Chapter 11 : Fluidisation

example 11.1 page no : 216

In [4]:
import math 

# Initialization of Variable
pi = 3.1428
d = 0.3/1000
mu = 2.21/100000
rho = 106.2             #density under operating condition
u = 2.1/100
rhos = 2600.            #density of particles
l = 3.25
g = 9.81
dt = 0.95               #fluidising diameter


#part 1
#calculation
a = u**2./d/g*d*rho*u/mu*(rhos-rho)/rho*l/dt
if a>100 :
    print "Bubbling fluidisation will occur as value is %.4f"%a

#part 2
Q = 2.04/100000
rhos = 2510.
rho = 800.
mu = 2.85/1000
l = 4.01
dt = 0.63
d = 0.1/1000
u = Q*4/pi/dt**2
a = u**2/d/g*d*rho*u/mu*(rhos-rho)/rho*l/dt
if a<100*10**-4:        #compare as value of a is much less than 100
    print "fluidisation occur in smooth mode as value is: %.4e"%a

Bubbling fluidisation will occur as value is 364.4332
fluidisation occur in smooth mode as value is: 1.0898e-07

example 11.2 page no ;218

In [5]:
import math 

# Initialization of Variable
d = 50./1000000
rhos = 1850.            #density of particle
rho = 880.              #density of hydrocarbon
mu = 2.75/1000          #viscosity of hydrocarbon
e = 0.45                #void fraction coeff.
g = 9.81
h = 1.37                #flow depth
c = 5.5/1000            #c = 1/K

#calculation
#part 1
u = c*e**3*d**2*g*(rhos-rho)/mu/(1-e)
print "The superficial linear flow rate in (m/s): %.3e"%u

#part 2
u = d**2*g*(rhos-rho)/18/mu
print "Terminal Settling Velocity in (m/s): %.4f"%u
Re = d*u*rho/mu
if Re<2 :
    print "Stoke law assumption is sustained with this velocity"

#part 3
P = g*(rhos-rho)*h*(1-e)
print "Pressure drop across fluidised bed in (N/m**2):",P

The superficial linear flow rate in (m/s): 7.883e-06
Terminal Settling Velocity in (m/s): 0.0005
Stoke law assumption is sustained with this velocity
Pressure drop across fluidised bed in (N/m**2): 7170.07995

example 11.3 page no : 221

In [6]:
import math 
from numpy import *
# Initialization of Variable
g = 9.81
rhos = 1980.            #density of ore
rho = 1.218             #density of air
e = 0.4
mu = 1.73/10**5
s = 0
wp = array([0, .08, .20, .40, .60, .80, .90, 1.00])        #weight percent
d = true_divide([0.4 ,0.5, 0.56, 0.62, 0.68, 0.76, 0.84, 0.94],1000)
dav = [0,0,0,0,0,0,0]
mf = [0,0,0,0,0,0,0]
a = [0,0,0,0,0,0,0]
#part 1
for i in range(7):
    dav[i] = d[i+1]/2+d[i]/2.                   #average dia
    mf[i] = wp[i+1]-wp[i]                       #mass fraction
    a[i] = mf[i]/dav[i]
    s = s+a[i]

db = 1/s                #d bar

#quadratic coeff. ax**2 +bx +c = 0
c = -(rhos-rho)*g
b = 150.*(1-e)/e**3/db**2*mu
a = 1.75*rho/e**3/db
y = poly1d([a,b,c],False)
U = roots(y)
print  "the linear air flow rate in (m/s): %.4f"%(abs(U[1]))

#part 2
d = 0.4/1000
a = 2*d**3/3/mu**2*rho*(rhos-rho)*g
a = math.log10(a)
print "log10(Re**2/rho/U**2*R) =  %.4f"%a

#using chart
Re = 10**1.853
u = Re*mu/rho/d
print  "speed required for smallest particle in (m/s): %.4f"%u

the linear air flow rate in (m/s): 0.2643
log10(Re**2/rho/U**2*R) =  3.5277
speed required for smallest particle in (m/s): 2.5313

example 11.4 page no : 224

In [7]:
import math 
from numpy import *

# Initialization of Variable
U = 2.032/10**4
pi = 3.1428
rho = 852
g = 9.81
mu = 1.92/1000
mf = 125/3600.          #mass flow rate

#calculation
#part 1
G = U*rho
A = mf/G
d = math.sqrt(4*A/pi)
print "the diameter of vessel will be in(m): %.4f"%d

#part 2
A = 0.201
e = 0.43
ms = 102.           #mass of solids
rhos = 1500.        #density of solid
L = ms/rhos/A
Lmf = L/(1-e)
print  "depth of bed in (m): %.4f"%Lmf 

#part 3
d1 = 0.2/1000
U = 2.*5.5/10**3*e**3*d1**2*(rhos-rho)*g/mu/(1-e)

#now euating for e
#a = e**3/(1-e)
a = U/5.5*10**3/(d1**2*(rhos-rho)*g/mu)
y = poly1d([1,0,a,-a],False)
e2 = roots(y)
L = Lmf*(1-e)/(1-e2[2])
print "depth of fluidised bed under operating condition in (m): %.4f"%L

the diameter of vessel will be in(m): 0.5052
depth of bed in (m): 0.5935
depth of fluidised bed under operating condition in (m): 0.6958

-c:45: ComplexWarning: Casting complex values to real discards the imaginary part

example 11.5 page no : 227

In [8]:
import math 

# Initialization of Variable
g = 9.81
pi = 3.1428
r = 0.51
e = 0.48            #void ratio
rhos = 2280.        #density of glass
rho = 1.204         #density of air
U = 0.015           #velocity of water entering bed
L = 7.32
gam = 1.4           #gamma
neta = 0.7          #efficiency
P4 = 1.013*10**5
P1 = P4
v1 = 1/1.204        #volume 1

#calculation
P3 = P4+g*(rhos-rho)*(1-e)*L
P2 = P3+0.1*85090
v2 = (P1*v1**gam/P2)**(1/gam)       #vlume 2
W = 1/neta*gam/(gam-1)*(P2*v2-P1*v1)    #work done
v3 = P2*v2/P3                       #volume 3
M = U*pi*r**2/v3                    #mass flow rate
P = M*W

# Results
print "The power supplies to the blower in (W): %.4f"%P

The power supplies to the blower in (W): 1948.7509

example 11.6 page no : 230

In [9]:
import math 

# Initialization of Variable
dt = 12.7/1000
d = 1.8/1000
Q = 2.306/10**6
pi = 3.1428

#calculation
#part 1
Sc = 4./dt
S = 6./d
f = (1+0.5*Sc/S)**2
U = Q*4/pi/dt**2        #velocity
Ua = f*U                #actual velocity
print "minimum fluidising velocity found using smaller glass column in (m/s): %.4f"%Ua

#part 2
dt = 1.5
Sc = 4./dt
f = (1+0.5*Sc/S)**2
Ua = f*U                #actual velocity
print "fluidising velocity found using larger glass column in (m/s): %.4f"%Ua

minimum fluidising velocity found using smaller glass column in (m/s): 0.0200
fluidising velocity found using larger glass column in (m/s): 0.0182

example 11.7 page no : 232

In [9]:
import math 

# Initialization of Variable
e = 0.4             #incipent to fluidisation

#calculation
#part 1
print "for Re<500"
print "the ratio of terminal velocity & minimmum fluidising velocity is"

a = 3.1*1.75/e**3

print math.sqrt(a)

#part 2
print "for Re>500"
print "the ratio of terminal velocity & minimmum fluidising velocity is"
a = 150.*(1-e)/18./e**3
print a

for Re<500
the ratio of terminal velocity & minimmum fluidising velocity is
9.20682491416
for Re>500
the ratio of terminal velocity & minimmum fluidising velocity is
78.125
In [ ]: