Chapter 24 : Sedimentation Centrifugation and Flotation¶

Example 24.1 Page no 350¶

In [1]:
from __future__ import division
print "Example 24.1 page no 350\n\n"
#glass sphere are settling in water at 20 deg C
#the slurry contains 60 wt% solids
# start by assuming a basis of 100 kg of slurry
m_f=40#mass of fluid,kg
rho_f=998#density of water,kg/m**3
V_f=m_f/rho_f#volume of the fluid,m**3
m_s=60#mass of solid,kg
rho_p=2467#density of glass,kg/m**3
V_s=m_s/rho_p#volume of glass,m**3
V = V_f + V_s#total volume,m**3
v_frac_f = V_f/V#volume fraction for the fluid particles
print "\n volume fraction fluid particles v_frac_f =%0.2f "%(v_frac_f)#
v_frac_p=1-v_frac_f#volume fraction for the glass particles
print "\n volume fraction for the glass particles v_frac_p=%0.2f  "%(v_frac_p)#
rho_m=round(v_frac_f*rho_f  + v_frac_p*rho_p)#bulk density of slurry
print "\n bulk density of slurry rho_m=%0.2f kg/m**3 "%(rho_m)#
b=10**(1.82*(1-v_frac_f))#dimensionless correction factor
g=9.807#gravitational acc.,m/s**2
D_p=0.0001554#diameter of particle,m
meu_f=0.001#viscosity of fluid
v_t = g*D_p**2*(rho_p-rho_f)*v_frac_f**2/(18*meu_f*b)#terminal velocity
print "\n terminal velocity v_t=%0.2f m/s"%(v_t)#
meu_m = meu_f*b#effective mixture viscosity
print "\n effective mixture viscosity meu_m=%0.2f kg/m.s"%(meu_m)#

Example 24.1 page no 350

volume fraction fluid particles v_frac_f =0.62

volume fraction for the glass particles v_frac_p=0.38

bulk density of slurry rho_m=1553.00 kg/m**3

terminal velocity v_t=0.00 m/s

effective mixture viscosity meu_m=0.00 kg/m.s


Example 24.2 Page no 352¶

In [2]:
from __future__ import division
print "Example 24.2 page no 352\n\n"
#refer to example 24.1
m_f=40#mass of fluid,kg
rho_f=998#density of water,kg/m**3
V_f=m_f/rho_f#volume of the fluid,m**3
m_s=60#mass of solid,kg
rho_p=2467#density of glass,kg/m**3
V_s=m_s/rho_p#volume of glass,m**3
V = V_f + V_s#total volume,m**3
v_frac_f = V_f/V#volume fraction for the fluid particles
print "\n volume fraction fluid particles v_frac_f =%0.2f "%(v_frac_f)#
v_frac_p=1-v_frac_f#volume fraction for the glass particles
print "\n volume fraction for the glass particles v_frac_p=%0.2f  "%(v_frac_p)#
rho_m=round(v_frac_f*rho_f  + v_frac_p*rho_p)#bulk density of slurry
print "\n bulk density of slurry rho_m=%0.2f kg/m**3 "%(rho_m)#
b=10**(1.82*(1-v_frac_f))#dimensionless correction factor
g=9.807#gravitational acc.,m/s**2
D_p=0.0001554#diameter of particle,m
meu_f=0.001#viscosity of fluid
v_t = g*D_p**2*(rho_p-rho_f)*v_frac_f**2/(18*meu_f*b)#terminal velocity
print "\n terminal velocity v_t=%0.2f m/s"%(v_t)#
meu_m = meu_f*b#effective mixture viscosity
print "\n effective mixture viscosity meu_m=%0.2f kg/m.s"%(meu_m)#
R_e=rho_m*v_t*D_p/(meu_m*v_frac_f)#reynolds no.
print "\n reynolds no R_e=%0.2f "%(R_e)#

Example 24.2 page no 352

volume fraction fluid particles v_frac_f =0.62

volume fraction for the glass particles v_frac_p=0.38

bulk density of slurry rho_m=1553.00 kg/m**3

terminal velocity v_t=0.00 m/s

effective mixture viscosity meu_m=0.00 kg/m.s

reynolds no R_e=0.12


Example 24.3 Page no 352¶

In [3]:
from __future__ import division
print "Example 24.3 page no 352\n\n"
#classification of small speherical particles of charcoal with a specific gravity of 2.2
#the particles are falling in a vertical tower against a rising current of air
#we have to calculate the minimum size of charcoal that will settle down to the bottom of the tower
rho =0.075#density of air,lb/ft**3
meu=1.23e-5#viscosity of air,lb/ft.s
#assume stokes law to apply
SG=2.2#specific gravity of charcoal
rho_w=62.4#density of water
rho_p=SG*rho_w#density of charcoal
v=15#velocity of air
g=32.2#grav. acc
D_p1=(18*meu*v/(g*rho_p))**0.5
K1 = D_p1*(g*rho*rho_p/meu**2)**(1/3)#settling factor
print "\n settling factor K1=%0.2f "%(K1)#
#from value of K,stokes law does not apply
#therefore,assume Intermediate range law applies
D_p =((v*rho**0.29*meu**0.43)/(0.153*(g*rho_p)**0.71))**(1/1.14)
print "\n particle diameter= D_p=%0.2e ft "%(D_p)#
K_n=(D_p/D_p1)*K1
print "\n final settling factor K_n=%0.2f"%(K_n)
#since the result is correct for the intermediate range

Example 24.3 page no 352

settling factor K1=11.26

particle diameter= D_p=2.18e-03 ft

final settling factor K_n=28.30


Exmple 24.4 Page no 354¶

In [4]:
from __future__ import division
print "Example 24.4 page no 354\n\n"
#a particle is spining in a 3 inch ID centrifuge
g=32.2
G=round(r*omega**2/g)
print "\n G=%0.2f "%(G)#

Example 24.4 page no 354

G=7.00


Example 24.5 Page no 357¶

In [5]:
from __future__ import division
from math import sqrt
print "Example 24.5 page no 357\n\n"
#a circular cylinder filled with water is rotated a uniform ,steady angular speed about it's central axis in rigid body motion
#since the cylinder is full the water will spill the moment the cylinder starts to spin ,spilling occur when omega > 0 rpm
# to determine the angular speed for 1/3 of the water to spill , consider the cylinder at rest when 1/3 of the water has already beem spilled
g=32.174#grav. acc
R = 0.25 #radius  of cylinder
z_st=2/3#the stationary height, ft
h = 2*(1-z_st)#increase in height is h/2,ft
omega=sqrt(4*g*(h/2)/R**2)

Example 24.5 page no 357



Example 26.6 Page no 392¶

In [6]:
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=%0.2f psf"%(P_drop)#

Example 26.6 page no 392

pressure drop P_drop=583.84 psf


Example 24.7 Page no 358¶

In [7]:
from __future__ import division
from math import sqrt
print "Example 24.7 page no 358\n\n"
#a cylindrical cup open to the atmosphere is filled with liquid to a height of 7 cm
#rotated around it's axis
#calculation of an angular velocity that will cause the liquid to start spilling
h=0.03#height,m
#applying eq. 24.22
g=9.807#grav. acc
omega=sqrt(2*h*g/(R**2))
omega=36.2#printing mistake in book
#calculation of pressure at point A and B that is P_a and P_b
z=.1#liquid height above point A and B,m
rho=1010#density of liquid,kg/m**3
P_a = rho*g*z
P_b=P_a#from symmetry P_a = P_b
print "\n pressure P_a=%0.2f Pa_gauge\n pressure P_b=%0.2f Pa_gauge"%(P_a,P_b)#
z_c=0.04#liquid height above point c,m
P_c=rho*g*z_c#pressure at point c
print "\n pressure P_c=%0.2f Pa_gauge"%(P_c)#
#to obtain the film thicknes,we have to find the original height
z_l=0.07#liquid height ,m
h_o=z_l-z_c#original height
r = 100*sqrt(2*h_o*g/(omega**2))#100 for centimeter
print "\n r=%0.2fcm "%(r)#
R=3
t_f=R-r#thikness of film
print "\n thickness film t_f=%0.2f m"%(t_f)##printing mistake in book

Example 24.7 page no 358

pressure P_a=990.51 Pa_gauge
pressure P_b=990.51 Pa_gauge

pressure P_c=396.20 Pa_gauge

r=2.12cm

thickness film t_f=0.88 m


Example 24.8 Page no 360¶

In [8]:
from __future__ import division
print "Example 24.7 page no 358\n\n"
#It is desired to separate quartz particles  from galena particles
SG_q = 2.65#specific gravity of quartz particle
SG_g=7.5#specific gravity of galena particles
rho_f=1000#density of water
rho_q=SG_q*rho_f#density of quartz paticles
rho_g=SG_g*rho_f#density of galena particle
#calculation of the settling veloctiy of the largest quartz particle with a diameter
D_q=9e-5#diameter of largest particle of quartz
g=9.807#grav. acc
meu_f=0.001#viscosity of water
K_q = D_q*(g*(rho_q-rho_f)*rho_f/(meu_f**2))**(1/3)#settling factor
print "\n settling factor K_q=%0.2f "%(K_q)#
#since K =2.27<3.3,stokes flow regime applies ,from the equation 23.36
v_q=g*D_q**2*(rho_q-rho_f)/(18*meu_f)#settling velocity of thelargest quartz particle
print "\n settling velocity (quartz) v_q=%0.2f m/s"%(v_q)#
#calculation of the settling velocity of the smallest galena partilce
d_g=4e-5#diameter of smallest galena particle
K_g = d_g*(g*(rho_g-rho_f)*rho_f/(meu_f**2))**(1/3)#settling factor
print "\n settling factor K_g=%0.2f "%(K_g)#
#since K = 1.6<3.3,stokes flow regime again applies
v_g=g*d_g**2*(rho_g-rho_f)/(18*meu_f)#settling velocity for galena particles
print "\n setling velocity v_g=%0.2f m/s"%(v_g)#
#to  obtain pure galena the upward velociy of the water must be equal to or greater than the settling veloctiy of the quartz particle
v_w=v_q#velocity of water
print "\n water velocity v_w=%0.2f m/s"%(v_w)#

Example 24.7 page no 358

settling factor K_q=2.28

settling velocity (quartz) v_q=0.01 m/s

settling factor K_g=1.60

setling velocity v_g=0.01 m/s

water velocity v_w=0.01 m/s


Example 24.9 Page no 361¶

In [9]:
from __future__ import division
from math import sqrt
print "\n Example 24.9 page no 361\n\n"
#refer to illustrative example 24.8
#we have to determine the size range of the galena in the top product
#to determine the size range of the galena product ,calculate the galena particle size that has a settling velocity equal to water velocity
#assume stokes law applies
v_w=0.0073#velocity of water
v_q=v_w#velocity of quartz particles
SG_q = 2.65#specific gravity of quartz particle
SG_g=7.5#specific gravity of galena particles
rho_f=1000#density of water
rho_q=SG_q*rho_f#density of quartz paticles
rho_g=SG_g*rho_f#density of galena particle
g=9.807#grav. acc
meu_f=0.001#viscosity of water
D = sqrt(18*meu_f*v_q/(g*(rho_g-rho_f)))
print "\n diameter D =%0.2e m"%(D)#
#check on the validity of stokes law by calculating the K factor
K = D*(g*(rho_g-rho_f)*rho_f/(meu_f**2))**(1/3)#settling factor
print "\n settling factor K=%0.2f "%(K)#
#since K =1.82<3.3 , the flow is in the stokes law range

 Example 24.9 page no 361

diameter D =4.54e-05 m

settling factor K=1.81


Example 24.10 Page no 362¶

In [10]:
from __future__ import division
from math import pi
print "Example 24.10 page no 362\n\n"
#air is being dried by bubbling through concentrated NaOH
q=4/60#flow rate of air,ft**3/min
D=2.5/12#diameter of tube
S=(pi/4)*D**2#cross sectional area
v=q/S#velocity of air,ft/s
meu=1.23e-5#viscosity of NaOH
rho=0.0775#density of air
g=32.2#grav. acc.
SG=1.34#specific gravity of NaOH
rho_w=62.4#density of water
rho_p=SG*rho_w#density of NaOH
D_p_max = (v*(rho**0.29)*(meu**0.43)/(0.153*(g*rho_p)**0.71))**(1/1.14)#assuming that the intermediate range applies ,maximum diamter of particle
print "\nD_p_max=%0.2e "%(D_p_max)#
#settling factor
K=D_p_max*(g*rho*rho_p/(meu**2))**(1/3)
print "\n settling factor K=%0.2f "%(K)#
#tus result for D_p_max is correct

Example 24.10 page no 362

D_p_max=5.01e-04

settling factor K=5.58