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6-76
CDP6-C
6-78
CDP6-D
6-80
CDP6-E
6-82
6-83
CDP6-F (a)
Mole balances:
dFA/dV = -rD-rU
V = 1425 ft3
POLYMATH Results
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
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(fu)/d(V) = ru
[2] d(fa)/d(V) = -ru-rd
[3] d(fd)/d(V) = rd
Explicit equations as entered by the user
[1] k1 = 15
CDP6-F (b)
Mole balances:
FA = FAo + rAV FD = rDV FU = rUV
6-84
POLYMATH Results
NLES Solution
Variable Value f(x) Ini Guess
ca 0.0023213 5.354E-08 0.00447
cd 0.0016225 -5.354E-08 0
cu 5.262E-04 -2.76E-15 0
cao 0.00447
E1 10000
ra -8.058E-05
NLES Report (safenewt)
Nonlinear equations
[1] f(ca) = ca-cao-ra*tau = 0
[2] f(cd) = cd-rd*tau = 0
[3] f(cu) = cu-ru*tau = 0
Explicit equations
[1] cao = 0.00447
[2] E1 = 10000
CDP6-G 3
6-86
CDP6-H
CDP6-I 2, 9-17
CDP6-J
CDP6-L
6-88
Solving we get: CF = 0.1gmol/l
Balance for B:
VCkCkCkVrCv FABBBo )()(0 2
423 +!=!=!
min60min07.01
min60)/1.0min.5.0/143.0min01.0(
1
)(
1
2111
3
2
42
!
!!!
+
+"
=
+
+
=lgmolmolllgmol
k
CkCk
CFA
B
#
#
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!=
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
Explicit equations
[1] vo = 10
[2] fao = 10
[3] W = 3
[4] fbo = 10
[5] rho = 0.0012
[6] k1 = 5
[7] k2 = 2
[18] r3 = rho*k3*cb*(cc^2)
[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
!
"
= = =
! !
CDP6-N 3, 6-21
CDP6-O 3, 6-25
