6-76
The concentration of B is highest at t = 100 sec, where its value is 0.1363 gmol/dm3
6-77
CDP6-C
6-78
6-79
CDP6-D
6-80
CDP6-E
6-81
6-82
6-83
CDP6-F (a)
Mole balances:
dFA/dV = -rD-rU
dFD/dV = rD
2
1D A
r k C=
k1 = 15 ft3/lbmol.s
2U A
r k C=
k2 = 0.015 s-1
Stoichiometry: CA = CTo(FA/FT), FT = FA + FD + FU
Cost = 60FB – 15FC -10FAO
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 2000 2000
fu 0 0 0.0237312 0.0237312
fd 0 0 0.0414537 0.0414537
fao 0.06705 0.06705 0.06705 0.06705
cao 0.00447 0.00447 0.00447 0.00447
Cost -0.6705 -0.6705 1.46684 1.4607519
rd 2.997E-04 2.319E-07 2.997E-04 2.319E-07
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(fu)/d(V) = ru
[1] k1 = 15
[2] k2 = 0.015
[3] fao = 0.06705
[4] ft = fu+fa+fd
[8] rd = k1*(ca^2)
variable name : V
initial value : 0
final value : 2000
CDP6-F (b)
Mole balances:
FA = FAo + rAV FD = rDV FU = rUV
6-84
1 1
exp 919.67
i
i io
E
k k
R T
! “! “
=#
$ %
$ %& ‘
POLYMATH Results
NLES Solution
Variable Value f(x) Ini Guess
ca 0.0023213 5.354E-08 0.00447
cu 5.262E-04 –2.76E-15 0
E1 10000
R 1.987
E2 2.0E+04
k1o 15
cost 0.6713603
k2 0.0085009
ru 1.973E-05
NLES Report (safenewt)
Nonlinear equations
[1] f(ca) = ca-cao-ra*tau = 0
[2] f(cd) = cd-rd*tau = 0
[1] cao = 0.00447
[2] E1 = 10000
[3] R = 1.987
[6] E2 = 20000
[7] k1o = 15
[11] cost = vo*(60*cd-15*cu-10*cao)
[15] ru = k2*ca
6-85
CDP6-G 3
6-86
6-87
CDP6-H
CDP6-I 2, 9-17
CDP6-J
CDP6-K 2, 9-13
CDP6-L
6-88
Solving we get: CF = 0.1gmol/l
Balance for B:
VCkCkCkVrCv FABBBo )()(0 2
423 +!=!=!
Balance for C:
VCkCkVrCv ABcco )()(0 13 !!=!=!
min60)/143.0min01.0/0907.0min07.0()( 11
13 lgmollgmolCkCkC ABC !+!=+= ““
#
lgmolCC/467.0=
Mole fraction of C =
583.0
8.0
467.0
1.0467.009.0143.0
467.0 ==
+++
=
+++ FCBA
C
CCCC
C
6-89
CDP6-M
4A + 5B 4C + 6D
2
1 1A A A B
r k C C!=
2A +1.5B E + 3D
2 2A A A B
r k C C!=
Rate laws:
-rA = r1A + r2A + (2/3)r4C –rB = 1.25r1A + 0.75r2A +r3B
-rC = -r1A + 2r3B + r4C –rD = -1.5r1A – 1.5r2A – r4C
Using these equations inpolymath to find the exiting molar flow rates.
POLYMATH Results
NLES Report (safenewt)
Nonlinear equations
[1] f(ca) = vo*ca-fao-ra*W = 0
[2] f(cb) = vo*cb-fbo-rb*W = 0
[3] f(cc) = vo*cc-rc*W = 0
[4] f(cd) = vo*cd-rd*W = 0
Explicit equations
[1] vo = 10
[2] fao = 10
[5] rho = 0.0012
[6] k1 = 5
[7] k2 = 2
[8] k3 = 10
[19] r4 = rho*k4*cc*(ca^(2/3))
[20] rf = 2*r3
[21] re = 0.5*r2+(5/6)*r4
[22] ra = -r1+r2-(2/3)*r4
6-90
(b)
3
3.6 10 0.336
10 9.989
E
AE
Ao A
F
Y
F F
!
“
= = =
! !
7
6
2.29 10 8.25 10
10 9.973
F
BF
F
Y
F F
!
!
“
= = = “
! !
CDP6-N 3, 6-21
CDP6-O 3, 6-25