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

PDf

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