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)#
```

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))
```

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"
```

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)#
```

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)))#
```