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FLOW IN PACKED BEDS
• Packed towers are finding applications in adsorption, absorption, ion-exchange, distillation, humidification, catalytic reactions, regenerative heaters etc.,
• Packing is to provide a good contact between the contacting phases.
• Based on the method of packing, Packings are classified as
(a) Random packings (b) Stacked packings
PACKED TOWERSFLUID FRICTION IN POROUS MEDIA
• The packings are made with clay, porcelain, plastics or metals.
Packing Void fraction, ε
Berl saddle 0.6 - 0.7
Intalox saddle 0.7 - 0.8
Rasching ring 0.6 - 0.7
Pall ring 0.9 - 0.95
Principal requirements of a tower packing
• It must be chemically inert to the fluids in the tower.
• It must be strong without excessive weight.
• It must contain adequate passages for the contacting streams without excessive pressure drop.
• It must provide good contact between the contacting phases.
• It should be reasonable in cost.
FLUID FRICTION IN POROUS MEDIA
• In this approach, the packed column is regarded as a bundle of crooked tubes of varying cross sectional area
• The theory developed for single st. tubes is used to develop the results of bundle of crooked tubes…..
– Laminar flow– Turbulent flow– Transition flow
Laminar flow in packed beds
• Porosity (void fraction) is given by
ε = (volume of voids in the bed / (total volume of bed )
• Superficial velocity ‘vs’ = (Q / Apipe)
• Interstitial velocity ‘vI’ = (Q / ε Apipe)
EMPTY TOWER VELOCITY
Velocity based on the area actually open to the flowing fluid
s
I
vv
• In a packed bed consider a set of crooked tubes (non-circular CSA)
• rH = (cross sectional area of channel) / (wetted perimeter of channel)
• Multiply and divide by LENGTH of bed
=rH = (A ε ) L / (wetted perimeter) L • =rH = ε (volume of bed) / (Total wetted surface area
of solids)
i.e., AREA AVAILABLE FOR FLOW = A ε
• To find wetted surface area…….
• Total wetted surface area of solids
= (no. of spherical particles) x (surface area of one particle)
and we know….
No. of particles = (volume of bed) (1- ε) / (volume of one particle) volume fraction
particleoneofareasurfaceparticleoneofvolbedvol
bedvolrH
.)1)(.(
).(
particleoneofareasurface
particleoneofvol.
1
61p
H
Dr
2
3
61 p
p
D
D
.Remod, noynoldsifiedso
Ieq
p
vDN Re,
sp vD
164
spp vDN
1
1Re,
ERGUN defined NRe,p without the constant term (4/6) for PACKED BED
232
)(min
eq
I
D
vLP
eqnPoiseuilleHagenflowarLafor
s
p
v
D
L2
164
32
3
2
2
)1(72
p
s
D
v
L
P
3
2
2
)1(150
p
s
D
v
L
P
By several experiments it has been found that the constant value should be 150
KOZNEY-CARMANN EQN.
Only if NRe,p < 10 (LAMINAR)
Turbulent flow in packed beds
22 I
eq
vL
factorfrictionforequationgeneraltheusewe
2
2
164
2
p
s
Dvf
L
P
3
2 )1(3
p
s
D
vf
By several expts it has been found that for turbulent flow, the ‘ 3f ’ should be replaced by a value 1.75
3
2 )1(75.1
p
s
D
v
L
P
BURK- PLUMMER EQN.
Only if NRe,p > 1000 (TURBULENT)
FOR TRANSITION REGION…
3
2
32
2 )1(75.1)1(150
p
s
p
s
D
v
D
v
L
P
ERGUN EQN.
if NRe,p between 10 and 1000
(TRANSITION)
PROB…..
• Calculate the pressure drop of air flowing at 30ºC and 1 atm pressure through a bed of 1.25 cm diameter spheres, at a rate of 60 kg/min. The bed is 125 cm diameter and 250 cm height. The porosity of the bed is 0.38. The viscosity of air is 0.0182 cP and the density is 0.001156 gm/cc.
Data:• Mass flow rate of Air = 60 kg/min = 1 kg/sec• Density of Air () = 0.001156 gm/cc = 1.156
kg/m3• Viscosity of Air () = 0.0182 cP = 0.0182 x 10-3
kg/(m.sec)• Bed porosity () = 0.38• Diameter of bed (D)= 125 cm = 1.25 m• Length of bed (L) = 250 cm = 2.5 m• Dia of particles (Dp)= 1.25 cm = 0.0125 m
• Volumetric flow rate = mass flow rate / density = 1 / 1.156 = 0.865 m3/sec
• Superficial velocity Vo = 0.865 / ( (/4) D2 ) = 0.865 / ( (/4) 1.252 ) = 0.705 m/sec
• NRe,P = 0.0125 x 0.705 x 1.156 / (0.0182 x 10-3 x ( 1- 0.38 ) ) = 903…….Transition region
• We shall use Ergun equation to find the pressure drop.
• p = 2492.92 N/m2
Pressure Drop in Regenerative Heater
• A regenerative heater is packed with a bed of 6 mm spheres. The cubes are poured into the cylindrical shell of the regenerator to a depth of 3.5 m such that the bed porosity was 0.44. If air flows through this bed entering at 25ºC and 7 atm abs and leaving at 200ºC, calculate the pressure drop across the bed when the flow rate is 500 kg/hr per square meter of empty bed cross section. Assume average viscosity as 0.025 cP and density as 6.8 kg/m3.
• Mass flow rate of Air / unit area = 500 kg/(hr.m2) = 0.139 kg/(sec.m2)
• Density of Air () = 6.8 kg/m3
• Viscosity of Air () = 0.025 cP = 0.025 x 10-3 kg/(m.sec)
• Bed porosity () = 0.44
• Length of bed (L) = 3.5 m
• Dia of particles (Dp)= 6 mm = 0.006 m
• Superficial velocity Vs = mass flow rate per unit area / density = 0.139 / 6.8 = 0.0204 m/sec
• NRe,p = 0.006 x 0.0204 x 6.8 / (0.025 x 10-3 x ( 1- 0.44 ) ) = 59.45
• We shall use Ergun equation to find the pressure drop.
• ∆P = 46.37 N/m2
Design of Packed Tower with Berl Saddle packing
• 7000 kg/hr of air, at a pressure of 7 atm abs and a temperature of 127oC is to be passed through a cylindrical tower packed with 2.5 cm Berl saddles. The height of the bed is 6 m. What minimum tower diameter is required, if the pressure drop through the bed is not to exceed 500 mm of mercury?
For Berl saddles, p = (1.65 x 105 Z Vs1.82 1.85 ) / Dp1.4
where p is the pressure drop in kgf/cm2, Z is the bed height in meter, is the density in g/cc, Dp is nominal diameter of Berl saddles in cm, Vs is the superficial linear velocity in m/sec.
Data:• Mass flow rate = 7000 kg/hr = 1.944 kg/sec• Height of bed (Z) = 6 m• Dp = 2.5 cm • 760 mm Hg = 1 kgf/cm2 = 1 atm• p = 500 mm Hg = (500/760) x 1 kgf/cm2 = 0.65 kgf/cm2
• Formula:• Ideal gas law: • PV = nRT• Formula given,• p = (1.65 x 105 Z Vs
1.82 1.85 ) / Dp1.4
Calculations:
= M(n/V) = M(P/RT) = 29 x 7 x 1.01325 x 105 / (8314 x (273 + 127) ) = 6.185 kg/m3 = 6.185 x 10-3 g/cc
• p = (1.65 x 105 Z Vs1.82 1.85 ) / Dp1.4
• 0.65 = (1.65 x 105 x 6 x Vs1.82 x (6.185 x 10-3 )1.85 ) / 2.51.4
• Vs1.82 = 0.02886
• Vs = 0.1432 m/sec
• Volumetric flow rate = mass flow rate/density = 1.944/6.185 = 0.3144 m3/sec
• Required Minimum Diameter (D) = 1.6719 m.
• Air flows thro a packed bed of powdery material of 1cm depth at a superficial gas velocity of 1cm/s. A manometer connected to the unit registers a pressure drop of 1cm of water. The bed has a porosity of 0.4. Assuming that Kozney-Carmann equation is valid for the range of study, estimate the particle size of the powder?
Density of air = 1.23kg/m3
viscosity of air = 1.8x10-5 kg/m-s
• Dp=1.24x10-4m
Flow Rate of Water through Ion-Exchange Column
Figure shows a water softener
in which water trickles by gravity
over a bed of spherical ion-exchange
resin particles, each 0.05 inch in
diameter. The bed has a porosity
of 0.33. Calculate the volumetric
flow rate of water. Assume laminar flow.
• Applying Bernoulli's equation from the top surface of the fluid to the outlet of the packed bed and ignoring the kinetic-energy term and the pressure drop through the support screen, which are both small, we find ………
• Since Laminar flow, apply Kozney-Carmann equation• vs = 0.01055 m/sec• = Q = 21cm3/sec
g(∆z) = hf
hf = ∆p/ρ=3.7376 J/kg
fhghvp
ghvp
222
21
21
1
2
1
2
1
• Water trickles by gravity over a bed of particles each 1mm dia in a bed of 6cm and height 2m. The water is fed from a reservoir whose dia is much larger than that of packed bed, with water maintained at a height of 0.1m above the top of the bed. The bed has a porosity of 0.31. calculate the volumetric flow rate of water if its viscosity is 1cP
Shape factor-Sphericity factor
• For non-spherical particles instead of diameter an equivalent diameter is defined.
• Sphericity Φs is defined as the surface-volume ratio for a sphere of dia Dp divided by the surface-volume ratio for the particle whose nominal size is Dp.
• Φs = (6/Dp) / (sp/vp)
• Therefore, actual dia to be used in Ergun eqn is = Φs Dp
• For a non-spherical particle, Ergun eqn is given by………
3
2
322
2 )1(75.1)1(150
ps
s
ps
s
D
v
D
v
L
P
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