Post on 06-Apr-2018
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(1.44)
The energy balance in a tubular reactor is given in eq. (1.44)
by:
For an adiabatic reactor, . Usually, there is a negligible
amount of work done on or by the reacting mixture.
(1.46)
0Q
0XFdTCTHdTCFWQ0A
T
TpR
0
Rx
n
1i
T
Tpii0AS
R0i
n1i
T
Tpii
T
TpR
0
Rx
0iR
dTCXdTCTH
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The circumflex denotes that the heat capacities are evaluated
at some mean temperature value between TR and T.
R
T
Tp
pTT
dTC
C R (1.48)
For the case of constant or mean heat capacities:
Rp
T
Tp
TTCdTCR
(1.47)
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In a similar fashion we can write the integral involving i andCpi in eq. (1.46) as
0ipiiTT
piiTTCdTC
0i
The circumflex denotes that the heat capacities are evaluated
at some mean temperature value between TR and T.
0i
T
Tpi
piTT
dTC
C 0i
n
1i0ipiiRpR
0
RxTTCXTTCTH
(1.49)
For constant or mean heat capacity, eq. (1.46) becomes:
(1.50)
(1.51)
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Eq. (1.51) can be solved for T
n
1i 0ipiiRpR
0
Rx TTC
XTTC
TH
n
1i0ipii
n
1ipiiRppR
0
RxTCTCXTCTXCXTH
Rp
n
1i0ipiiR
0
Rxp
n
1ipii
TCXTCTHXCXCT
p
n
1ipii
Rp
n
1i0ipiiR
0
Rx
CXC
TCXTCTHX
T
(1.52)
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This equation can be combined with the differential mole
balance derived from eq. (1.1):
(1.1) T,XrF
dX
dV
A
0Ato obtain the temperature, conversion, and concentration
profiles along the length of the reactor.
One way of accomplishing this combination is to use eq.
(1.52) to construct a table of T as a function of X.
Once we have T as a function of X, we can obtain k(T) as a
function of X and hence rA as a function of X alone.
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The elementary reversible gas-phase reaction is carried out in
a PFR in which pressure drop is neglected and pure A entersthe reactor:
A BMole balance:
Rate law:
with
and
dXr
FdV
A
0A
C
B
AAK
CCkr
T1T1REexpkk 11
T
1
T
1
R
HexpTKK
1
0
Rx
1CC
(a)
(b)
(c)
(d)
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Stoichiometry: Gas, = 0, P = P0
TT
X1CC 00AA
CB
AAKCCkr
T
TXCC 0
0AB
(1.46) n
1i
T
Tpii
T
TpR
0Rx
0iR
dTCXdTCTH
(e)
(f)
(g)
Eqs. (a) through (g) and (1.52) can easily be solved using
either Simpson's rule or an ODE solver.
Combine:
Energy balance (eq. 1.46):
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Normal butane, C2H4, is to be isomerized to isobutane in a plug-
flow reactor. The reaction is to be carried out adiabatically in theliquid phase under high pressure using essentially trace amounts of
a liquid catalyst which gives a specific reaction rate of 3 1.1 h1 at
360 K. Calculate the PFR volume necessary to process 100,000
gal/day (163 kg mol/h) of a mixture 90 mol % n-butane and 10 mol
% i-pentane, which is considered an inert. The feed enters at 330 K.
Additional information:
HRx = -6900 J/mol* butaneCpn-B = 141 J/mol.K
Cpi-B= 141 J/mol.KCpi-P = 161 J/mol.K
Activation energy = 65.7 kJ/mol
KC = 3.03 at 60C
CA0
= 9.3 g mol/dm3 = 9.3 kg mol/m3
EXAMPLE 1.2
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SOLUTION
Reaction:
Mole balance:
Rate law:
with
dXr
FdV
A
0A
C
B
AA
K
CCkr
T
1
T
1
R
Eexpkk
1
1
(a)
(b)
(c)
T1
360
1
31.8
700,65exp1.31k
T
1
360
17906exp1.31k
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T
1
T
1
R
HexpTKK
1
0
Rx
1CC
(d)
T1
333
1
31.8
6900exp03.3K
C
T133313.830exp03.3KCStoichiometry (liquid phase, v = v0):
X1CC0AA
XCC 0AB
(e)
(f)
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C
0A0AA
K
XCX1Ckr
Combine:
XK
111kCr
C
0AA
(g)
(h)
(1.46)
n1i
T
Tpii0A0A
T
TpR
0
Rx
0iR
dTCFXFdTCTH
Energy balance (eq. 1.46):
1
F
F
0A
0A
A
with:
0
F
F
0A
0B
B
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0K.molJ141141CCC pApBp 1111.0
9.0
1.0
F
F
0A
0I
I
(i)
Eq. (46) becomes:
n1i
0ipiiR
0
RxTTCXTH
n1i
pii
R
0
Rx
0
C
XTHTT
with: 161111.0141CCCCpIIpBBpAA
n
1ipii
= 158.9 J/mol . K
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Eq. (i) becomes:
X43.43330T (j)
molJ900,6THR
0
Rx
At equilibrium rA = 0, equation (b) becomes:
C
e
0AAK
X1kC0r
C
C
eK1
KX (k)We will now integrate Equation (a) with Simpson's rule, but
first we have to calculate (FA0/-rA) as a function of X.
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For example for X = 0.1
K34.3341.043.43330T 1
h77.534.334
1
360
17906exp1.31k
00.334.334
1
333
1
3.830exp03.3KC
75.0
0.31
0.3
K1
KX
C
C
e
h.m
kmol83.511.0
0.3
111
m
kmol3.9
h
77.5r
33A
33A0A m830.2h.mkmol83.51
htotalkmol163totalmolbutanemol9.0
r
F
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X T k KC Xe - rA FA0/(-rA)
0 330.00 4.22 3.10 0.76 39.28 3.735
0.1 334.34 5.77 3.00 0.75 51.83 2.830
0.2 338.69 7.81 2.91 0.74 67.62 2.1700.3 343.03 10.49 2.82 0.74 87.19 1.682
0.4 347.37 14.00 2.73 0.73 111.12 1.320
0.5 351.72 18.54 2.65 0.73 139.93 1.048
0.6 356.06 24.39 2.58 0.72 174.02 0.843
0.7 360.40 31.87 2.51 0.71 213.62 0.687
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0
5
10
15
20
25
30
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
FA0
/(-rA
)
X
1 2 3 4 5 76
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2
yy
2
yy
2
yy
2
yy1.0Area 43322110
2
yy
2
yy
2
yy
2
yy43766554
2y
yyyyyy2
y1.0Area 7
654321
0
3m13.3
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When the problem is to find the temperature profile along
the reactor, then the algorithm is as follows.
Mole balance:
Rate law:
Energy balance:
0A
A
F
r
dV
dX AA
Ckr (1.1)
For no work done on the system, , and adiabatic
operation, , the energy balance is written as eq. (1.46)
0WS
0Q
0XdTCTHdTC TT
pR
0
Rx
T
Tpii
R0i
(1.46)
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Differentiating eqs. (1.46) with respect to V yields
dVdX
THdV
dT
C R0
Rxpii
0dV
dXdTC
dV
dTCX
T
Tpp
R
dV
dXdTCTH
dV
dTCXC
T
TpR
0
Rxppii
R
dV
dX
CXC
dTCTH
dV
dT
ppii
T
TpR
0
Rx
R
(1.53)
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Combining eqs. (1.1) and (1.53) yields
ppii0A
T
TpR
0
RxA
CXCF
dTCTHr
dV
dTR
(1.54)
Here we have two equations, (1.1) and (1.54), that must be
solved simultaneously.
One method that can be used to solve the problem is Runge-Kutta method.
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xz,xy,xfdx
dy
xz,xy,xgdx
dz
00yxy
00zxz
nnn1 z,y,xf.hk nnn1 z,y,xg.hl 2
lz,
2
ky,
2
hxf.hk 1
n
1
nn2 2l
z,2
ky,
2
hxg.hl 1
n
1
nn2
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3n3nn4lz,ky,hxf.hk
3nnn4lz,ky,hxg.hl
4321 kk2k2k61k 4321 ll2l2l61l hxx
n1nkyy
n1nlzz
n1n
2l
z,2
ky,
2
hxf.hk 2
n
2
nn3 2l
z,2
ky,
2
hxg.hl 2
n
2
nn3
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One of the key steps in producing acetic anhydride is the
vapor-phase cracking of acetone to ketene and methane:
CH3COCH3 CH2CO + CH4This reaction is first-order with respect to acetone and the
specific reaction rate can be expressed by:
where k is in reciprocal seconds and T is in kelvin.
In this design it is desired to feed 8000 kg of acetone per hour
to a tubular reactor. The reactor consists of a bank of 1000 1-inch schedule 40 tubes. The reactor is operated adiabatically.
The inlet temperature and pressure are at 1035 K and 162 kPa
(1.6 atm), respectively. Plot the conversion and temperature
along the length of the reactor.
EXAMPLE 1.3
T3422234.34kln
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SOLUTION
Let A = CH3COCH3, B = CH2C0, and C = CH4.
Rewriting the reaction symbolically gives us:
A B + CMole balance:
Rate law:
0A
A
F
r
dV
dX AA
Ckr (a)
(b)
A B C Total
Initial NA0 0 0 NA0
Reaction NA0 X
Final NA0 (1 X) NA0 X NA0 X NA0 (1 + X)
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0
0A
0AV
NC
V
X1N
V
N
C
0AA
A
If the gas mixture is assumed to behave as ideal gas:
RTN
PV
RTN
PV
t00t
0
0
0
0
00A
0A
0
00t
t VT
TX1V
T
T
N
X1NV
T
T
N
NV
T
T
X1
X1C
T
T
X1V
X1NC 00A0
0
0A
A
(c)
(d)
(e)
Combining (d) and (e) gives us:
(f)
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Combining eqs. (b) and (f) yields
T
T
X1
X1kC
r00A
A (g)
while combining eqs. (a) and (g) yields
TTX1 X1vkTTX1F X1kCdVdX 00
0
0A
0A (h)For no work done on the system, , and adiabatic
operation, , the energy balance leads to eq. (1.54)
0WS
0Q
ppii0A
T
TpR
0
RxA
CXCF
dTCTHr
dV
dTR
(1.54)
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Calculation of mole balance parameters:
smol3.38hkmol9.137kmolkg58
hkg8000F0A
3330A
0Ammol8.18
mkmol0188.0
K1035K.kmol
m.kPa31.8
kPa162RTPC
sm037.2C
Fv
3
0A
0A
0
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Calculation of energy balance parameters:
a. The standard heat of reaction :R0Rx TH molkJ67.216TH
acetoneR
0
Rx
molkJ09.61THketeneR
0
Rx
molkJ81.74THmethaneR
0
Rx
67.21681.7409.61TH R0Rx molJ80770molkJ77.80
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b. Cp :K.molJT1086.45T183.063.26C
26
pA
K.molJT1095.30T0945.004.20C
26
pB
K.molJT1071.18T077.039.13C
26
pC
8.663.2604.2039.13
ABC
0115.0183.00945.0077.0ABC
ABC
6108.3 2
pTTC
26
T108.3T0115.08.6
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c. i Cpi :
A
= 1
B = C = 0i Cpi = A CpA + B CpB + A CpB = CpA
The two equations are solved simultaneously using 2-
dimensional Runge-Kutta method.
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0.00
0.05
0.10
0.15
0.20
0.25
0 1 2 3 4
X
V (m3)
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900
925
950
975
1000
1025
1050
0 1 2 3 4
T
(K)
V (m
3
)