Sec. R4.

1 Sperical Packed-Bed Reactors 168

R4.1 Sperical Packed-Bed Reactors

Another advantage of spherical reactors is that they are the most economical
shape for high pressures. As a first approximation we will assume that the fluid
moves down through the reactor in plug flow. Consequently, because of the
increase in cross-sectional area, Ac , as the fluid enters the sphere, the superfi-
cial velocity, G  ṁ
Ac , will decrease. From the Ergun equation [Equation
(4-22)],

dP G( 1   ) 150( 1   )
------   ---------------------3- -------------------------------  1.75 G (4-22)
dz gc Dp Dp

we know that by decreasing G, the pressure drop will be 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 exit (Fig-
ures R4-1 and R4-2). Here L is the location of the screen from the center of the
Feed

z0 = 0

R L

z=L

L'

zf = L + L'

Catalyst

+z axis

Screens

Products

Figure R4-1 Schematic drawing of the Figure R4-2 Coordinate system and
inside of a spherical reactor. variables used with a spherical reactor. The
initial and final integration values are shown as
z0 and zf

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 R4-2:
Ac   [ R 2  ( z  L )2 ] (R4-1)

Spherical reactor 1 1
catalyst weight W  c( 1   )V  c( 1   ) R2 z  --- ( z  L )3  --- L3 (R4-2)
3 3
 169 Chap.

By using these formulas and the standard pressure drop algorithm, one can
solve a variety of spherical reactor problems. Note that Equations (R4-1) and
(R4-2) 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, z f  L  L .

Example R4–1 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 mol/s, respectively. In the
reformer, dehydrogenation reactions such as
Paraffin → Olefin  H2

occur. The reaction is first order in paraffin. Assume that pure paraffin enters the
reactor 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 tubular 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 reac-
tor, 173,870 kg.
r A  k C A
r A  B( r A )  C ( 1   )( r A )  C ( 1   )k C A

Additional information:
3
0  0.032 kg/dm
DP  0.02 dm   0.4
k  0.02 dm3/kg cat  s   1.5  106 kg/dm  s
L  L  27 dm c  2.6 kg/dm3

Solution
We begin by performing a mole balance over the cylindrical core of thickness z
shown in Figure RE4-1.1.
Following 1. Mole balance:
the algorithm
In  Out  Generation  0
FA  FA  r A Ac z  0
z z  z

Dividing by z and taking the limit as z ⎯⎯→ 0 yields

dFA
---------  r A Ac
dz
In terms of conversion
 Sec. R4.1 Sperical Packed-Bed Reactors 170

FA

z
∆z
z+∆z

FA

Figure RE4-1.1 Spherical reactor.

dX r A Ac
-------  ----------------- (RE4-1.1)
dz FA0

2. Rate law:

r A  kC A  k C A c ( 1   ) (RE4-1.2)

3. Stoichiometry. Gas, isothermal (T  T0 ):

⎛ 1X ⎞
C A  C A0 ⎜ ----------------⎟ y (RE4-1.3)
⎝ 1  X ⎠

  yA0   1  ( 1  1  1 )  1 (RE4-1.4)

where
P
y  ----- (RE4-1.5)
P0

Note that yA0 ( y with a subscript) represents the mole fraction and y alone represents
the pressure ratio, P / P0 .
The variation in the dimensionless pressure, y, is given by incorporating the
variable y in Equation (4-24):

dy 
-----   -------0- ( 1  X ) (RE4-1.6)
dz P0 y

The units of 0 for this problem are kPa/dm3.

The equations in
boxes are the key
G( 1   ) 150( 1   )
equations used 0  -----------------------3 -------------------------------  1.75G (RE4-1.7)
in the ODE solver 0 gc Dp  Dp
program
171 Chap.

ṁ
G  ----- (RE4-1.8)
Ac

For a spherical reactor

Ac   [ R 2  ( z  L )2 ] (RE4-1.9)

1 1
W  c( 1   ) R2 z  --- ( z  L )3  --- L3 (RE4-1.10)
3 3

Parameter evaluation:
Recall that gc  1 for metric units.

G( 1  0.4 )
0  ---------------------------------------------------------------------------
( 0.032 kg/dm3 )( 0.02 dm )( 0.4 )3
(RE4-1.11)
150( 1  0.4 )( 1.5  106 kg/dm  s )
 ------------------------------------------------------------------------------------  1.75G
0.02 dm

⎛ kPa/dm ⎞
0  [ ( 98.87 s1 )G  ( 25, 630 dm2/kg )G2 ]  ⎜ 0.01 ------------------------
-⎟ (RE4-1.12)
⎝ kg/dm2  s2⎠

The last term in brackets converts ( kg
dm2  s ) to (kPa/dm). Recalling other param-
eters, ṁ  44 kg/s, L  27 dm, R  30 dm, and cat  2.6 kg/dm3.
Table RE4-1.1 shows the POLYMATH input used to solve the above equa-
tions. The MATLAB program is given as a living example problem on the
CD-ROM.
TABLE R4-1.1 POLYMATH PROGRAM

Equations Initial Values

d(X)/d(z)=–ra*Ac/Fao 0
d(y)/d(z)=–beta/Po/y*(1+X) 1
Fao=440
Po=2000
CaO=0.32
R=30
phi=0.4
kprime-0..02
L=27
rhocat=2.6
m=44
Ca=CaO*(1–X)*y/(1+X)
Ac=3.1416*(R^2–(z–L)^2)
V=3.1416*(z*R^2–1/3*(z–L)^3–1/3*L^3)
S=m/Ac
ra=–kprime*Ca*rhocat*(1–phi)
 Sec. R4.1 Sperical Packed-Bed Reactors 172

TABLE R4-1.1 POLYMATH PROGRAM (CONTINUED)

beta=(98.87*G+25630*G^2)*0.01
W=rhocat*(1–phi)*V
z0 = 0, zf = 54

Figure RE4-1.2 Pressure and conversion for: 1, tubular PBR; 2, spherical PBR.

For the spherical reactor, the conversion and the pressure at the exit are
A comparison
between reactors X  0.81 P  1980 kPa
If similar calculations are performed for the tubular PBR, one finds that for the same
catalyst weight the conversion and pressure at the exit are
X  0.71 P  308 kPa
Figure RE4-1.2 shows how conversion, X, and dimensionless pressure, y, vary with
catalyst weight in each reactor. Here X1 and y1 represent the tubular reactor and X2
and y2 represent the spherical reactor. In addition to the higher conversion, the
spherical reactor has the economic benefit of reducing the pumping and compres-
sion cost because of higher pressure at the exit.
Because the pressure drop in the spherical reactor is very small, one could
increase the reactant flow rate significantly and still maintain adequate pressure at
the exit. In fact, Amoco uses a reactor with similar specifications to process 60,000
barrels of petroleum naphtha per day.