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9-1
Solutions for Chapter 9 – Unsteady State
Nonisothermal Reactor Design
P9-1 Individualized solution
P9-2 (a) Example 9-1
P9-2 (b) Example 9-2
To show that no explosion occurred without cooling failure.
9-2
P9-2 (c) Example 9-3
Decreasing the electric heating rate (Tedot in polymath program from example 9-3) by a factor of
P9-2 (d) Example 9-4
Decreasing the coolant rate to 10 kg/s gives a weak cooling effect and the maximum temperature
P9-2 (e) Example 9-5
P9-2 (f) Example 9-6
Using the code from Example 9-5, we could produce the following graphs either by changing TO
9-4
P9-2 (g) Example 9-7
The temperature trajectory changes significantly. Instead of a maximum in temperature, there is
now a minimum.
P9-2 (h) Example RE9-1
P9-2 (i) Example RE9-2
9-6
P9-2 (j) No solution will be given
P9-3
See Polymath program P9-3.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 2.9 2.9
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(t) = -dH*k/Cpa
Explicit equations as entered by the user
P9-4
9-8
POLYMATH Results
No Title 08-11-2005, Rev5.1.233
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 100 100
Na 1.0E-09 1.0E-09 326.10947 312.51175
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(Na)/d(t) = Fao+Vra
[2] d(T)/d(t) = ((Vra*dHr)-Fao*Cpa*(T-To))/(Na*Cpa)
Explicit equations as entered by the user
[1] Fao = 100
9-9
P9-5 (a)
POLYMATH Results
08-11-2005, Rev5.1.233
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 120 120
Na 500 0.1396707 500 0.1396707
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Na)/d(t) = ra*V
Explicit equations as entered by the user
[1] Fbo = if(t<50)then(10)else(0)
P9-5 (b)
This is the same as part (a) except the energy balance.
Energy balance:
9-11
P9-5 (c)
This is the same as part (b) except the reaction is now reversible.
See Polymath program P9-5-c.pol
P9-6 (a)
P9-6 (b)
9-13
P9-7
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 10 10
X 0 0 0.2504829 0.2504829
ODE Report (RKF45)
Differential equations as entered by the user
Explicit equations as entered by the user
[1] k1 = .002*exp((100000/8.314)*(1/373-1/T))
9-15
P9-8 (a)
Use Polymath to solve the differential equations developed from the unsteady state heat and mass
balances.
See Polymath program P9-8-a.pol
POLYMATH Results
No Title 08-11-2005, Rev5.1.233
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 300 300
Cc 0.1 0.1 205.63558 205.63558
Cs 300 43.080524 300 43.080524
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg
Explicit equations as entered by the user
[1] Iprime = (0.0038*T*exp(21.6-6700/T))/(1+exp(153-48000/T))
9-17
P9-8 (b)
When we change the initial temperature we find that the outlet concentration of species C has a
maximum at T0 = 300
P9-8 (c)
Cc can be maximized with respect to T0 (inlet temp), Ta (coolant/heating temperature), and heat
P9-9
First order liquid phase, CSTR
First solve the steady state problem for the heat exchange area A for normal operaiton T = 358K.
Steady state solution:
9-18
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 10 10
Ca 180 4.1007565 180 4.1007565
9-19
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ((180-Ca)/tau)+ra
Explicit equations as entered by the user
[1] tau = .4
P9-10 (a)
( ) ( )
[ ]
!"##$=Rxii HTTCpX /
0
(1)
X=1 T=Tf gives
Insert (1) in (2)
( )( )
000 TT
H
Cp
CCC
Rx
ii
AAA !
"!
#
$!=%
(3)
For X= 1, Tf = T expression (1) becomes
Remember that CB0 is not constant here. Similar to CA, use equations (1) and (4)
"
%
For developing –rA use
Insert (7) and (8) in (9)
!"
