8-1
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
( )
X=
$
k
1+
$
k=
$
Ae“E RT
1+Ae“E RT
One equation, two unknowns
Adiabatic energy balance
In two equations and two unknowns
In Polymath form the solution
f X
( )
=X“
#
Ae“E RT
1+Ae“E RT
f T
( )
=T0“$HRx X
CPA
Enter X, A, E, R,
CPA
, T0 and ΔHRx to find τ and from that you can find V.
P8-2 (b) Example 8-2
Helium would have no effect on calculation
%Error =“#CPT“TR
( )
” #HRx
o+#CPT“TR
( )
[ ]
=1270
23,210 $100 =5.47%
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
Cao 9.3 9.3 9.3 9.3
T 340 340 363.39881 363.39881
Kc 2.8783812 2.4595708 2.8783812 2.4595708
Xe 0.7421605 0.7109468 0.7421605 0.7109468
rate 79.470998 79.470998 110.4184 85.208593
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
[3] T = 340+43.3*X
[4] Kc = 3.03*exp(-830.3*((T-333)/(T*333)))
PFR
T
CSTR has the same trend.
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
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
Cao 9.3 9.3 9.3 9.3
Kc 3.6518653 2.7812058 3.6518653 3.6255777
Xe 0.7850325 0.7355341 0.7850325 0.7838108
dHrx –6900 -6900 –6900 -6900
m 50 50 50 50
Cpc 75 75 75 75
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra/Fao
[2] d(T)/d(V) = ((ra*dHrx)-Ua*(T-Ta))/Cpo/Fao
[1] Cao = 9.3
[2] Fao = .9*163*.1
[3] Kc = 3.03*exp(-830.3*((T-333)/(T*330)))
[4] k = 31.1*exp(7906*(T-360)/(T*360))
[5] Xe = Kc/(1+Kc)
[6] ra = -k*Cao*(1-(1+1/Kc)*X)
[11] m = 50
[12] Cpc = 75
P8-2 (e) Example 8-5
At V = 0, Ta = 995.15 and gives a counter current entering temperature of 1250 K.
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
T 1035 972.39417 1035 1034.4748
Ta 995.15 986.00676 1249.999 1249.999
Fao 0.0376 0.0376 0.0376 0.0376
Cpa 163 163 163 163
To 1035 1035 1035 1035
Cpc 34.5 34.5 34.5 34.5
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra/Fao
[2] d(T)/d(V) = (Ua*(Ta-T)+ra*dHrx)/(Fao*(Cpa+X*delCp))
Explicit equations as entered by the user
[1] Fao = .0376
[2] Cpa = 163
[3] delCp = -9
[4] Cao = 18.8
[5] To = 1035
[6] dHrx = 80770+delCp*(T-298)
P8-2 (f) Example 8-6
Energy balance will remain the same
XEB =2“10#3T#300
( )
for
2A“2B
Xe=Ke
1+Ke
8-5
P8-2 (g) Example 8-7
Both Xe and XEB will change. The slope of energy balance will decrease by a factor of 3.
Also Xe will be more temperature sensitive
Ke=Keexp “HRx
1
#1
$
&
‘
)
The dotted line in the plot below shows an increase in –ΔHRx
1st Order
2nd Order
T
X
8-6
P8-2 (h) Example 8-8
(1) CA0 will decrease but this will have no effect
“
=401.1 ft3
(3) In the energy balance the slope of the energy balance of X vs. T will be greater
“#iCP
C=35 +18.65
( )
18
( )
+4$1.67
( )
19.5
( )
=35 +335.7 +130.2
=501 BTU
kmol°R
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
tau 0.1229
A 1.696E+13
NLES Report (safenewt)
Nonlinear equations
Basecase
Change QM
Less Conversions
8-8
Explicit equations
[1] tau = 0.1229
[2] A = 16.96*10^12
[3] E = 32400
Vary the heat exchanger area to find the effect on conversion.
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
k2a 553.05566 553.05566 1.48E+07 3.716E+06
Cto 0.1 0.1 0.1 0.1
Ft 100 77.521631 100 77.521631
Cb 0 0 0.0415941 0.0415941
Cc 0 0 0.016986 0.016986
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(V) = r1a+r2a
[2] d(Fb)/d(V) = -r1a
[3] d(Fc)/d(V) = -r2a/2
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
[4] Ft = Fa+Fb+Fc
[10] r2a = -k2a*Ca^2
[11] Fto = 100
[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
UA =60,000 J m3•s•K
intersection on the graph.
UA =700 J m3•s•K
only three steady states T = 300 (about), T = 350 (about) and one are a very high temperature off
the scale of the R (T) and G(T) plot.
In all cases
SBC
remains low at 0.05, meaning that the reaction has neared completion to form
“
=0.001
only the lower steady state at T = 316 about and other off scale
SBC
= 0.05
“
=0.00001
,
SBC
= 5
P8-2 (l) Example PRS P8-4.1
“
~1
dPP
0
“
2=
“
1
dP
1
P
01
dP2
P
02
=1
1 2
( )
1
2
( )
#
$
%
&
‘
( =1
No effect for turbulent flow if both dP and P changed at the same time.
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
W 0 0 4500 4500
Ta 320 320 334.77131 334.77131
y 1 0.3044056 1 0.3044056
alpha 2.0E-04 2.0E-04 2.0E-04 2.0E-04
To 350 350 350 350
Uarho 0.5 0.5 0.5 0.5
Mc 200 200 200 200
Cpmc 18 18 18 18
8-10
thetaI 1 1 1 1
CpI 40 40 40 40
Ea 2.5E+04 2.5E+04 2.5E+04 2.5E+04
Kc 66.01082 0.8247864 66.01082 31.551036
ka 0.046809 0.046809 11.205249 0.1177827
xe 0.8024634 0.3122841 0.8024634 0.7374305
sumcp 80 80 80 80
Ca 0.1060606 0.0137198 0.1060606 0.0137198
Cc 0 0 0.0724316 0.0355685
ra –5.265E-04 -0.0143957 –1.745E-05 -1.745E-05
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ta)/d(W) = Uarho*(T-Ta)/(Mc*Cpmc)
[2] d(y)/d(W) = -alpha/2*(T/To)/y
[3] d(T)/d(W) = (Uarho*(Ta-T)+(-ra)*(-Hr))/(Fao*sumcp)
Explicit equations as entered by the user
[1] alpha = .0002
[2] To = 350
[3] Uarho = 0.5
[4] Mc = 200
[5] Cpmc = 18
[8] thetaI = 1
[9] CpI = 40
[10] CpA = 20
[11] thetaB = 1
[12] CpB = 20
[13] Cto = 0.3
[14] Ea = 25000
[15] Kc = 1000*(exp(Hr/1.987*(1/303-1/T)))
[16] ka = .004*exp(Ea/1.987*(1/310-1/T))
[20] sumcp = (thetaI*CpI+CpA+thetaB*CpB)
[21] Ca = Cao*(1-X)*y*To/T
[22] Cb = Cao*(1-X)*y*To/T
[23] Cc = Cao*2*X*y*To/T
P8-2 (n)
(1) The concentration of A near the wall is lower than in the center because the velocity profile is
parabolic. This means near the walls the velocity is much lower and therefore the time space near
8-11
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
Inlet flow rate of A
6
0
0107
830
101.581.0 !
“=
““
==
A
AA
A
MF
v
#
m3/s
Inlet flow rate of B
6
00 103525.2 !
“=““=AB vv
m3/s
Inlet total flowrate
666
000 1049103510142 !!! “=“+“=+= BA vvv
m3/s
Inlet concentration of A
8.2040
1049
1.0
6
.0
0
0=
!
== “
v
F
CA
A
mol/m3
Inlet concentration of B
5.39682
10001035
63
6
0
0
0=
!!
=== ““
“
v
F
C
BBB
B
#
mol/m3
(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
● Specific reaction rate
28.1
0=k
/3600 1/s @300K
● Reaction rate
!
#
$
&‘=)
11
(exp
0TTR
E
kk
or
!
#
$
&‘=RT
E
Ak exp
AA kCr =!
● Thermal conductivity of the reaction mixture, ke = 0.559 W/m·K
● Average density of the reaction mixture, ρ, rho = 1000 kg/m3
● Heat capacity of the reaction mixture, Cp = 4180 J/kg·K
● Overall heat transfer coefficient, Uk = 1300 J/m2·s·K
2. Size of the Tubular Reactor
● Reactor radius, Ra = 0.1 m
8-12
2. Femlab screen shots
(1) Domain
(2) Constants and scalar expressions
– Constants
– Scalar expressions
(3) Subdomain Settings
– Physics
8-13
(Energy balance)
(Cooling Jacket)
(Source Term)
F = Taz-2*pi*Ra*Uk*(T-Ta)/(CpJ*mJ)
– Initial values
(Mass balance) cA(t0) = cA0
(Energy Balance) T(t0) = T0
8-14
– Boundary Conditions
@ r = 0, Axial symmetry
@ inlet, cA = cA0 (for mass balance)
T = T0 (for energy balance)
@ outlet, Convective flux
(4) Results
(Concentration, cA)
(Temperature, T)
8-15
Second order reaction
8-16
[2] Constants & Scalar expressions
(1) Constants
(2) Scalar expressions
8-17
[3] Subdomain Settings
(Mass balance)
(Energy balance)
(Cooling Jacket)
(Source Term)
F = Taz-2*pi*Ra*Uk*(T-Ta)/(CpJ*mJ)
8-18
(2) Initial values
(Mass balance) cA(t0) = cA0
(3) Boundary Values
@ r = 0, Axial symmetry
@ inlet, cA = cA0 (for mass balance)
T = T0 (for energy balance)
8-19
(Temperature, T)
P8-2 (o) Individualized solution
P8-3 Solution is in the decoding algorithm available in the beginning of this manual.
P8-4
NH 4NO3l
( )
“2H2O g
( )
+N2O g
( )
Al
( )
“2W g
( )
+B g
( )
From Rate Data
Mole Balance
V=FA0X
“r
A
XMB =“r
AV
FA0
=
kM
V#V
FA0
=kM
FA0
Energy Balance
FA0HA0+FW0HW0“FAHAg
( )
“FWHWg
( )
“FBHBg
( )
=0
FA0HA0+FA0#WHW0“FA01“X
( )
HAg
( )
“FA0#W+2FA0X
( )
HWg
( )
“FA0XHBg
( )
=0
HAg,T
( )
=HAl,T
( )
+$HVap
$HRx =2HWg
( )
+HBg
( )
“HAl
( )
HA0“HAg
( )
+#WHW0“HWg
( )
( )
“2HWg
( )
+HBg
( )
“HAl
( )
$HRx
6 7 4 4 4 4 8 4 4 4 4
” $HVap
%
&
‘
‘
(
)
*
*
X=0
HAl,T
( )
“HA0
( )
CPAT“660
( )
6 7 4 4 8 4 4
+1“X
( )
$HVap +#WHS500°F
( )
“HW200°F
( )
+CPST“500
( )
[ ]
=“$HRx X
XE=CPT“660
( )
+#WHS500°F
( )
“HW200°F
( )
+CPST“500
( )
[ ]
“$HRx
“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
HW500°F
( )
=1,202 BTU
lb =21,636 BTU
lbmol