Chapter 25 : Porous Media and Packed Beds

Example 25.1 Page no 370

In [1]:
from __future__ import division
print "Example 25.1 page no 370\n\n"
#calculation of efffective particle diameter for a set of packing
V=0.2#packing volume
n=100#no. of particle assume
V_p=V*1000/n#the volume of single particle,mm**2#
S_p=2.18#average surface area of particle,mm**2
a_p=S_p/V_p#specific surface area of particle ,(mm)**-1
D_p = 6/a_p#effective diameter of particle,mm
print "\n effective partcle diameter D_p=%.2f mm "%(D_p)#
Example 25.1 page no 370



 effective partcle diameter D_p=5.50 mm 

Example 25.2 Page no 371

In [2]:
from __future__ import division
print "Example 25.2 page no 371\n\n"
#refer to example 25.1
V=0.2#packing volume
n=100#no. of particle assume
V_p=V*1000/n#the volume of single particle,mm**2#
S_p=2.18#average surface area of particle,mm**2
a_p=S_p/V_p#specific surface area of particle ,(mm)**-1
D_p = 6/a_p#effective diameter of particle,mm
D_p=5.50#round off value for accurate answer
rho=0.235#density of fluid,g/cm**3
meu=2e-4#viscosity,g/cm.s
v=10#interstitial velocity ,cm
R_e=round((D_p/v)*rho*v/meu)#reynolds no
print "\n Reynolds no R_e=%.2f "%(R_e)#
#from R_e value we can conclude that the flow of fluid would be in the turbulent region
Example 25.2 page no 371



 Reynolds no R_e=6463.00 

Example 25.3 Page no 372

In [3]:
from __future__ import division
from math import pi
print "Example 25.3 page no 372\n\n"
#air flows across a packed bed 
d_p=1.5#diamter of cylinderical particles,cm
h=2.5#height ,cm
V_p=pi*d_p**2*h/(4)#volume of the cylinderical particles
S_p=pi*d_p*h + 2*(pi*d_p**2/4)#cylinderical particle surface area,cm**2
a_p=S_p/V_p#particle specific surface 
print "\n particle specific surface a_p =%.2f cm**-1 "%(a_p)#
d_p_e=6/a_p#effective particle diameter
print "\n effective particle diameter d_p_e=%.2f cm"%(d_p_e)#
Example 25.3 page no 372



 particle specific surface a_p =3.47 cm**-1 

 effective particle diameter d_p_e=1.73 cm

Example 25.4 Page no 373

In [4]:
from __future__ import division
print "\nExample 25.4 page no 373\n\n"
#a absorber bed consists of cube particles 
L=3/4#edge length of particle
V_p=L**3#volume of particle 
S_p=6*L**2#surface area of particle
a_p=6*L**2/L**3#specific particle surface area
print "\n specific particle surface area a_p=%.2f in**-1"%(a_p)#
d_p_e = L#effective particle diameter = edge length
print "\n effective particle diameter d_p_e=%.2f in"%(d_p_e)
Example 25.4 page no 373



 specific particle surface area a_p=8.00 in**-1

 effective particle diameter d_p_e=0.75 in

Example 25.5 Page no 373

In [5]:
from __future__ import division
from math import pi
print "Example 25.5 page no 373\n\n"
#gas(propane) flows through a catalyst tower
Mw=44.1#molecular weight
P=4320#pressre at the bottom of the catalyst bed,psf
R=10.73#gas constant
T=960#temperature,Rankine
rho=P*Mw/(R*T*144)#density of propane
L=50#height of bed,ft
D=20#diameter of bed,ft
V=pi*D**2*L/4#bed volume
theta=10#contact time,s
e=0.4#bed porosity
q=V*e/theta#volumetric flow rate
v_s=4*q/(pi*D**2)#superficial  velocity
print "\n superficial velocity v_s=%.2f ft/s"%(v_s)#
v_i=v_s/e#interstitial velocity
print "\n interstitial velocity v_i=%.2f ft/s"%(v_i)#
rho_s=77.28#ultimate density(spheres )
rho_b=(1-e)*rho_s#bulk density
print "\n bulk density rho_b=%.2f lb/ft**3"%(rho_b)#
d_p=0.0833#diameter of particles
a_p=6/d_p#specific surface area
print "\n specific surface area a_p=%.2f ft**-1"%(a_p)#
a_b=a_p*(1-e)#bed specific surface
print "\n bed specific surface a_b=%.2f ft**-1"%(a_b)
Example 25.5 page no 373



 superficial velocity v_s=2.00 ft/s

 interstitial velocity v_i=5.00 ft/s

 bulk density rho_b=46.37 lb/ft**3

 specific surface area a_p=72.03 ft**-1

 bed specific surface a_b=43.22 ft**-1

Example 25.6 Page no 375

In [6]:
from __future__ import division
print "Example 25.6 page no 375\n\n"
#refer to example 25.5
d_p=0.0833#diameter of particles,ft
e=0.4#bed porosity
D_h=2/3*(e/(1-e))*d_p#hydraulic diameter
r_h=D_h/4#hydrulic radius
print "\n hydraulic diameter D_h=%.2f ft\n hydrulic radius r_h=%.3f ft"%(D_h,r_h)# 
Example 25.6 page no 375



 hydraulic diameter D_h=0.04 ft
 hydrulic radius r_h=0.009 ft