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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
The new T0 of 20 ˚F (497 ˚R) gives a new ΔHRn and T. With T=497+89.8X the polymath
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
10 gives a conversion of 9.72 % at the onset temperature. For a decrease by a factor 10 the
P9-2 (d) Example 9-4
Decreasing the coolant rate to 10 kg/s gives a weak cooling effect and the maximum temperature
in the reactor becomes 315 K. An increase of the coolant rate to 1000 kg/s gives a Tmax of 312 K.
P9-2 (e) Example 9-5
9-3
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
9-5
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
T 970 970 1.119E+23 1.119E+23
dH -336 -336 -336 -336
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(t) = -dH*k/Cpa
9-7
Explicit equations as entered by the user
[1] dH = -336
P9-4
9-8
See Polymath program P9-4.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 0 100 100
Na 1.0E-09 1.0E-09 326.10947 312.51175
T 400 400 801.376 800
Cpa 20 20 20 20
Vra -1.9E-10 -101.76832 -1.9E-10 -100
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
[2] dHr = -8000
[3] Cpa = 20
9-9
P9-5 (a)
9-10
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 120 120
vo 1 0 1 0
V 50 50 100 100
ra 0 -0.3365379 0 -1.969E-05
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Na)/d(t) = ra*V
[2] d(Nb)/d(t) = ra*V+Fbo
[3] d(Nc)/d(t) = -ra*V
Explicit equations as entered by the user
[1] Fbo = if(t<50)then(10)else(0)
[2] Nao = 500
[3] Cbo = 10
[4] k = .01*exp((10000/1.987)*(1/300-1/T))
P9-5 (b)
This is the same as part (a) except the energy balance.
9-11
See Polymath program P9-5-b.pol
P9-5 (c)
This is the same as part (b) except the reaction is now reversible.
9-12
P9-6 (a)
P9-6 (b)
9-13
P9-7
9-14
See Polymath program P9-7.pol
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
T 373 373 562.91803 562.91802
Ca 0.1 0.0749517 0.1 0.0749517
k2 3.0E-05 3.0E-05 366.75159 366.75083
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(t) = -ra/.1
[2] d(T)/d(t) = ((40000+(10*(T-298)))*(-ra)*(1/.1))/(56.25-(10*X))
Explicit equations as entered by the user
[1] k1 = .002*exp((100000/8.314)*(1/373-1/T))
[2] Ca = .1*(1-X)
[3] Cb = .1*(1.25-X)
[4] k2 = .00003*exp((150000/8.314)*(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
T 278 278 333.55016 333.55016
mu 0.0426159 2.912E-04 0.1725264 2.912E-04
Q 0 0 0 0
Cps 74 74 74 74
Cpc 74 74 74 74
rs -0.005327 -30.908876 -0.005327 -0.074831
9-16
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg
[2] d(Cs)/d(t) = -rg/Ycs
Explicit equations as entered by the user
[1] Iprime = (0.0038*T*exp(21.6-6700/T))/(1+exp(153-48000/T))
[2] Km = 5.0
[3] mu1max = 0.5
[4] Ycs = 0.8
[5] mu = mu1max*Iprime*(Cs/(Km+Cs))
[6] Q = 0
[7] Cps = 74
[8] Hrxn = -20000
9-17
P9-8 (b)
When we change the initial temperature we find that the outlet concentration of species C has a
P9-8 (c)
Cc can be maximized with respect to T0 (inlet temp), Ta (coolant/heating temperature), and heat
exchanger area. Therefore, if we are to find the optimal heat exchanger area the inlet temperature
P9-9
First order liquid phase, CSTR
First solve the steady state problem for the heat exchange area A for normal operaiton T = 358K.
9-18
See Polymath program P9-9.pol
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
tau 0.4 0.4 0.4 0.4
k 1.1 1.1 107.23721 107.23721
Na 0.9 0.0205038 0.9 0.0205038
9-19
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ((180-Ca)/tau)+ra
[2] d(Cb)/d(t) = -ra-(Cb/tau)
Explicit equations as entered by the user
[1] tau = .4
[2] A = 22.696
[3] k = 1.1*exp(11409*(1/313-1/T))
[7] Nb = Cb/V
P9-10 (a)
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Remember that CB0 is not constant here. Similar to CA, use equations (1) and (4)
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