one stream (FA0) fed to a CSTR (V) with the same
volume as the 2 parallel CSTR's (V1 + V2)
Why should one use a parallel configuration?
2.1.3 Isothermal PFR design
Fj
Tubular reactors
Tubular reactors are frequently used in the
Fj0
chemical process industry
For modeling purposes, often plug flow is assumed (no radial
velocity and temperature profile; no dispersion)
From SU 2, general description of a PFR
(differential form):
݀ܺ
= െ‫ݎ‬஺
‫ܨ‬஺଴
ܸ݀
݀ܺ
‫ܨ‬஺଴
= ܸ݀
െ‫ݎ‬஺
Plug flow (high Re, turbulent (idealised),
flat velocity profile; uniform velocity
profile
௑
݀ܺ
ܸ = ‫ܨ‬஺଴ න
െ‫ݎ‬஺
଴
PFR (2nd order irreversible (in A) liquid phase reaction)
௑
݀ܺ
ܸ = ‫ܨ‬஺଴ න
െ‫ݎ‬஺
Step 1) Mole balance
െ‫ݎ‬஺ = ݇‫ܥ‬஺ଶ
Step 2) Rate law
2A Æ B
AƽB
଴
Liquid phase reaction (v = v0, constant density reaction)
‫ܥ‬஺ = ‫ܥ‬஺଴ 1 െ ܺ
Step 3) Stoichiometry
Isothermal operation (k = constant)
௑
‫ܨ‬஺଴
݀ܺ
ܸ=
ଶ න 1െܺ ଶ
݇‫ܥ‬஺଴
଴
Step 4) Combine
௑
‫ܨ‬஺଴
݀ܺ
ܸ=
ଶ න 1െܺ ଶ
݇‫ܥ‬஺଴
଴
Step 5) Evaluate
Integration:
‫ܨ‬஺଴
ܺ
ܸ=
ଶ 1െܺ
݇‫ܥ‬஺଴
‫ܨ‬஺଴
ܸ=
ଶ
݇‫ܥ‬஺଴
1
1െܺ
௑
଴
‫ݒ‬଴
ܺ
ܸ=
݇‫ܥ‬஺଴ 1 െ ܺ
߬݇‫ܥ‬஺
‫ܽܦ‬ଶ
ܺ=
=
1 + ߬݇‫ܥ‬஺ 1 + ‫ܽܦ‬ଶ
PFR
2A Æ B
AƽB
(2nd order irreversible (in A) gas phase reaction)
௑
݀ܺ
ܸ = ‫ܨ‬஺଴ න
െ‫ݎ‬஺
Step 1) Mole balance
െ‫ݎ‬஺ = ݇‫ܥ‬஺ଶ
Step 2) Rate law
଴
Gas phase reaction:
‫ܥ‬஺଴ 4௝ + Q௝ ܺ
Step 3) Stoichiometry ‫ܥ‬௝ =
1 + Hܺ
ܲ
ܲ଴
ܶ଴
ܶ
(Eq. 4-25
(3-46))
For species A: 4A = 1; QA = - 1; and if isothermal and isobaric:
‫ܥ‬஺଴ 1 െ ܺ
‫ܥ‬஺ =
1 + Hܺ
௑
݀ܺ
ܸ = ‫ܨ‬஺଴ න
െ‫ݎ‬஺
െ‫ݎ‬஺ = ݇‫ܥ‬஺ଶ
‫ܥ‬஺଴ 1 െ ܺ
‫ܥ‬஺ =
1 + Hܺ
Mole balance
Rate law
Stoichiometry
଴
௑
‫ܨ‬஺଴
1 + Hܺ ଶ
ܸ=
ଶ න 1 െ ܺ ଶ ݀ܺ
݇‫ܥ‬஺଴
Step 4) Combine
଴
Step 5) Evaluate:
ଶܺ
‫ݒ‬଴
1
+
H
ܸ=
2H 1 + H ln 1 െ ܺ + Hଶ ܺ +
݇‫ܥ‬஺଴
1െܺ
ଶ
‫ݒ‬଴
1
+
H
ܺ
ܸ=
2H 1 + H ln 1 െ ܺ + Hଶ ܺ +
݇‫ܥ‬஺଴
1െܺ
ߝ = ߜ‫ݕ‬஺଴
(as previously defined)
For an on species A
normalised reaction
ߜ = ෍ Q௜
1
ܱܵ2 + ܱଶ ՜ ܱܵ3
2
1
‫ܣ‬+ ‫ ܤ‬՜‫ܥ‬
2
1
0.9
0.8
0.7
ߜ = െ1 + െ
X (-)
0.6
1
1
+1= െ
2
2
0.5
‫ ܱܥ‬+ ‫ܪ‬ଶ ܱ ՜ ‫ܱܥ‬ଶ + ‫ܪ‬ଶ
0.4
‫ܣ‬+‫ ܤ‬՜‫ܥ‬+‫ܦ‬
0.3
0.2
0.1
Here H = G (if yA0 = 1, hence pure feed of A)
Isothermal, isobaric,
2nd order in A gas
phase reaction in a
PFR
ߜ = െ1 + െ1 + 1 + 1 = 0
0
0
5
10
V (m3)
15
20
‫ܥ‬ଶ ‫ ଺ܪ‬՜ ‫ܥ‬ଶ ‫ܪ‬ସ + ‫ܪ‬ଶ
‫ܣ‬՜‫ܤ‬+‫ܥ‬
ߜ = െ1 + 1 + 1 = 1
X as a function of V (m3) for a reaction with ((v0/kCA0) = 2.0 m3) for different H.
Can you qualitatively understand the trends?
There is an alternative way of solving the problems,
then the previously shown analytical method, which
is more generally applicable and more effective in
complex problems (complex kinetics, pressure drop,
non-isothermal, multiple reactions …)
This is by using Polymath, which is a more effective way of solving
problems in this course, specifically also when non-isothermal,
multiple reactions, or complex reaction kinetics problems are of
interest
This will be further discussed in Chapter 6
2.1.4 Pressure drop
Pressure drop
Up till now, the reaction was assumed to take place isobaric (P =
constant), but in practice P can change significant in a reactor
In well mixed reactors, pressure gradients are normally not
experienced, due to the mixing (Ci, T and P constant)
In tubular flow reactors, pressure gradients can occur, specifically when
packed bed reactors are considered
2 practical examples; 1) flow through a packed bed and 2) flow through
a cylindrical pipe will be discussed
Pressure drop over a packed bed reactor (gas phase reaction)
gas
gas
Often used for catalysed gas phase reactions
The packed (or fixed) bed can be seen as a collection of packed
particles, through which the gas is transported (gas flows, solid is fixed)
Isobaric catalysed gas phase
reaction; 2nd order in A (isothermal):
݀ܺ ݇‫ܥ‬஺଴ 1 െ ܺ
=
ܹ݀
‫ݒ‬଴ 1 + ߝܺ
ଶ
non- isobaric catalysed gas phase
reaction; 2nd order in A (isothermal):
݀ܺ ݇‫ܥ‬஺଴ 1 െ ܺ
=
ܹ݀
‫ݒ‬଴ 1 + ߝܺ
ଶ
ܲ
ܲ଴
ଶ
݀ܺ
= ݂1(ܺ, ܲ)
ܹ݀
Pressure drop over a packed bed reactor
gas
gas
Most frequently used for the description of pressure in a packed bed
reactor is the Ergun equation:
݀ܲ
1
1െI
=െ
݀‫ݖ‬
ߩ݃௖ ‫ܦ‬௣ Iଷ
P
z
U
P
Dp
I
G
150(1 െ I)ߤ
‫ ܩ‬+ 1.75‫ ܩ‬ଶ
‫ܦ‬௣
= pressure
= length along the bed
Term 1 (dominant Term 2 (dominant
at low G, low Re) at high G, high Re)
= gas density
= gas viscosity
= particle diameter of bed material
= porosity of the bed; (1- I) = volume solids/total bed volume
= superficial mass flux (kg/m2 s)
(gc is a conversion factor in the case non-SI units are used)
Pressure drop over a packed bed reactor
G
= superficial mass flux (kg/m2 s);
this means mass flux based on the empty pipe
G [kg/m2 s] = U [kg/m3] u [m/s];
usuperficial
u based on empty pipe
uinterstitial
usuperficial
What is normally larger? Usuperficial or Uinterstitial?
Pressure drop over a packed bed reactor
For steady state:
From SU 1.4:
In terms of U:
with Ergun:
݉ሶ ଴ = ݉ሶ
[kg/s]
ߩ଴ ‫ݒ‬଴ = ߩ‫ݒ‬
[kg/m3·m3/s]
‫ݒ = ݒ‬଴
ܲ଴
ܲ
ܶ
ܶ଴
‫்ܨ‬
‫்ܨ‬଴
ߩ = ߩ଴
ܲ
ܲ଴
ܶ଴
ܶ
‫்ܨ‬଴
‫்ܨ‬
݀ܲ
1
1െI
=െ
݀‫ݖ‬
ߩ଴ ݃௖ ‫ܦ‬௣ Iଷ
150 1 െ I ߤ
‫ ܩ‬+ 1.75‫ ܩ‬ଶ
‫ܦ‬௣
ܲ
ܲ଴
ܶ଴
ܶ
‫்ܨ‬଴
‫்ܨ‬
݀ܲ
1
1െI
=െ
݀‫ݖ‬
ߩ଴ ݃௖ ‫ܦ‬௣ Iଷ
150 1 െ I ߤ
‫ ܩ‬+ 1.75‫ ܩ‬ଶ
‫ܦ‬௣
ܲ
ܲ଴
ܶ଴
ܶ
‫்ܨ‬଴
‫்ܨ‬
= E0 (only a function of bed geometry and inlet conditions)
In terms of weight of catalyst (preferred for catalytic reactions) :
Ac
(m2)
ܹ = 1 െ I ‫ܣ‬௖ ‫ߩ ݖ‬௖
z (m)
݀ܲ
ߚ଴
ܲ଴
=െ
ܹ݀
‫ܣ‬௖ 1 െ I ߩ௖ ܲ
ܶ
ܶ଴
Uc = catalyst density (kg/m3)
‫்ܨ‬
‫்ܨ‬଴
݀ܲ
ߚ଴
ܲ଴
=െ
ܹ݀
‫ܣ‬௖ 1 െ I ߩ௖ ܲ
Resulting in:
݀‫݌‬
ߙ ܶ
=െ
ܹ݀
ʹ‫ܶ ݕ‬଴
ܶ
ܶ଴
‫்ܨ‬
‫்ܨ‬଴
‫்ܨ‬
‫்ܨ‬଴
2ߚ଴
Where p = P/P0 and : ߙ =
‫ܣ‬௖ ߩ௖ 1 െ I ܲ଴
In terms of H:
݀‫݌‬
ߙ
=െ
1 + ߝܺ
ܹ݀
ʹ‫ݕ‬
Isothermal operation:
݀‫݌‬
= ݂ଶ ܺ, ܲ
ܹ݀
From previous:
݀ܺ
= ݂1(ܺ, ܲ)
ܹ݀
ܶ
ܶ଴
2 ODE's that have
to be solved
simultaneously
݀‫݌‬
ߙ
=െ
1 + ߝܺ
ܹ݀
ʹ‫ݕ‬
ܶ
ܶ଴
If the reaction is carried out isothermally and H = 0, the differential
equation can be solved analytically
݀‫݌‬
ߙ
=െ
ܹ݀
ʹ‫ݕ‬
ܲ
2ߚ଴ ‫ݖ‬
‫=݌‬
= 1െ
ܲ଴
ܲ଴
ଵ
ଶ
1.2
Typical shape:
P/P0
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
1.2
z (m)
Can you qualitatively understand the trend?
Pressure drop over an empty pipe (non catalytic reaction)
The pressure drop over a pipe is normally small (and significant smaller than a
packed reactor); the pressure drop can be determined with the Fanning equation,
and results in:
ଶ
ଵ
ܲ
‫=݌‬
= 1 െ ߙ௉ ܹ ଶ
ܲ଴
Ͷ݂‫ܩ‬
ߙ௉ =
‫ܣ‬௖ ߩ଴ ܲ଴ ‫ܦ‬
where:
1.02
1
Empty pipe (ID =
D = 2.54 cm)
P/P0
0.98
0.96
0.94
0.92
0.9
0.88
0.86
0.84
Packed Bed (ID = 2.54 cm),
filled with spherical catalyst
with Dp = 1 mm
0.82
0
0.2
0.4
0.6
0.8
1
z (m)
1.2
For further learning on the influence of pressure drop in a packed bed reactor,
study examples 5.4, 5.5, 5.6, 5.7 (4.4, 4.5 and 4.6) intensively
Recommendation: initially convert all units to standard SI units and proceed
with calculations
Recall Slide 23 (SU 2.1)
There is an alternative way of solving the problems,
then the previously shown analytical method, which
is more generally applicable and more effective in
complex problems (complex kinetics, pressure drop,
non-isothermal, multiple reactions …)
This is by using Polymath, which is a more effective way of solving
problems in this course, specifically also when non-isothermal,
multiple reactions, or complex reaction kinetics problems are of
interest
This will be further discussed in Chapter 6
Exercise
1. Due to environmental concerns, Sasol has strongly invested in the development
of a natural gas line from Mozambique to their operational. Natural gas is hereby
seen as a cleaner feedstock for the production of fuels and chemicals than coal.
The natural gas pipe line system stretches over several 100 km’s and consists of
several recompression units to transport the gas over the long distance.
Why do you think is natural gas a cleaner alternative fuel than coal?
2. Multiple recompression units are needed since the transport of gas is associated
with significant pressure losses. If one inspects the gas velocity in the gas line
between 2 recompression units, what will happen to the gas velocity over the
length of the pipe? Motivate your answer!
a) The gas velocity will increase
b) The gas velocity will remain the same
c)
The gas velocity will decrease
Study Unit 2.2:
Isothermal reactor design: Moles and Molar Flow Rates
Fogler (Chapter 6)
SU 1: Fogler, Ch. 1-4, more or less introductory (setting the scene, learning the jargon)
1.1 Ch 1: Mole balances
1.2 Ch 2: Conversion and Reactor Sizing
1.3 Ch 3: Rate Laws
1.4 Ch 4: Stoichiometry
SU 2: Fogler, Ch. 5-7, reactor design and analysis; the chemical engineer's bread and
butter (still introductory …..)
2.1 Ch 5: Isothermal Reactor Design: Conversion
2.2 Ch 6: Isothermal Reactor Design: Moles and Molar Flow rates
2.3 Ch 7: Collection and Analysis of Rate Data
SU 3 + P: Fogler, Ch. 8 and 11, and Project, somewhat more to practice … (still introductory …..)
3.1 Ch 8: Multiple Reactions
3.2 Ch 11:Nonisothermal Reactor Design - The Steady State Energy Balance and
Adiabatic PFR Applications
2.2. Study Unit 2.2
2.2.1 Mole Balances
2.2.2 Membrane reactors
2.2.3 Unsteady-State operation of Stirred Reactors
2.2.4 Semibatch Reactors
2.2.1 Mole Balances
Previously, reactor equations were mainly written in terms of conversion (As
detailed in Chapter 5).
For relative simple cases (single reaction, isothermal), this approach works well,
although time consuming.
For more complex cases (if multiple reactions occur, non-isothermal), another
approach, where the mole balances of each species are expressed in terms of
number of moles, Ni (batch) or molar flow rates, Fi (continuous) is often preferred.
This approach (of course) also works for simple cases, and is as such more
generally applicable, and a confident use of this method, makes problem solving (a
lot) easier.
Chapter 6
Chapter 5
PFR
(2nd order irreversible (in A) gas phase reaction)
2A Æ B
AƽB
‫ݒ‬଴
1 + H ଶܺ
ଶ
ܸ=
2H 1 + H ln 1 െ ܺ + H ܺ +
݇‫ܥ‬஺଴
1െܺ
For an on species A
normalised reaction
ߝ = ߜ‫ݕ‬஺଴ (as previously defined)
1
ܱܵ2 + ܱଶ ՜ ܱܵ3
2
1
‫ܣ‬+ ‫ ܤ‬՜‫ܥ‬
2
1
0.9
0.8
0.7
ߜ = െ1 + െ
X (-)
0.6
1
1
+1= െ
2
2
0.5
‫ ܱܥ‬+ ‫ܪ‬ଶ ܱ ՜ ‫ܱܥ‬ଶ + ‫ܪ‬ଶ
0.4
‫ܣ‬+‫ܤ‬՜‫ܥ‬+‫ܦ‬
0.3
0.2
0.1
Here H = G (if yA0 = 1, hence pure feed of A)
Isothermal, isobaric,
2nd order in A gas
phase reaction in a
PFR
ߜ = െ1 + െ1 + 1 + 1 = 0
0
0
5
10
V (m3)
15
20
‫ܥ‬ଶ ‫ ଺ܪ‬՜ ‫ܥ‬ଶ ‫ܪ‬ସ + ‫ܪ‬ଶ
‫ܣ‬՜‫ܤ‬+‫ܥ‬
ߜ = െ1 + 1 + 1 = 1
X as a function of V (m3) for a reaction with ((v0/kCA0) = 2.0 m3) for different H.
Can you qualitatively understand the trends?
From SU 2.1 slide 22
Fj0
Fj
'V
Fj
Fj
lim 'V Æ 0
From SU 1.1, derivation of molar balance on a PFR
PFR (2nd order (in A) gas phase reaction, isothermal, isobaric)
݀‫ܨ‬௝
= ‫ݎ‬௝
ܸ݀
ʹ‫ ܣ‬՜ ‫ܤ‬
݀‫ܨ‬஺
= ‫ݎ‬஺
ܸ݀
‫ ܣ‬՜ 0.ͷ‫ܤ‬
ߜ = െ1 + 0.5 = െ0.5
െ‫ݎ‬஺ = ݇‫ܥ‬஺ଶ
݀‫ܨ‬஻
= െ0.5 ‫ݎ‬஺
ܸ݀
The previous example was worked out with (v0/kCA0) = 2.0 m3 and in order to
compare CA0 = chosen as 1.0 mol/m3, k = 1.0 m3/mol s and v0 = 2.0 m3/s, so that H =
yA0 G = -0.5, and that FA0 = CA0 v = 2.0 mol/s
Pure A = fed (yA = 1) so that CT0 = CA0 = 1.0 mol/m3 (isothermal, isobaric)
So that:
ிಲ
and
‫Ͳܶܥ = ܣݕ Ͳܶܥ = ܣܥ‬
ி்
‫ܨ = ܶܨ‬஺ + ‫࡮ܨ‬
Polymath code:
d(FA) / d(V) = rA
FA(0) = 2
d(FB) / d(V) =-0.5* rA
FB(0) = 0
V(0) = 0
V(f) = 20
DE’s and BCs to describe flow of A and B
Initial and final condition
rA = -k*CA^2
k=1
Reaction kinetic equation and kinetic constant(s)
FT = FA+FB
CT0 = 1
CA = CT0*FA/FT
Determination of CA (at any point in the reactor),
X = 1-FA/FA0
FA0 = 2
Definition of conversion
The following is obtained from running the Polymath code (dotted line)
1
0.9
0.8
0.7
X (-)
0.6
0.5
‫ݒ‬଴
1 + H ଶܺ
ଶ
ܸ=
2H 1 + H ln 1 െ ܺ + H ܺ +
݇‫ܥ‬஺଴
1െܺ
0.4
Slide 17
0.3
Results from solving mole-balances
directly (using polymath)
0.2
0.1
0
0
5
10
V (m3)
15
20
This approach is (of course) also applicable for single reaction, and will generate
the same results as the approach where the conversion was written in terms of
volume, see Example 6-1 on this, and try to reproduce Fig E6-1.1 using Polymath.
Summary of approach:
The reaction 2 NOCl Æ 2 NO + Cl2 is considered (isothermal, isobaric), which
can be written as A Æ B + ½ C kinetic rate law: -rA = kA CA2
݀‫ܨ‬஺
= ‫ݎ‬஺
ܸ݀
‫ܨ‬஺ (ܸ = 0) = ‫ܨ‬஺଴
݀‫ܨ‬஻
= ‫ݎ‬஻
ܸ݀
݀‫ܨ‬஼
= ‫ݎ‬஼
ܸ݀
Stoichiometry
Concentration calculation
(total moles and
concentration)
‫ܨ‬஻ (ܸ = 0) = ‫ܨ‬஻଴
‫ݎ‬஻ = െ‫ݎ‬஺
‫ܨ = ்ܨ‬஺ + ‫ܨ‬஻ + ‫ܨ‬஼
‫ܨ‬஼ (ܸ = 0) = ‫ܨ‬஼଴
1
‫ݎ‬஼ = െ ‫ݎ‬஺
2
‫ܨ‬௝
‫ܥ‬௝ = ‫்ܥ‬଴
‫்ܨ‬
A Æ B + ½ C kinetic rate law: -rA = kA CA2
35
30
Total
A
B
C
Fj (Pmol/s)
25
20
15
10
5
0
0.E+00
2.E-06
4.E-06
6.E-06
V (dm3)
8.E-06
1.E-05
2.2.2 Membrane reactors
Up till now, we considered reactions in which the total mass flow in equals the
total mass flow out (at steady state)
݉ሶ
݉ሶ ଴ = ݉ሶ
݉ሶ ଴
Is this always (desired) the case?
Does the mass rate in the reactive phase always be the same?
A membrane reactor is a reactor where this is not the case
•
Membrane reactors (§6.4 (4.9))
The principles are demonstrated here, while the examples (in the book) can be
followed for further illustration
Membrane reactors (§ 6.4 (4.9))
Butadiene can be prepared by the gas-phase catalytic dehydrogenation of 1-Butene:
C4H8 œ C4H6 + H2.
o
c) Calculate the equilibrium constant at 25 C for this reaction (4)
o
In practice, the reaction is carried out at elevated temperatures (500 – 700 C).
d) Motivate the choice of elevated temperature. Give two arguments (restrict your answer to not more than
30 words per argument) (4)
In order to suppress side reactions, the 1-butene is strongly diluted with steam before it passes into the
reactor. Therefore the feed to the reactor consists of a mixture of 8.3 mol% 1-butene and 91.7 mol% steam.
Steam does not react in this case.
For the following problems you can assume that the enthalpy of reaction is not a function of temperature,
and that the reaction attains equilibrium.
o
e) Determine the conversion of 1-butene if the reactor is operated at 600 C and 2.0 bar. Motivate
assumptions you make (8)
f) Estimate the temperature at which the reactor must be operated in order to convert 70% of the 1-butene at
a reactor pressure of 2.0 bar. Motivate assumptions you make (4)
From CEMI 313 exam (July 2013)
Butadiene can be prepared by the gas-phase catalytic dehydrogenation of 1-Butene:
C4H8 œ C4H6 + H2.
'h0rxn,298 = 110 kJ/mol
q
o
e) Determine the conversion of 1-butene if the reactor is operated at 600 C and 2.0 bar. Motivate
assumptions you make (8)
Answer: 47%
How can I increase conversion (from a thermodynamic perspective)
How can I increase conversion (from a reaction kinetic perspective)
Selective removal of 1 of the reactants (e.g. H2), can increase the conversion (and
facilitate the reaction and separation in 1 apparatus)
Membranes (wikipedia):
"A membrane is a thin, film-like structure that separates two
fluids. It acts as a selective barrier, allowing some particles or
chemicals to pass through, but not others"
Illustration of membrane reactors to increase conversion (yield) in
equilibrium limited reactions
C4H8 Ù C4H6 + H2
A Ù B +C
Xeq (600oC, 2 bar) = 0.47
݀‫ܨ‬஺
= ‫ݎ‬஺
ܸ݀
C4H6
C4H8
݀‫ܨ‬஻
= ‫ݎ‬஻
ܸ݀
H2
ܴ஼ = ݇஼ᇱ ܽ ‫ܥ‬௖ െ ‫ܥ‬஼ௌ
ܴ஼ = ݇஼ᇱ ܽ‫ܥ‬௖
(If sweep gas rate is high)
ௗி಴
= ‫ݎ‬஼ - RC
ௗ௏
CC
Reactor side
membrane
Sweep gas
H2 transported through
the membrane
H2
CCS
Potential applications of (inorganic) membrane reactors
High-temperature membrane reactors: potential and problems, G. Saracco, H.W.J.P. Neomagus, G.F. Versteeg,
W.P.M. van Swaaij, Chemical Engineering Science 54 (1999) 1997-2017
2.2.3 Unsteady-State operation of Stirred Reactors
Unsteady-State Operation of Stirred Reactors
Continuous reactors were up-till-now only discussed at steady-state
In reality, continuous processes operated very seldom at steady state
Can you identify a few causes why not?
In this section, the focus is on non-steady state operation of stirred reactions
In what case you can even not avoid unsteady state operation?
In this section, we discuss
•
Start-up of a CSTR
•
Semi-batch stirred reactors
Startup of a CSTR
General Mole balance (Eq. 1.3):
݀ܰ௝
‫ܨ‬௝଴ െ ‫ܨ‬௝ + ‫ܩ‬௝ =
݀‫ݐ‬
For steady-state operation:
‫ܨ‬௝଴ െ ‫ܨ‬௝ + ‫ݎ‬௝ ܸ = 0
For species A (unsteady state):
݀ܰ஺
‫ܨ‬஺଴ െ ‫ܨ‬஺ + ‫ݎ‬஺ ܸ =
݀‫ݐ‬
We inspect a specific case in which the volume remains
constant (V = V0), for a liquid reaction (without density
effects) (v = v0), in which initially no A is in the reactor,
working isothermally (even at start-up)
(This is an idealised example, e.g. liquid reaction that takes
place diluted in a solvent, where initially only solvent is
present in the reaction). This example can illustrate the
effect of start-up, avoiding excessive difficult mathematics
݀ܰ஺
‫ܨ‬஺଴ െ ‫ܨ‬஺ + ‫ݎ‬஺ ܸ =
݀‫ݐ‬
At t = 0: CA = 0 (Only Solvent in reactor)
After t = 0, the reactor is fed with a constant
liquid flow (v = v0) containing a constant
concentration of A (CA0)
Consequently, the concentration of A will
initially accumulate in the reactor
Reminding that Fj = Cj v gives (v=v0, V=V0):
݀‫ܥ‬஺ ܸ0
‫ܥ‬஺଴ ‫ݒ‬଴ െ ‫ܥ‬஺ ‫ݒ‬଴ + ‫ݎ‬஺ ܸ0 =
݀‫ݐ‬
ܸ଴ ܸ଴ ݀‫ܥ‬஺
‫ܥ‬஺଴ െ ‫ܥ‬஺ + ‫ݎ‬஺ =
‫ݒ‬଴ ‫ݒ‬଴ ݀‫ݐ‬
With space time,
W = V0/v0
‫ܥ‬஺଴ െ ‫ܥ‬஺ + ‫ݎ‬஺
=
݀‫ܥ‬஺
݀‫ݐ‬
‫ܥ‬஺଴ െ ‫ܥ‬஺ + ‫ݎ‬஺
=
݀‫ܥ‬஺
݀‫ݐ‬
For a first order reaction (-rA = k CA):
‫ܥ‬஺଴ െ ‫ܥ‬஺ െ ݇‫ܥ‬஺
=
݀‫ܥ‬஺
݀‫ݐ‬
Which can be re-written to the 1st order DE:
݀‫ܥ‬஺
1+݇
‫ܥ‬஺଴
+
= ‫ܥ‬஺
W
݀‫ݐ‬
W
Liquid reaction:
v=v0
And solved with the initial condition CA (t=0) = 0
V=V0
T= T0
‫Ͳܣܥ‬
1 + ݇W
Feed Concentration of A: CA0
‫ݐ‬
‫ܥ‬஺ =
1 െ ݁‫ ݌ݔ‬െ
W
1 + ݇W
Exit concentration of A: CA
‫Ͳܣܥ‬
‫ܥ‬஺ =
1 + ݇W
1 + ݇W
1 െ ݁‫ ݌ݔ‬െ
‫ݐ‬
W
Example with the following constants:
10 mol/L
k
-1
v0
1.00E-04 s
3
0.01 m /s
V
W
kW
100 m3
10000 s
1
For a space time of
10 000 s, the time to
reach steady state is
about 25 000 s (in
this example)
In general,
6
Steady state value, ‫ܥ‬஺ =
5
‫ܥ‬஺଴
1 + ݇W
4
CA (mol/L)
CA0
3
2
1
ts = 4.6W/(1+kW)
ts = time to reach 99% of
steady state value
0
0
5000
10000
15000
t (s)
20000
25000
30000
Also here, the use of Polymath is easier: If one takes the initial derived DE (slide 25):
‫ܥ‬஺଴ െ ‫ܥ‬஺ + ‫ݎ‬஺
=
݀‫ܥ‬஺
݀‫ݐ‬
݀‫ܥ‬஺
= (‫ܥ‬஺଴ െ ‫ܥ‬஺ + ‫ݎ‬஺ )/
݀‫ݐ‬
, which can be
rewritten as:
d(CA) / d(t) = (CA0-CA+rA*tau )/tau
CA(0) = 0
6
CA (mol/L)
5
t(0) = 0
t(f) = 30000
4
3
2
1
CA0 = 10 # mol/L
tau = 10000 # s
rA = -k*CA
k = 1e-4 # s-1
0
0
5000
10000
15000
20000
25000
30000
t (s)
CA0
10 mol/L
k
-1
v0
1.00E-04 s
3
0.01 m /s
V
W
kW
100 m3
10000 s
1
Will give the same result, without doing
excessive mathematics
2.2.4 Semibatch Reactors
Semi batch reactors
A semi-batch reactor, is a reactor that is carried out batch-wise, but some of the
components are added (b) or disappear (c)
Main reason: Selectivity increase, example for the following reaction scheme:
A and B react to the desired product D, but also to the undesired product U
A+BÆD
rD = -kDCA2CB (desired reaction)
A+BÆU
rU = -kUCACB2 (undesired reaction)
If I work semi-batch, what is better to do? Add B to A, or A to B?
If the reaction is strongly exothermic, a (slow) feed of one of the reactants can also
be desired
Also, in non-steady state processes, the use of Polymath for
problem-solving is strongly recommended
CEMI323 Class Test 1
2010
11 Aug.
a)
Overall mass balance of all species:
- [ out ] + [ gen. ] = [ acc. ] 3 (1 punt)
You have two (2) hours to complete this test. Answer all the questions.
[ in ]
Question 1 [15]
U 0v0 0 0
1.1)
dV
dt
Derive from first principles the design equation (integral form) for a batch reactor in
terms of conversion. Take note: This question will be marked negative.
for U
v0
d ( UV )
dt
U0
3 (1 punt)
Thus after integration:
1.2)
Consider the semi-batch reactor as given in Figure 1 for the elementary liquid-phase
reaction of A + B C.
V
V0 X 0 t
b)
Mole balance on species A:
B
[ in ] - [ out ]
A
Figure 1 – Semibatch reactor with constant molar feed.
The reactor is initially charged at time t = 0 min. with pure A. The reactor volume is
V0 at t = 0 min. Thereafter, only species B is fed slowly to A in the well mixed
reactor. Take note as B is continually fed to the reactor, the reactor volume increase,
V(t). The system is a constant-density system but the reactor volume change as time
progress.
a)
b)
c)
Show clearly that from a overall mass balance, the reactor volume is:
V(t) = V0 + vot
Show clearly that from a mole balance on species A:
v
dC A
rA 0 C A
dt
V
Show clearly that from a mole balance on species B:
v (C C B )
dC B
rB 0 B 0
dt
V
+ [ gen. ]
=
[ acc. ]
dN A
dt
VdC A C AdV
dt
dt
0 0 rAV ( t )
3 (1 punt)
d C AV
dt
3 (1 punt)
rAV
And from question 1.2 (a) dV/dt = v0
Thus
VdC A
dt
v 0C A VrA
dC A
dt
rA 3 (1 punt)
v0
CA
V
c)
Mole balance on specials B:
[ in ] - [ out ]
+ [ gen. ]
FB 0 0 rBV ( t )
=
[ acc.]
dN B
dt
dN B
rBV FB 0
dt
dC B
dV
dN B d (VC B )
CB
V
dt
dt
dt
dt
rBV FB 0 rBV v 0C B 0
dC B
dt
rB 33 (2 punte)
33 (2 punte)
3 (1 punt)
v 0 (C B 0 C B )
V
Exercise: Work this problem out using Polymath and compare the 2 methods (define
here values for constants here, such as CA0, CB0, v0, V0 etc.)
Study Unit 2.3:
Collection and Analysis of Rate Data (Ch 7)
Fogler (Chapter 7)
SU 1: Fogler, Ch. 1-4, more or less introductory (setting the scene, learning the jargon)
1.1 Ch 1: Mole balances
1.2 Ch 2: Conversion and Reactor Sizing
1.3 Ch 3: Rate Laws
1.4 Ch 4: Stoichiometry
SU 2: Fogler, Ch. 5-7, reactor design and analysis; the chemical engineer's bread and
butter (still introductory …..)
2.1 Ch 5: Isothermal Reactor Design: Conversion
2.2 Ch 6: Isothermal Reactor Design: Moles and Molar Flow rates
2.3 Ch 7: Collection and Analysis of Rate Data
SU 3 + P: Fogler, Ch. 8 and 11, and Project, somewhat more to practice … (still introductory …..)
3.1 Ch 8: Multiple Reactions
3.2 Ch 11:Nonisothermal Reactor Design - The Steady State Energy Balance and
Adiabatic PFR Applications
2.3. Study Unit 2.3
2.3.1 Background and introduction
2.3.2 The method of excess
2.3.3 The integral method
2.3.4 The differential method of analysis
2.3.5 Nonlinear regression
2.3.6 Reaction-Rate data from differential reactors
2.3.1 Background and introduction
•
•
•
•
•
•
In the design of reactor, the kinetic rate (-ri) equation is an important aspect.
It was already discussed that this equation can have various (mathematical)
forms.
Normally, the rate equation is obtained under well controlled experimental
conditions in laboratory (small scale) equipment.
Often batch reactors and differentially operating flow reactors are used for this
purpose.
In this section, the various reactor systems and specifically methods of analysis
are discussed.
Homogeneous systems are often analysed by batch methods, while for
heterogeneous reactions differential flow reactors are preferred.
The idea
Chemistry
A few grams
(batch)
Pilot plant operation
A few kilograms per hour
(continuous)
Large scale chemical operation
A few hundred kilotons per year
(continuous)
Chemical Engineering
Stage of development
A rate equation is a function of T and can be a function of all species that are
present in the reaction system (SU 1.3):
െ‫ݎ‬஺ = ݇஺ (ܶ)݂ ‫ܥ‬஺ , ‫ܥ‬஻ … … . .
Reaction rate analyses are normally performed isothermally, as to solely find the
influence of concentration.
The temperature dependent variables (A, E) can then be obtained by comparing
isothermal results performed at different temperature (Arrhenius plot)
݇஺ ܶ = ‫ܣ‬
ln ݇஺
ିா
݁ ோ்
‫ܧ‬
= ݈݊ ‫ ܣ‬െ
ܴܶ
ln ݇஺
2.3.2 The method of excess
െ‫ݎ‬஺ = ݇஺ (ܶ)݂ ‫ܥ‬஺ , ‫ܥ‬஻ … … . .
The kinetic rate equation can be dependent on more then one species
(e.g. for a second order reaction, -rA = k CA CB).
For an easier interpretation, one can chose on of the reactant in
excess, so that this concentration is negligibly changes over time.
If D is to be found, have B in large excess, and if E has to be found
have A in large excess.
If B in large excess:
where
If A in large excess:
where
Once D and E known, kA can be determined by
2.3.3 The integral method
Constant volume batch reactor analysis
If a constant batch reactor is used, the reaction parameters can be
obtained by following the concentration.
General mole balance for a batch reactor:
݀ܰ஺
= ‫ݎ‬஺ ܸ
݀‫ݐ‬
݀ ‫ܥ‬஺ ܸ
= ‫ݎ‬஺ ܸ
݀‫ݐ‬
V = constant
݀‫ܥ‬஺
= ‫ݎ‬஺
݀‫ݐ‬
Since rA = f (CA), the reaction rate parameters can such be obtained,
e.g. for a 1st order in A:
݀‫ܥ‬஺
= െ݇ ‫ܥ‬஺
݀‫ݐ‬
And for an arbitrary order (D):
݀‫ܥ‬஺
= െ݇‫ܥ‬஺ఈ
݀‫ݐ‬
Since kinetic parameters are also obtained from chemical reactors
(although on a controlled way and on a small scale), the general
chemical reactor theory is also applicable to these test reactors.
Since often the exact form of the kinetic rate equation is not known,
a trial-and-error procedure can be followed, e.g. for a power-ratelaw, the various integral rate laws can be determined for the
different powers (D)
An D can be guessed and the CA, t curve can be predicted from
which k can be determined (fitted)
Batch reactor
If D = 0,
If data fits, the kinetic rate
constant (k) can be obtained
from the respective figure (as
the negative slope of the curve)
and:
Slope = - k
Batch reactor
If D = 1,
If data fits, the kinetic rate
constant (k) can be obtained
from the respective figure (as
the slope of the curve)
and:
Slope = k
Batch reactor
If D = 2,
If data fits, the kinetic rate
constant (k) can be obtained
from the respective figure (as
the slope of the curve)
and:
Slope = k
Batch reactor
So in the integral method, we evaluate (guess) an order and see if it
fits. If it does not fit, e.g. if for a 2nd order reaction, not a straight line
is obtained between 1/CA and t, another value for D should be tried
(so this can be a tedious trial and error procedure)
2.3.4 The differential method of analysis
The integral method should be tested per case (e.g. per D), while in the
differential method a more general approach can be applied.
Consider general power law kinetics:
After a mathematical operation, this can be rewritten to:
So, a plot of (ln(–dCA/dt)) vs. ln(CA)
results in a straight line with the slope of
Dand kA can be obtained from:
ln
ln
How to obtain d(CA)/dt?
Experimental data is resulting in CA as a function of A. So, how can
now d(CA)/dt be obtained?
This can be done on various ways:
• Graphical differentiation (old school method)
• Numerical differentiation
• Differentiation of a polynomial fit
The graphical and numerical methods are clearly explained in
Example 7-2 (Example 5-1)
3
CA (mol/m )
0
50.0
50
38.0
100
30.6
150
25.6
200
22.2
250
19.5
300
17.4
t (min)
60
CA (mol m-3)
50
40
30
20
10
0
0
50
100
150
t (min))
200
250
300
350
Differentiation of a polynomial fit
60
CA (mol m-3)
50
y = 3.629E-04x 2 - 2.111E-01x + 4.892E+01
R² = 9.930E-01
40
30
20
10
0
0
50
100
150
200
250
300
t (min))
‫ܥ‬஺ = 0.0003629 ‫ ݐ‬ଶ െ 0.2111 ‫ ݐ‬+ 48.29
݀‫ܥ‬஺
= 0.0007258 ‫ ݐ‬െ 0.2111
݀‫ݐ‬
350
Once d(CA)/dt has be obtained (here the finite difference method was
used), it can be plotted against CA (logarithmic scale)
1
CA (mol/m ) - dCA/dt
0
50.0
0.286
0.194
50
38.0
100
30.6
0.124
0.084
150
25.6
200
22.2
0.061
250
19.5
0.048
0.036
300
17.4
t (min)
1.0
- dCA/dt (mol m-3 min-1)
3
10.0
0.1
0.01
CA (mol m-3)
100.0
And ln(d(CA)/dt) can be plotted against ln(CA)
CA (mol/m3) - dCA/dt
0
50.0
0.286
50
38.0
0.194
100
30.6
0.124
150
25.6
0.084
200
22.2
0.061
250
19.5
0.048
300
17.4
0.036
t (min)
ln CA
3.912
3.638
3.421
3.243
3.100
2.970
2.856
ln (- dCA/dt)
-1.252
-1.640
-2.087
-2.477
-2.797
-3.037
-3.324
D = 2.0
kA = 1.3·10-4 mol m-3 min-1
e-8.9712
െ‫ = ܣݎ‬1.27 ȉ 10ିସ ‫ܥ‬஺ଶ.଴଴
[mol m-3 min-1]
ln (-dCA/dt (mol m-3 min-1))
0.0
2.0
2.5
3.0
3.5
-0.5
-1.0
-1.5
y = 1.9959x - 8.9712
R² = 0.9942
-2.0
-2.5
-3.0
-3.5
ln(CA (mol m-3))
4.0
4.5
2.3.5 Nonlinear regression
Using the previous example, one can also analytically integrate the
rate equation first and then fit the data.
If
•
•
•
•
•
with: CA(t = 0) = CA0 , then:
Guess D and k (e.g. 1.5 and 1.00·10-4 mol m-3 min-1)
Determine CA for each t (CA,calc)
Determine “error” between exp. and calc. value (CA,exp – CA, calc)2
Sum the “error”
Minimise the sum of errors
Can be done with Excel (solver routine)
Initial guess:
D
k
1.5
1.00E-04 mol m-3 min-1
EXP
CALC
2
3
3
t (min) CA (mol/m ) CA (mol/m ) (exp-calc)
0
50.0
50.0
0.0000
50
38.0
48.3 105.6378
100
30.6
46.6 257.3940
150
25.6
45.1 379.8836
200
22.2
43.6 458.5586
250
19.5
42.2 515.6862
300
17.4
40.9 550.8553
ɇ;ĞdžƉͲĐĂůĐͿ2
After minimising error
changing D and k:
(compare to previously found 2.0 and 1.3 ȉ 10ିସ )
D
k
2.04
-3
-1
1.11E-04 mol m min
EXP
CALC
2
t (min) CA (mol/m ) CA (mol/m3) (exp-calc)
0
50.0
50.0
0.0000
50
38.0
37.9
0.0038
100
30.6
30.6
0.0003
150
25.6
25.7
0.0090
200
22.2
22.2
0.0022
250
19.5
19.5
0.0003
300
17.4
17.4
0.0000
3
2268.02
ɇ;ĞdžƉͲĐĂůĐͿ2
0.02
Off course other software (Polymath, Matlab) can also be used
With the current computational power of computers, an approach
where a differential equation (dCA/dt) is solved (numerically)
simultaneously with a solver (fitting) routine to obtain the various
kinetic parameters (e.g. k, D etc.) is preferred, specifically also when
multiple reactions occur
This can easily be carried out by e.g. Matlab (but falls outside the
scope of this course)
2.3.6 Reaction-Rate data from differential reactors
This approach is often used for heterogeneous (e.g. catalytic) reactions.
In a differential reactor, the conditions are such chosen that the conversion of
the various reactants is small (< 10%).
This is done to keep the concentration of the reactants approximately
constant (e.g. if CA0 = 1.0 mol dm-3, and X = 5% then CA = 0.95 mol dm-3,
which is almost the same as the inlet concentration), so that the reaction
conditions can be approximated by the inlet (feed) conditions
The advantage of the method is that the reaction rate can be obtained directly
by relative straightforward calculations (without having to assume a model)
So the reaction rate can be directly determined from the
concentration of the formed product (here P), and the selected flow
rate and catalyst weight
The inlet concentrations can be varied and related to the obtained
reaction rates and a rate equation can be found
Study Example 7-4 (5-5) for more detail
Study Unit 3.1:
Multiple reactions
Fogler (Chapter 8)
SU 1: Fogler, Ch. 1-4, more or less introductory (setting the scene, learning the jargon)
1.1 Ch 1: Mole balances
1.2 Ch 2: Conversion and Reactor Sizing
1.3 Ch 3: Rate Laws
1.4 Ch 4: Stoichiometry
SU 2: Fogler, Ch. 5-7, reactor design and analysis; the chemical engineer's bread and
butter (still introductory …..)
2.1 Ch 5: Isothermal Reactor Design: Conversion
2.2 Ch 6: Isothermal Reactor Design: Moles and Molar Flow rates
2.3 Ch 7: Collection and Analysis of Rate Data
SU 3 + P: Fogler, Ch. 8 and 11, and Project, somewhat more to practice … (still introductory …..)
3.1 Ch 8: Multiple Reactions
3.2 Ch 11:Nonisothermal Reactor Design - The Steady State Energy Balance and
Adiabatic PFR Applications
3.1. Study Unit 3.1
3.1.1 Introduction and definitions
3.1.2 Algorithm for multiple reactions
3.1.3 Reactions in series, parallel and complex reactions
3.1.4 Reactor selection and operating conditions
3.1.5 Membrane reactors to improve selectivity in
multiple reactions
3.1.1 Introduction and definitions
Up to date only reaction systems in which 1 reaction
occurs were studied.
In reality, this is seldom the case, and this study units
will discuss the consequence of multiple reactions
In general, reaction systems consists of parallel
reactions, series reactions, of a combination of the 2
(complex reaction systems) and independent reactions
Parallel reactions
In a parallel reaction, the reactant (e.g. A) can react to different
products (e.g. B and C), according to different rate laws:
B
A
C
An example is the partial oxidation of methane to formaldehyde
(desired) and CO2 (undesired). In general important in partial
oxidation reactions
CH4 + O2
CH4 + 2O2
= CH2O + H2O (desired)
= CO2 + 2 H2O (undesired)
Series (consecutive) reactions
In a series reaction, the reactant (e.g. A) reacts to a product (e.g. B),
that can undergo a consecutive reaction to another product (e.g. C)
A
B
C
An example is the chlorination of benzene (C6H6) to mono-chlorobenzene (C6H5Cl) and further alkylation to di-chloro-benzene (C6H4Cl2):
C6H6 + Cl2 = C6H5Cl + HCl
C6H5Cl + Cl2 = C6H4Cl2 + HCl
Complex reactions
In a complex reaction, both series and parallel reactions occur
simultaneously:
An example is the partial oxidation of methane, in which both
formaldehyde and methane can react to CO2
CH4 + O2
CH2O + O2
CH4 + 2O2
= CH2O + H2O (desired)
= CO2 + H2O (undesired)
= CO2 + 2 H2O (undesired)
B
A
C
Independent reactions
Are reactions that occur simultaneously, but reactants and products
do not react with each other:
An example is the cracking of crude oil, in which multiple
reactions occur, but 2 of them are:
C15H32 = C12H26 + C3H6
C8H18 = C6H14 + C2H4
Reactions occur independently, when reactants and products do
not interfere each other
Selectivity and Yield (2 important process parameters)
Desired Reaction:
kD
A o
D
Undesired Reaction:
kU
A o
U
Instantaneous
Selectivity
Yield
S DU
rD
rU
YD
rD
rA
Overall
~
S DU
~
YD
FD
FAO FA
FD
FU
ND
N AO N A
Reactors and reaction conditions are often designed to maximise
the yield of the desired product
kD
A o
D
kU
A o
U
Remark: The definitions of Fogler are not used throughout the scientific
community and are even a bit outdated.
More conventionally (specifically in research papers), the following definitions
are used for Selectivity and Yield:
Selectivity ‫= ܦ ݋ݐ‬
Yield ‫= ܦ ݂݋‬
௠௢௟௘௦ ௢௙ ஽ ௙௢௥௠௘ௗ
௠௢௟௘௦ ௢௙ ஽ ௙௢௥௠௘ௗ ௜௙ ௔௟௟ ࢉ࢕࢔࢜ࢋ࢚࢘ࢋࢊ ஺ ௥௘௔௖௧௦ ௧௢ ஽
௠௢௟௘௦ ௢௙ ஽ ௙௢௥௠௘ௗ
௠௢௟௘௦ ௢௙ ஽ ௙௢௥௠௘ௗ ௜௙ ௔௟௟ ஺ ௜௦ ௖௢௡௩௘௥௧௘ௗ ௧௢ ஽
The advantage of these definitions is that both values are between 0 and 1 (0
and 100%) and that:
Yield = Conversion x Selectivity
Take home message: In analyzing papers reporting on yield and selectivity,
critically review how these are defined.
We further continue with the definitions as stated in the book
Exercise: 12 mol A/s is fed to a reactor, and the following reactions take place:
A Æ 2B
AÆC
The effluent stream contains 2.0 mol/s A and 1.0 mol/s C. Determine the
overall selectivity and yield of B (desired component) with the definitions of
Fogler, and the definitions given in the previous slide.
Optimisation strategy
What should be the criterion for designing the reactor ?
Is it necessary that reactor operates such that minimum amount of undesired products
are formed ?
D
A
Reactor
System
D, A
U
S
E
P
A
R
A
T
O
R
S
E
P
A
R
A
T
O
R
Total Cost
Cost
A, D
U
Will be further developed in 4th year design project
Reactant Conversion
3.1.2 Algorithm for multiple reactions
Design Equation for Reactors
Gas-Phase
Batch
Semi-Batch
CSTR
PFR
PBR
Liquid Phase
NOTE the design equations
are EXACTLY as it were for
single reaction
Net Rate of Reaction
Sum up the rates of formation for each reaction in order
to obtain the net rate of formation.
If N reactions are taking place
k
3C D
Reaction 1: A B o
1A
Reaction 2: A 2C o 3E
k2 A
Reaction 3:
k3 B
2 B 3E o
4F
rA
r1 A r2 A r3 A ... rqA
N
¦r
iA
i 1
rB
r1B r2 B r3 B ... rqB
N
¦r
iB
i 1
...
Reaction q: A 1 B ko G
NA
2
rj
N
¦r
ij
i 1
species
reaction number
Example: Stoichiometry & Rate Laws
Consider the following set of reactions (and associated rate equations), which
occur in the reduction of NO:
k1
4 NH 3 6 NO o
5 N 2 6 H 2O
r1NO
k2
2 NO o
N 2 O2
k3
N 2 2O2 o
2 NO2
k1NO C NH 3 C NO
r2 N 2
r3O2
1.5
k 2 N 2 C NO
2
k3O2 C N 2 CO2
2
Write the rate law for each species in each reaction and then write the net
rates of formation of NO, O2, and N2.
2
5
k1
NO NH 3 o N 2 H 2O
3
6
k2
2 NO o
N 2 O2
1
k3
O2 N 2 o
NO2
2
r1NO
k1NO C NH 3 C NO
r2 N 2
r3O2
1 .5
k 2 N 2 C NO
2
k3O2 C N 2 CO2
2
Example: Stoichiometry & Rate Laws
2
5
k1
NO NH 3 o N 2 H 2O
3
6
r1NO
Reaction 1:
r1NH 3
2
3
r1NO
1
r1N 2
5
6
r1H 2O
r1NH 3
2
(r1NO )
3
2
1.5
k1NO C NH 3 C NO
3
r1N 2
5
(r1NO )
6
5
1. 5
k1NO C NH 3 C NO
6
r1H 2O
r1NO
k1NO C NH 3 C NO
1. 5
1
k1NO C NH 3 C NO
1.5
Example: Stoichiometry & Rate Laws
Similar, for reaction 2, we have
k2
2 NO o
N 2 O2
r2 N 2
k 2 N 2 C NO
k3
N 2 2O2 o
2 NO2
k3O2 C N 2 CO2
2r2 N 2
r3 N 2
1
(r3O2 )
2
r3 NO2
r3O2
2r2O2
2k 2 N 2 C NO
2
2
For reaction 3:
r3O2
r2 NO
1
2
k3O2 C N 2 CO2
2
k3O2 C N 2 CO2
2
then write the net rates of formation of NO, O2, and N2.
2
Example: Stoichiometry & Rate Laws
N
¦r
rj
ij
i 1
rNO
3
¦ riNO
5
2
k1
NO NH 3 o N 2 H 2O r1NO
3
6
k1NO C NH 3 C NO
k2
2 NO o
N 2 O2
r2 N 2
1
k3
O2 N 2 o
NO2
2
r1NO r2 NO 0
i 1
rN 2
rO2
3
¦r
i 1
3
iN 2
¦ riO2
i 1
k1NO C NH 3 C NO
r1N 2 r2 N 2 r3 N 2
r1O2 r2O2 r3O2
1.5
r3O2
2k 2 N 2 C NO
1.5
k 2 N 2 C NO
2
k3O2 C N 2 CO2
2
2
5
1
1.5
2
2
k1NO C NH 3 C NO k 2 N 2 C NO k3O2 C N 2 CO2
6
2
2
k 2 N 2 C NO k3O2 C N 2 CO2
2
Rates for the other components can also be generated
3.1.3 Reactions in series, parallel and
complex reactions
An illustration of how series and parallel reactions effect conversion and selectivity
follows, on the hand of a liquid phase reaction, carried out in a PFR, in which
different reaction schemes are discussed. (The description of CSTRs can be followed
from the book)
Intermezzo: A 1st order reaction that is carried out in a PFR, in which no volume
changes take place can be analysed as follows:
݀‫ܨ‬஺
= ‫ݎ‬஺
ܸ݀
1st order reaction
Since FA = CA v0 and V = W v0 :
݀‫ܨ‬஺
= െ݇ ‫ܥ‬஺
ܸ݀
݀‫ܥ‬஺
= െ݇ ‫ܥ‬஺
݀W
A nice way to follow the concentration of A as a
function of the residence time in the reactor
1st order irreversible liquid-phase reaction (A ÆB) in a PFR; CA0 = 1000 mol/m3
d(CA) / d(tau) = rA #mol/m3/s
CA(0) = 1000 #mol/m3
d(CB) / d(tau) = -rA #mol/m3/s
CB(0) = 0
AÆB
-rAB = kAB CA
(Analytical solution:
CA = CA0·e-kW
kAB = 3.0·10-4 s-1
tau(0) = 0
tau(f) = 7200 #s
1200
1000
rA = -kAB*CA
kAB = 0.0003 # s-1
A
CA (mol/m3)
800
600
400
B
200
0
0
1000
2000
3000
4000
W (s)
5000
6000
7000
8000
1st order irreversible liquid-phase parallel reaction (A ÆB, A Æ C) in a PFR; CA0 =
1000 mol/m3
d(CA) / d(tau) = rAB + rAC
CA(0) = 1000
d(CB) / d(tau) = -rAB
CB(0) = 0
d(CC) / d(tau) = -rAC
CC(0) = 0
A
B
-rAB = kAB CA
C
-rAC = kAC CA
kAB = 3.0·10-4 s-1
kAC = 1.0·10-4 s-1
1200
10
9
1000
8
rAB = -kAB*CA
kAB = 0.0003
rAC = -kAC*CA
kAC = 0.0001
CA (mol/m3)
800
B
A
600
7
6
5
4
400
3
C
200
2
1
0
0
1000
2000
3000
4000
W (s)
5000
6000
7000
0
8000
Selectivity (FB/FC)
tau(0) = 0
tau(f) = 7200
1st order irreversible liquid-phase series reaction (A ÆB Æ C) in a PFR; CA0 = 1000
mol/m3
d(CA) / d(tau) = rAB
CA(0) = 1000
d(CB) / d(tau) = -rAB+rBC
CB(0) = 0
d(CC) / d(tau) = -rBC
CC(0) = 0
A
B
-rAB = kAB CA
C
kAB = 3.0·10-4 s-1
-rBC = kBC CB
kBC = 3.0·10-4 s-1
10
1200
9
1000
8
7
rAB = -kAB*CA
kAB = 0.0003
rBC = -kBC*CB
kBC = 0.0003
CA (mol/m3)
800
C
A
600
6
5
4
400
B
3
2
200
1
0
0
0
1000
2000
3000
4000
W (s)
5000
6000
7000
8000
Selectivity (FB/FC)
tau(0) = 0
tau(f) = 7200 #s
1st order irreversible liquid-phase series and parallel reaction (A ÆB Æ C, and A Æ C)
in a PFR; CA0 = 1000 mol/m3
A
B
-rAB = kAB CA
C
-rBC = kBC CB
-rAC = kAC CA
kAB = 3.0·10-4 s-1
kBC = 3.0·10-4 s-1
kBC = 1.0·10-4 s-1
1200
tau(0) = 0
tau(f) = 7200
9
1000
8
A
800
CA (mol/m3)
rAB = -kAB*CA
kAB = 0.0003
rBC = -kBC*CB
kBC = 0.0003
rAC = -kAC*CA
kAC = 0.0001
10
7
C
6
600
5
4
400
B
200
3
2
1
0
0
1000
2000
3000
4000
W (s)
5000
6000
7000
0
8000
Selectivity (FB/FC)
d(CA) / d(tau) = rAB + rAC
CA(0) = 1000
d(CB) / d(tau) = -rAB+rBC
CB(0) = 0
d(CC) / d(tau) = -rAC-rBC
CC(0) = 0
For the complex reduction of NO by NH3 (as discussed previously)
Consider the following set of reactions (and associated rate equations), which
occur in the reduction of NO:
k1
4 NH 3 6 NO o
5 N 2 6 H 2O
r1NO
k2
N 2 O2
2 NO o
k1NO C NH 3 C NO
r2 N 2
k3
N 2 2O2 o
2 NO2
r3O2
k 2 N 2 C NO
2
k3O2 C N 2 CO2
2
2
5
k1
NO NH 3 o N 2 H 2O r1NO
3
6
k1NO C NH 3 C NO
k2
2 NO o
N 2 O2
r2 N 2
1
k3
O2 N 2 o
NO2
2
r3O2
1.5
1.5
k 2 N 2 C NO
2
k3O2 C N 2 CO2
2
2
5
k1
NO NH 3 o N 2 H 2O r1NO
3
6
k1NO C NH 3 C NO
k2
2 NO o
N 2 O2
r2 N 2
1
k3
O2 N 2 o
NO2
2
r3O2
1.5
k 2 N 2 C NO
2
k3O2 C N 2 CO2
2
In an isobaric and isothermal PFR (P = 5.0 bar, T = 301 K), NO is reduced in a
nitrogen stream, by injecting NH3. The feed consists of NO and NH3 only, and NH3
is fed with an excess of 50% (according to reaction 1). The total flow rate of the
stream is 1.0 mol/s.
6 components (NO, NH3, N2, H2O, O2 and NO2), so 6 DE's
From given data: FNO(0) = 0.5 mol/s; FNH3(0) = 0.5 mol/s; Fother (0) = 0
2
5
k1
NO NH 3 o N 2 H 2O r1NO
3
6
k1NO C NH 3 C NO
k2
N 2 O2
2 NO o
r2 N 2
1
k3
O2 N 2 o
NO2
2
r3O2
1.5
k 2 N 2 C NO
2
k3O2 C N 2 CO2
2
Write the reaction equations in terms of the given reactions (i.e. for NO and NH3):
‫ݎ‬ேை = ‫ݎ‬ଵேை െ 2 ‫ݎ‬ேమ
2
‫ݎ‬ேுయ = ‫ݎ‬ଵேை
3
d(FNO) / d(V) = RNO #mol/s
FNO(0) = 0.5
d(FNH3) / d(V) = RNH3 #mol/s
FNH3(0) = 0.5
d(FN2) / d(V) = RN2 #mol/s
FN2(0) = 0
d(FH2O) / d(V) = RH2O
FH2O(0) = 0
d(FO2) / d(V) = RO2
FO2(0) = 0
d(FNO2) / d(V) = RNO2
FNO2(0) = 0
V(0) = 0
V(f) = 500
R2NO = -2*R2N2
R2O2 = R2N2
6 DE's
(1 per species)
Initial and final
condition
R1NO = -k1NO*CNH3*CNO^1.5
R2N2 = k2N2*CNO^2
R3O2 = -k3O2*CN2*CO2^2
R1NH3 = (2/3)*R1NO
R1N2 = -(5/6)*R1NO
R1H2O = -R1NO
Kinetic rate
equations (3)
Disappearance rate
of Reaction 1
Disappearance rate
of Reaction 2/3
R3N2 = (1/2)*R3O2
R3NO2 = -R3O2
RNO = R1NO+R2NO
RNH3 = R1NH3
RN2 = R1N2+R2N2+R3N2
RH2O = R1H2O
RO2 = R2O2+R3O2
RNO2 = R3NO2
Disappearance
rate per species
CNO = CT0*FNO/FT
CN2 = CT0*FN2/FT
CNH3 = CT0*FNH3/FT
CO2 = CT0*FO2/FT
Concentration
of species
FT = FNO+FNH3+FN2+FH2O+FO2+FNO2
CT0 = 0.20 # mol/dm3
k1NO = 4.3
k2N2 = 2.7
k3O2 = 5.8
Reaction rate
constants (k1NO has
changed compared to
book)
2
5
k1
NO NH 3 o N 2 H 2O r1NO
3
6
k1NO C NH 3 C NO
k2
N 2 O2
2 NO o
r2 N 2
1
k3
O2 N 2 o
NO2
2
r3O2
1.5
k 2 N 2 C NO
2
k3O2 C N 2 CO2
2
0.6
1.05
1
0.5
0.95
0.3
NH3
NO
N2
H2O
O2
NO2
0.9
Total
0.85
0.2
0.8
0.1
0.75
0.7
0
0
100
200
300
V (dm3)
400
500
FT (mol/s)
Fi (mol/s)
0.4
3.1.4 Reactor selection and
operating conditions
Parallel reactions
S DU
D(esired)
ாವ
ఈ
ିோ்
‫ݎ‬஽ = ‫ܣ‬஽ ݁
ȉ ‫ܥ‬஺ భ
U(ndesired)
ாೆ
ఈ
ିோ்
‫ݎ‬௎ = ‫ܣ‬௎ ݁
ȉ ‫ܥ‬஺ మ
rD
rU
A
For this reaction system:
If D1 > D2, keep concentration of A high: Use PFR and avoid dilution, high
pressure (gas-phase reactions)
If D2 > D1, keep concentration of A low: Use CSTR and promote dilution, low
pressure (gas-phase reactions)
If ED > EU, carry out reaction at high temperature
If EU > ED, carry out reaction at low temperature
Section 8.6
Consider the following reaction scheme:
A+BÆD
-r1A = k1A·CA2·CB
k1A = 2 dm6/mol2 s
A+BÆU
-r2A = k2A·CA·CB2
k2A = 3 dm6/mol2 s
Under what conditions should you carry out the reaction, as to optimise the production
of the desired product (D)?
A+BÆD
A+BÆU
-r1A = k1A·CA2·CB
-r2A = k2A·CA·CB2
k1A = 2 dm6/mol2 s
k2A = 3 dm6/mol2 s
If this reaction was carried out in a PFR (FA0 = FB0 = 4.0 mol/s, CT0 = 0.8 mol/m3), the
following profiles can be obtained:
d(FA) / d(V) = r1A+r2A
FA(0) = 4
d(FB) / d(V) = r1A+r2A
FB(0) = 4
d(FD) / d(V) = -r1A
FD(0) = 0
d(FU) / d(V) = -r2A
FU(0) = 0
4.5
4
3.5
r1A = -k1A*CA^2*CB
r2A = -k2A*CA*CB^2
Fi (mols/)
3
V(0) = 0
V(f) = 50
FA, FB
2.5
FU
2
1.5
k1A = 2
k2A = 3
CT0 = 0.8
1
FD
0.5
0
CA = CT0*FA/FT
CB = CT0*FB/FT
FT = FA+FB+FD+FU
0
5
10
15
20
25
V (dm3)
30
35
40
45
50
A+BÆD
A+BÆU
-r1A = k1A·CA2·CB
-r2A = k2A·CA·CB2
k1A = 2 dm6/mol2 s
k2A = 3 dm6/mol2 s
Overall Selectivity (FD/FU) and oevrall yield (FD/(FA0-FA))
0.8
Overall Selectivity (FD/FU)
0.7
0.6
0.5
Overall Yield (FD/(FA0 - FA)
0.4
0.3
0.2
0.1
0
0
5
10
15
20
25
V (dm3)
30
35
40
45
50
3.1.5 Membrane reactors to improve
selectivity in multiple reactions
Selectivity can be increased by dosing reactant (B) through a porous
medium (membrane), application d) of Figure 1
High-temperature membrane reactors: potential and problems, G. Saracco, H.W.J.P. Neomagus, G.F. Versteeg,
W.P.M. van Swaaij, Chemical Engineering Science 54 (1999) 1997-2017
A+BÆD
A+BÆU
-r1A = k1A·CA2·CB
-r2A = k2A·CA·CB2
k1A = 2 dm6/mol2 s
k2A = 3 dm6/mol2 s
So instead of feeding B at once (at the feed), B is gradually dosed over the length of the
reactor.
If the same amount of B is fed (compared to PFR) the feeding rate is (FBtot/VPFR), and
the amount fed to dV of the reactor is (FBtot/VPFR)dV
In this case, FBtot = 4.0 mol/s (as in the PFR) and VPFR = 50 dm3, so FBtot/VPFR =
0.080 mol/dm3 s
d(FA) / d(V) = r1A+r2A
FA(0) = 4
d(FB) / d(V) = r1A+r2A+Rb
FB(0) = 0
d(FD) / d(V) = -r1A
FD(0) = 0
d(FU) / d(V) = -r2A
FU(0) = 0
A+BÆ D
A+BÆ U
CT0 = 0.8
CA = CT0*FA/FT
CB = CT0*FB/FT
FT = FA+FB+FD+FU
Rb = 4/50 # mol/m3 s
4
FA
3.5
3
Fi (mols/)
k1A = 2
k2A = 3
k1A = 2 dm6/mol2 s
k2A = 3 dm6/mol2 s
4.5
V(0) = 0
V(f) = 50
r1A = -k1A*CA^2*CB
r2A = -k2A*CA*CB^2
-r1A = k1A·CA2·CB
-r2A = k2A·CA·CB2
2.5
2
1.5
FD
1
FB
0.5
FU
0
0
5
10
15
20
25
V (dm3)
30
35
40
45
50
Comparison of PFR and
Membrane Reactor (M)
10
A+BÆ D
A+BÆ U
-r1A = k1A·CA2·CB
-r2A = k2A·CA·CB2
k1A = 2 dm6/mol2 s
k2A = 3 dm6/mol2 s
1.2
Selectivity MR
9
Overall Selectivity (FD/FU)
8
Yield MR
7
0.8
6
5
0.6
4
Yield PFR
0.4
3
2
0.2
Selectivity PFR
1
0
0
0
5
10
15
20
25
V (dm3)
30
35
40
45
50
Overall Yield (= FD/(FA0 - FA)
1
Study Unit 3.2:
Nonisothermal Reactor Design – The steady-State Energy
Balance and Adiabatic PFR Applications
Fogler (Chapter 11)
SU 1: Fogler, Ch. 1-4, more or less introductory (setting the scene, learning the jargon)
1.1 Ch 1: Mole balances
1.2 Ch 2: Conversion and Reactor Sizing
1.3 Ch 3: Rate Laws
1.4 Ch 4: Stoichiometry
SU 2: Fogler, Ch. 5-7, reactor design and analysis; the chemical engineer's bread and
butter (still introductory …..)
2.1 Ch 5: Isothermal Reactor Design: Conversion
2.2 Ch 6: Isothermal Reactor Design: Moles and Molar Flow rates
2.3 Ch 7: Collection and Analysis of Rate Data
SU 3 + P: Fogler, Ch. 8 and 11, and Project, somewhat more to practice … (still introductory …..)
3.1 Ch 8: Multiple Reactions
3.2 Ch 11:Nonisothermal Reactor Design - The Steady State Energy Balance and
Adiabatic PFR Applications
3.2. Study Unit 3.2
3.2.1 Introduction and rationale
3.2.2 The energy balance
3.2.3 Overview of energy balances
3.2.4 The user friendly Energy Balance
3.2.5 Adiabatic operation
3.2.6 Adiabatic Equilibrium conversion
3.2.1 Introduction and rationale
From SU1
Coal
Coal
CO
Gasification H2
H2O
O2
Entrained flow reactor
(dp ~ 50 Pm)
The conversion of coal into syngas is
carried out in different form of reactors, all
over the world.
The choice of reactor is often related to the
type of coal used.
Most (coal gasification)
reactors do not work under
isothermal conditions!
Fluidised bed reactor
(dp ~ 5 mm)
Fixed bed reactor
(dp ~ 50 mm)
Why do we need an Energy Balance (EB)?
Every reaction proceeds with release or absorption of
heat.
The amount of heat released or absorbed depends on
– the nature of reacting system
– the amount of material reacting
– temperature and pressure of reacting system
and can be calculated from heat of reaction ('HRxn)
Most industrial reactors will require heat input or heat
removal, hence, we need energy balances to describe this.
Up to now, we discussed only the “mass balance”.
How about the “energy balance” or the heat effects?
dE
dt
n
n
i 1
i 1
Q W ¦ Fi Ei in ¦ Fi Ei out
The basis is "old" material; a combination
of material that was discussed in
Thermodynamics and Process Principles II
From CEMI222, sign convention for work
'U Q W
1st law for closed
systems (neglecting
kinetic and potential
energy)
Important note (compared to Process Principles II).
In this module the following equation will be given:
'U Q W
This is only a matter of definition: In Felder and
Rousseau, work is defined as work done by the
system; in Koretsky work is defined as work done
on (into) the system (Koretsky’s definition more
preferred lately)
CEMI222-Chapter 2, Slide 28
Fogler follows the Felder approach by defining the work
as DONE by the system (work out = positive)
Why Energy Balance ?
A
A & B, X = 0.7
Ao B
Consider an exothermic 1st order liquid phase reaction, operated adiabatically in a PFR:
Arrhenius Equation
E
rA
FA0
dX
Mole balance
dV
Rate law rA
v
Stoichiometry FA
CA
kC A
k
Ae
RT
dX
dV
k (1 X )
v0
dX
dV
ª9
E § 1 1 ·º (1 X )
k1 exp « ¨¨ ¸¸»
9
T1 T ¹¼ v9
0
¬ R ©9
v0
C Av0
C A0 (1 X )
Need relationships : X
T
V
3.2.2 The energy balance
From Thermodynamics (closed systems)
GQ
First law
To a closed (mass) system:
'E GQ GW
The change in total energy of the system
GW
The work done by the system
to the surroundings
The heat flow to the system
From Thermodynamics
(open systems)
GQ
Rate of flow of heat to the system
from the surroundings
dE
dt
Fin
Fout
Hin
Hout
GW
n
n
i 1
i 1
Q W ¦ Fi Ei in ¦ Fi Ei out
Rate of accumulation of
energy in the system
Rate of work done by the
system on the surroundings
Rate of energy leaving the system
by mass flow out the system
Rate of energy added to the system
by mass flow into the system
General Form of Energy Balance
Rate of flow
of heat to the
system from
the
surrounding
–
Rate of work
done by the
system on
the
surroundings
+
Rate of energy
Rate of energy
added to the
leaving the
system by
– system by
mass flow into
mass flow out
the system
of the system
=
W s
Fin Ein
Fout Eout
Control Volume
Q
Q W Fin Ein Fout Eout
dEˆ
dt
Rate of
accumulation
of the energy
within the
system
Evaluation of the Work (W) term
W s
F1 , E1
F1 , E1
F2 , E2
F2 , E2
...
...
Q
Fn , En
dE
dt
Fn , En
n
n
i 1
i 1
Q W ¦ Fi Ei in ¦ Fi Ei out
In a chemically reacting systems, there are usually two types of work that need
to be accounted for – (i) Shaft Work (e.g. work done by impellers in a CSTR
and batch reactor) and (ii) Flow Work
W
W s Rate of Flow Work
Rate of flow work is rate of work to
get mass into and out of the system
molar volume
Rate of Flow Work
n
n
i 1
i 1
¦ Fi PVˆ i in ¦ Fi PVˆi
out
Koretsky: v
~
Fogler: V
Evaluation of the E term
W s
F1 , E1
F2 , E2
F2 , E2
...
...
Q
Fn , En
dE
dt
Ei
F1 , E1
n
~
~
Q W s ¦ Fi ( Ei PVi ) in ¦ Fi ( Ei PVi ) out
n
i 1
u 2i
Ui g z ...
2
2
Fn , En
u
U i !! i gzi other
2
i 1
Energy Ei is the sum of internal, kinetic, potential
and any other type of energies.
? Ei # U i
For a majority of reactors, only internal energy is important
Energy Balance Equation in terms of Enthalpy
dE
dt
n
n
~
Q Ws ¦ Fi (U i PVi ) in ¦ Fi (U i PVi ) out
~
i 1
i 1
~
U i PVi Enthalpy!!, function of T
Hi
dE
dt
n
n
i 1
i 1
Q W s ¦ Fi H i in ¦ Fi H i out
n
n
i 1
i 1
Q W s ¦ Fi 0 H i 0 ¦ Fi H i
0
Steady state
The starting point for analysing T-effects in reactors
Fogler:
dE
dt
n
n
Q W s ¦ Fi H i in ¦ Fi H i out
§ dU ·
Koretsky: ¨ dt ¸
©
¹ sys
i 1
i 1
¦ n h ¦ n h
in
in
in
out
out
W
Q
out
S
Eq 11-9
3.2.3 Overview of energy balances
3.2.4 The user friendly Energy Balance
Here’s what we’ll do with Enthalpy terms
• Express Hi in terms of Enthalpy of Formation (Hio )
and Heat Capacity (Cpi)
• Express Fi in terms of conversion (for single reaction)
or rates of reaction
• Define Heat of Reaction ('HRxn)
• Define ' Cp
Enthalpy Relationships: Single Reaction System
FA , H A
FA0 , H A0
FB 0 , H B 0
T0
...
T
b
c
d
A B o C D
a
a
a
FB , H B
...
FI , H I
FI 0 , H I 0
If A is the limiting reactant
n
In
¦ H i 0 Fi 0
Out
i 1
H A0 FA0
H A FA
H B 0 FB 0
H B FB
H C 0 FC 0
H C FC
H D 0 FD 0
H D FD
H I 0 FI 0
H I FI
n
FA
i 1
FB
¦ H i Fi
FC
Fd
FI
FA0 (1 X )
b
X)
a
c
FA0 (T C X)
a
d
FA 0 (T D X)
a
FA0 (T I ) FI 0
FA 0 (T B Fi in terms of FA0 and X
from stoichiometry
Enthalpy Relationships: Single Reaction System
n
n
¦ F H (T ) ¦ F H (T )
i0
i0
i
0
i 1
i
i 1
FA0 {[ H A0 (T0 ) H A (T )] T B [ H B 0 (T0 ) H B (T )]
T C [ H C 0 (T0 ) H C (T )] T D [ H D 0 (T0 ) H D (T )] T I [ H I 0 (T0 ) H I (T )]}
[
d
c
b
H D (T ) H C (T ) H B (T ) H A (T )]FA0 ˜ X
a
a
a
'HRxn
n
n
¦ F H (T ) ¦ F H (T )
i0
i 1
i0
i
0
i 1
i
ª n
º
FA0 «¦T i [ H i 0 (T0 ) H i (T )]» >FA0 ˜ X ˜ [ 'H Rxn ] @
¬i 1
¼
Next, we will evaluate the different terms of RHS
Expressing Hi(T) in terms of Hi0 and Cpi
n
ª n
º
FA0 «¦T i [ H i 0 (T0 ) H i (T )]» >FA0 ˜ X ˜ [ 'H Rxn ] @
¬i 1
¼
n
¦ F H (T ) ¦ F H (T )
i0
i0
i
0
i 1
i
i 1
Enthalpy at any given temperature is related to enthalpy at a reference
T
temperature and heat capacity
o
H i (Tref ) H i (T )
³ C dT
pi
T Tref
H io (Tref )
Heat of formation of species " i" at Tref
Cp either taken
constant or f(T)
Hi is available from
Chemical Engg handbook
>H i 0 (T0 ) H i (T )@
Therefore,
T0
³ Cp dT
i
T
and,
T0
n
¦T ˜ >H (T ) H (T )@ ³ T ˜ Cp dT
i
i 1
i0
0
For no phase change
i
i
T
i
Heat of Reaction ('HRxn)
n
ª n
º
FA0 «¦T i [ H i 0 (T0 ) H i (T )]» >FA0 ˜ X ˜ [ 'H Rxn ] @
¬i 1
¼
n
¦ F H (T ) ¦ F H (T )
i0
i0
i
0
i 1
i
i 1
Heat of reaction is defined as:
'H Rxn (T )
[
d
c
b
H D (T ) H C (T ) H B (T ) H A (T )]
a
a
a
Enthalpy at any given temperature is:
T
H i (T )
H io (Tref ) ³ C dT
pi
T Tref
?
'H Rxn (T )
where, 'Cp
'H
o
Rxn
(Tref ) ³
T
T Tref
'Cp dT
d
c
b
Cp D CpC Cp B Cp A
a
a
a
General Form of Energy Balance
W s
F1 , E1
F2 , E2
F2 , E2
...
...
Q
Fn , En
Q W F1 , E1
n
¦F E
i
i 1
i in
Fn , En
n
¦ Fi Ei out
i 1
dEˆ
dt
Energy Balance Equation in terms of Enthalpy
n
n
ˆ
d
E
Q W s ¦ Fi H i in ¦ Fi H i out
dt
i 1
i 1
Energy Balance Equation in terms of Conversion
º
ª n T
Q W s FA0 «¦ ³ T i Cpi dT ]» >FA0 ˜ X ˜ 'H Rxn (T )@
»¼
«¬ i 1 T0
dEˆ
dt
º
ª n T
Q Ws FA0 «¦ ³ T i Cpi dT ]» >FA0 ˜ X ˜ 'H Rxn (T )@
»¼
«¬ i 1 T0
dEˆ
dt
If the heat capacities can be assumed constant:
>
@
n
ª
º
0
Q W s FA0 «¦ T i Cpi T Ti 0 » 'H Rxn
TR 'C p T TR FA0 ˜ X
¬i 1
¼
dEˆ
dt
At steady-state:
>
@
n
ª
º
0
Q Ws FA0 «¦ T i Cpi T Ti 0 » 'H Rxn
TR 'C p T TR FA0 ˜ X
¬i 1
¼
0
In non-stirred reactors, there is no shaft work, and in stirred reactors,
the shaft work is almost always negligible (Ws ~ 0)
3.2.5 Adiabatic operation
General heat balance (based on constant Cp's), at steady state:
>
@
n
ª
º
0
Q W s FA0 «¦ T i Cpi T Ti 0 » 'H Rxn
TR 'C p T TR FA0 ˜ X
¬i 1
¼
0
Adiabatic (Q = 0) and neglect shaft work (Ws ~ 0)
>
@
ªn
º
0
FA0 «¦ T i Cpi T Ti 0 » 'H Rxn
TR 'C p T TR FA0 ˜ X
¬i 1
¼
0
n
In terms of conversion:
¦ 4 C (T T )
i
X
>
'H
i 1
$
RX
pi
i0
(TR ) 'C p (T TR )
@
If 'Cp (T-TR) term much smaller than 'H0RX term (often the case), a linear
relation between X and T can be expected (for adiabatic operation!)
Example 11.3 (on another way)
Normal butane (n-C4H10, A) is isomerised to isobutane (i-C4H10, B) in a PFR that
operates adiabatically. The reversible reaction is carried out in the liquid phase and
is first order in A with a reaction constant of 31.1 h-1 at 360 K. Calculate the PFR
and CSTR volumes to process a feed of 163 kmol/h (feed is 90% A, with 10% inert
i-pentane, C). The feed enters at T = 330 K. What volume is needed for 70%
conversion of n-C4H10
Further:
'H0Rx = -6900 J/mol
CpA = 141 J/mol K
Activation energy = 65.7 kJ/mol CpB = 141 J/mol K
KC (T = 60 oC) = 3.03
CpC = 161 J/mol K
CA0 = 9.3 kmol/m3
What is the equilibrium conversion at 60 oC (333 K)?
Answer: 75 %
The conversion is close to the equilibrium conversion, which requires relative large
amounts of reactor volume
Example 11.3 (on another way)
Normal butane (n-C4H10, A) is isomerised to isobutane (i-C4H10, B) in a PFR that
operates adiabatically. The reversible reaction is carried out in the liquid phase and
is first order in A with a reaction constant of 31.1 h-1 at 360 K. Calculate the PFR
and CSTR volumes to process a feed of 163 kmol/h (feed is 90% A, with 10% inert
i-pentane, C) for 70% conversion. The feed enters at T = 330 K. What volume is
needed for 70% conversion of n-C4H10
ο‫ܥ‬௣ = 0, Æ
Further: 'H0Rx = -6900 J/mol
CpA = 141 J/mol K
Activation energy = 65.7 kJ/mol CpB = 141 J/mol K οHRx = οH0Rx = const.
KC (T = 60 oC) = 3.03
CpC = 161 J/mol K
CA0 = 9.3 kmol/m3
What do you expect for the following dependencies?
T
T
X
X
V
-rA
V
V
Example 11.3 (description of DE's)
Normal butane (n-C4H10, A) is isomerised to isobutane (i-C4H10, B) in a PFR that operates adiabatically.
The reversible reaction is carried out in the liquid phase and is first order in A with a reaction constant of
31.1 h-1 at 360 K. Calculate the PFR and CSTR volumes to process a feed of 163 kmol/h (feed is 90% A,
with 10% inert i-pentane, C); The feed enters at T = 330 K.
A Ù B; -rA = k (CA –CB/KC)
k = f(T)
݀‫ܨ‬஺
= ‫ݎ‬஺
ܸ݀
݀‫ܨ‬஻
= െ‫ݎ‬஺
ܸ݀
݀‫ܨ‬஼
=0
ܸ݀
݀ܶ ‫ݎ‬஺ ο‫ܪ‬ோ௫
=
ܸ݀ σ ‫ܨ‬௜ ‫ܥ‬௉೔
FA(V=0) = FA0 (=0.9·163 = 146.7 kmol/hr)
FB(V=0) = FB0 (= 0)
FC(V=0) = FC0 (= 0.1·163 = 16.3 kmol/hr)
T(V=0) = T0 (= 330 K)
οHRx = const.
KC = f(T)
d(FA) / d(V) = rA
FA(0) = 146.7
d(FB) / d(V) = -rA
FB(0) = 0
d(FC) / d(V) = 0
FC(0) = 16.3
mol balances
d(T) / d(V) = rA*deltaHrx/(CpA*FA+CpB*FB+CpC*FC)
T(0) = 330
energy balance
V(0) = 0
V(f) = 4
FA0 = 146.7 # mol/s
CA0 = 9.3 # kmol/m3
CpA = 141 # kJ/kmol K
CpB = 141 # kJ/kmol K
CpC = 161 # kJ/kmol K
deltaHrx = -6900 # J/mol
CA = CA0*(1-X)
CB = CA0*X
thermodynamic
constants
CA(X), CB(X)
R = 8.314 # J/mol K
DHrx = -6900 # J/mol
T2 = 60+273.15
KC0 = 3.03
KC = KC0*exp((DHrx/R)*((1/T2)-(1/T)))
E = 65700 # J/mol
T1 = 360
k1 = 31.1 # hr-1
k = k1*exp((E/R)*((1/T1)-(1/T)))
rA = -k*(CA-CB/KC)
X = 1-FA/FA0
KC (T)
kinetic constants
k1 (T)
reaction rate
Example 11.3 (Polymath code)
Example 11.3 (reaction rate and T as function of V)
365
70
360
60
355
350
T (K)
40
345
30
340
20
335
330
10
325
0
0
0.5
1
1.5
2
V (m3)
2.5
3
3.5
4
-rA (kmol/m3 hr)
50
Example 11.3 (Temperature and reaction rate as f(X))
365
70
360
60
50
350
40
345
30
340
20
335
330
10
325
0
0
0.1
0.2
0.3
0.4
X (-)
0.5
0.6
0.7
0.8
-rA (kmol/m3 hr)
Temperature (K)
355
Example 11.3 (Temperature and conversion as f(V))
1
365
0.9
360
Equilibrium
conversion
355
0.8
0.7
0.6
0.5
345
0.4
340
0.3
335
0.2
330
0.1
325
0
0
0.5
1
1.5
2
V (m3)
2.5
3
3.5
4
X (-)
T (K)
350
Example 11.3 (Levenspiel plot)
30
25
FA0/-rA (m3)
20
VCSTR ~ 17 m3
VPFR ~ 2.6 m3
15
10
5
0
0
0.1
0.2
0.3
0.4
X (-)
0.5
0.6
0.7
0.8
3.2.6 Adiabatic Equilibrium conversion
As shown in the previous example, the temperature has an influence on the
equilibrium conversion, and therefore (inter-staged) cooling/heating is used to
increase yield in equilibrium-limited reactions.
As an example (Example 8-6, in SI-units), the solid-catalysed liquid phase reaction
A Ù B is used, which is carried out in an adiabatic reactor. Pure A is fed to the
reactor at T = 300 K. The reaction kinetics follow an elementary rate law.
Additional information:
HA0 = -167.5 kJ/mol;
HB0 = -251.2 kJ/mol
CPA = 209.3 J/mol K
CPB = 209.3 J/mol K
Ke = 100 000 (at T = 298 K)
We will analyse the reaction in an X-T figure
X
Qualitatively,
• how will the equilibrium conversion (Xe) vary with T?
• How will the Temperature relate to the conversion (XEB)?
T
Equilibrium Conversion (Xe as f(T))
As an example (Example 11-4, in SI-units), the solid-catalysed liquid phase reaction
A Ù B is used, which is carried out in an adiabatic reactor. Pure A is fed to the
reactor at T = 300 K. The reaction kinetics follow an elementary rate law.
Additional information:
HA0 = -167.5 kJ/mol;
HB0 = -251.2 kJ/mol
CPA = 209.3 J/mol K
CPB = 209.3 J/mol K
Ke = 100 000 (at T = 298 K)
From Stoichiomitry, liquid reaction (no density changes), pure A feed: CA = CA0(1-X)
CB = CA0 X
At equilibrium:
‫ܥ‬஺ ‫ܥ‬஺଴ (1 െ ܺ݁) (1 െ ܺ݁)
‫ܭ‬௘ =
=
=
‫ܥ‬஺଴ ܺ݁
ܺ݁
‫ܥ‬஻
Temperature dependency of Ke (van 't Hoff)
‫ܭ‬௘ (ܶ) = ‫ܭ‬௘ (ܶଵ )݁
‫ܭ‬௘ (ܶ)
ܺ௘ =
1 + ‫ܭ‬௘ (ܶ)
బ
οுೃೣ
ோ
ଵ ଵ
ି
்భ ்
‫ܭ‬௘ (ܶ) = ‫ܭ‬௘ (ܶଵ )݁
బ
οுೃೣ
ோ
‫ܭ‬௘ (ܶ)
ܺ௘ =
1 + ‫ܭ‬௘ (ܶ)
ଵ ଵ
்భ ି்
As an example (Example 11-4, in SI-units), the solid-catalysed liquid phase reaction
A Ù B is used, which is carried out in an adiabatic reactor. Pure A is fed to the
reactor at T = 300 K. The reaction kinetics follow an elementary rate law.
Additional information:
HA0 = -167.5 kJ/mol;
HB0 = -251.2 kJ/mol Æ 'H0Rx = -83700 J/mol
CPA = 209.3 J/mol K
CPB = 209.3 J/mol K
Ke = 100 000 (at T = 298 K)
Ke (-)
Xe (-)
298
350
400
425
450
475
500
550
600
650
100000
659.772
18.085
4.113
1.103
0.339
0.118
0.019
0.004
0.001
1.000
0.998
0.948
0.804
0.524
0.253
0.105
0.018
0.004
0.001
1.2
1.0
Equilibrium conversion (Xe)
T (K)
0.8
0.6
0.4
0.2
0.0
300
350
400
450
Temperature (K)
500
550
600
Conversion (XEB as f(T))
As an example (Example 11-4, in SI-units), the solid-catalysed liquid phase reaction
A Ù B is used, which is carried out in an adiabatic reactor. Pure A is fed to the
reactor at T = 300 K. The reaction kinetics follow an elementary rate law.
Additional information:
HA0 = -167.5 kJ/mol;
HB0 = -251.2 kJ/mol Æ 'H0Rx = -83700 J/mol
CPA = 209.3 J/mol K
CPB = 209.3 J/mol K
Ke = 100 000 (at T = 298 K)
Adiabatic (Q = 0) and neglect shaft work (Ws ~ 0)
>
@
ªn
º
0
FA0 «¦ T i Cpi T Ti 0 » 'H Rxn
TR 'C p T TR FA0 ˜ X
¬i 1
¼
From Slide 31; Analysis of
adiabatic operation
0
n
In terms of conversion:
¦ 4 C (T T )
i
X
>
'H
i 1
$
RX
pi
i0
(TR ) 'C p (T TR )
@
If 'Cp (T-TR) term much smaller than 'H0RX term (often the case), a linear
relation between X and T can be expected (for adiabatic operation!)
Conversion (XEB as f(T))
n
¦ 4 C (T T )
i
X
Since only pure A (no B) is fed to the
reactor, CPA = CPB:
X
X
y=
>
'H
i 1
$
RX
pi
(TR ) 'C p (T TR )
C pA (T T0 )
$
'H RX
C pA
'H
$
RX
mx
i0
T
C pAT0
$
'H RX
+c
@
Conversion (XEB and Xe as f(T))
1.2
conversion (X)
1.0
Equilibrium
conversion
0.8
0.6
0.4
0.2
X – T relation according
to energy balance
0.0
300
350
400
450
Temperature (K)
500
550
600
As an example (Example 11-4, in SI-units), the solid-catalysed liquid phase reaction
A Ù B is used, which is carried out in an adiabatic reactor. Pure A is fed to the
reactor at T = 300 K. The reaction kinetics follow an elementary rate law.
Additional information:
HA0 = -167.5 kJ/mol;
HB0 = -251.2 kJ/mol Æ 'H0Rx = -83700 J/mol
CPA = 209.3 J/mol K
CPA = 209.3 J/mol K
Ke = 100 000 (at T = 298 K)
1.2
A reactor scheme could be selected in which
the reaction mixture is cooled after the first
reactor and than send to a second one.
The exit from the second reactor could than
again be send to a third one etc.
Reactor staging with interstage cooling
Reactor
conversion (X)
1.0
0.8
0.6
0.4
0.2
0.0
300
Reactor
350
400
450
Temperature (K)
500
550
Reactor
600
1.2
1.0
Example 11-4, with
interstage cooling (to
350 oC) after 95% of
the equilibrium
conversion is reached
Conversion (-)
0.8
0.6
Energy Balance line after 2nd cooling
0.4
Energy balance line after 1st cooling
Energy Balance line
0.2
Equilibrium curve
0.0
300
Reactor
350
400
450
Temperature (K)
550
600
Reactor
Reactor
T1 = 459 K
X1 = 0.397
500
T1 = 439 K
X1 = 0.620
T1 = 421 K
X1 = 0.798
For endothermic reactions, there is a similar approach, which is detailed in
Section 11.6.3
Reactor staging with interstage heating
Section 11.7 is reading material