# Chapter 9 - CENTRIFUGAL SEPARATIONS¶

## Page 482 Example 9.1¶

In :
from __future__ import division
from math import sqrt
d_particle = 5#          #particle size is in um
p = 1000#               #density of water in kg/m**3
ps = 2800#              #density of solids in kg/m**3
viscosity = 10**(-3)#    #viscosity is in Ns/m**2
uo = ((d_particle*10**(-6))**2)*(ps-p)*9.81/(18*viscosity)#
print"\n Terminal falling velocity of particles of diameter = %.2f m/sec"%(uo*10**5)#
Q = 0.25#              #volumetric flow rate is in m**3/sec
print"\n E = %.2f*10**(4) m**2"%((Q/uo)*10**(-4))#

print"\n For coal-in-oil mixture"
uo1 = 0.04/(Q/uo)#
print"\n uo = %.2f*10**-6 m/sec"%(uo1*10**6)#

d = sqrt((18*0.01*uo1)/((1300-850)*9.81))#
print"\n d = %d um"%((d/3)*10**6)#

 Terminal falling velocity of particles of diameter = 2.45 m/sec

E = 1.02*10**(4) m**2

For coal-in-oil mixture

uo = 3.92*10**-6 m/sec

d = 4 um


## Page 488 Example 9.3¶

In :
from sympy import symbols, solve
from __future__ import division
from math import sqrt,pi
from scipy import log
#In the filter press
# V**2 + 2(AL/v)V = 2(-deltaP)A**2*t/(ruv)

l = 0.025#         #l is in meters
L = 0.003#         #L is in meters
deltaP = 350#      #it is in N/m**2
t = 3600#          #t is in secs

# x = v/ru
x = symbols('x')#
x1 = solve(0.025**2 + 2*0.003*0.025 - 2*3.5*10**(5)*3600*x)
print"\n the value of ru/v = %.2f*10**12"%((1/x1)*10**(-12))#

#In the centrifuge
R = 0.15#             #R is in meters
H = 0.20#             #H is in meters
V = 0.00225#          #V is in m**3
r = symbols('r')
r1 = solve(pi*(R**2 - r**2)*H-V)
print"\n Value of ro = %.3f   mm"%(r1/2)#

#(R**2 -r**2)(1+2L/R)+2r**2ln(r/R) = 2vtpw**2/ru(R**2-ro**2)
t = symbols('t')#
t1 = solve((R**2 - r1**2)*(1+2*(L/R))+2*(r1**2)*log(r1/R)-2*t*1000*408.4**(2)/(3.25*10**12)*(R**2-(r1/2)**2))#
print" \n time required = %.3f   secs"%abs(t1)#

 the value of ru/v = 3.25*10**12

Value of ro = 0.069   mm