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Bring Chapter 3 exercises to Calculation Session:
3.64, 3.66, and 3.113.
Lecture 13 (Monday) 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?
Lecture 14 (today): 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.
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
N2 + 3H2 2NH3
Calculate equilibrium curves.
2
total
0
4
220rxn
0rxn
)1(27)2(16ln
PP
XXXRS
HT eqn 3.180, p. 141
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
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
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.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.
CH4 + H2O(g) CO + 3H2
Use a series of adiabatic reactors and interstage heaters.
arbitrary; set by upper limit for steel.equivalent unit for Reactor 1
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 eqn 3.171 on p. 138
Best linear fit for 500 to 800C:C2020
1slope
adiabatic temperature change
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
Substitute expression for Kinto the equation for T. KRS
HTln0
rxn
0rxn
Constraint set bythermodynamics
Constraint set byreactor materials
Constraint set by kinetics;rxn too slow below 500C.
Heater 1Reactor 1
Heater 2Reactor 2
