Chemical Reaction Engineering

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Chemical Reaction Engineering
Lecture 7
Lecturer ???
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This course focuses on non-isothermal reactors.
Up to now, we discussed only the mass
balance. How about the energy balance or the
heat effects?
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Isothermal and non-isothermal

• The major difference between the design of
  isothermal reactors and that of the
  non-isothermal reactors lies in the method of
  evaluating the design equation when temperature
  varies along the length of a PFR or when heat is
  removed from a CSTR.

Nonisothermal
Energy balance
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Why Energy balance?
B, X 0.7
A

• Consider an exothermic reaction and operated
  adiabatically in a PFR

Arrhenius Equation
Mole balance
Rate law
Stoichiometry
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Thermodynamics

• First law
• To a closed (mass) system

The work done by the system to the surroundings
The change in total energy of the system
The heat flow to the system
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An open system (for example, CSTR)
Rate of flow of heat to the system from the
surroundings
Rate of energy leaving the system by mass flow
out the system
Rate of accumulation of energy in the system
Rate of work done by the system on the
surroundings
Rate of energy added to the system by mass flow
into the system
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The work term
Rate of work done by the system on the
surroundings

• The work term is usually saparated into flow
  work and other work.
• Flow work
• the work that is necessary to get the mass into
  and out of the system
• for example, when shear stresses are absent

P pressure Vi specific volume
Flow work other work (shaft work)
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Such as electric or magnetic energy, light etc.
Potential energy
Kinetic energy
Internal energy
Usually
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Enthalpy!!, function of T
unit (cal / mole)
Steady state
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Heat of reaction at temperature T (outlet)
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(Heat of reaction)
when NO phase change
(if phase change is involved, p 433)
Enthalpy of formation of species i at reference
temperature (25 C)
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Its value can be found in the Handbook of
Chemistry and Physics.
constant or mean heat capacities
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Constant or mean heat capacities
What about variable heat capacities?
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Heat capacities are strong functions of
temperature over a wide temperature range
Therefore
integrating
Heat capacity as a function of temperature
Similar procedure
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Highly temperature sensitive heat capacities
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Calculate the heat of reaction for the synthesis
of ammonia from hydrogen and nitrogen at 150 ºC
in kcal/mol of N2 reacted and in kJ/mol of H2
reacted.
(At 298 K)
constant or mean heat capacities
Negative sign exothermic reaction
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Back to our energy balance equation (at steady
state) ...
Next issue
Highly temperature sensitive heat capacities
Constant or mean heat capacities
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Heat added to the reactor

• The heat flow to the reactor is usually in terms
  of the overall heat-transfer coefficient, U, the
  heat-exchange area, A, and the difference between
  the ambient temperature, Ta, and the reaction
  temperature, T.

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CSTR with a heat exchanger
The rate of heat transfer from the exchanger to
the reactor
Energy balance on heat exchanger
Rate of energy in by flow
Rate of energy out by flow
Rate of heat transfer from exchanger to reactor
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Eliminate Ta2
Taylor series expanding and neglect the second
order terms
when Ta1 Ta2Ta
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Tubular reactors (PFR/PBR)
Integrate the heat flux equation along the length
of the reactor to obtain the total heat added to
the reactor
a heat-exchange area per unit volume of reactor
Heat transfer to a PFR
For a tubulat reactor of diameter D, a 4 / D
For a PBR
Heat transfer to a PBR
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Non-isothermal operation
Constant or mean heat capacities
Xenergr balance
CSTR, PFR, PBR, Batch
Temperature
Adiabatic exothermic reactions
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Non-isothermal CSTR
Design equation ( From mass balance)
coupling
Energy balance
With heat exchanger
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CSTR
e.g.
Relationships bewteen V, X , and T
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Non-isothermal CSTR rxn example
P. 445
FA0
T 58F
T 58F
FB0
FM0
T 75F
Operated adiabatically
V 300 gal
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CSTR
e.g.
Relationships bewteen V, X , and T
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Mole balance
and
Rate laws
Stoichimetry
Energy balance
(adiabatic)
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Solve the two equations simultaneously and we
obtain X f (T)
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Similar to the previous system. However, a
cooling coil is now used. It has 40 ft2 of
cooling surface and the cooling water flow rate
inside the coil is sufficiently large that a
constant coolant temperature of 85 F can be
maintained. A typical overall heat transfer
coefficient for such a coil is 100 Btu/h ft2 F.
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Solve the two equations simultaneously and we
obtain X f (T)
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General energy balance equation
Adiabatic Tubular Reactor
Usually, there is negligible amount of work done
on or by the reacting mixture. The work term can
be neglected in tubular reactor design.
Adiabatic operation of PFR
Mole balance of PFR
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PFR/PBR
e.g.
Relationships bewteen V, X , and T
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Something about thermodynamics ...
k
For a reversible reaction
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Normal butane is to be isomerized to isobutane in
a PFR.
The reaction is to be carried out adiabatically
in the liquid phase under high pressure
using essentially trace amount of a liquid
catalyst which gives a specific reaction rate of
3.1 h-1 at 360 K. Calculate the PFR volume
necessary to process 100000 gal/day of a mixture
of 90 mol n-butane and 10 mol i-pentane
(inert). The feed enters at 330 K.
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Stoichiometry
Mole balance (PFR)
Rate law
Energy balance
0
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These equations can be solved simultaneously
X
Xe
V
T
V
At equilibrium -rA 0
-rA
Max. reaction rate _at_ somewhere in the tube
V
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Steady-state tubular reactor with heat exchanger
Heat is added or removed through the cylindrical
walls of the reactor
Differentiating w.r.t. V
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The change of temperature with volume (i.e.
distance) down the reactor
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Back to our general equation for energy balance
X is not always the most convenient parameter...
Differentiating w.r.t. V
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PFR/PBR
CSTR (integrating w.r.t. V)
Using (-rA) instead of X
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Vapor-phase cracking of acetone to ketene and
methane
The reaction is first order with respect to
acetone and
unit k (s-1) and T(K)
It is desired to feed 8000 kg of acetone per hour
to a tubular reactor. The reactor consists of a
bank of 1000 1-inch schedule 40 tubes. Consider
two cases here 1. The reactor is operated
adiabatically. 2. The reactor is surrounded by a
heat exchanger where the heat-transfer
coefficient is 110 J/m2 s K, and the ambient
temperature is 1150 K. The inlet temperature and
pressure are the same for both cases at 1035 K
and 162 kPa, respectively. Plot the conversion
and temperature along the length of the reactor.
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Stoichiometry
Mole balance (PFR)
Rate law
And then, energy balance to find X f (T)
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Case 1, Adiabatic operation
Highly temperature sensitive heat capacities
where in general
only A at V 0 (t 0)
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All these can be solved together.
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X
T
Vc
V
Adiabatic endothermic reaction and the reaction
virtually dies out after Vc.
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Case 1, Operation of a PFR with heat exchanger
Highly temperature sensitive heat capacities
where in general
only A at V 0 (t 0)
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a heat-transfer area per unit volume of pipe
v0 was total initial feeding rate (adiabatic
case). When heat exchanger was applied, it should
be based on each tube
...
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X
T
V
Adiabatic endothermic reaction and the supply of
heat from the heat exchanger increases the
conversion.
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Equilibrium conversion

• The highest conversion that can be achieved in
  reversible reactions is the equilibrium
  conversion.
• For reversible reactions, the equilbrium
  conversion is uaually calculated first.
• The equilbrium conversion increases with
  increasing temperature for endothermic reactions.
• The equilbrium conversion decreases with
  increasing temperature for exothermic reactions.

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Exothermic reactions
The inlet temperature increases, the adiabatic
equilibrium conversion decreases
To determine the maximum conversion that can be
achieved in an exothermic reaction carried out
adiabatically Find the intersection of the
equilibrium conversion as a function of
temperature with temperature-conversion
relationships from energy balance
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Adiabatic equilibrium temperature example
For the elementary solid-catalyzed liquid-phase
reaction
Make a plot of equilbrium conversion as a
function of time. Determine the adiabatic
equilibrium temperature and conversion when pure
A is fed to the reactor at a temperature of 300
K.
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equilibrium
Rate law
-rA 0
The Vant Hoff equation
Only thermodynamics! nothing to do with the
energy balance
Xe f (T)
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Now, the reaction is carried out adiabatically
(energy balance)
From thermodynamics
XEB
From energy balance
How to increase the conversion?
T
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cooling process (with interstage cooling)
final conversion
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Interstage cooling example
Following the previous example, what conversion
could be achieved if two interstage coolers were
available that had the capacity to cool the exit
stream to 350 K? Also determine the heat duty of
each exchanger for a molar feed rate of A of 40
mol/s. Assume that 95 of equilibrium conversion
is achieved in each reactor. The feed temperature
to the first reactor is 300 K.
From previous work, we found that the adiabatic
equilibrium conversion
95 of equilibrium conversion
Temperature leaving the first reactor
cooling to 350 K
Cooling duty ?
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(Energy balance in the heat exchanger)
No reaction take place in the exchanger
being absorted by the coolant stream
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Th2 460 K
Th1 350 K
Reactant mixture
Heat Exchanger
Coolant
Tc2 400 K
Tc1 270 K
From transport phenomena
A 3.16 m2
The condition entering the second reactor
energy balance
95 of equilibrium conversion
following similar procedure again (page 476)
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Endothermic reactions
The equilibrium conversion increases with
increaing temperature
XEB
final conversion
heating process (with interstage heating)
T
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Optimum feed temperature
For a reversible and exothermic reaction
From thermodynamics
XEB
0.75
T0 500
0.33
T0 600
0.15
T0 350
600
500
350
T
At a very low feed temperature, the specific
reaction rate will be so small that virtually all
of the reactant will pass through the reactor
without reacting. There is an optimum inlet
temperature!
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Non-adiabatic reactor operation oxidation of
sulfur dioxide example
for X gt 0.05
The feed to an SO2 converter is 7900 lb mol/h and
consists of 11 SO2, 10 O2, and 79 inerts. The
converter consists of 4631 tubes packed with
catalyst, each 20 ft long. The tubes are 3 in.
o.d. and 2.782 in i.d. The tubes will be cooled
by a boiling liquid at 805 F, so the coolant
temperature is constant over this value. The
entering pressure is 2 atm. For inlet
temperatures of 740 and 940 F, plot the
conversion, temperature, equilbrium conversion,
and reaction rate profile down the reactor.
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From thermochemical tables, the equilibrium
constant at any temperature T is
T in R, Kp in atm-1/2
From the data of Eklund (1956), the rate
constants
T in R
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Stoichiometry
Mole balance (PFR)
Rate law
per tube
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Operation of a tubular with heat changer
Highly temperature sensitive heat capacities
4 parameter (X,W,P, and T), we need more
relations!
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Ergun equation (lecture 2)
Viscosity is also a function of temperature, but
we ignore here.
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Mole balance
Energy balance
Momentum balance
Three O.D.E.s are solved simultaneously.
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Multiple steady states
When more than one intersection occurs of the
energy and mole balance curves, there will be
multiple steady states at which the reactor may
operate.
Energy balance
Set ?Cp 0
Assume no shaft work
CSTR
Cp0
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For Ta lt T0
Heat-generated term G(T)
For CSTR
Heat-removed term R(T)
function of T ... i.e. mole balance, energy
balance ...
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Temperature ignition-extinction curve
The points of intersection of R(T) and G(T) give
the temperature at which the reactor can operate
at steady state and they are termed Ts.
Upper steady state
Ts, steady-state temperature
Unstable steady states
Lower steady state
ignition temperature
extinction temperature
T0, entering temperature
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• Ignition temperature is very important, since
  once the entering temperature exceeds this value,
  transition to the upper steady state will occur
  
• undesirable
• dangerous

Runaway reactions
Bifurcation analysis is usually applied to
analyze whether or not multiple steady states are
possible. (refer to page 498)
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Non-isothermal multiple reactions
We have discussed the energy balance for a single
reaction, for example (PFR)
When q multiple reactions and m species are
involved
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For example, series reactions take place in a PFR

Energy balance
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Non-isothermal multiple reactions in a CSTR
reactor
Single reaction
CSTR
When q multiple reactions and m species are
involved
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Parallel reactions with heat effects (PFR example)
Gas-phase reactions occur in a PFR
A, 100 mol /s, 150 C, 0.1 mol/dm3
Determine the temperature and flow rate profile
down the reactor.
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Mole balance (PFR)
Rate laws for each species
Convert C to F
(PFR)
A
B
C
Energy balance
All these O.D.E.s are solved together
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Series reactions with heat effects (CSTR example)
Liquid-phase elementary reactions occur in a CSTR

take place in a 10 dm3 CSTR. What are the
effluent concentrations for a volumetric feed
rate of 10 dm3/min at a concentration of A of 0.3
mol/dm3
283 K
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Mole balance (CSTR)
Rate laws for each species
Convert F to C
(CSTR)
A
B
C
Energy balance
All these O.D.E.s are solved together
Multi-steady state (page 507)