Chapter 13 : Centrifugal Separation Operations¶

example 13.1 page no : 259¶

In :
import math

# Initialization of Variable
rho = 998.
g = 9.81
pi = 3.1428
omega = 2*pi*1055./60           #angular rotation
ld = 1.55/100.                   #liq. depth
l = 10.25/100.

#calculation
#part1
a = r*omega**2/g
print "ratio of cetrifugal force & gravitational force is: %.4f"%a

#part2
V = pi*(r**2-ri**2)*l
sigma = (omega**2*V)/(g*math.log(r/ri))
print "equivalent to gravity settling tank of crossectional area of in (m**2): %.4f"%sigma
ratio of cetrifugal force & gravitational force is: 31.7517
equivalent to gravity settling tank of crossectional area of in (m**2): 0.2358

example 13.2 page no : 261¶

In :
import math

# Initialization of Variable
sigma = 55.*10**6           #maximum stress
d = 35.2/100
rhos = 8890.                #density of bronze
rho = 1105.                 #density of solution
t = 80./1000                #thickness
tau = 4.325/1000.
pi = 3.1428

#calculation
#part1
def fround(x,n):
# fround(x,n)
# Round the floating point numbers x to n decimal places
# x may be a vector or matrix# n is the integer number of places to round to
y = round(x*10**n)/10**n
return y

omega = math.sqrt((sigma*tau*2/d)/(.5*rho*(d**2/4-ri**2)+rhos*tau*d/2))
N = 60*omega/2/pi
print "The maximum safe speed allowed in rpm: %.4f"%N

#part2
P = .5*rho*(d**2./4-ri**2)*omega**2
P = fround(P/10**4,1)*10.**4
#print (P,"the power in N/m**2:")
print 'the power in N/m**2: %3.2e'%( P)
a = rho*omega**2*d/2
a = fround(a/10**6,1)*10**6
The maximum safe speed allowed in rpm: 2560.1495
the power in N/m**2: 8.65e+05

example 13.3 page no : 262¶

In :
import math

# Initialization of Variable
rhos = 1425.        #density of organic pigment
rho = 998.          #density of water
pi = 3.1428
omega = 360*2*pi/60.
mu = 1.25/1000.
t = 360.
r = 0.165+0.01
ro = 0.165

#calculation
d = math.sqrt(18*mu*math.log(r/ro)/t/(rhos-rho)/omega**2)
print 'the minimum diameter in organic pigment in m: %3.1e'%( d)
the minimum diameter in organic pigment in m: 2.5e-06

example 13.4 page no : 263¶

In :
import math

# Initialization of Variable
rhos = 1455.            #density of crystals
rho = 998.              #density of wliquid
g = 9.81
pi = 3.1428
mu = 1.013/1000
omega = 2*pi*60000/60.
l = 0.5
d = 2*10.**-6.            #dia of particles
t = 38.5/1000           #thickness of liquid

#calculation
ri = r-t
V = pi*l*(r**2-ri**2)
Q = d**2*(rhos-rho)/18/mu*omega**2*V/math.log(r/ri)
print "the maximum volumetric flow rate in (m**3/s): %.4f"%Q
the maximum volumetric flow rate in (m**3/s): 0.0104

example 13.5 pageno : 265¶

In :
import math

# Initialization of Variable
rhoc = 867.         #density of cream
rhom = 1034.        #density of skimmem milk
rm = 78.2/1000.     #radius of skimmed milk
rc = 65.5/1000.     #radius of cream

#calculation
r = math.sqrt((rhom*rm**2-rhoc*rc**2)/(rhom-rhoc))

# results
print "distance of xis of rotation of cream milk interface in (m): %.4f"%r
distance of xis of rotation of cream milk interface in (m): 0.1249

example 13.6 page no : 266¶

In :
import math

# Initialization of Variable
rho = 1.210             #density of air
mu = 1.78/10**5
g = 9.81
rhos = 2655.            #density of ore
pi = 3.1428
d = 0.095
dp = 2.*10**-6           #particle diameter
dt = 0.333              #dia of cyclone separator
h = 1.28

#calculation
U = dp**2*g*(rhos-rho)/18/mu
Q = 0.2*(pi*d**2/4)**2*d*g/U/pi/h/dt
print "volumetric flow rate in(m**3/s): %.4f"%Q
volumetric flow rate in(m**3/s): 0.0215

example 13.7 page no : 268¶

In :
import math
from numpy import linspace
# Initialization of Variable
b = 4.46*10**4
c = 1.98*10**4
s = 0.
def intregrate():
s = 0.
for i in range(10889):
d = linspace(0,10000,10889)
y = (1-math.exp(-b*d[i])*c*(1-math.exp(-c*d[i])))*0.69
s = s+y
a = y
return a

a = intregrate()

print "overall efficiency of cyclone separator in %",a*100
overall efficiency of cyclone separator in % 69.0
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