Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
11-1#
Solutions)for)Chapter)11)–)Non-isothermal)Reactor)Design-The)
Steady-State)Energy)Balance)and)Adiabatic)PFR)Applications)
)
P11-1)(a))Example)Table)11-2)
(i)#T0#and#CPA#have#the#greatest#effect,#FA0#and#k0#have#the#least#effect#on#the#temperature#profiles.#
(v)#Individualized#answer#
P11-1)(b))Example)11-3)
(i)#No#solution#will#be#provided#
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 0.2603491 0.2603491
ODE)Report)(RKF45)
Differential equations as entered by the user
11-2#
[4] Kc = 3.03*exp(-830.3*((T-333)/(T*333)))
[5] k = 31.1*exp(7906*(T-360)/(T*360))
[6] Xe = Kc/(1+Kc)
X#
0.26#
0.54#
0.68#
0.66#
0.65#
0.62#
0.59#
0.55#
0.48#
(viii)#We#see#that#there#is#a#maximum#in#the#exit#conversion#corresponding#to#T0#=#350K.##Therefore,#the#
#
)
P11-1)(c))Aspen#Problem#
#
X#
11-3#
P)11-1)(d))
(i)#
#
(ii)#At#high#To,#the#graph#becomes#asymptotic#to#the#X-axis,#that#is#the#conversion#approaches#0.#
At#low#To,#the#conversion#approaches#1.#
#
⇒XEB =
2CP
A
T−T0
( )
ΔHRx
#
$
%&
'
(
#
#
(iv)#
It#is#observed#that#at#the#same#inlet#temperature,#the#equilibrium#conversion#on#the#addition#of#inerts#is#
11-4#
P11-1)(e))Example)11-5)
#
Q=
mCPΔT
Q=220kcal /s
⇒
m=220×103
18(400 −270)
=854.7mol /s
#
####For#the2nd#case:#
###Hot#Stream:##460K#!#350K#
###Cold#Stream:#270K#!##400K#
###LMTD#=#69.52#oC#
###
⇒A=220 ×103
100 ×69.52
=31.64m2
#
)
P11-1)(f))Example)11-6)
(i)#The#optimum#feed#temperature#is#480#K.#
#
(vi)#Individualized#solution#
#
)
P11-2))
rA=k(CA−
CB
KC
)=0
rA=k(CA0(1−X)−
CA0X
KC
)
rA=0,#then
(1−Xe)=
Xe
KC
KC=
Xe
1−Xe
)
11-5#
P11-2)Continued#
Since&KC(T2)=KC(T
1)exp[ΔHRx
R(1
T
1
−1
T2
)]
We(have(
Xe2
1−Xe2
=
Xe1
1−Xe1
exp[ΔHRx
R(1
T
1
−1
T2
)]
0.8
1−0.8 =0.5
1−0.5 exp[ΔHRx
R(1
(227+273) −1
(127+273))]
ΔHRx
=−23.05&kJ /mol
)
#
)
P11-3)(a))True.#Endothermic#&#adiabatic#so#T#decreases,#for#endothermic#reaction,#Xe#decreases#as#T#
decreases.#Once#X#reaches#Xe,#T#doesn’t#change#and#therefore#Xe#reaches#a#plateau.#))
P11-3)(b)) False.#If#the#reaction#is#endothermic#and#heated#along#the#length,#T#will#decrease#and#then#
#
)
P11-4)))
A+B→C
#
Since#the#feed#is#equimolar,#CA0#=#CB0#=#.1#mols/dm3#
CA#=#CA0(1-X)#
CB#=#CB0(1-X)#
11-6#
P11-4)Continued#
−rA=k C A0
2(1−X)2=.01k(1−X)2
#
)
P11-4)(a)
For#the#PFR,#FA0#=#CA0v0#=#(.1)(2)#=#.2#mols/dm3#
See#Polymath#program#P11-4-a.pol.#
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca0
0.1
0.1
0.1
0.1
2
Fa0
0.2
0.2
0.2
0.2
3
k
0.01
0.01
7.85772
7.85772
4
ra
-0.0001
-0.0018941
-4.3E-20
-4.3E-20
5
T
300.
300.
500.
500.
6
T0
300.
300.
300.
300.
7
V
0
0
3.441E+09
3.441E+09
8
X
0
0
1.
1.
Explicit equations
1
Ca0 = .1
2
Fa0 = .2
3
T0 = 300
4
T = T0 + 200 * X
5
k = .01*exp((10000 / 2) * (1 / 300 - 1 / T))
6
ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2)
The#conversion#and#temperature#both#increase#down#the#reactor#volume#because#the#reaction#is#
exothermic.#They#increase#faster#because#the#reaction#rate#increases#as#the#temperature#increases.#
#
11-7#
P11-4)(b)
T#=#T0#+#200X#
P11-4)(c))
T#and#therefore#k#is#constant#under#isothermal#condition)
V=FA0
dX
−rA
∫
−rA=k CA0
2(1−X)2
V=FA0
kCA0
2
dX
(1−X)2
∫
V=FA0
kCA0
2(1
1−X−1)
X=1
FA0
kVCA0
2+1
Qr=FA0X(−ΔHR(T0)) =−FA0ΔHR(T0)
FA0
kVCA0
2+1
=−(0.2)(−6000)
(0.2)
(0.01)V(0.01)2+1
=1200
200,000
V+1
)
P11-4)(d)
rA=k(CACB−
Cc
KC
)=0
rA=k(CA0
2(1−X)2−
CA0X
KC
)
Xe=
2+1
KCCA0
−[(2+1
KCCA0
)2−4]0.5
2
See#Polymath#program#P11-4-d.pol.#
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca0
0.1
0.1
0.1
0.1
2
dHrx
-6000.
-6000.
-6000.
-6000.
3
Fa0
0.2
0.2
0.2
0.2
4
k
0.01
0.01
0.010583
0.010583
5
Kc
22.29685
22.11524
22.29685
22.11524
6
ra
-0.0001
-0.0001047
-0.0001
-0.0001047
7
T
300.
300.
301.0234
301.0234
8
T0
300.
300.
300.
300.
9
V
0
0
10.
10.
10
X
0
0
0.005117
0.005117
11
Xe
0.518003
0.5166571
0.518003
0.5166571
Differential equations
1
d(X)/d(V) = -ra / Fa0
Explicit equations
1
Ca0 = .1
2
Fa0 = .2
3
T0 = 300
4
T = T0 + 200 * X
5
k = .01*exp((10000 / 2) * (1 / 300 - 1 / T))
6
dHrx = -6000
7
Kc = 10*exp((dHrx/8.314)*(1/450-1/T))
8
ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2-Ca0*X/Kc)
9
Xe = 0.5*(2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)
#
11-9#
P11-4)(d))Continued
The#conversion#and#temperature#are#both#lower#than#in#part#(a),#because#the#reaction#rate#is#lower#due#
)
P11-5)(a))
#
P11-5)(a))Continued
)
#
#
P11-5)(b)#Individualized#solution)
)
P11-5)(c))#
V=FA0
dX
−rA
∫
−rA=k C A0(1−X) / (1+X)
V=
FA0
k CA0
1+X
1−X
∫dX
11-11#
P11-5)(c))Continued#
Qr=FA0X(−ΔHR(T0)),%so%X= Qr
FA0(−ΔHR(T0))
V=v0
k(2ln 1
1−Qr
FA0(−ΔHR(T0))
−Qr
FA0(−ΔHR(T0))),where%FA0=P
0v0
RT0
#
Qr#vs#V#can#be#plotted#according#to#the#above#equation.#
#
#
)
0
5
10
15
20
25
30
0 10 20 30 40 50
Qr
V
Qr*vs.*V
11-12#
P11-5)(d))&)(e)#continued#
#
#
#
#
11-13#
P11-5)(d))&)(e)#continued#
)
)
)
)
P11-5)(d))&)(e)#continued#
)
#
#
#
#
#
P11-5)(f))Individualized#Solution#
)
#
P11-6)(a))
A→B+C
CA=CT
FA
F
T
θ
I=FI
FA
#
CT=CA+CI
F
T=FA+FI
#
CA01 =CA0+CI0
( )
FA0
FA0+FI0
CA01 =CA0+CI0
θ
I+1
#
