Lecture 19
Chemical Reaction Engineering (CRE) is the
field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in
which they take place.
Web Lecture 19
Class Lecture 17 – Tuesday 3/8/2011
 Energy Balance Fundamentals
 Adibatic reactors
2
Let’s calculate the volume necessary to achieve a conversion, X,
in a PFR for a first-order, exothermic and adiabatic reaction.
The temperature profile might look something like this:
T
3
k
V
X
V
V
dX  rA

dV FA0
 E  1 1 
rA   ki exp   C A
 R  T1 T 
C A  C A0 1  X 
4
 E  1 1 
k i exp   
R  T1 T 
dX


C A0 1  X 
dV
FA 0
 E  1 1 
ki exp   
R  T1 T 
dX


C A0 1  X 
dV
FA0
We cannot solve this Equation because we don’t have X either
as a function of V or T.
We need another Equation. That Equation is:
The Energy Balance
5
1. Adiabatic CSTR, PFR, Batch or PBR
Cˆ P  0
WS  0
X EB
 C T  T 


X
i
C˜ PA T  T0 
T  T0 

0
H o Rx

6
Pi
H Rx
o
H
 Rx X EB
 C
i
Pi
Exothermic
Endothermic
T0
T
T
T0
0
7
XEB
0
XEB
2. CSTR with heat exchanger: UA(Ta-T) and a large coolant
flow rate
X EB

C
m
Ta
8
UA

 T T a   iCPi T T 0 
FA 0


H o Rx
T
3. PFR/PBR with heat exchange
Ta
FA0
T0
Coolant
3A. PFR in terms of conversion
Qg
9
Qr
Qg  Qr
dT rAH Rx T   Ua T  Ta 


dV
FA0   iCPi  C p X  FA0   iCPi  C p X 
3B. PBR in terms of conversion
Ua

rA H Rx T  
T  Ta 

b
dT

dW
FA 0
 C
i
Pi
 C p X

3C. PBR in terms of molar flow rates
dT

dW
10
Ua

rA H Rx T  
T  Ta 
b
 FiCP
i
3D. PFR in terms of molar flow rates
dT rA H Rx T   Ua T  Ta  Qg  Qr


dV
 FiCPi
 FiCPi
4. Batch
dT rAV H Rx   UA T  Ta 

dt
 N iCPi
11
5. For Semibatch or unsteady CSTR

n
dT

dt
Q  WS   Fi 0 CPi T  Ti 0    H Rx T     rAV 
i 1
n
N C
i 1
i
Pi
6. For multiple reactions in a PFR (q reactions and m species)
q
dT

dV
 r H
ij
i1
Rxij
 UaT  Ta 
m
FC
i
j 1
12
Pj
Let’s look where these User Friendly Equations came from.

Reactor with no Spatial Variations
Reactor
Reactor with no Spatial Variations
Q
Reactor
Rate of flow of
heat to the
system from
the
surroundings
Q
(J/s)
Reactor with no Spatial Variations
Q
W
Reactor
Rate of flow of
heat to the
system from
the
surroundings
-
Rate of work
done by the
system on the
surroundings
Q
-
W
(J/s)
(J/s)
Reactor with no Spatial Variations
Q
Fin
W
Reactor
Ein
Rate of flow of
heat to the
system from
the
surroundings
Q
(J/s)
-
Rate of work
done by the
system on the
surroundings
+
Rate of energy
added to the
system by mass
flow into the
system
-
W
+
Fin Ein
(J/s)
(J/s)
Reactor with no Spatial Variations
Q
Fin
W
Fout
Reactor
Ein
Rate of flow of
heat to the
system from
the
surroundings
Q
(J/s)
-
Rate of work
done by the
system on the
surroundings
-
W
(J/s)
Eout
+
Rate of energy
added to the
system by mass
flow into the
system
-
Rate of energy
leaving system
by mass flow
out of the
system
+
Fin Ein
-
Fout Eout
(J/s)
(J/s)
Reactor with no Spatial Variations
Q
Fin
=
Rate of flow of
heat to the
system from
the
surroundings
=
Q
dEˆ sys
dt
(J/s)
Fout
Reactor
Ein
Rate of
accumulation
of energy
within the
system
W
(J/s)
-
Rate of work
done by the
system on the
surroundings
-
W
(J/s)
Eout
+
Rate of energy
added to the
system by mass
flow into the
system
-
Rate of energy
leaving system
by mass flow
out of the
system
+
Fin Ein
-
Fout Eout
(J/s)
(J/s)


e.g., FA 0  

H i in

e.g., H A 0 
Q
Fi in

W
S
Fi out

e.g., FA 

H i out
e.g., H 
A

Energy Balance on an open system: schematic.
dE system


1
Q  WS   Fi 0 E i 0 in   Fi E i out 
dt
19
OK folks, here is what we are going to do to put the above equation into
a usable form.
1. Replace Ui by Ui=Hi-PVi
2. Express Hi in terms of heat capacities
3. Express Fi in terms of either conversion or rates of reaction
4. Define ΔHRx
5. Define ΔCP
6. Manipulate so that the overall energy balance is in terms of
the User Friendly Equations.
20
Assumptions:
=0
=0
E i  U i  P Ei  KEi Other energies small compared to internal
  flow work  shaft work
W
~
~
flow work    Fi 0 P0 Vi 0   Fi PVi
Recall:
Hi  Ui  PV˜i
21
 ~ m3 
 V 

mol

Substituting for W
F U
i0
i0
  FU
i i Q  
0 i 0   Fi PVi  WS 
  Fi 0 PV

Hi 0
dEsys
F
U i 0  PV
0 i0 
   Fi U i  PVi   Q  WS  dt
i0
F
i0
22
Hi
H i 0   Fi H i  Q  WS 
Steady State:
dEsys
dt
 W
   F H  F H  0
Q
S
i0 i0
i i
dEsys
dt
General Energy Balance:
 W
   F H  F H 
Q
S
i0 i0
i i
dE system
dt
For Steady State Operation:
 W
   F H  F H  0
Q
S
i0 i0
i i
23
F H
i0
i0
 FA 0 i Hi0
H Rx
F H
i
i
 FA0   i  i X H i  FA0  i H i  FA0 X i H i
Q  WS  FA0   i  H i 0  H i   FA0 X H Rx   0
24
For No Phase Changes
Hi T   H TR  T CPidT
R
Enthalpy of formation at temperature TR
Constant Heat Capacities
0
i
T
Hi T  Hi0 TR  CPi T  TR 
Hi0  Hi  CPi T  T0 
0

H


H
 i i  i i  iCPi T  TR 
Heat of reaction at temperature T
25
 H   H   C T  T 
i
i
i
0
i
i
Pi
R
HR T  HR TR  Cˆ P T  TR 
d ˆ
c ˆ
b ˆ
ˆ
ˆ
ˆ

C


C

C

C

C

C
 i Pi
P
PD
PC
PB
PA
a
a
a
Substituting back into the Energy Balance
Q  WS  FA0 X  H R TR   Cˆ P T  TR    FA0  iCPi T  Ti 0   0

26
Adiabatic (Q=0) and no Work (WS  0)
H Rx
d
c
b
 HD  HC  HB  HA
a
a
a
d
c
b
C P  C PD  C PC  C PB  C PA
a
a
a
27
Q  WS  FA0   i  H i 0  H i   FA0 X H Rx   0
Substituting back into the Energy Balance
Q  WS  FA0 X  H R TR   Cˆ P T  TR    FA0  iCPi T  Ti 0   0
28

Adiabatic (Q=0) and no Work (WS  0)


X H R TR   Cˆ P T  TR 
X H R T 
T  T0 
 T0 
~
~
ˆ
  i C Pi  XC P
  i C Pi  XCˆ P
T
Exothermic
T0
29
X
A↔B
1) Mole balance:
2) Rate Laws:
dX
rA

dV
FA 0

CB 
rA  k C A 

k
C 

C P  0
30
 E  1 1 
k  k1 exp   
 R  T1 T 
 H 0X  1 1 
  
k C  k C 2 exp
 k  T2 T 
A↔B
3) Stoichiometry:
C A  C A 0 1  X 
CB  CA0X
4) Energy Balance:
 H 0X X
T  T0 
 iCPi
31
First need to calculate the maximum conversion which is at the
Adiabatic Equilibrium.
A↔B
XC
 H X0 X
T  T0 
iCPi
Adiabatic equilibrium conversion
T
X eq
32
KC

1 K C
0

H


H
 i i  i i  iCPi T  TR 
HR T  HR TR  Cˆ P T  TR 
d ˆ
c ˆ
bˆ
ˆ
ˆ
iCPi  CP  a CPD  a CPC  a CPB  Cˆ PA

33
H Rx
d
c
b
 HD  HC  HB  HA
a
a
a
d
c
b
C P  C PD  C PC  C PB  C PA
a
a
a
34
Q  WS  FA0   i  H i 0  H i   FA0 X H Rx   0
Substituting back into the Energy Balance
Q  WS  FA0 X  H R TR   Cˆ P T  TR    FA0  iCPi T  Ti 0   0
35




X HR TR  Cˆ P T  TR 
X H R T0 
T  T0 
 T0 
iC˜ Pi  XCˆ P
iC˜ Pi  XCˆ P
T
T0
36
X
A↔B
1) Mole balance:
2) Rate Laws:
dX
rA

dV
FA 0

CB 
rA  k C A 

K
C 

C P  0
3) Stoichiometry:
37
0
 H Rx
K C  K C 2 exp
 R
C A  C A 0 1  X 
CB  CA0X
 E  1 1 
k  k1 exp   
 R  T1 T 
 1 1 
  
 T2 T 
A↔B
4) Combine:

X 
rA  kCA01 X 

K

C 
5) Energy Balance:

HoRx X
T  T0 
iCPi
First need to calculate the maximum conversion for adiabatic
operation which is at the Adiabatic Equilibrium Conversion.
 –r  0, solving for
At equilibrium
A
38
Xe 
KC
1 K C
A↔B
 H X0 X
T  T0 
iCPi
Xe
Adiabatic equilibrium conversion and
temperature
T
X eq
40
KC

1 K C
We can now form a table. Set X, then calculate T, -VA,
and FA0/-rA, increment X, then plot FA0/-rA vs. X:
FA0/-rA
41
X
42