5_-_Flow_Through_Packed_Beds (1)

Flow Through Packed Beds of Solids

Instructor: Dr. Mohammad Al-Harahsheh
Department of Chemical Engineering
Jordan University of Science and Technology

Introduction

There are numerous physical processes of interest to chemical engineers that involve the flow of fluids
through porous solids or beds of particles. Some examples include:

 Oil and Gas Production - oil and gas deposits usually occur in porous structures such as

sands, sandstones, limestones, and dolomites. To be recovered, the oil or gas must flow through
these porous structures to a well hole. In the later stages of reservoir production, the oil or gas
must often be displaced from the porous solid by water or miscible solvents.

 Hydrology - underground water sources usually occur in stratified layers of sand called

aquifers. The recovery of water for drinking or irrigation, the movement of trace pollutants into
aquifers, salt water encroachment into the fresh water aquifers, and the dissolution of underground
structures such as salt domes are some of the problems in hydrology dealing with flow through
porous materials.

 Biophysics - the flow of blood and gases in the lung, and the flow of blood in the kidney are

examples of life processes involving flow in porous structures.

 Filtration - solids can be separated from a suspension by a porous medium that retains the

solids but allows the liquid to pass. Examples include water treatment by sand beds and the
recovery of solid products by various types of filters.

Introduction

 Reactor Engineering - a common type of chemical reactor consists of a fixed bed of solid

catalyst particles through which the reactants and the products flow. In some cases the bed of
particles moves slowly countercurrent to the reacting stream. Reactors of the latter type are
called moving bed reactors.

 Packed Columns - in operations involving the transfer of components from one phase to a

second phase, it is advantageous to have a large interfacial area for the transfer process. This
is often achieved by contacting the phases in a column filled with solid particles. An example
is a packed column for the absorption of gases that flow upward through the packing (solid
particles), into a liquid that flows downward over the packing. Other examples include
packed columns for distillation, drying of gases by silica gel or molecular sieves,
chromatography, and ion exchange.

 Fluidized Beds - if a gas flow upward through a bed of particles with sufficiently high

velocity, the drag force on the particles balances the gravitational force on the particles and
the bed expands into a well-mixed, fluid-like phase. Industrial applications of fluidized beds
include catalytic cracking of heavy crude oil fractions, ore roasting, extraction of oil from
shale and bituminous sand, and combustion.

Packed Towers for Absorption and Distillation

 Used for continuous countercurrent

contacting of gas and liquid in absorption
as well as vapor-liquid contacting in
distillation.

Consist of: 1. Cylindrical column
Operation: 2. Distributing space at the bottom
3. Distributing device at the top
4. Gas outlet at the top
5. Liquid outlet at the bottom
6. Packing material

Gas enters the distributing space at the bottom below the
packed section and rises upward to contact the
descending liquid
Why packing? To provide large area of intimate contact between liquid and gas

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 3-4

Clean gas out Packed Towers

Mist Eliminator
Liquid Sprays

Packing

Dirty gas in

Liquid outlet ChE 362 Unit Operations Chapter 3-5

JUST Department of Chemical Engineering

Industrial Applications

• Fixed bed catalytic reactors
• Absorption
• Distillation
• Adsorption
• Filtration

Packed Bed Scrubber Flash video

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 3-6

Darcy’s Law and Permeability

Experimental work (1830)

Vo  1, Vo  P Vo  Q
L A

Darcy’s law:

Vo k  P  (1)

L

k: a constant, depends on the physical
properties of bed and fluid.

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 3-7

Linear relation between the velocity and the pressure drop → laminar flow

→ the resistance to flow arises mainly from viscous drag:

(1) → V o  k P P (2)
=B

L L

: the viscosity of the fluid

B: the permeability coefficient for the bed  indication of the ease of passing
a fluid through a bed of particles.

If: Vo cm/s (2) B  cm2.cp  Darcy
 cp s.atm
atm
P cm

L

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 3-8

The Channel Model

 The most common methods of calculating the pressure drop through a
packed bed are based on estimates of the total drag on the boundaries of the
tortuous channels through the bed.

 The actual channels are irregular in shape, have a variable cross section and
orientation, and are highly interconnected.

 To calculate an equivalent channel diameter (Deq), it assumed that the bed
has a set of uniform circular channels whose total surface area and void
volume match those of the bed.

V

Deq

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 3-9

 Tortuosity:

Tortuosity is a measure of how
much a pathway deviates from a
straight line.

The path that fluid takes through a granular
material is governed by how individual pore
spaces are connected.

The greater the tortuosity the lower the
permeability because viscous
resistance is cumulative along the
length of the pathway.

Geometric Relations

1. Sphericity, s:

s  Dp2 / 1 D3p  6 / Dp (3)
sp 6 sp / vp

/ vp

2. Specific surface of a particle, av:

av  sp = (surface of a particle) ? (total surface of particles in bed)
vp (volume of a particle) (volume of particles)
=

(3) av  6 (4)
sDp
(5)
For a spherical particle (s=1):
Chapter 3-11
av  6
Dp

JUST Department of Chemical Engineering ChE 362 Unit Operations

3. Voidage, 
  volume of voids = cross sectional area of voids
total volume of particles and voids cross sectional area of bed

4. Total surface area in bed/total volume in bed, a:

a = total surface area of particles
total volume of bed

=  total surface area of particles   volume of particles 
 volume of particles   total volume of bed 

a  av 1  (6)
(7)
(4) into (6): a  6 1 

s Dp

Dp  sDp

eff

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 3-12

5. Hydraulic radius, rH
Flow through noncircular conduits:

Empirically, it is found that f vs. Re correlations for circular conduits

matches the data for noncircular conduits if D is replaced with equivalent
hydraulic diameter DH.

rH  cross-sectional area available for flow  DH
wetted perimeter 4

For the special case of circular duct:  DH  D

rH  cross-sectional area available for flow   D2 / 4  D
wetted perimeter D 4

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 3-13

6. Equivalent channel diameter, Deq

rH  cross-sectional area available for flow  voide volume available for flow
wetted perimeter total wetted surface of solids

 volume of voids/volume of bed   (8)

wetted surface/volume of bed a (7) a  6 1 

 rH  s Dp (9) s Dp

61  Deq

For the special case of circular channel: (9)

rH  cross-sectional area available for flow   De2q / 4  Deq  Deq  4rH
wetted perimeter  Deq 4

 Deq  2 s Dp  (10)
3 
1 

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 1-14

7. Average velocity in the channel,V

V Q V
cross sectional area of voids
Deq
V : average interstitial (local) velocity
Chapter 1-15
Vo  cross sectional Q of entire bed
area

Vo : superficial or empty-tower velocity

Vo V  cross sectional area of voids  V 
 cross sectional area of entire bed 

 V  Vo (11)



JUST Department of Chemical Engineering ChE 362 Unit Operations

Laminar Flow

From “Fluid Mechanics” and for laminar flow in circular and straight pipes:

P  32  V ?

L D2

?

What are the proper velocity and diameter to account for the fact that
channels are tortuous and not straight and parallel ?

V

Deq

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 1-17

Laminar Flow V  Vo (11)

P  32  V 

L D2

Deq  2 s D p  (10)
3 
1 

 P  72Vo1 2 A correction factor 1 is added to account for
L the tortuous path through the bed
s2D2p3

 12.1experiment 2
2 P 150Vo 1 
P 721Vo 1  L  s2D2p3 (12)
 L 
s2D2p3

Kozeny-Carman Equation ReP  Vo DP 1

1   

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 1-18

Turbulent Flow

From “Fluid Mechanics” and for turbulent flow in circular and straight pipes:

P  2 f V 2 V  Vo (11) f : friction factor

LD 

Deq  2 s Dp  (10)
3 
1 

 P  2 f   Vo 2 31   3 f 2Vo2 1 2 is added to account for the
L   s Dp 3 tortuous path through the bed
   2s Dp

P  1.75Vo2 1  (13) Burke-Plummer Equation
L s Dp  3
ReP  Vo DP  1000

1   

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 1-19

Ergun Equation (The entire range of flow rates)

P  150Vo 1  2  1.75Vo2 1  (14) P  k1Vo  k2Vo2
L sDp 3 L
s2D2p3 ↓↓

??

Valid for spheres, cylinders, and granular packings (: 0.3 – 0.6)

Dp  Ds  n 1 Dp/Dt  for spheres  for cylinders
0 0.34 0.34
  xi Dp 0.1 0.38 0.35
0.2 0.42 0.39
i1 i 0.3 0.46 0.45
0.4 0.50 0.53
Dt: tower or bed diameter 0.5 0.55 0.60

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 1-20

Example 1:

A packed tower is composed of cylinders having a diameter D = 0.02 m and a
length h = D. b = 962 kg/m3 p = 1600 kg/m3

Calculate: , Dpeff, a

Solution:

b  p (1 )    0.399

Dp  sDp and av  6  Dp 6
sDp av
eff eff

sp 2 D2  Dh hD 6 6 6
vp 4 D av 6/D
av  = D2 h av   Dp  D  0.02 m

eff

4

a 6 1   180.3 m2
Dp m3

eff

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 1-21

Example 2:

Calculate P

Solution: Dt = 0.61 m Spherical particles
L = 2.44 m DP = 12.7 mm

 = 0.38

P  150Vo1 2  1.75Vo2 1 
L s2D2p3 sDp 3
Air m  0.358 kg/s

( = 1.221 kg/m3 ,  = 1.9×10-5 Pa.s)

m  0.358 kg/s = AcV o    Dt2 V  V o  1 m/s
4
o

 P 124.2 1913.99  P  0.0497 105Pa
L↓
↓
? ? Re=816.14

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 1-22

Example

 A gas absorption tower of diameter (2 m) contains ceramic Raschig rings
randomly packed to a height of 5 m. Air containing a small proportion of
sulphur dioxide passes upwards through the absorption tower at a flow
rate of (6 m3/s). The viscosity and density of the gas may be taken as
1.80 x 10-5 Pa.s and 1.2 kg/m3 respectively. Details of the packing are:

 Ceramic Raschig rings, Surface area per unit volume of packed bed, a =
190 m2/m3,Voidage of randomly packed bed = 0.71

1. Calculate the diameter, Dsv, of a sphere with the same surface-volume
ratio as the Raschig rings.

2. Calculate the frictional pressure drop across the packing in the tower.

3. Discuss how this pressure drop will vary with flow rate of the gas within

±10% of the quoted flow rate.

4. Discuss how the pressure drop across the packing would vary with gas

pressure and temperature.

JUST Department of Chemical Engineering ChE 362 Unit Operations Chapter 1-23

Packings Raschig Lessing ring
rings
 Classes of packings: Pall ring
Berl saddle Nutter ring
 Broken solids: cheapest; size
from10-100 mm to the size of Pall ring
column

 Shaped packings: better liquid
flow and higher effective
surface

 Grids: easy to fabricate; used
in cooling towers

 Structured packings:

 High efficiency
 But of High cost

 Packing should be non
porous to prevent crystal
formation in the pores
when packing dries

Packings

Grids and structured packings

Grids

Structured packing