Download Review: SS Nonisothermal Reactors - University of Illinois Urbana

Review: Multiple Steady States in
CSTR
1
L16-1
XA,EB
XA,MB
0.8
XA
0.6
0.4
0.2
0
0
100
200
300
400
500
600
T (K)
• Plot of XA,EB vs T and XA,MB vs T
• Intersections are the T and XA that satisfy both mass balance (MB) &
energy balance (EB) equations
• Each intersection is a steady state (temperature & conversion)
• Multiple sets of conditions are possible for the same reaction in the same
reactor with the same inlet conditions!
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-2
Review: Heat Removal Term R(T) & T0
Heat removed: R(T)
Heat generated G(T)
 r V 
Cp0 1     T  TC   HRX  A 
 FA0 
R(T) line has slope of CP0(1+)
  UA Cp0FA0 Tc   Ta  T0
1 
R(T)
=∞
=0
R(T)
Increase 
Increase T0
T
When T0 increases, slope stays
same & line shifts to right
For Ta < T0
Ta
T0
T
When  increases from lowering
FA0 or increasing heat exchange,
slope and x-intercept moves
Ta<T0: x-intercept shifts left as ↑
Ta>T0: x-intercept shifts right as ↑
=0, then TC=T0 =∞, then TC=Ta
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-3
Review: CSTR Stability
G(T)
Tc 
R(T)
 Ta  T0   UA Cp0FA0
1 
R(T) > G(T) → T
falls to T=SS1
G(T) > R(T)
→ T rises to
T=SS1
G(T) > R(T)
→ T rises to
T=SS3
2
Heat removed: R(T)
1
3
R(T) > G(T)
→T falls to T=SS3
Heat generated G(T)
 rA V 

 FA0 
Cp0 1     T  TC   HRX 
T
• Magnitude of G(T) to R(T) curve determines if reactor T will rise or fall
• G(T) = R(T) intersection, equal rate of heat generation & removal, no
change in T
• G(T) > R(T) (G(T) line above R(T) on graph): rate of heat generation > heat
removal, so reactor heats up until a steady state is reached
• R(T) > G(T) (R(T) line above G(T) on graph): rate of heat generation < heat
removal, so reactor cools off until a steady state is reached
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-4
L16: Unsteady State Nonisothermal
Reactor Design
Q
Goal: develop EB for unsteady state reactor
An open system (for example, CSTR)
Fin
Fout
Hin
Hout
W
dEsys
dt

Q

Rate of
rate of heat
accumulation
flow from
=
of energy in
surroundings
system
to system
dEsys
dt
W
n

 FE
i i
i1
in

Rate of work
Rate of energy
done by
added to
+
system on
system by
surroundings
mass flow in
 0 steady state
dEsys
dt
n
 FE
i i
i1
out
Rate of energy
leaving system
by mass flow
out
 0 unsteady state
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-5
Change in System Energy with Time
dEsys
dt
n
 Q  W   FE
i i
i1
n
in
  FE
i i
i1
out
n
Energy of system is the sum of products of each species E
sys   NiEi
specific energy Ei & the moles of each species:
i1
n
Ei  Ui & Ui  Hi  PVi  so:  Esys   Ni Hi  PVi  Differentiate wrt time
i1

dEsys
dt
dEsys n dHi n dNi d  n

d n



N

H

P
N
V
 i
 i
 i i
   Ni Hi  PVi  

dt
dt
dt
dt
dt i1
 i1

i1
i1

Total V

dEsys
dt
n
dHi n dNi d
  Ni
  Hi
 PV 
dt i1 dt dt
i1
n
n
Q  WS   FH
i i
i1
n
  FH
i i
in
i1
0
For well-mixed reactor
with constant PV- variation
out
dHi n dNi
  Ni
  Hi
dt i1 dt
i1
Total Energy Balance for unsteady state, constant PV
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-6
Well-Mixed Reactors, Constant PV
Total Energy Balance for unsteady state
n
n
n
dHi n dNi d
Q  WS   FH
  Ni
  Hi
 PV  = 0
i i   FH
i i
dt i1 dt dt
i1
in i1
out i1
Special case: well-mixed reactors (e.g., batch, CSTR or semibatch) with
constant PV- variation in total P or V can be neglected
n
n
 Q  WS   FH
i i
i1
n
  FH
i i
in
i1
out
dHi n dNi
  Ni
  Hi
dt i1 dt
i1
Total Energy Balance for unsteady state, constant PV
T
dHi
dHi
dT
Evaluate
recalling that Hi  H RX  TR    Cpi dT so
Cpi
dt
dt
dt
T
R
n
n
 Q  WS   FH
i i
i1
n
  FH
i i
in
i1
out
dT n dNi
  NiCpi
  Hi
dt i1 dt
i1
Need to put dNi/dt into terms that can be measured
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-7
EB for Well-Mixed Reactors, PV=0
n
n
Q  WS   FH
i i
i1
n
  FH
i i
in
i1
out
dT n dNi
  NiCpi
  Hi
dt i1 dt
i1
From the mass balance:
Accumulation = In - Out + Gen
dNi
dNi
 Fi0  F i  irA V Substitute
 Fi0  F i  i  rA  V 
dt
dt
n
n
n
dT n
 Q  WS   FH
  NiCpi
  Hi Fi0  F i  irA V 
i i   FH
i i
dt i1
i1
i1
in i1
out
n
n
n
n
n
dT n
 Q  WS   Fi0Hi0   FH
  HiFi0   HiFi    iHi  rA V 
i i   NiCpi
dt i1
i1
i1
i1
i1
i1
Add SFiHi to both sides of equation:
n
n
n
dT n
 Q  WS   Fi0Hi0   NiCpi
  HiFi0    iHi  rA V 
dt i1
i1
i1
i1
Substitute ΣiHi =H°RX(T):
n
n
H°RX(T)
dT n
 Q  WS   Fi0Hi0   NiCpi
  HiFi0  H RX  T  rA V 
dt i1
i1
i1
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-8
Simplified EB for Well-Mixed Reactors
n
n
dT n
  HF
Solve for dT/dt: Q  WS   Fi0Hi0   NiCpi
i i0  H RX  T  rA V 
dt i1
i1
i1
Bring SFi0Hi and H°RX(T) terms to other side of equation:
n
n
n
dT
 Q  WS   Fi0Hi0   HF
i i0  H RX  T  rA V    NiCpi
dt
i1
i1
i1
Factor SFi0Hi0 and SFi0Hi terms and divide by SNiCpi :
n
Energy balance for
unsteady state reactor
with phase change:
Q  WS   Fi0 Hi  Hi0   H RX  T  rA V 
i1
n
 NiCpi

dT
dt
i1
n
Energy balance for Q  WS   Fi0Cpi  T  Ti0   H RX  T rA V  dT
i1

unsteady state reactor
n
dt
 NiCpi
without phase change:
i1
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-9
Unsteady State EB, Liquid-Phase Rxns
n
Q  WS   Fi0 Cpi  T  Ti0   H RX  T  rA V 
i1
n
 NiCpi

dT
dt
i1
For liquid-phase reactions, often Cp = SiCpi is so small it can be neglected
When Cp can be neglected, then:
n
 NiCpi  NA0 Cps where Cps  SiCpi is the heat capacity of the solution
i1
If the feed is well-mixed, it is convenient to use:
SFi0 Cpi  FA0 Cps
Plug these equations and Ti0 = T0 into the EB gives:
Q  WS  FA 0 Cps  T  Ti0   H RX  T  rA V  dT

NA0 Cps
dt
This equation for the EB is simultaneously solved with the mass
balance (design eq) for unsteady state, nonisothermal reactor design
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-10
Nonisothermal Batch Reactor Design
n
Q  WS   Fi0Cpi  T  Ti0   H RX  T rA V 
i1
0
n
 NiCpi

dT
dt
i1
No flow, so:

Q  WS  H RX  T  rA V 
n
 NiCpi

dT
dt
i1
Put the energy balance
in terms of XA:
Ni  NA0  i   i X A 
Ni0
where i 
&  iCpi  CP
NA 0
Q  WS  H RX  T  rA V 
dT


n
dt


NA0   iCpi  Cp X A 
i1

Solve with the batch reactor design equation using an ODE solver (Polymath)
dX A
NA0
 rA V
dt
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Adiabatic Nonisothermal Batch
Reactor
Design
0 0
L16-11
Q  WS  H RX  T  rA V  dT

n
dt


NA0   iCpi  Cp X A 
i1

In the case of no stirring work and adiabatic operation, WS =0 & Q  0


H RX  T  rA V 
n

NA0   iCpi  Cp X A 
i1


dT Substitute:
H RX  T  rA V 
dT


dt  iCpi  Cps
NA0 Cps  Cp X A  dt
dT
dt
dX A

H
T


H
T


C
T

T
an
d
-N
 rA V
Substitute:
RX  
RX  R 
p
R
A0
dt
dX A
dT
   H RX  TR   Cp  T  TR   NA 0
 NA0 CpS  Cp X A 


dt
dt
dX
dT
   H RX  TR   Cp  T  TR   A  CpS  Cp X A 

 dt
dt
Rearrange:
 H RX  T  rA V   NA0 CpS  Cp X A 
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
dX A
dT
Solve for how XA   H



T


C
T

T

C


C
X
RX  R 
p
R 
p A
 pS

dt
dt
changes with T
L16-12
   H RX  TR   Cp  T  TR   dX A  CpS  Cp X A  dT


Get like terms together:

XA

X A0 0 CpS
T
dX A
dT
 
 Cp X A T H RX  TR   Cp  T  TR 
0
Integrate & solve for XA:
 XA
 T  T0 

 XA
  H RX  TR   Cp  T  TR  
Cp
S

 T  T0 
 HRX  T 
Cp
S
Solving for T:
 H RX  T0   X A
 H RX  T0   X A


T  T0  
 T  T0  
n
Cp  X A Cp
s
 iCp  X A Cp
i1
Heat capacity of soln (calculate Cps if not given)
i
Solve with the batch reactor design equation using an ODE solver (Polymath)
XA
t  NA0 
0
dX A
rA V
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-13
1st
A
order, liquid-phase, exothermic reaction A→B is run in a batch reactor.
The reactor is well-insulated, so no heat is lost to the surroundings. To control
the temperature, an inert liquid C is added to the reaction. The flow rate of C
is adjusted to keep T constant at 100 °F. What is the flow rate of C after 2h?
TC0 = 80 °F
V0= 50 ft3
H°RX=-25000 Btu/lb mol
k(100 °F)= 1.2 x 10-4 s-1
Cp, (all components)= 0.5 Btu/lb mol °F
CA0= 0.5 lb mol/ft3
1. Solve design eq for comp as function of t
2. Solve EB for FC0 using that info & T=100 °F
This is essentially a semi-batch reactor since only C is fed into the reactor
dX A
dNA
 rA V would complicate
Design eq:
 rA V Note, using NA0
d
t
dt
the calculation because V depends on t
Rate eq: -rA = kCA
dNA
dNA


kC
V

 kNA Rearrange and integrate for NA
Combine:
A
dt
dt
 NA 
dNA t

ln
 
  kdt

  kt  NA  NA0 exp  kt 
 NA0 
NA0 NA
0
NA
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-14
1st
A
order, liquid-phase, exothermic reaction A→B is run in a batch reactor.
The reactor is well-insulated, so no heat is lost to the surroundings. To control
the temperature, an inert liquid C is added to the reaction. The flow rate of C
is adjusted to keep T constant at 100 °F. What is the flow rate of C after 2h?
Use EB to find how the flow rate of C depends on the rxn (solve EB for FC0)
0 0 n
Q  WS   Fi0 Cpi  T  Ti0   H RX  T  rA V 
dT
i1

n
dt
 NiCpi
0
i1
n
n
i1
i1
 0    Fi0Cpi  T  Ti0   H RX  T rA V    Fi0 Cpi  T  Ti0   H RX  T  rA V 
C is the only species that flows, so:  FC0 CpC  T  Ti0   H RX  T  rA V 
rAV = -kCAV = -kNA where NA  NA0 exp  kt 

 FC0 CpC  T  Ti0   H RX  T  kNA0 ekt
Isolate FC0:  FC0 

H RX  T  kNA0 ekt
CpC  T  Ti0 


Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-15
1st
A
order, liquid-phase, exothermic reaction A→B is run in a batch reactor.
The reactor is well-insulated, so no heat is lost to the surroundings. To control
the temperature, an inert liquid C is added to the reaction. The flow rate of C
is adjusted to keep T constant at 100 °F. What is the flow rate of C after 2h?
H°RX=-25000 Btu/lb mol
Cp, (all components)= 0.5 Btu/lb mol °F
TC0 = 80 °F
V0= 50 ft3
k(100 °F)= 1.2 x 10-4 s-1
CA0= 0.5 lb mol/ft3
FC0 

H RX  T  kNA 0 ekt
CpC  T  Ti0 

NA0  CA0 V0  0.5
At 2h (7200s):
FC0
lb mol
ft 3


 1.210
Btu 

4

1
 1.2  10 s  25 lb mol  e
25,000
lb mol 


Btu
0.5
100 F  80 F
lb mol  F

 FC0  3.16


50ft 3  25 lb mol

4 s1 7200s 



lb mol
s
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-16
Instead of feeding coolant to the reactor, a solvent with a low boiling point is
added (component D). The solvent has a heat of vaporization of 1000 Btu/lb
mol, and initially 25 lb mol of A are placed in the tank. The reactor is wellinsulated. What is the rate of solvent evaporation after 2 h if T is constant at 100
°F? Additional info: k(100 °F)= 1.2 x 10-4 s-1 H°RX=-25000 Btu/lb mol
Still a semibatch reactor, where D is removed from the reactor
Use EB that accounts for a phase change:
0 0 n
Q  WS   Fi0 Hi0  Hi   H RX  T rA V 
Q̇ =0
dT
i1

ẆS=0
n
dt
 NiCpi
0
i1
Clicker Question: Does dT/dt = 0?
a) Yes
b) No
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-17
Instead of feeding coolant to the reactor, a solvent with a low boiling point is
added (component D). The solvent has a heat of vaporization of 1000 Btu/lb
mol, and initially 25 lb mol of A are placed in the tank. What is the rate of
solvent evaporation after 2 h?
Additional info: k(100 °F)= 1.2 x 10-4 s-1
H°RX=-25000 Btu/lb mol
Still a semibatch reactor, where D is removed from the reactor
Use EB that accounts for a phase change:
0 0 n
Q  WS   Fi0 Hi0  Hi   H RX  T rA V 
Q̇ =0
dT
i1

ẆS=0
n
dt
 NiCpi
dT/dt = 0
0
i1
n
  Fi0 Hi0  Hi   H RX  T  rA V 
i1

D is the only species that ‘flows’,
and rAV = -kNA0(exp[-kt]), so:

 FD0 Hi0  Hi   H RX  T  kNA 0 ekt  F 
D0
Hi0-Hi = heat of vap
25,000
FD0 

H RX  T  kNA0 ekt
Hi0  Hi 

Btu  0.00012
 25 lb mol e 0.00012 s 7200s  F  0.0316 lb mol

lbmol 
s
D0

s
1000Btu lb mol
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
A liquid phase exothermic reaction A →B is carried out at 358K in a 0.2 m3 CSTR. L16-18
The coolant temperature is 273K and the heat transfer coefficient (U) is 7200
J/min·m2·K. What is the heat exchange area required for steady state operation?
Using this heat exchange area, plot T vs t for reactor start-up.
CPA =CPS=20 J/g•K ẆS=0 CA0= 180 g/dm3 u0= 500 dm3/min
T0= 313 K
r= 900 g/dm3
H°RX(T) = -2500 J/g
E=94852 J/mol·K
k(313K)= 1.1 min-1
a) Heat exchange area for steady state operation:
SS operation
means that T is
constant, so:
n
dT

dt
Q  WS   Fi0Cpi  T  Ti0   H RX  T rA V 
i1
n
0
 NiCpi
n
i1
 Q  WS   Fi0 Cpi  T  Ti0   H RX  T  rA V 
i1
Q̇ =UA(Ta-T), ẆS=0, and A is only species that flows
 UA  Ta  T   FA0 CpA  T  TA0   H RX  T  rA V 
Plug in rA = -kCA and solve for A
 UA  Ta  T   FA0 CpA  T  TA0   H RX  T  kC A V 
A
H RX  T  kC A V   FA0 CpA  T  TA0 
U  Ta  T 
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
A liquid phase exothermic reaction A →B is carried out at 358K in a 0.2 m3 CSTR. L16-19
The coolant temperature is 273K and the heat transfer coefficient (U) is 7200
J/min·m2·K. What is the heat exchange area required for steady state operation?
Using this heat exchange area, plot T vs t for reactor start-up.
CPA =CPS=20 J/g•K ẆS=0 CA0= 180 g/dm3 u0= 500 dm3/min
T0= 313 K
r= 900 g/dm3
H°RX(T) = -2500 J/g
E=94852 J/mol·K
k(313K)= 1.1 min-1
Use material balance to determine steady state value of CA
FA0  FA  rA V  0  C A0u0  CAu0  kC A V  0  CA0u0  CAu0  kCA V

1.1
1 
 94852 J mol  1
C A0u0
107.4
exp 

 C A k  358K  =

k
358K





min
 8.314 J mol  K  313 358  
u0  kV
min
CA 

180 g dm3 500 dm3 min



500 dm3 min  107.4min1 200dm3
Solve EB for A:
A
 C A  4.1g dm3
H RX  T  kC A V   FA0 CpA  T  TA0 
g 
dm3 

FA0   180
500
  90000 g min
3  
min 

dm  
U  Ta  T 
3
3  1000dm
V=0.2m 
3


m

3
  200dm

Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
A liquid phase exothermic reaction A →B is carried out at 358K in a 0.2 m3 CSTR. L16-20
The coolant temperature is 273K and the heat transfer coefficient (U) is 7200
J/min·m2·K. What is the heat exchange area required for steady state operation?
Using this heat exchange area, plot T vs t for reactor start-up.
CPA =CPS=20 J/g•K ẆS=0 CA0= 180 g/dm3 u0= 500 dm3/min
T0= 313 K
r= 900 g/dm3
H°RX(T) = -2500 J/g
E=94852 J/mol·K
k(313K)= 1.1 min-1
Solve for heat exchange area at SS:
A
H RX  T  kC A V   FA 0 CpA  T  TA 0 
U  Ta  T 
FA0=90,000 g/min
H°RX(T) = -2500 J/g
U= 7200 J/min·m2·K
V=200 dm3
k= 107.4 min-1 CA= 4.1 g/dm3 CPA =20 J/g•K
TA0= 313 K
T= 358K
Ta=273K
J   107.4 
g 
g 
J 

3 
  2500   
4.1
200dm

90000
20

 

  358K  313K 
3
g
m
in
m
i
n
g

K


dm 


 
A 
J


7
200

  273K  358K 
2

min m  K 
A=227.4 m2
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Will use Polymath to plot T vs t for CSTR start-up (unsteady-state). Need
equations for dCA/dt, dT/dt, and k.
dNA
dC A C A0 C A



 rA
 FA0  FA  rA V
Mass balance:
dt
t
t
dt
L16-21
dT UA  Ta  T   FA0CpA  T  Ti0   H RX rA V 

n
dt
 Ni0Cps
i1
CPs is in terms of mass (J/g·K), so FA0 & Ni0 must also be in terms of mass
FA0=90,000 g/min
Substitute ṁi0 for Ni0, & use r for the solution to calculate:
g 
3

m

200d
m
9
0
0
 mi0  180,000g
mi0  Vr
i0

3

dm 
rA=-kCA
Amount of gas leaving reactor (L7)
1.1
1 
 94852J mol  1
1.1
1 
11408.7  1
k=
exp 


k=
exp





 K
min
min
 313 T  

 8.314 J mol  K  313 T  
U= 7200 J/min·m2·K
A=227.4m2 CA0=180g/dm3
HRX=-2500 J/g Ni0=mi0 V=200 dm3 CPs=20 J/g˙K
t=V/u0 Ta=273K
u0=500 dm3 T0=313
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-22
Will use Polymath to plot T vs t for CSTR start-up (unsteady-state). Need
equations for dCA/dt, dT/dt, and k.
1.1
1 
11408.7  1
dC A C A0 C A
k=
exp

rA=-kCA



 rA



min
dt
t
t
 313 T  
 K
dT UA  Ta  T   FA0CpA  T  Ti0   H RX rA V 
mi0  180,000g

n
dt
 Ni0Cps
F =90,000 g/min
i1
U= 7200 J/min·m2·K
A=227.4m2 CA0=180g/dm3
HRX=-2500 J/g Ni0=mi0 V=200 dm3 CPs=20 J/g˙K
A0
t=V/u0 Ta=273K
u0=500 dm3 T0=313
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
T (K)
L16-23
t (min)
Reaches steady state at ~12 minutes
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-24
The elementary, liquid phase, exothermic reaction A →B is carried out in a 2 m3
CSTR that is equipped with a heat jacket. Pure A enters the reactor at 60 mol/min and
a temperature of 310K. The coolant in the heat jacket is kept at 280 K. Provide all
equations, all constants, the initial time, and the final time that must be entered into
Polymath in order to plot temperature vs time for the first 20 min of reactor start-up.
ΔHRX(TR) = -10,000 cal/mol CPA = CPS =15 cal/mol·K
CpB=15 cal/mol·K
ẆS=0
E = 20,000 cal/mol k = 1 min-1 at 400 K
UA= 3200 cal/min•K
u0= 300 L/min
Need equations for how T changes with time, CA changes with time, & k changes with T.
n
dT

dt
Q  WS   Fi0 Cpi  T  Ti0   H RX  T  rA V 
i1
n

WS  0
Q  UA T - T
a

 NiCpi
i1
HRX  T   H RX  TR   CP  T  TR  CP  CPB  CPA  Cp  15  15
b
a
cal
0
mol  K
cal


H
T


10,000
 H RX  T   H RX  TR   0
RX  
mol
n
F C
i 1
i0
pi
n
 FA 0 Cp,A  FB0 Cp,B FB0  0   Fi0 Cpi  FA0 Cp,A


i 1
n
 NiCpi  NA0 Cps
i1
 
UA T - T  FA 0 Cp,A  T  Ti0   H
T r V
a
RX R A
Combine with EB: dT 
dt
NA0 Cps
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L16-25
The elementary, liquid phase, exothermic reaction A →B is carried out in a 2 m3
CSTR that is equipped with a heat jacket. Pure A enters the reactor at 60 mol/min and
a temperature of 310K. The coolant in the heat jacket is kept at 280 K. Provide all
equations, all constants, the initial time, and the final time that must be entered into
Polymath in order to plot temperature vs time for the first 20 min of reactor start-up.
ΔHRX(TR) = -10,000 cal/mol CPA = CPS =15 cal/mol·K
CpB=15 cal/mol·K
ẆS=0
E = 20,000 cal/mol k = 1 min-1 at 400 K
UA= 3200 cal/min•K
u0= 300 L/min
Need equations for how T changes with time, CA changes with time, & k changes with T.
cal
UA T - T  FA 0 Cp,A  T  Ti0   H
T r V
H RX  T   10,000
dT
a
RX
R A
mol



UA  3200
FA0  60
FA0
Ta  280
CPA  15
dt
t  f   20
NA0 Cps
Ti0  310
FA 0 V
NA 0

u0  NA0 
u0
V
t 0  0
 
rA  kC A
u0  300
CPS  15
 20,000  1
1 
k  1 exp 
 400  T  
1.987





V  2m3 1000L 1m3  V  2000
Use the mass balance to get eq for CA(t)
dC A C A0 C A
dNA
dNA



 kC A


C
u

C
u

k
C
V
 FA0  FA  rA V
A0 0
A 0
A
dt
t
t
dt
dt
t  V u0 C A0  FA 0 u0
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.