Chapter 7 - LIQUID FILTRATION

Page 394 Example 7.1

In [5]:
from __future__ import division
from numpy import mat
from sympy import symbols,solve
#In the leaf filter filtration is at const pressure from the start
#V**2 + 2ALV/v = 2(-deltaP)A**2t/(ruv)

#In the filter press,a volume V1 of filtrate is obtained under const rate conditions in time t1,and filtration is then carried out at constant pressure.
#V1**2 + 2ALV1/v = 2(-deltaP)A**2t1/(ruv) 
#and (V**2 − V1**2 ) + 2AL/υ(V − V1) = 2(−P)A**2/rμυ(t − t1)

#for the leaf filter
t2 = 300#          #t2 is in secs
V2 = 2.5*10**(-4)#   #V2 is in m**3
t3 = 600#           #t3 is in secs
V3 = 4*10**(-4)#     #V3 is in m**3
A = 0.05#           #A is in m**2
deltaP = -7.13*10**(4)# #it is in N/m**2
#putting these values in above eq

a = [[2*7.13*10**(4)*0.05**(2)*300,-2*0.05*2.5*10**(-4)],[2*7.13*10**(4)*0.05**(2)*600,-2*0.05*4*10**(-4)]]
b = [[(2.5*10**(-4))**2],[(4*10**(-4))**2]]
a=mat(a)
b=mat(b)
x=(a**-1*b)
y = [1/x[0],x[1]]
print"\n L/υ=%f*10**(-3)   and rμυ = %.3f  *10**(11)"%(y[1]*10**3,y[0]*10**(-11))#

#for the filter press
V1 = symbols('V1')
s = solve(V1**2 + (2.16*y[1]*V1)-(4*10**(5)*2.16**2)/y[0]*180)
print"\n the value of V1 = %.3f  m**3"%(s[1])

#For a constant pressure period (t - t1)=900secs
#Calculting the total volume of filtrate
V = symbols('V')
d = solve((V**2-3.33*10**(-4))+(1.512*10**(-2)*(V-1.825*10**(-2))-5.235*10**(-6)*900))#
print"\n The value of V = %.3f m**3"%(d[1])

f = (4*10**(5)*(2.16)**2)/(7.13*10**(11)*(6.15*10**(-2) + 2.16*3.5*10**(-3)))#
print"\n The final rate of filtration is %.2f*10**(-5) m**3/sec"%(f*10**(5))#

# Assuming  viscosity of the filtrate is the same as that of the wash-water
rw_400 = (0.25)*f#
print"\n Rate of washing at 400 kN/m2 = %.1f*10**(-6) m**3/sec"%(rw_400*10**(6))#

rw_275 = rw_400*(275/400)#
print"\n Rate of washing at 275 kN/m**2 = %.1f*10**(-6) m**3/sec"%(rw_275*10**6)#
print"\n Thus the amount of wash-water passing in 600s = %.3f m**3"%(600*rw_275)#

 L/υ=3.500000*10**(-3)   and rμυ = 7.130  *10**(11)

 the value of V1 = 0.018  m**3

 The value of V = 0.066 m**3

 The final rate of filtration is 3.79*10**(-5) m**3/sec

 Rate of washing at 400 kN/m2 = 9.5*10**(-6) m**3/sec

 Rate of washing at 275 kN/m**2 = 6.5*10**(-6) m**3/sec

 Thus the amount of wash-water passing in 600s = 0.004 m**3

Page 398 Example 7.2

In [7]:
from sympy import symbols,solve
#The slurry contains 100kg whiting/m**3 of water
print"\n Volume of 100 kg whiting = %.3f   m**3"%(100/3000)#
print"\n Volume of cake = %.3f   m**3"%(0.0333/0.6)#
print"\n Volume of liquid in cake = %.3f   m**3"%(0.05556*0.4)#
print"\n Volume of filtrate = %.3f m**3"%((1-0.0222))#
print"\n volume of cake/volume of filtrate v = %.3f  "%(0.0556/0.978)#
A = 10**(-4)#                 #area in sq meters
deltaP = -1.65*10**(5)#       #P is in pascals
l = 0.01#                    #length is in meters
vol_flow_rate = 2*10**(-8)#           #Volume flow rate is in m**3/sec
u = 10**(-3)#                 #vicosity is in Ns/m**2

r = symbols('r')
r1 = solve((10**4)*(2*10**(-8)*r)-1.65*10**(5)/(10**(-5)))[0]
print"\n r = %.2f*10**(13)/m**2"%(r1*10**(-13))#

def optimum():
    Lopt = 1.161*10**(-3)*(900)**(0.5)#  #t = 900 secs
    return Lopt
print"\n optimum frame thickness = %.1f mm"%(2*optimum()*1000)#

#total cycle time = 1.015L**2 + 900
#rate of cake production R = L/(1.015L**2 + 900)
 
L = symbols('L')
L1 = solve(1.025*10**(6)*L**2 + 900 - 2.050*10**(6)*L**2)
print"\n Frame thickness = %.2f mm"%(2*L1[0]*10**3)#

 Volume of 100 kg whiting = 0.033   m**3

 Volume of cake = 0.056   m**3

 Volume of liquid in cake = 0.022   m**3

 Volume of filtrate = 0.978 m**3

 volume of cake/volume of filtrate v = 0.057  

 r = 8.25*10**(13)/m**2

 optimum frame thickness = 69.7 mm

 Frame thickness = -59.26 mm

Page 404 Example 7.3

In [11]:
from sympy import symbols,solve
from numpy import mat
from sympy.mpmath import quad
V = 0.094#         #volume in m**3
deltaP = -3530#    #P is in kN/m**2

#At t = 1105 secs
V1 = 0.166#       #V is in m**3
deltaP1 = -5890#   #P is in kN/m**2

a = [[2.21*10**(6), -0.094],[6.51*10**(6),-0.166]]
b = [[0.0088],[0.0276]]
a=mat(a)
b=mat(b)
x = (a**-1)*b
y = [x[1],x[0]]
print"\n LA/v =%f       A**2/rμv = %.3f  *10**(-7)"%(y[0],y[1]*10**7)#
print"\n For the full size plant:"
print"\n LA/v = %.3f         A**2/rμv=%f*10**(-7)"%(10*y[0],y[1]*10**8)#


#Solving LHS of the integral
LHS = quad(lambda b:b+0.154+2.31,[0,1])#
#Equating LHS = RHS
t = LHS/(3.46*10**(-3))#
print"\n t = %d secs"%(t)#
print"\n deltaP = %d kN/m**2"%((1+0.154)/(4.64*10**(-7)*857))#

 LA/v =0.015130       A**2/rμv = 0.046  *10**(-7)

 For the full size plant:

 LA/v = 0.151         A**2/rμv=0.462543*10**(-7)

 t = 856 secs

 deltaP = 2902 kN/m**2

Page 423 Example 7.4

In [13]:
from sympy import symbols,solve
from numpy import mat,sqrt,pi


a = [[2*84300*0.02**(2)*60,-2*0.02*0.0003],[2*84300*0.02**(2)*120,-2*0.02*0.00044]]
b = [[0.0003**2],[0.00044**2]]
a=mat(a)
b=mat(b)
x = (a**-1)*b
y = [x[1],1/x[0]]#
print"\n L/v = %.3f           ruv = %.3f  *10**(10)"%(y[0],y[1]*10**(-10))#
print"\n Area of filtering surface = %.3f   m**2"%(4*(pi))#
print"\n Bulk volume of cake deposited =%.3f m**3/revolution"%(4*(pi)*0.005)#

V = sqrt(1*10**(-6)*143**2)#
print"\n V = %.3f m**3"%(V)#

t =symbols('t')
t1 = solve(0.141**2 +2*2.19*10**(-3)*0.141-2*84300*(4*(pi))**(2)*t/(3.48*10**10))[0]
print"\n t = %.3f   secs"%(t1)#
print"\n time for 1 revolution =%.1f secs"%(t1/0.4)#
print"\n speed = %.3fHz"%(0.4/t1)#
print"\n rate of filtrate production w = %.2f kg/sec"%(143/67.3)
print"\n mass of slurry S =%.1f kg/sec"%(1.66*2.11)#

 L/v = 0.002           ruv = 3.503  *10**(10)

 Area of filtering surface = 12.566   m**2

 Bulk volume of cake deposited =0.063 m**3/revolution

 V = 0.143 m**3

 t = 26.793   secs

 time for 1 revolution =67.0 secs

 speed = 0.015Hz

 rate of filtrate production w = 2.12 kg/sec

 mass of slurry S =3.5 kg/sec

Page 424 Example 7.5

In [38]:
from sympy import symbols,solve
from numpy import mat,pi
from scipy import sqrt
V1 = 0.00025#               #V is in m**3
t = 300#                    #t is in secs
a = mat([[7.14*10**(-6),2.86*10**(-4)],[11.42*10**(-6),2.86*10**(-4)]])
b = mat([[1.2*10**(6)],[1*10**(6)]])
x = (a**-1)*b#

#for the plate and frame filter
B1 = x[0]/(2*2.2**2*413*10**3)#
B2 = x[1]/(2.2*413*1000)#

print"\nrμv = %d\n"%(x[0])#
print"\n rμl = %d"%(x[1])#
print"\n B1= %.3f      B2= %.3f  "%(B1,B2)#
print"\n the filtration time for maximum throughput is:"

t1 = 21.6*10**3#
t0= t1 +B2*sqrt(t1/B1)[0]

print"\n t = %.3f   secs"%abs(t0)#
V = sqrt(t1/B1)[0]
print"\n V= %.3f   m**3"%abs(V)#
fr=((V/(t1+t0))/10**-6)[0,0]
print"\nMean rate of filtration is: %.2f *10**-6 m**3/s"%abs(fr)

rμv = -46728971962


 rμl = 5362394614

 B1= -11688.555      B2= 5901.821  

 the filtration time for maximum throughput is:

 t = 23041.859   secs

 V= 1.359   m**3

Mean rate of filtration is: 30.94 *10**-6 m**3/s

Page 426 Example 7.6

In [48]:
from math import pi, sqrt,ceil
from sympy import symbols,solve

A=0.6*0.6*pi#       #in m**2
rate=1.25*10**-4#     # in m**3/s

v_w=0.2/(3*10**3)#
v_f=10**-3-v_w#

v=v_w/v_f#
v_rate=rate*v#
w=360*0.2#

t=v_rate*w/A#
print"\nThickness of cake produced is : %.1f mm"%(t/10**-4)#
K = symbols('K')
K1 = solve((1.25*10**(-4)*360)**2-K*(6.5*10**(4)*(0.36*(pi))**(2)*72))[0]
print"\n The value of K is %.2f*10**(-10)"%(K1*10**(10))#

#Filter press
#Using a filter press with n frames of thickness b m the total time, for one complete cycle of the press =(tf+120n+240),where tf is the time during which filtration is occurring
#overall rate of filtration = Vf/(tf + 120n + 240)

# Vf = 0.3**(2)*n*b/0.143
#tf = 2.064*10**5 b**2

b = symbols('b')
b1 = solve(b**2 - 0.0458*b - 0.001162)#
print"\n The thickness is %.4f m"%(b1[0])#

def number_of_plates():
    
    n = (0.030 + 25.8*b1[1]**2)/(0.629*b1[1]-0.015)#
    return n
n = number_of_plates()#
print"\n The minimum number of plates required is %d"%(ceil(n))#

d = symbols('d')
d1 = solve(ceil(n)*(0.629*d-0.015)-0.030-25.8*d**2)[0]
print"\n The sizes of frames which will give exactly the required rate of filtration when six are used are %.3f   mm"%(d1*10**3)#
print"\n\n\n  Thus any frame thickness between 47 and 99 mm will be satisfactory. In practice,50 mm (2 in) frames would probably be used."

Thickness of cake produced is : 5.7 mm

 The value of K is 3.38*10**(-10)

 The thickness is -0.0182 m

 The minimum number of plates required is 6

 The sizes of frames which will give exactly the required rate of filtration when six are used are 46.716   mm



  Thus any frame thickness between 47 and 99 mm will be satisfactory. In practice,50 mm (2 in) frames would probably be used.

Page 430 Example 7.7

In [50]:
from __future__ import division
t1 = 600/(1/4)*(11/32)#
print"\n t1 = %dsecs"%(t1)#
#Substituting gives 
deltaT = (19200/3)*(784-121+34)/2048#
print"\n t -t1 = %d secs"%(deltaT)# 
Cycle_time = 180+900+t1+deltaT#
print"\n cycle time = %d secs"%(Cycle_time)#
Increase = (0.000214 - 0.000196)/(0.000196)*100#
print"\n Increase in filtration rate is %.1f per cent"%(Increase)#
 

 t1 = 825secs

 t -t1 = 2178 secs

 cycle time = 4083 secs

 Increase in filtration rate is 9.2 per cent