# Chapter 26 : Fluidization¶

## Example 26.2 Page no 382¶

In [1]:
from __future__ import division
from math import sqrt,pi
print "Example 26.2 page no 384\n\n"
#a water softner unit consists of a large diameter tank ,the bottom of tank is connected to a vertical ion exchange pipe
h_f=1.25#total fluid height
h_l=h_f
g=32.174#grav. acc
e=0.25# bed porosity
d_p=0.00417#ion exchange resin particle diameter,ft
L=1#pipe length ,ft
#assume turbulent flow ,apply burke purmer equation
v_s=sqrt(g*h_f*e**3*d_p/(1.75*(1-e)*L))#superficial velocity
print "\n superficial velocity v_s=%.2f ft/s"%(v_s)#
meu=6.76e-4#absolute viscosity of water
rho=62.4#density of water
#check for turbulent flow
R_e=d_p*v_s*rho/((1-e)*meu)
print "\n R_e=%.2f"%(R_e)#
#since reynold no is low the calculation is not valid
#assume laminar flow and use Blake-Kozeny equation 26.9
v_s_t=rho*g*h_f*e**3*d_p**2/(150*meu*((1-e)**2)*L)#superficial velocity
print "\n superficial velocity v_s_t=%.2f ft/s"%(v_s_t)#
#check the porous medium reynolds no
R_e_t=v_s_t*d_p*rho/((1-e)*meu)
print "\n reynolds no R_e_t=%.2f "%(R_e_t)#
#since reynolds no R_e < 10,the flow is therfor laminar
D=0.167#diameter of pipe
S=(pi/4)*D**2#empty cross sectional area
q=v_s_t*S#volumetric flow rate
print "\n vol. flow rate q=%.2e ft**3/s"%(q)#

Example 26.2 page no 384

superficial velocity v_s=0.04 ft/s

R_e=22.93

superficial velocity v_s_t=0.01 ft/s

reynolds no R_e_t=6.14

vol. flow rate q=2.62e-04 ft**3/s


## Example 26.3 Page no 384¶

In [2]:
from __future__ import division
from math import sqrt,pi
print "Example 26.3 page no 384\n\n"
#refer to Example 26.2
#a water softner unit consists of a large diameter tank ,the bottom of tank is connected to a vertical ion exchange pipe
h_f=1.25#total fluid height
h_l=h_f
g=32.174#grav. acc
e=0.25# bed porosity
d_p=0.00417#ion exchange resin particle diameter,ft
L=1#pipe length ,ft
#assume turbulent flow ,apply burke purmer equation
v_s=sqrt(g*h_f*e**3*d_p/(1.75*(1-e)*L))#superficial velocity
print "\n superficial velocity v_s=%.2f ft/s"%(v_s)#
meu=6.76e-4#absolute viscosity of water
rho=62.4#density of water
#check for turbulent flow
R_e=d_p*v_s*rho/((1-e)*meu)
print "\n R_e=%.2f"%(R_e)#
#since reynold no is low the calculation is not valid
#assume laminar flow and use Blake-Kozeny equation 26.9
v_s_t=rho*g*h_f*e**3*d_p**2/(150*meu*((1-e)**2)*L)#superficial velocity
print "\n superficial velocity v_s_t=%.2f ft/s"%(v_s_t)#
#check the porous medium reynolds no
R_e_t=v_s_t*d_p*rho/((1-e)*meu)
print "\n reynolds no R_e_t=%.2f "%(R_e_t)#
#since reynolds no R_e < 10,the flow is therfor laminar
#calculation of the  pressure drop due to friction and the pressure drop across the resin bed
k=e**3*d_p**2/(150*(1-e)**2)#packed bed permeability
P_drop_fr=rho*h_f#friction pressure drop across resin bed,psf
print "\n fricion pressure drop P_drop_fr=%.2f psf"%(P_drop_fr)#
z_d=-1#length from point 2 to 3,ft
P_drop_r=rho*(z_d+h_f)#pressure drop across the resi bed
print "\n pressure drop across across the resin bed P_drop_r=%.2f psf"%(P_drop_r)#


Example 26.3 page no 384

superficial velocity v_s=0.04 ft/s

R_e=22.93

superficial velocity v_s_t=0.01 ft/s

reynolds no R_e_t=6.14

fricion pressure drop P_drop_fr=78.00 psf

pressure drop across across the resin bed P_drop_r=15.60 psf


## Example 26.4 Page no 387¶

In [3]:
from __future__ import division
from math import sqrt,pi
print "\nExample 26.4 page no 387\n\n"
#air is used to fluidize a bed of speherical particles
D=0.2#bed diameter,m
d_p=7.4e-5#diameter of 200 mesh particles from table 23.2,m
rho_s=2200#ultimate solid density
rho_f=1.2#density of air
meu=1.89e-5#viscosity of air
g=9.807#grav. constant
e=0.45#bed porosity
L_mf=0.3#length at minimum fluidization
#assume laminar flow
#applying equation 26.29
v_mf=(1-e)*g*rho_s*d_p**2/(150*e**3*meu)#minimum fluidizaton veloctiy
print "\n min. fluidization velocity v_mf=%.2f m/s"%(v_mf)#
#check the flow regime
R_e=v_mf*d_p/(meu*(1-e))
print "\n Reynolds no R_e=%.2f "%(R_e)#
#since R_e= 1.79 <10,flow is laminar
m_dot=pi*v_mf*D**2*rho_f/4#mass flow rate
print "\n mass flow rate m_dot =%.2f kg/s"%(m_dot)#
P_fr=round((1-e)*rho_s*g*L_mf)#gas pressure drop across the bed
print "\n gas pressure drop P_fr=%.2f Pa"%(P_fr)#

Example 26.4 page no 387

min. fluidization velocity v_mf=0.25 m/s

Reynolds no R_e=1.79

mass flow rate m_dot =0.01 kg/s

gas pressure drop P_fr=3560.00 Pa


## Example 26.5 Page no 389¶

In [4]:
from __future__ import division

print "Example 26.5 page no 389\n\n"
#air  flowing through a 10 ft packed bed
V_o=4.65#superficial velocity,ft/s
meu_g=1.3e-5#viscosity of air
rho_g=0.67#density of air,lb/ft**3
e=0.89#void volume
g_c=32.2#grav. constant
L=10#length of packed bed
d_p=0.007815#effective particle diameter
P_drop = ((150*V_o*meu_g/(g_c*d_p**2))*((1-e)**2/e**3) + (1.75*rho_g*V_o**2/(g_c*d_p))*((1-e)**2/e**3))*L#pressure drop
print "\n pressure drop P_rop=%.2f lb/ft**2"%(P_drop)##calculation error in book


Example 26.5 page no 389

pressure drop P_rop=18.08 lb/ft**2


## Example 26.6 Page no 392¶

In [5]:
from __future__ import division

print "Example 26.6 page no 392\n\n"
#a bed of pulverized is to be fluidized with liquid oil
D=4#diameter of bed ,ft
d_p=0.00137#particle diameter ,ft
rho_s=84#coal particle density ,lb/ft**3
rho_f=55#oil density,lb/ft**3
e_mf=0.38#void fraction
L_mf=8#bed height at minimum fluidization,ft
P_drop=(rho_s-rho_f)*(1-e_mf)*L_mf +rho_f*L_mf

print "\npressure drop P_drop=%.2f psf"%(P_drop)#

Example 26.6 page no 392

pressure drop P_drop=583.84 psf


## Example 26.7 Page no 393¶

In [6]:
from __future__ import division
print "Example 26.7 page no 393\n\n"
#refer to example 26.6
D=4#diameter of bed ,ft
d_p=0.00137#particle diameter ,ft
rho_s=84#coal particle density ,lb/ft**3
rho_f=55#oil density,lb/ft**3
meu_f=3.13e-4#viscosity of oil
e_mf=0.38#void fraction
L_mf=8#bed height at minimum fluidization,ft
L_f=10#bed height,ft
e=1-L_mf*(1-e_mf)/L_f#bed voidage
g=32.174#grav acc
v_s=(d_p**2)*g*(e**3)*(rho_s-rho_f)/(150*meu_f*(1-e)) #superficial velocity
print "\n superficial velocity v_s=%.2f ft/s"%(v_s)#
q=(pi/4)*D**2*v_s#volumetric flow rate
print "\n vol. floe rate q=%.2f ft**3/s"%(q)#
#check on the laminar flow assumption
meu_f=0.01
R_e=d_p*v_s*rho_f/(meu_f*(1-e))
print "\n reynolds no R_e=%.2f"%(R_e)#
print "\n since R_e is less than 10 ,flow is laminar"

Example 26.7 page no 393

superficial velocity v_s=0.01 ft/s

vol. floe rate q=0.12 ft**3/s

reynolds no R_e=0.15

since R_e is less than 10 ,flow is laminar


## Example 26.8 Page no 393¶

In [7]:
from __future__ import division
print " Example 26.8 page no 393\n\n"
#refer to example 25.6
#obtain the porous medium friction factor usingthe burke -plummer equation
#/since the flow is turbulent ,eq.26.6 applies
f_pm=1.75#porous medium friction facot
v_s=2#superficial velocity
e=.4#porosity
L=50#length of bed
d_p=0.0833#particle diameter
g=32.174#grav. acc
print "\n head loss h_f=%.2f ft of propane "%(h_f)#
#applying bernoulli eq. between the entrance and gas exit
P2=4320#pressure at the bottom of the catalyst bed
rho_f=0.0128#density of fluid
z_d=-50#length from point 2 to 3,z2-z1
P1 = P2 + rho_f*(z_d-h_f)# absolute pressure of the inlet gas
print "\n pressure P1=%.2f psf"%(P1)#
#since flow is turbulent , permeablity of the medium k can not be calculated

 Example 26.8 page no 393

head loss h_f=1224.30 ft of propane

pressure P1=4303.69 psf


## Example 26.9 Page no 394¶

In [8]:
from __future__ import division
print "Example 26.9 page no 394\n\n"
#turbulent flow of water through a carbon bed
d_p=0.001#particle diameter
meu=0.001#viscosity of water
e=0.25#porosity
R_e=1000#R_e is >1000 for turbulent flow,for minimum pressure drop
rho=1000#density of water,kg/m**3
v_s=R_e*meu*(1-e)/(d_p*rho)#superficial velocity
print "\n superficial velocity v_s=%.2f m/s"%(v_s)#
phi_s=1#spehercity
L=0.5#length of bed,m
P_drop = 1.75*rho*L*v_s**2*(1-e)/(phi_s*d_p*(e**3))#presssure drop
print "\npressure drop P_drop=%.2e Pa"%(P_drop)#

Example 26.9 page no 394

superficial velocity v_s=0.75 m/s

pressure drop P_drop=2.36e+07 Pa


## Example 26.10 Page no 395¶

In [9]:
from __future__ import division
from math import sqrt,pi
print "Example 26.10 page no 395\n\n"
#a bed of 200 mesh particles is fluidized with air
d_b=0.2#diameter of bed,m
d_p=7.4e-5#particle diameter
L_mf=0.3#bed height at minimum fludization
e_mf=0.45#bed porosity at min. fluidization
L_o=L_mf*(1-e_mf)#the zero porosity bed height
print "\n zero porosity bed height L_o=%.2f m"%(L_o)#
rho_s=2200#density of particles
rho_f=1.2#density of fluid
g=9.807#grav. acc
meu_f=1.89e-5#viscosity of fluid
#assuming laminar flow ,use equation  26.9
v_mf =(e_mf**3)*(g*(rho_s-rho_f)*(d_p**2))/(150*(1-e_mf)*meu_f)#velocity at minimum fluidization
print "\n velocity at min. fluidization v_mf=%.2f m/s"%(v_mf)#
v_t=0.35#terminal velocity from example 26.3
e=0.91#value of e porosity from eq26.9
L_f=L_o/(1-e)#expanded bed height L_f
m=rho_s*pi*d_b**2*L_o#bed inventory
print "\n expanded bed height L_f=%.2f m\n bed inventory m=%.2f kg"%(L_f,m)#

Example 26.10 page no 395

zero porosity bed height L_o=0.17 m

velocity at min. fluidization v_mf=0.01 m/s

expanded bed height L_f=1.83 m
bed inventory m=45.62 kg


## Example 26.11 Page no 396¶

In [10]:
from __future__ import division
print "\n Example 26.11 page no 396\n\n"
#refer to illustrative example 26.9
d_p=7.4e-5#particle diameter
L_mf=0.3#bed height at minimum fludization
e_mf=0.45#bed porosity at min. fluidization
L_o=L_mf*(1-e_mf)#the zero porosity bed height
print "\n zero porosity bed height L_o=%.2f m"%(L_o)#
rho_s=2200#density of particles
rho_f=1.2#density of fluid
g=9.807#grav. acc
meu_f=1.89e-5#viscosity of fluid
#assuming laminar flow ,use equation  26.9
v_mf =(e_mf**3)*(g*(rho_s-rho_f)*(d_p**2))/(150*(1-e_mf)*meu_f)#velocity at minimum fluidization
print "\n velocity at min. fluidization v_mf=%.2f m/s"%(v_mf)#
F_mf=v_mf**2/(g*d_p)#fluidization mode
print "\n fluidization mode F_mf=%.2f "%(F_mf)#
#from value of F_mf ,fluidization is smoth,F_mf =0.66<0.13

 Example 26.11 page no 396

zero porosity bed height L_o=0.17 m

velocity at min. fluidization v_mf=0.01 m/s

fluidization mode F_mf=0.07