The algorithm
We can evaluate the combine step either 1) Analytically 2) Graphically 3) Numerically, or 4) Using software
Eq, (4-33) is valid or e x 4 1 oalyif E = 0
162 isothermal Reactor Design
Solution
1. biffereatial mole balance:
Sec. 4.4 Pressure Drop in Reactors 163
Evaluating the pmsure drop parameters
Separating variables to form the integrals yields
Integrating gives us
Solving for W, we obtain
1 - [ I - (3aFAO/2kr)((l $-E) In[ 1/(1 - x ) ] - Ex)]u3
W = a (E4-6.11)
6. Parameter evaluation per tube (i.e., divide feed rates by 1000):
Ethylene:
Oxygen:
I = inerts = N,:
FAo = 3 X
FBo = 1.5 X
lb mob's = 1.08 lb movh lb moYs = 0.54 lb mom
0.79 mol N, F, = 1.5 X 10-4 lb molVs X 0.21 mol 0, Fi = 5.64 X lop4 Ib moUs = 2.03 ib m o h Summing: F , = F A o
+
FBo+
Fx = 3.65 ib moyhE = yAoG = (0.3)(1 -
3
- 1) = -0.15PA0 = yaoPo = 3.0 atm
lb mol
X 3 atm X 0.63 = 0.0266
-
h Ib cat k' = kPA0(i)2/3 = 0.0141 lb
atm.lb catch
1 - [ 1 -(3aFAo/2k'){(l -0.15) ln[l/(l -0.6)]-(-0.!5)(0.6)}12/3
w = -
a
For 60% conversion, Equation (E4-6.11) becomes 1 - (1 - 1.303aFA0/k')2'3 W =
ci (E4-6.12)
In order to calculate a,
2PO
= A,(1 - +)P,PO
we need the superficial mass velocity, G . The mass flow rates of each entering species are:
Neglecting pressure drop results in poor design (here 53%
vs. 60%
164 Isothermal Reactor Design Chap. 4
lb mol lb
hAo = 1.08
-
X 28-
= 30.24 Ibh-
- 17.28 Ibh mIo = 2.03
-
X 28-
= 56.84 Ibhh lb mol
lb mol h
Ib mol lb
h lb mol
b m o l x 32
-
lbh B o = 0.54
-
The total mass flow rate is
lb hTo = 104.4 - h
This is essentially the same superficial mass velocity, temperature, and pres- sure as in Example 4-5. Consequently, we can use the value of Po calculated in Example 4-5.
Po = 0.0775 atm - ft
280 - - (2)(0.0775) a t d f t a =
A c ( l - +)pcPo (0.01414 ft2)(0.55)(120 Ib cat/ft3)(10 atmj
- 0.0166 Ib cat
_ -
Substituting into Equation (E4-6.12) yields
1
lb mol l - l l - ' o.0266 G h W = L
0.0166Ab cat
= 45.4 lb of catalyst per tube or 45,400 lb of catalyst total
This catalyst weight corresponds to a pressure drop of approximately 5 atm.
If we had neglected pressure drop, the result would have been
1
w=--;
(1+&)h---EX 1FAo[ k 1 - x
(1-0.15) In --(-0.15)(0.6) 1 - 0.6 0.0266
L
= 35.3 Ib of catalyst per tube (neglecting pressure drop) ,
and we would have had insufficient catalyst to achieve the desired conpersion.
Substituting this catalyst weight (Le., 35,300 Ib total) into Equation (E4-6.10) gives a conversion of only 53%.
i ,~.-
See. 4.4 Pressure Drop in Reactors "3
Example 4- 7 Pressure Drop with Reaction-Numerical Solutiort
Rework Example 4-6 for the case where volume change is nor rlegiccted i n t h . Ergun equation and the two coupled differential equations
of conversion and preasure u itii ~ d t i ! ) \t sscight are solved
1
I SolutionRather than rederive everything starting horn the etry, and pressure drop equations, we will use th 4-6 Combining Equationc (E4 6.1) and (E4-6.8) g I
Program examples POLYMATH, MatLab can be loaded from the CD-ROM (see the Introduction)
Next. we let
E3- / .3)
(E1-7.4 j
For the reaction conditions descnbed in Example 4-6, we habe the bomindarJ condi- tioris W = 0, X = 0, and y = 1.0 and the parameter values a = C: s " ! ~ b / l b cat,
E = -0.15, k' = 0.0266 Ib mol/h.Ib cat, and FAo = 1.08 Ib mol/h.
'4 large number of ordinary differential equation solver software pacKages (i.e., ODE solvers) wfiich are extremely user friendly have become available. We shall use POLYMATH4 to solve the examples in the pnnted text. However, the CD-ROM contains an example that uses ASPEN, as v~ell as all the MATLAB and PlOLY- MATH solution programs to the example prcprams. With POLYMATH one simply enters Equations (E4-7.3) and (E4-7.4) and the corresponding parameter value into the computer (Table E4-7.1 1 with the l ~ l l l d ~ (rather, boundary) conditions and they are solved and displayed as shown in Figure E4-7.1,
We note that neglecting EX in the Ergun equation in Example 4-6 ( E X = -0.09) to obtain an analyticai solution resulted in less than a 10% error.
Developed by Professor M. Cutlip of the University of Connecticut. and Professor M.
Shacham of Ben Gurion University. Available from the CACHE Corporation, P.O.
Box 7939, Austin, TX 78713.
166 1,sothermal Reactor Design Chap. 4
TABLE EX-7.1. POLYMATH SCREEN SHOWING EQUATIONS TYPED IN AND READY TO BE SOLVED.
Equations Initial Values
d(y)/d(w)=-alpha*(l+eps*x) /2/Y d[x)/d(w) =rate/faO
faO=1.00 alpha=0.0166 eps=-O. 15 kprime=O. 0266 f=(l+eps*x) / y
rate=kprime*( (l-x)/(l+ePS*X))*Y
f = 60 w0 = a , w
1 0
3‘000
T
Scale:
0.000 12.00a 24. ooo 36.000 48.000 60, ooc
U
Figure E4-7.1 Reaction rate profile down the PBR.
However, larger errors will result if large values of EX are neglected! By taking into account the change in the volumetric flow rate (Le., E = -0.15) in the pressure drop term, we see that 44.0 lb of catalyst is required per tube as opposed to 45.4 lb when E was neglected in the analytical solution, Equation (E4-7.4). Why was less catalyst required when E was not neglected in Equation (E4-7.4)? The reasonis that the numerical solution accounts for the fact that the pressure drop will be less because E is negative.
Sec. 4.4 Pressure Drop in Reactors 167
Volumetric flow rate increases
with increasing pressure drop
Effect of added
It is also interesting to learn what happens to the volumetric flow rate along the length of the reactor. Recalling Equation (3-44),
Po T - u , ( l + E X ) ( T / T , ) p To P / P o
u = U0(1+EX)
-
- --
(3-44)We let f be the ratio of the volumetric flow rate, u, to the entering volumetric flow rate, u o , at any point down the reactor. For isothermal operation Equation (3-44) becomes
(E4-7.5)
Rgure E4-7.2 shows X, y (i.e., y = P/Po), and f down the length of the reactor. We see that both the conversion and the volumetric flow increase along the length of the reactor while the pressure decreases. For gas-phase reactions with orders greater than zero, this decrease in pressure will cause the reaction rate to be less than in the case of no pressure drop.
4.000
i
3.200
2.400
1.600
0.800
0.000
W
Figure E4-7.2 Output in graphical form from POLYMATH
We note from Figure W-7.2 that the catalyst weight necessary to raise the conversion the last 1% from 65% to 66% (3.5 lb) is 8.5 times more than that (0.41 lb) required to raise the conversion 1% at the reactor's entrance.
Also, during the last 5% increase in conversion, the pressure decreases from 3.8 atm to 2.3 atm.
conversion on
168 Isothermal Reactor Design Chap. 4
4.4.3 Spherical Packed-Bed Reactors
When small catalyst pellets are required, the pressure drop can be signif- icant. In Exa~iiple 4-6 we saw that significant design flaws can result if pressure drop is neglecced or if steps are not taken to minimize pressure drop. One type of reactor that minimizes pressure drop and is also inexpensive to build is the spherical reactor, shown in Figure 4-8. In this reactor, called an ultraformer, dehydrogenation reactions such as
paraffin
--+
aromatic t 3H2 are carried out.Figure 4-8 Sphencal Ultraformer Reactor. (Courtesy of Amoco Petroleum Products.) This reactor I S one in a series of SIX used by Amoco for reforming petroleum naphtha. Photo by K. R Renicker, Sr.
Another advantage of spherical reactors js that they are the most eco- nomical shape for high pressures. As a first approximation we will assume that the fluid moves down through the reactor in plug Bow. Consequently, because
Sec. 4.4 Pressure Drop in Reactors 169
Spherical reactor catalyst weight
of the increase in cross-sectional area, A,, as the fluid enters the sphere, the superficial velocity, G = riz/A,, will decrease. From the Ergun equation [Equation (4-%2)],
(4-22) we h o w that by decreasing G, the pressure drop w i l l he reduced significantly, resulting in higher conversions.
Because the cross-sectional area of the reactor is small near the inlet and outlet, the presence of catalyst there would cause substantial pressure drop;
thereby reducing the efficiency of the spherical reactor. To solve this problem, screens to hold the catalyst are placed near the reactor entrance and exii. (Fig- ures 4-9 and 4-10). Were L is the location of the screen from the center of the
Feed I I
talyst
'i'
+ L'
Products +z axis
Figure 4-9 Schematic drawing of the inside Figure 4-10 Coordinate system and of a sphenca! reactor. variables used with a spherical reactor. The
initial and final integration values are slhown as zo and z,.
reactor. We can use elementary geometry and integral calculus to derive the following expressions for cross-sectional area and catalyst weight as a function of the variables defined in Figure 4-10:
A , = T [ R * - ( z - L ) * ] (4-38)
By using these formulas and the standard pressure drop algorithm, one can solve a variety of spherical reactor prablems. Note that Equations (4-38) and
170 Isothermal Reactor Design Chap. 4
(4-39) make use of L and not L'. Thus, one does not need to adjust these for- mulas to treat spherical reactors that have different amounts of empty space at the entrance and exit (i.e., L # L ' ) . Only the upper limit of integration needs to be changed, zf = L
+
L'.
Example 4-8 Dehydrogenation Reactions in a Spherical Reactor
Reforming reactors are used to increase the octane number of petroleum. In a reforming process 20,000 barrels of petroleum are to be processed per day. The cor- responding mass and molar feed rates are 44 kg/s and 440 molls, respectively. In the reformer, dehydrogenation reactions such as
paraffin
+
olefin -k H,occur. The reaction is first-order in paraffin. Assume that pure paraffin enters the reac- tor at a pressure of 2000 kPa and a corresponding concentration of 0.32 mol/dm3.
Compare the pressure'drop and conversion when this reaction is carried out in a tubu- lar packed bed 2.4 m in diameter and 25 m in length with that of a spherical packed bed 6 m in diameter. The catalyst weight is the same in each reactor, 173,870 kg.
- T i = k' CA
- T A = pB(-Ti) = pc(l-$)(-?$.) = p c ( l - $ ) k ' C ~ Additional information:
po = 0.032 kg/dm3 D , = 0.02 dm L = L' = 27 dm
= 0.4 kg/dm.s kt = 0.02 dm3kg cat
.
I.'. = 1.5 xpc = 2.6 kg/dm3 Solution
We begin by performing a mole balance over the cylindrical core of thickness Az shown in Figure E4-8.1.
Figure E4-8.1 Spherical reactor.