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Solutions for Chapter 8 – Steady-State Nonisothermal Reactor Design
P8-1 Individualized solution
P8-2 (a) Example 8-1
For CSTR
V=FA0X
"r
A
=X
#
0k1"X
( )
P8-2 (b) Example 8-2
Helium would have no effect on calculation
8-2
P8-2 (c) Example 8-3
V = 0.8 m3
See Polymath program P8-2-c.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 0.8 0.8
X 0 0 0.5403882 0.5403882
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra/Fao
Explicit equations as entered by the user
[1] Cao = 9.3
[2] Fao = .9*163
PFR
T
330
340
350
370
390
420
450
500
600
X
0.26
0.54
0.68
0.66
0.65
0.62
0.59
0.55
0.48
P8-2 (d) Example 8-4
Counter-Current: Guess Ta at V = 0 to be 330 and it will give an entering coolant temperature of
310 K.
8-3
See Polymath program P8-2-d.pol.
POLYMATH Results
No Title 08-17-2005, Rev5.1.233
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 5 5
X 0 0 0.7797801 0.7797801
T 310 310 344.71423 310.83085
Ta 330.7 310.16835 335.79958 310.16835
ODE Report (RKF45)
Differential equations as entered by the user
Explicit equations as entered by the user
[1] Cao = 9.3
[2] Fao = .9*163*.1
P8-2 (e) Example 8-5
8-4
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 0.001 0.001
X 0 0 0.3512403 0.3512403
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra/Fao
Explicit equations as entered by the user
[1] Fao = .0376
[2] Cpa = 163
P8-2 (f) Example 8-6
Energy balance will remain the same
XEB =2"10#3T#300
( )
P8-2 (g) Example 8-7
Both Xe and XEB will change. The slope of energy balance will decrease by a factor of 3.
X
8-6
P8-2 (h) Example 8-8
(1) CA0 will decrease but this will have no effect
(2) τ will decrease
X
T
X
M
XM
,T
8-7
P8-2 (i) Example 8-9
Change CP = 29 and –ΔH = 38700
POLYMATH Results
NLES Solution
Variable Value f(x) Ini Guess
X 0.7109354 2.444E-11 0.367
k 20.01167
NLES Report (safenewt)
Basecase
8-8
Explicit equations
[1] tau = 0.1229
P8-2 (j)
α = 1.05 dm3
See Polymath program P8-2-j.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 1 1
Fa 100 2.738E-06 100 2.738E-06
Fb 0 0 55.04326 55.04326
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(V) = r1a+r2a
Explicit equations as entered by the user
[1] k1a = 10*exp(4000*(1/300-1/T))
[2] k2a = 0.09*exp(9000*(1/300-1/T))
[3] Cto = 0.1
[12] alpha = 1.05
8-9
P8-2 (k) Example 8-11
Vary UA
UA =70,000 J m3•s•K
only the lower steady state exists at T = 318 K
SBC =0.05
Vary
T0
T0=275
, very little effect.
Vary τ
P8-2 (l) Example PRS P8-4.1
"
~1
dPP
0
P8-2 (m) Example T8-3
mc = 200 g/s
See Polymath program P8-2-m.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
8-10
thetaI 1 1 1 1
CpI 40 40 40 40
CpA 20 20 20 20
thetaB 1 1 1 1
CpB 20 20 20 20
ODE Report (RKF45)
Differential equations as entered by the user
Explicit equations as entered by the user
[1] alpha = .0002
[2] To = 350
[12] CpB = 20
[13] Cto = 0.3
[14] Ea = 25000
[15] Kc = 1000*(exp(Hr/1.987*(1/303-1/T)))
P8-2 (n)
(1) The concentration of A near the wall is lower than in the center because the velocity profile is
Below is the FEMLAB solution.
1. Parameters in simulation on the tubular reactor from Example 8-12 (First Order
reaction):
Reaction:
CBA !+
A- propylene oxide; B- water; C- propylene glycol
(1) operating parameters
Reactants
Feed rate of A FA0 = 0.1 mol/s
Inlet flow rate of A
6
3
0
0107
830
101.581.0 !
!
"=
""
==
AA
A
MF
v
#
m3/s
(2) properties of reactants
● Heat of reaction, ∆HRx, dHrx = -525676+286098+154911.6=-84666.4 J/mol
● Activation energy, E = 75362 J/mol
● Pre-exponential factor, A = 16.96×1012 /3600 1/s
(3) properties of coolant
2. Size of the Tubular Reactor
8-12
2. Femlab screen shots
(1) Domain
(2) Constants and scalar expressions
- Constants
- Scalar expressions
(3) Subdomain Settings
- Physics
(mass balance)
8-13
(Energy balance)
(Cooling Jacket)
8-14
- Boundary Conditions
@ r = 0, Axial symmetry
(4) Results
(Concentration, cA)
8-15
Second order reaction
[1] Domain
[2] Constants & Scalar expressions
(1) Constants
(2) Scalar expressions
8-17
[3] Subdomain Settings
(1) Physics
(Mass balance)
(Energy balance)
(Cooling Jacket)
(2) Initial values
(3) Boundary Values
@ r = 0, Axial symmetry
[4] Results
(Concentration, cA)
8-19
(Temperature, T)
P8-2 (o) Individualized solution
P8-4
NH 4NO3l
( )
"2H2O g
( )
+N2O g
( )
Al
( )
"2W g
( )
+B g
( )
Mole Balance
V=FA0X
"r
A
XMB ="r
AV
FA0
=
kM
V#V
FA0
=kM
FA0
Energy Balance
HA0"HAg
( )
+#WHW0"HWg
( )
( )
"2HWg
( )
+HBg
( )
"HAl
( )
$HRx
6 7 4 4 4 4 8 4 4 4 4
" $HVap
%
&
'
'
(
)
*
*
X=0
"W=F
W
FA
=0.17
( )
18
( )
0.83
( )
80
( )
=0.9103
CPA=0.38 BTU
lb°R#80 lb
mol =30.4 BTU
lbmol°R
CPS=0.47 BTU
lb°R#18 lb
mol =8.46 BTU
lbmol°R
