engrd 2190 – lecture 14conduct reaction in a series of adiabatic reactors and heat exchangers....
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Concept: Process Analysis by Mathematical Modeling –Energy Balances
Context: Processes with Endothermic Chemical Reactions –Reactor Design
Defining Question: For endothermic reactions, conversion is high at low temperatures (thermodynamics)and reaction rate is high at high temperatures (kinetics).
How to balance thermodynamics and kinetics?
EngrD 2190 – Lecture 14
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Homework 3 due today at noon.Write team code and names of all contributing team members on all solutions. Indicate this week’s Team Coordinator.
Prelim 1: Sunday 10/4, 7:30-9:30 p.m., 407 Willard Straight (Memorial Room)Covers Chapter 2 and mass balances (formal and informal).Covers through Lecture 10, Homework 3, Calculation Session 3. Open book, open notes, open exercise solutions.Bring a calculator. Graphing calculators are allowed. Laptops only fordigital textbook and material stored on laptop. Must be approved pre-prelim.
Special TA Office Hours Saturday Afternoon – Zoom + in-personDetails this evening.
Practice Exercises for Prelim 1. Optional - Solutions are posted.Process Design with real chemicals: 2.18Process Design with qualitative, informal mass balances: 3.123 and 3.132Formal Mass Balances: 3.20, 3.25, and 3.45Informal Mass Balances: 3.41
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Average starting salaryfor Cornell ChemE B.S. 2020
was $84,000
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Lecture 13 RecapWhat is the adiabatic temperature rise?
2CO + O2 2CO2
25°C T = ?
q adiabatic
fractional conversion, X0 1
temperature
25C
T ?needed only this point;T = Tadiabatic at X = 1.ignored path
Today we calculate T for X < 1. We calculate the path.
We need a relation between T and X.
And we need to know how far the reaction proceeds. What is X at equilibrium as a function of T?
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Recap of textbook pp. 139-145.
92.2 kJ/mol exothermic
198.8 J/(molK)order increases
P0 = 1 bar 1 atm
Must express partial pressures in terms of fractional conversion, X.
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N2 + 3H2 2NH3
mols initially: 1 3 0 total = 4
mols later: 1X 3(1X) 2X total = 2(2X)
Choose X and calculate K. Use K to calculate T.
Check at X = 0. Okay.Check at X = 1. Okay.
2
total
0
4
22
3
totaltotal
202
total
3eq,Heq,N
202eq,NH
)1(27)2(16
)2(2)1(3
)2(21
)(2)(
22
3
PP
XXX
PXXP
XX
PPX
X
PP
PPK
RTSTH
RTG
e
e
/)(
/
0rxn
0rxn
0rxn
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N2 + 3H2 2NH3
Calculate equilibrium curves.
2
total
0
4
220rxn
0rxn
)1(27)2(16ln
PP
XXXRS
HT eqn 3.180, p. 141
200°C: high conversion,but slow reaction.
500°C: fast reaction,but low conversion.
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N2 + 3H2 2NH3
Calculate the adiabatic temperature path.
Like eqn 3.189, p. 143but with average CP’s)25(2)25)(3(
)300)(3(
322
22
NHP,HP,NP,0rxn
HP,NP,
TCTCCH
TCCX
average CP’s
Shomate eqn
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N2 + 3H2 2NH3
Conduct reaction in a series of adiabatic reactors and heat exchangers.
Arbitrary limit 0.05 below equilibrium line.
Adiabatic temperature rise in reactor 1.
0.24
515CX = 0.24
Cool reactor 1 effluent to 300C.
Adiabatic temperature rise in reactor 2.
435C
435CX = 0.38
515C
0.38
reaction rate is zero onthe equilibrium line
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At present, most chemical products start with crude oil.crude oil hydrocarbon building blocks fuels, lubricants, polymers, plastics,
In the future, chemical products will start with coal or methane.coal and/or CH4 syngas dimethyl ether fuels, lubricants, polymers, plastics,
CH4 + H2O(g) CO + 3H2
)11(31)C25( 0OH f,
0CH f,
0H f,
0CO f,
0rxn 242
HHHHH
kJ/mol1.206
)8.2418.74(035.110
endothermic; must supply heat to drive reaction.
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CH4 + H2O(g) CO + 3H2
Use a series of adiabatic reactors and interstage heaters.
arbitrary; set by upper limit for steel.equivalent unitfor Reactor 1
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Energy balance on fictitious energy combiner and splitter.q1 + q3 = q2 + q4
25
800OHP,CHP,1 )(
24TdCXCXq
)( 0rxn2 HXq > 0 for an endothermic reaction.
final
24
800OHP,CHP,3 ))1()1(
T
TdCXCXq
> 0
final
2
25HP,COP,4 )3(
T
TdCXCXq
> 0
> 0
Substitute into energy balance, substitute Shomate equations,and solve for X.
See analogous equations 3.167-3.170on p. 138.
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Best linear fit for 500C to 800C:C2020
1slope
adiabatic temperature change
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CH4 + H2O(g) CO + 3H2
)11(31)C25( 0OH f,
0CH f,
0H f,
0CO f,
0rxn 242
SSSSS
K)J/(mol6.214
)83.18826.186(68.130367.197
order decreases; rxn favored by high T.
)C25( need 0rxn S
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Substitute expression for Kinto the equation for T. KRS
HTln0
rxn
0rxn
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Constraint set bythermodynamics
Constraint set byreactor materials
Constraint set by kinetics;rxn too slow below 500C.
Heater 1Reactor 1
Heater 2Reactor 2