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KGT 002 Kemisk Reactionsteknik I, 5p KGT 002 Chemical Reaction Engineering I, 5p
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KGT 002 Kemisk Reactionsteknik I, 5p KGT 002 Chemical Reaction Engineering I, 5p
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Instructors Prof. Jonas Hedlund - Lectures Dr. Göran Olofsson - Lectures Lic.Eng. Mattias Grahn - Tutorial sessions M.Sc. Alessandra Mosca – Tutorial sessions and Laboratory exercise
KGT 002 Kemisk Reaktionsteknik I, 5p KGT 002 Chemical Reaction Engineering I, 5p
Welcome General Course Topics: 1.Numerical methods and Material balances 2.Conversion and simple reactor design equations 3.Rate laws and stoichiometry 4.Isothermal reactor design 5.Pressure drop and unsteady-state reactor operation 6.Selectivity and multiple reactions 7.Nonisothermal reactor design 8.Multiple steady-states and bifurcation analysis Course Literature: Elements of Chemical Reaction Engineering, 4th Edition H. Scott Fogler Full IT support: http://www.ltu.se/web/pub/jsp/polopoly.jsp?d=1592 Parts of Course:
“It is primarily a knowledge of chemical kinetics and reactor design that distinguishes the chemical engineer from other engineers.” – H. Scott Fogler, Professor of Chemical Engineering, University of Michigan.
KGT 002 Kemisk Reactionsteknik I, 5p KGT 002 Chemical Reaction Engineering I, 5p
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• • • • •
Lectures. Tutorial sessions (räkneövningar). Assignment problems Laboratory exercise. Exam.
Introduction to Chemical Reaction Engineering
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Introduction to Chemical Reaction Engineering Date 19/1 20/1 23/1 27/1 27/1 31/1 31/1 1/2 2/2 3/2 6/2 7/2 7/2 10/2 13/2 14/2 15/2 17/2 20/2 21/2 24/2 24/2 27/2 28/2 2/3 6/3 7/3 10/3
Pass 3 4 2 1 3-4 2 3 3 3 4 2 1-2 3 3-4 1 2-3 3 4 2 3 1 3-4 1 2 3-4 2 4 3
Activity, Teacher 1. Introduction and numerical methods, Jonas 1. Material balances, Jonas 1. Material balances, Jonas Tutorial Session, Topic 1, Alessandra Computer Lab – Matlab tutorial, Mattias Tutorial Session, Topic 1, Alessandra 2. Conversion and design equations, Jonas Tutorial Session, Topic 2, Alessandra 3. Rate laws and stoichiometry, Jonas Tutorial Session, Topic 3, Mattias 4. Isothermal reactor design, Jonas Computer Lab – Matlab tutorial, Mattias Tutorial Session, Topic 4, Mattias Laboratory exercise, group 1&2, Alessandra Tutorial Session, Topic 4, Mattias Laboratory exercise, group 3&4, Alessandra 5. Pressure drop and unsteady-state, Göran Tutorial Session, Topic 5, Mattias 6. Selectivity and multiple reactions, Göran Tutorial Session, Topic 6, Mattias 7. Nonisothermal reactor design, Fredrik Laboratory exercise, group 5&6, Alessandra 7. Nonisothermal reactor design, Göran 7. Nonisothermal reactor design,Göran Tutorial Session, Topic 7, Mattias 8. Multiple steady-states, Göran Tutorial Session, Topic 8, Mattias Tutorial Session, exam consultation, Mattias
Literature Lecture notes Lecture notes 1.1-1.4+additional Problems Matlab tutorial Problems 2.1-2.5 Problems 3.1-3.4 Problems 4.1-4.3 Matlab tutorial Problems Lab. comp. Problems Lab. comp. 4.4-4.7+additional Problems 6.1-6.6 Problems 8.1-8.3,8.7 Lab. comp. 8.1-8.3,8.7 9.1-9.2,8.4 Problems 8.6 Problems
Chemical Reaction Engineering
Typical questions that CRE may answer: • Is a batch or a continuous process preferred? • What kind of reactor should be used? • How large reactor is necessary to produce wanted amount? • What factors influence the yield? • How is the process optimized? • How should the process be controlled? • How is the reactor affected by a change in the operating conditions?
Assignments: Assignment #
1 2 3 4
Topic
1-2 3-4 5-6 7-8
KGT002 Chemical Reaction Engineering I, 5p gives the fundamentals. KGT003 Chemical Reaction Engineering II, 5p, gives deeper understanding.
Introduction to Chemical Reaction Engineering
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Some chemicals, plant sizes and waste produced: Category
Plant capacity
Introduction to Chemical Reaction Engineering
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Chemical reactors:
Price
Waste/product
Petroleum 106-108 tons/y Refining Commodity 104-106 Chemicals Fine Chemicals 102-104
$0.1/lb
0.1
0.1-2
1-3
2-10
2-10
Foods
1-50
Materials
0-∞
Pharmaceuticals 10-103
10-∞
Thermodynamics
Fluid flow
Kinetics
CHEMICAL REACTOR
Design
10-100
CHEMICAL PROCESS
Chemical processes
Mass transfer Heat transfer
Process control
Economics
A generic flow diagram of a chemical process:
Raw Materials
Separation
Chemical
Separation
Products
Process
Reactor(s)
Process
By-products
Chemical process: Methane
Natural Gas
Syngas
Methanol
Acetic acid
1. Numerical Methods and material balances
Acetylsalicylic acid
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1. Numerical Methods and material balances
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b
Identical solution using Matlab:
a
In the command window, write the following (press return after each command): clear X = 0:pi/100:pi; Y=sin(X); Z = trapz(X,Y)
Integration: S= ∫ f ( x )dx Analytical solution S=F(b)-F(a), where F is a primitive to f. Numerical solution, trapeze formula S ≈ h(0.5*fa + f1 + f2 + … + 0.5*fb) The function values are calculated at equal distance h. Example
The Matlab will answer: Z =1.9998
π
S= ∫ sin( x)dx Analytical solution S=-cos(π)+cos(0)=2
Ordinary differential equations (ODE)
0
Example (first order ODE): Numerical solution, with h=π/4 X
SINx 0 0,785398 1,570796 2,356194 3,141593
y′ = 0.5( x + y ) − 1
TRAPZ 0 0,707107 1 0,707107 3,23E-15
0 0,707106781 1 0,707106781 1,61557E-15 Sum*h =1,90
Suppose that we are interested of the solution for x=4. A few solution curves:
S=1,90 With h=π/100, S=1,9998
y(0) = 1 (boundary condition, bivillkor), selects only one of these solutions Analytical solution (will not be considered here):
y = e 0. 5 x − x y(4) = 3.389056 Eulers method:
1. Numerical Methods and material balances
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1. Numerical Methods and material balances
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Matlab solution: 1. Start at the boundary condition n=0 (xn,yn), in this case (0,1). 2. Calculate xn+1 = xn + h and yn+1 = yn + hy´n 3. Increase n (n=n+1) and go to 2 until x=4. For h=1: n
yn y´n xn 0 0 1 1 1 0,5 2 2 0,25 3 3 0,375 4 4 1,0625 Reducing h to half, also reduces the error to about half:
-0,5 -0,25 0,125 0,6875 1,53125
h y(4) y(4)Analytical - y(4)Euler 1 1.065 2.3 0.5 1.96 1.4 0.25 2.58 0.8 0.001 3.385 0.004 It is possible to use Richardson extrapolations to improve the result, but it is also possible to use Excel or similar with small h.
1. Material Balances
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clear [X,Y] = ode45('diffeq1',[0,4],1); plot(X,Y) 2. Save another m-file with the name “diffeq1.m” and the content: function dydx = diffeq1(x,y); dydx=0.5*(x+y)-1; 3. Write “example1” in the command window and press “Enter”. Matlab will now run the m-file with the name example1, which will run the built in ODE solver ode45 (also a m-file). Ode45 will create a matrix (the solution) with X values in the first column and corresponding Y values in the second. The plot command will draw a figure of the solution:
1. Material Balances
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If no generation or consumption occurs, i.e. no chemical reaction:
1. Material Balances Before we start designing reactors, we have to consider Material Balances (MB). MB:s can be very useful in order to calculate data which can not be measured directly. MB:s are also an essential part in reactor design.
Material balances are made over a selected system boundary:
Input flow streams
1. Save an m-file with the name “example1.m” and the content:
Output flow streams
System
Accumulation = input - output If the system is at steady state as well (accumulation=0): Input = output
Choice of a Basis for Calculations: System boundary
Start by choosing basis for calculations, it could be: • A unit of mass/mole • A unit of volume • A unit of time
Material balance:
Input to system
+
Production within system
=
output from system
+
Accumulation within system
Units can be moles/time, moles or kg. Use the most convenient units.
(solid or liquid systems) (gas systems) (continuous processes)
• Arrival at the same solution independent of basis • Poorly chosen basis can make calculations unnecessarily difficult. Problem analysis: Problem analysis increase the clarity (överskådlighet) of a MB problem and thus simplify the solution. Analyse the problem by determine the degrees of freedom: Degrees of freedom = (number of independent flow variables) (number of independent balance equations) – (number of specified independent flow variables) – (number of help equations).
Material Balance on a system can be for: • Total mass (no generation or consumption if no nuclear processes) • Total moles (Very useful in CRE) • Mass of a chemical compound • Moles of a chemical compound (Very useful in CRE) • Mass of an atomic species • Moles of an atomic species (Very useful in CRE)
Degrees of freedom = 0 for a correctly specified problem.
1. Material Balances
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Example 1: Titanium dioxide TiO2 is a white hiding pigment used heavily in the paint and paper industries. A pigment plant is to produce 2000 kg h-1 dry TiO2 product. An intermediate stream consisting of TiO2 precipitate suspended in an aqueous salt solution, is to be purified of salt so that the final product contains, on a water-free basis, at most 100 ppm (mass basis) of salt. The salt removal is to be accomplished by washing the precipitate with water. If the raw pigment stream contains 40% TiO2, 20% salt and the rest water (all mass %) and if the washed pigment is, upon settling, projected to consist of about 50% (mass) TiO2 solids, what will the composition of the waste wash water stream be? Determine degrees of freedom for this problem, is it solvable?
1. Material Balances
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Example 2: Propane is burned with 25% excess air. How many moles of air per second are needed to produce 100 moles of exhaust gas per second? Calculate the degrees of freedom and solve with three different bases for calculation. C3H8 + 5O2 → 3CO2 + 4H2O Example 3: Solubility of barium nitrate at 100°C is 34kg/100kg water, at 0°C it is 5kg/100kg water. If you start with 100kg barium nitrate in a saturated solution at 100°C, how much precipitates after cooling to 0°C? Calculate the degrees of freedom to verify that the problem is correctly specified.
Bypass flow:
Possible system boundaries
Reactor
Example 4: A process removes radioactive strontium (Sr90) from milk by filtration through a bed of CaHPO4. The process removes all Sr but also unfortunately removes 97% Ca from the milk. Milk containing 4.85⋅10-14 kg Sr/m3 and 10-3 kg Ca/m3 is fed to the process. If the milk must contain at least 5⋅10-5 kg Ca/m3, what will the Sr concentration be? Process: Milk
1. Material Balances
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1. Material Balances
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Example 6: A stoichiometric mixture of H2 and N2 will be produced for ammonia synthesis (75% H2 ,25% N2) by letting generator gas (78% N2, 20% CO, 2% CO2) react with water gas (50% H2, 50% CO). The reaction shall be carried out such that all CO is consumed because CO poisons the ammonia synthesis catalyst. The generator gas and water gas will react with steam so that CO and H2 react by the reaction:
Recirculation and purge:
Fresh Feed
CaHPO4 bed
Reactor
Product
Separation Purge
Recycle Example 5: Methanol synthesis is based on the reaction between CO2 and H2: CO2 + 3H2 → CH3OH + H2O
CO + H20→ CO2 + H2 (water gas shift reaction) All water is consumed and CO2 is removed in a scrubber. Calculate the ratio between the feeds of generator gas and water gas and between generator gas and steam.
Process:
CO2 and H2 are fed stoichiometrically to the process and the feed also contain 0.5 vol.% inert gas. 60% of CO2 and H2 are converted in the reactor to CH3OH. The concentration of inert in the reactor feed is 2 vol.%. Calculate (a) mol gas recycled per mol fresh feed and (b) mol gas purged per mol fresh feed.
78% N2 20% CO 2% CO2
Process:
50% H2 50% CO
1
Process
2
5
3 H 2O CO2 H2 I
F
B
Reactor
C
D R CO2 ,H2 ,I
A
Separation P
CH3OH, H2O CO2 ,H2 ,I
4
CO2
75% H2 25% N2
1. Material Balances
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1. Material Balances
The rate of reaction
rA is dependent on:
A chemical reaction occurs when molecular species lose their identity and take a new form:
• • • •
Types of reactions: • Decomposition • Combination • Isomerization
A→B+C B+C→A A→B
(sönderdelning) (kombinering) (isomerisering)
species concentration temperature pressure type of catalyst
Observe! rA may not be the same at every position in the reactor. Forms of the reaction rate equation (note that rA < 0 and k > 0):
Reaction occurs at a certain “rate” or speed;
- rA = kCA
If A → 2B (decomposition)
- rA = kCA2
rA
rate of formation of A per unit time per unit volume of reactor
=
rB
- rA =
mol A
k1 ⋅ C A 1+ k 2 ⋅ CA
0
s ⋅ (m3 reactor)
dC A is a very confusing definition of reaction rate and dt
should be used with great care (or not at all). It is only valid for a constant volume batch reactor (consider for instance a steady state reactor). The definitions above can be used for all kinds of reactors.
The General Mole Balance then - 2rA = rB (-2*(rate of formation of a) = rate of formation of b)
Fj0
Fj
System
Rate can be expressed based on different units of size:
Mole balance on species j at any time t;
rA = rate of formation of A per volume reactor rA´ = rate of formation of A per mass of catalyst
Input to system
rA´´ = rate of formation of A per surface area of catalyst
+
Production within system
=
output from system
+
Accumulation within system
moles time
In + prod = out + acc
1. Material Balances
FjO + Pj = Fj +
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1. Material Balances
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Batch Reactor
dN j dt
where Nj is number of moles of j in system at time t.
• Fj0 = Fj = 0 • ∫ rjdV = rj ⋅ V (No spatial variation if well mixed) V
Can express Pj in terms of reaction rate:
Then
Pj = rj ⋅ V
dN j
moles
dt
moles =
time
time ⋅ volume
⋅
Volume
= rj ⋅ V
Usually V is constant Continuous Stirred Tank Reactor (CSTR)
Only true if rj is constant throughout system volume V, i.e. if concentration, temperature, etc. are constant.
Usually operated at steady state: dN j dt
If rj varies with V:
=0
Also perfectly mixed (no spatial variation):
M
Pj = ∑ r ji Vi i =1
∫ rjdV = rj ⋅ V
If system divided into M infinitesimal volume elements (dV) then
V
Then
Pj = ∫ r j dV V
Thus the general mole balance equation for any reactor becomes: FjO – Fj + ∫ rjdV = V
dN j dt
FjO – Fj + rj ⋅ V = 0
V=
Fj0 − Fj
Fj0
− rj
The CSTR is modelled so that the concentration in the outlet stream is the same as inside the tank. Fj = Cj ⋅ υ υ is volumetric flow rate (volume/time) rj = function of Cj etc.
Fj
Conc. Cj0
Cj Position
1. Material Balances
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1. Material Balances
Tubular Reactor dFj dV Fj0
Fj
Plug flow, no mixing
= rj
If the reactor has constant cross-sectional area (A); dV = A dz
Operated at steady-state: dN j
dFj dz
=0
dt then
END OF TOPIC 1
FjO – Fj + ∫ rjdV = 0 V
Modelled so that concentration and temperature are constant radially but varies axially throughout the reactor. Mole balance for an infinitesimally thin section of the reactor with volume dV:
Fj
Fj0
Fj + dFj
Fj
dV dZ Input to system
+
Production within system
Material balance over dV: Fj + rj dV = (Fj + d Fj)
= A ⋅ rj
=
output from system
+
Accumulation within system
moles time
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