# Chapter 23 : Particle Dynamics¶

## Example 23.1 Page no 323¶

In [9]:
from math import sqrt
from __future__ import division
print "Example 23.1 Page no 323\n\n"
#calculation of aerodynamic diameter for the following particles
d_es=1.4#equivalent dia of solid sphere,micrometer
sg_s=2#specific gravity of solid sphere
d_eh=2.8#equivalent diameter of hollow sphere, mirometer
sg_h=0.51#specific gravity of hollow sphere
d_pa1=d_es*sqrt(sg_s)#aerodynamic dia for solid sphere
d_pa2=round(d_eh*sqrt(sg_h))#aerodynamic dia for hollow sphere
print "\n d_pa1=%0.2f micron\nd_pa2=%0.2f micron"%(d_pa1,d_pa2)#

Example 23.1 Page no 323

d_pa1=1.98 micron
d_pa2=2.00 micron


## Example 23.2 Page no 323¶

In [10]:
from math import sqrt
from __future__ import division
print "Example 23.2 Page no 323\n\n"
#calculation of aerodynamic diameter of irregular saped sphere
d_e=1.3#eq. diameter,micron
sg=2.35
d_pa=d_e*sqrt(sg)#aerodynamic diameter
print "\n aerodynamic diameter d_pa=%0.2f micron"%(d_pa)#

Example 23.2 Page no 323

aerodynamic diameter d_pa=1.99 micron


## Example 23.3 Page no 335¶

In [11]:
from math import exp
from __future__ import division
print "Example 23.3 Page no 335\n\n"
#calculation of cunningham correction factor
dp=0.4#particle diameter
lemda=6.53e-2
A=1.257 + 0.40*exp(-1.10*dp/(2*lemda))
C= 1 + 2*A*lemda/dp#cunningham correction factor(CCF)
print "CCF C=%0.2f "%(C)#

Example 23.3 Page no 335

CCF C=1.41


## Example 23.4 Page no 336¶

In [12]:
from math import exp
from __future__ import division
print "Example 23.4 Page no 336\n\n"
#three different diameter sized fly ash particls settle through air
#we have to calculate the particle terminal velocity and determine how far each will fall in 30 seconds
#assume the particles are speherical
SG=2.31#specific gravity of fly ash
rho_w=62.4#density of water
rho_p=SG*rho_w#density of particles
#properties of air
R=0.7302#gas constant
T=698#temperature,R
P=1#pressure ,atm
Mw=29#mol. wt of air
rho_a=P*Mw/(R*T)#density of air,lb/ft**3
meu=1.41e-5#viscosity of air,lb/ft.s
g=32.174#grav. acc
D1=0.4#diameter of particle 1,microns
D2=40#diameter of particle 2,microns
D3=400#diameter of particle 3,microns
K1=(D1/(25400*12))*(g*rho_p*rho_a/(meu**2))**(1/3)#dimensionless constant for particle 1
K2=(D2/(25400*12))*(g*rho_p*rho_a/(meu**2))**(1/3)#dimensionless constant for particle 2
K3=(D3/(25400*12))*(g*rho_p*rho_a/(meu**2))**(1/3)#dimensionless constant for particle 3
print "\n dimensionless constant K1=%0.2f \n K2=%0.2f \n K3=%0.2f "%(K1,K2,K3)#
#first we determine which fluid particle dynamic law applies for the above values of K
#for particle 1,strokes law applies
#for particle 2,strokes law applies
#for particle 3,intermediate law applies
#terminal settling velocity for each particle
v1=(D1/(25400*12))**2*g*rho_p/(18*meu)
print "\n terminal settling velocity for particle 1 v1=%0.2f ft/s"%(v1)#
v2=(D2/(25400*12))**2*g*rho_p/(18*meu)
print "\n terminal settling velocity v2=%0.2f ft/s"%(v2)#
v3=(D3/(25400*12))**1.14*0.153*g**0.71*rho_p**0.71/(rho_a**0.29*meu**0.43)
print "\n terminal settling velocity v3=%0.2f ft/s "%(v3)#
#calculation of how far x,the fly ash particles will fall in 30 seconds
t=30#time,sec
x2=v2*t#distance travel by 2 particle
x3=v3*t#distance travel by 3 particle
print "\n distance by 2 particle x2=%0.2f ft\n distance by 3 particle x3=%0.2f ft"%(x2,x3)#
#for 1 particle K1 and v1 value are without the CCF.With the correction factor lemda=6.53e-8,gives
lemda=6.53e-8#correction factor
y=-1.10*(D1/(25400*12))/(2*lemda)
A =1.257 + 0.40*exp(y)
C=1 + 2*A*lemda/(D1/(25400*12))#cunningham correction factor(ccf)
#now equation 23.36 can be employed
v1_corrected=v1*C#corrected velocity of 1 particle
x1=v1_corrected*t#distance travel by 1 particle
print "\n distance travel by 1 particle x1=%0.2f ft"%(x1)#

Example 23.4 Page no 336

dimensionless constant K1=0.01
K2=1.44
K3=14.42

terminal settling velocity for particle 1 v1=0.00 ft/s

terminal settling velocity v2=0.31 ft/s

terminal settling velocity v3=8.90 ft/s

distance by 2 particle x2=9.44 ft
distance by 3 particle x3=266.88 ft

distance travel by 1 particle x1=0.00 ft


## Example 23.5 Page no 338¶

In [13]:
from math import sqrt
from __future__ import division
print "\n Example 23.5 Page no 338\n\n"
#refer to example 23.5
#we have to calculate size of a flyash particle that will settle with a velocity of 1.384 ft/s
SG=2.31#specific gravity of fly ash
rho_w=62.4#density of water
rho_p=SG*rho_w#density of particles
#properties of air
R=0.7302#gas constant
T=698#temperature,R
P=1#pressure ,atm
Mw=29#mol. wt of air
rho_a=P*Mw/(R*T)#density of air,lb/ft**3
meu=1.41e-5#viscosity of air,lb/ft.s
g=32.174#grav. acc
v=1.384#velocity at which particle settle down,ft/s
W= v**3*rho_a**2/(g*rho_p*meu)#dimensionless constant
print "\n dimensionless constant W=%0.2f "%(W)#
#since W < 0.2222 stokes' law applies
D_p=sqrt(18*meu*v/(g*rho_p))#diameter of particle
print "\n diameter of particle D_p=%0.2e ft"%(D_p)#


 Example 23.5 Page no 338

dimensionless constant W=0.13

diameter of particle D_p=2.75e-04 ft


## Example 23.7 Page no 340¶

In [14]:
from math import exp
from __future__ import division
print "\n Example 23.7 Page no 340\n\n"
# In a plant manufacturing ivory soap detergent explodes one windy day
#we have to calculate the distance from the plant where the soap particles will start to deposit and where they will cease to deposit
#the smallest particle wll travel the greatest distance while the largest will travel the least distance
#for the minimumdistance ,we use largest particle
D_l=3.28e-3#largest diameter,ft
g=32.174#grav. acc.
SG=0.8#specific gravity of soap particle
rho_w=62.4
rho_p=SG*rho_w#density of particle
rho_a=0.0752#density of given atmosphere,lb/ft**3
meu=1.18e-5#viscosity
K_l = D_l*(g*(rho_p-rho_a)*rho_p/(meu**2))**(1/3)#dimensionless constant
print "\n dimensionless constant K_l=%0.2e "%(K_l)#
#value of K indicates the intermediate range applies
#the settling velocity is given by
v_l=0.153*g**0.71*D_l**1.14*rho_p**0.71/(meu**0.43*rho_a**0.29)
print "\n settling velocity v_l=%0.2f ft/s"%(v_l)#
H=400#vertical height blowen by particle,ft
t_l=H/v_l#descent time
v_w=20#wind velocity in miles/h
L=t_l*v_w*(5280/3600)#horizontal distance travelled by particles
print "\n descent time t_l=%0.2f second\n horizontal distance L=%0.2f ft"%(t_l,L)#
#for the minimum distance we use smallest particle
D_s=6.89e-6#diameter of smallest particle,ft
K_s=D_s*(g*(rho_p-rho_a)*rho_a/(meu**2))**(1/3)
print "\n dimensionless constant K_s=%0.2e "%(K_s)#
#velocity is in the stokes regime and is given by
v_s=g*D_s**2*rho_p/(18*meu)
print "\n settling velocity v_s=%0.2e ft/s"%(v_s)#
t_s=H/v_s#descent time
L_s=t_s*v_w*(5280/3600)#horizontal distance travelled
print "\n descent time t_s=%0.2f s\nhorizontal distance travelled by smallest particle L_s=%0.2e ft"%(t_s,L_s)#
m=100*2000#mass of particles
V_act=m/rho_p#actual volume of particles
e=0.5#void fraction
V_b=V_act/e#bulk volume
print "\n actual volume V_act=%0.2f ft**3\nbulk volume V_b=%0.2f "%(V_act,V_b)#
L_d=L_s-L#length of drop area
print "\n L_d=%0.2e "%(L_d)#
W=100#width ,ft
A_d=L_d*W#deposition area
H_d=V_b/A_d#deposition height
print "\n deposition height H_d=%0.2e ft"%(H_d)#
#deposition height can be ,at bestt, described asa sprinkling

 Example 23.7 Page no 340

dimensionless constant K_l=2.73e+02

settling velocity v_l=11.85 ft/s

descent time t_l=33.74 second
horizontal distance L=989.79 ft

dimensionless constant K_s=6.57e-02

settling velocity v_s=3.59e-04 ft/s

descent time t_s=1114285.36 s
horizontal distance travelled by smallest particle L_s=3.27e+07 ft

actual volume V_act=4006.41 ft**3
bulk volume V_b=8012.82

L_d=3.27e+07

deposition height H_d=2.45e-06 ft


## Example 23.8 Page no 342¶

In [15]:
from math import pi,sqrt
from __future__ import division
print "Example 23.8 Page no 342\n\n"
#a small sphere is observed to fall through caster oil
v_t=0.042#terminal velocity of particle
meu_f=0.9#viscosity of oil
rho_f=970#density of oil
g=9.807#grav. acc.
D_p=0.006#diameter of particle
rho_p=(18*meu_f*v_t)/(g*D_p**2)  + rho_f
print "\n density of particle rho_p=%0.2f kg/m**3"%(rho_p)#
neu_f=9.28e-4#dynamic viscosity of fluid
R_e=D_p*v_t/neu_f#reynolds no
print "\n reynolds no R_e=%0.2f "%(R_e)#
#since R_e < 0.3
#calculation of the settling criterion factor ,K
K=D_p*(g*rho_f*(rho_p-rho_f)/(meu_f**2))**(1/3)#the settling criterion factor
print "\n K=%0.2f "%(K)#
#since K <3.3, stokes law applies
#the drag coeff. C_d
C_d=24/R_e
print "\n drag coeff C_d=%0.2f "%(C_d)#
F_d=3*pi*meu_f*D_p*v_t#drag force
print "\n drag force F_d=%0.2f N"%(F_d)#
F_b=(pi/6)*D_p**3*rho_f*g#buoyancy force
print "\n buoyancy force F_b=%0.2f N"%(F_b)#
#Consider the case when same sphere is dropped in water
rho_w=1000#density of water,kg/m**3
meu_w=0.001#viscosity of water,kg/m.s
#the particle will move faster because of the lower viscosity of water ,stokes law will almost definietly not apply
K_w=D_p*(g*rho_w*(rho_p-rho_w)/(meu_w**2))**(1/3)#the settling criterion factor
print "\n k_w settling factor =%0.2f "%(K_w)#
#since K_w = 158 > 43.6,the flow is in the Newton's law regime
#employ eq. 23.31 but include the (buoyant) density ratio factor
v_t_w=1.75*sqrt((rho_p-rho_w)/(rho_w)*g*D_p)#terminal velocity
print "\n terminal velocity in water v_t_w=%0.2f m/s"%(v_t_w)#

Example 23.8 Page no 342

density of particle rho_p=2897.19 kg/m**3

reynolds no R_e=0.27

K=1.70

drag coeff C_d=88.38

drag force F_d=0.00 N

buoyancy force F_b=0.00 N

k_w settling factor =158.99

terminal velocity in water v_t_w=0.58 m/s


## Example 23.9 Page no 344¶

In [16]:
from __future__ import division
print "Example 23.9 Page no 344\n\n"
#the bottom of a ship,moving in water
rho=1000#density of water
v=12#velocity of boat,m/s
L=20#length,m
W=5#width ,m
meu=1e-3#viscosity
R_e=rho*v*L/meu#reynolds no
print "Reynolds no R_e=%0.2e "%(R_e)#
#from reynolds no  flow is turbulent
C_d=0.031/(R_e**(1/7))#coeff. discharge\
print "\ncoeff. discharge C_d=%0.2e "%(C_d)#
#calculation of the drag on area LW
F_d=(1/2)*C_d*rho*v**2*L*W#drag force
print "\n drag force F_d=%0.2e N"%(F_d)#

Example 23.9 Page no 344

Reynolds no R_e=2.40e+08

coeff. discharge C_d=1.97e-03

drag force F_d=1.42e+04 N