# Chapter 3 - MOTION OF PARTICLES IN A FLUID¶

## Page 159 Example 3.1¶

In [2]:
d = 0.00040#   #Diameter of the particle in m
p1 = 820#      #Density of the fluid in kg/m**3
meu = 0.01#    #Viscosity of the fluid in N s/m**2
p2 = 7870#    #Density of steel in kg/m**3
g = 9.81#     #Acceleartion due to gravity in m/sec**2
#Computation of terminal velocity of a spherical particle
def galileo_number():

x = (2*d**3*(p2-p1)*p1*g)/(3*(meu)**2)# #x = (Ro/pu**2)*Re**2
return x

x = galileo_number()#
print"\n The value of (Ro/pu**2)/Re**2 is  %.3f  \n"%(x)#

#From table 3.4 log(x) corresponds to log(Re)=0.222
Re = 1.667#
def teminal_velocity():
u=(Re*meu)/(p1*d)#
return u

u = teminal_velocity()#
print"\n The terminal velocity of the steel ball is %.3f m/sec or %.0f mm/sec"%(u,u*1000)#

 The value of (Ro/pu**2)/Re**2 is  24.197

The terminal velocity of the steel ball is 0.051 m/sec or 51 mm/sec


## Page 159 Example 3.2¶

In [4]:
from __future__ import division
from math import sqrt
u_water = 5*10**(-3)#  #The flow velocity of the water in m/sec
p_galena = 7500#      #The density of galena is in kg/m**3
p_limestone = 2700#   #The density of limestone is in kg/m**3
viscosity = 0.001#    #The viscosity of water in N s/m**2

#calculating maximum value of reynold's number considering 5mm particle size
Re_max = (u_water*1000*0.0001)/(viscosity)#
print"\n The maximum permissible value of Re is %.3f  "%(Re_max)#

#maximum particle size of galena which will be carried away by water
d = sqrt((u_water*(18*viscosity))/((7500-1000)*9.81))#
print"\nmaximum particle size of galena which will be carried away by water is %.1f um"%(d*10**(6))#

#maximum particle size of limestone which will be carried away by water
d1 = sqrt((u_water*(18*viscosity))/((2700-1000)*9.81))#
print"\nmaximum particle size of limestone which will be carried away by water is %.1f um"%(d1*10**(6))#

#From the given data 43% galena and 74% limestone will be removed .
#Given that in the feed there is 20% galena and 80% limestone
#Assuming 100g feed
print"\n\nIn the overflow:"
print"\nAmount of galena is %.3f  g"%((20*0.43))#
print"\nAmount of limestone is %.3f  g"%((80*0.74))#
print"\nconcentration of galena is %.1f per cent"%((20*0.43*100)/(20*0.43+80*0.74))#
print"\n\concentration of galena is %.3f  per cent"%((80*0.74*100)/(20*0.43+80*0.74))#
print"\n\nIn the underflow:"
print"\nconcentration of galena is %.1f percent"%((20*(1-0.43)*100)/(20*(1-0.43)+80*0.26))
print"\nconcentration of limestone is %.1f per cent"%((80*0.26*100)/(20*0.57+80*0.26))

 The maximum permissible value of Re is 0.500

maximum particle size of galena which will be carried away by water is 37.6 um

maximum particle size of limestone which will be carried away by water is 73.5 um

In the overflow:

Amount of galena is 8.600  g

Amount of limestone is 59.200  g

concentration of galena is 12.7 per cent

\concentration of galena is 87.316  per cent

In the underflow:

concentration of galena is 35.4 percent

concentration of limestone is 64.6 per cent


## Page 167 Example 3.3¶

In [15]:
from math import pi,sqrt
from __future__ import division
min_area =6*10**(-6)#    #minimum area of mica plates in m**2
max_area =6*10**(-4)#    #maximum area of mica plates in m**2
p_oil = 820#            #density of the oil in kg/m**3
Viscosity = 0.01#      #Viscosity is in N s/m**2
p_mica = 3000#         #Density of mica in kg/m**3

print"\n    smallest particles \t\t\t largest particles"
print"\nA:   %.3f  m**2 \t\t\t\t %.3f  m**2 "%(min_area,max_area)#
print"\ndp:  %.3f  m \t\t\t\t\t %.3f  m"%(sqrt(4*min_area/(pi)),sqrt(4*max_area/(pi)))#
print"\ndp**3:%f*10**(-8)m**3\t\t\t%.3f  *10**(-5)m**3"%(sqrt(4*min_area/(pi))**(3)*10**(8),sqrt(4*max_area/(pi))**(3)*10**5)#
print"\nv:   %.3f  *10**9m**3 \t\t\t\t%.3f  *10**7m**3"%(0.285*sqrt(4*min_area/(pi))**(3)*10**(9),sqrt(4*max_area/(pi))**(3)*(10**7)*0.0285)#
print"\nk:   %.3f \t\t\t\t\t%.3f  "%(0.285,0.0285)#

x1 = (4*0.285/(pi*0.01**2))*(3000-820)*(820*2.103*10**(-8)*9.81)#
x2 = (4*0.0285/(pi*0.01**2))*(3000-820)*(820*2.103*10**(-5)*9.81)#
print"\n(Ro/pu**2)Reo**2 is %d for the smallest particles and %d for the largest particles"%(x1,x2)#
#From table 3.4 Re for smallest particle is 34.9 and that for the largest particle is 361
u1 = 34.9*0.01/(820*2.76*10**(-3))#
print"\nTerminal velocity for the smallest particle is %.3f m/sec"%(u1)#
u2 = 361*0.01/(820*2.76*10**(-2))#
print"\nTerminal velocity for the largest particle is %.3f m/sec"%(u2)#
print"\n\n Thus it is seen that all the particles settle at approximately the same velocity"

    smallest particles 			 largest particles

A:   0.000  m**2 				 0.001  m**2

dp:  0.003  m 					 0.028  m

dp**3:2.111505*10**(-8)m**3			2.112  *10**(-5)m**3

v:   6.018  *10**9m**3 				6.018  *10**7m**3

k:   0.285 					0.029

(Ro/pu**2)Reo**2 is 1338 for the smallest particles and 133823 for the largest particles

Terminal velocity for the smallest particle is 0.154 m/sec

Terminal velocity for the largest particle is 0.160 m/sec

Thus it is seen that all the particles settle at approximately the same velocity


## Page 181 Example 3.4¶

In [16]:
from __future__ import division
from math import log,cos,sin
v_particle = 6#  #velocity of the particle in m/sec
v_water = 1.2#   #veloicity of the water in m/sec
v_rel = v_particle - v_water##relative velocity of particles relative to the fluid in m/sec
Re1 = 6*10**(-3)*v_rel*1000/(1*10**(-3))##Re1 is the reynold's no.
print"\nReynold no. is %d"%(Re1)#

#When the particle has been retarded to a velocity such that Re=500
ydot = (v_rel*500)/Re1#
print"\nParticle velocity is %.3f m/sec\n"%(ydot)#
c = 0.33/(6*10**(-3))*(1000/2500)#
f = sqrt((3*6*10**(-3)*(2500-1000)*9.81)/1000)#
def Fa(t):
y = (-1/22)*(log(cos(0.517*22*t) + 4.8/0.517*sin(0.517*22*t)))#
return y

def deriv(t) :
yd= -0.083+(0.517*(9.28*cos(11.37*t) - sin(11.37*t))/(cos(11.37*t) + 9.28*sin(11.37*t)))#
return yd
def double_deriv(t):
ydd = -0.517*(11.37)**2*(9.28*cos(11.37*t) - sin(11.37*t))/(cos(11.37*t) + 9.28*sin(11.37*t))#
return ydd

told = 0#
while 1:
tnew = told - deriv(told)/double_deriv(told)#
if (tnew == told) :
y = Fa(told)#
d = y#
print"\nThe distance moved with speed less than 0.083m/sec is %.3fm"%(d)#
t=told#
print"\n The time taken by particle to move this distance is %.3fsec"%(t)#
break#

told = tnew#

print"\nThe distance moved by the particle relative to the walls of the plant %.3fm"%(1.2*t - d)#

Reynold no. is 28800

Particle velocity is 0.083 m/sec

The distance moved with speed less than 0.083m/sec is -0.101m

The time taken by particle to move this distance is 0.115sec

The distance moved by the particle relative to the walls of the plant 0.239m


## Page 183 Example 3.5¶

In [17]:
#rate of dissolution of salt
def dissolution(d):
x = (3*10**(-6))-(2*10**(-4)*3.406*10**(5)*d**2)#
return x

#rate of falling of the particle in stokes law region
def rate_h(d):
y = 3.406*d**(2)/(-3*10**(-6)-68.1*d**2)##y is in m/sec
return y
print"\n By trial and error the solution for d is 0.88 mm"
print"\n The rate of dissolution is %.3f   "%(dissolution(8.8*10*(-4)))#
print"\n The rate of falling of the particle is %.3f   m/sec"%(rate_h(8.8*10**(-4)))#

 By trial and error the solution for d is 0.88 mm

The rate of dissolution is -8440340.480

The rate of falling of the particle is -0.047   m/sec