8-61
P8-16 (a)
G(T)=“HRX
X=
“
k
1+
“
k,k=6.6 #10$3exp E
R
1
350 $1
T
%
&
‘ (
)
*
+
,
–
.
/
0
Graphical method
POLYMATH Results
02-22-2006, Rev5.1.233
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 1000 1000
T 350 350 450 450
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(t) = 0.1
Explicit equations as entered by the user
[1] RT = 150*(T-350)
– Equation
POLYMATH Results
02-22-2006, Rev5.1.233
NLE Solution
Variable Value f(x) Ini Guess
T 399.9425 3.181E-09 300
NLE Report (safenewt)
Nonlinear equations
[1] f(T) = RT-GT = 0
Explicit equations
8-63
P8-16 (b)
First, we must plot G(T) and R(T) for many different T0’s on the same plot. From this we must
generate data that we use to plot Ts vs To.
380
390
400
P8-16 (c)
For high conversion, the feed stream must be pre-heated to at least 404 K. At this temperature, X
P8-16 (d)
For a temperature of 369.2 K, the conversion is 0.935
P8-17
The energy balance for a CSTR:
0
( ) ( ) A
Rx
A
r V
G T H
F
! “
#
=#$ % &
‘ (
0
( ) A pS r
R T F C UAT UAT= + !
P8-18 (a)
[ ]
0 0 0
Mol Balance :
/
A A
A A B e
F X v C X
V
r k C C K
= =
! !
8-65
P8-18 (b)
0
3600 9
10* 40
A pA
UA
F C
!
= = =
P8-18 (c)
P8-18 (d)
The next plot shows how to find the ignition and extinction temperatures. The ignition
temperature is 358 K and the extinction temperature is 208 K.
8-67
P8-18 (f)
P8-18 (g)
At the maximum conversion G(t) will also be at its maximal value. This occurs at approximately
T = 404 K. G(404 K) = 73520 cal. For there top be a steady state at this temperature, R(T) =
G(T). See Polymath program P8-18-g.pol.
P8-18 (h) Individualized solution
P8-18 (i)
The adiabatic blowout flow rate occurs at
0.0041s
!
=
See Polymath program P8-18-i.pol.
P8-18 (j)
Lowing T0 or Ta or increasing UA will help keep the reaction running at the lower steady state.
P8-19
Given the first order, irreversible, liquid phase reaction:
A B!
1.0 / min/UA cal C= °
CTa
o
100=
0.5 / minPure A Feed g mol=
2 / /
A B
p p
C C cal g mol C= = !°
Simplifying,
1
1
11
k
X
k
k
!
!
!
= =
++
8-69
P8-19 (a)
(a)
R(T) = 4(T-(Ta+To)/2) = 4T-200-2To = 27.5 cal/gmolA (From Figure P8-19)
T = 163oC (From Figure P8-19) and the marginal To = 212oC
or
P8-19 (b)
(b)
Ts=180 oC
8-70
P8-19 (c)
To = 209 oC. R(T) = 4T-618. From Figure P8-19, G(T) ~ 90. Therefore X = 0.9
(c)
P8-19 (d)
(d)
From Figure P8-19, T = 148oC,
TT
Toa
c
1
+
+
=
“
“
8-71
P8-19 (e) Individualized solution
P8-20 (a)
The following are the explanations for the unexpected conversion and temperature profiles
Case 1: Broken preheater or ineffective catalyst
P8-20 (b)
The following are the explanations for the unexpected conversion and temperature profiles
Case 1: Broken preheater or ineffective heat exchanger
P8-21
Below is the FEMLAB solution.
1. Parameters in simulation on the tubular reactor in Problem 8-6:
Reaction:
CBA !+
(1) operating parameters
Reactants
Inlet concentration of A
100
0=
A
C
mol/m3
(2) properties of reactants
● Heat of reaction, ∆HRx, dHrx = -41+20+15=-6 kcal/mol=-25100 J/mol
8-72
Assumption:
● Thermal conductivity of the reaction mixture, ke = 0.68 W/m·K
(needed in the mass balance and the energy balance)
2. Size of the Tubular Reactor
(1) Volume of reactor sized by a PFR = 0.317 m3
3. Femlab Screen Shots
(1) Domain
(2) Constants and Scalar Expressions
– Constants
– Scalar Expressions
8-73
(3) Subdomain Settings
– Physics
(Mass balance)
(Energy balance)
– Initial Values
(Mass balance) cA(t0) = cA0
(Energy Balance) T(t0) = T0
– Boundary Conditions
(4) Results
(Concentration, cA)
8-75
P8-22 (a)
0
:
A
A
Liquid Phase A B C
r
dX
dV F
+!
“
=
The energy Balance:
0
( ) ( )( )
A B
a A R
A p p
Ua T T r H
dT
dV F C C
!+! !“
=# $
+
% &
8-76
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
V
0
0
1000.
1000.
2
T
300.
300.
434.7779
348.2031
3
X
0
0
0.9620102
0.9620102
4
Ta
300.
300.
300.
300.
Differential equations
1
d(T)/d(V) = (U * a * (Ta – T) + (-ra) * (-Dhr1)) / (fao * (cpa + cpb))
2
d(X)/d(V) = -ra / fao
Explicit equations
1
Ta = 300
2
R = 1.988
3
E = 10000
8-77
P8-22 (b)
Gas Phase:
A“B+C
dX
dW =“rA
FA0
The energy Balance:
( ) ( )( )
5
a A R
Ua T T r H
dT
dW F C
U
!+! !“
=
=
See Polymath program P8-22-b.pol.
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
W
0
0
50.
50.
8-78
7
kr
0.2
0.0622149
0.2
0.0622149
8
Uarho
5.
5.
5.
5.
Differential equations
1
d(X)/d(W) = –rA / v0 / CA0
2
d(T)/d(W) = (Uarho * (Ta – T) + rA * 20000) / v0 / CA0 / 40
Explicit equations
1
T0 = 400
8-79
P8-23
First note that ΔCP = 0 for both reactions. This means that ΔHRx(T) = ΔHRx° for both reactions.
Now start with the differential energy balance for a PFR;
dT
dV =Ua(Ta“T)+r
ij #HRxij
( )
$
FjCPj
$=Ua(Ta“T)+r
1A#HRx1A
( )
+r2B#HRx2B
( )
FjCPj
$
dV =
We can then solve for r1A from this information.
r
1A=Ua(T“Ta)“r
2B#HRx 2B
( )
#HRx1A
=Ua(T“Ta)+2k2DCBCC#HRx 2B
( )
#HRx1A
r
1A=10(500 “325) +2#0.4 0.2
( )
0.5
( )
5000
( )
“50000 =“0.043
8-80
P8-24 (b)
P8-24 (c)
solution
P8-24 (d) No solution will be given