# Chapter 11 : Flow through porous media¶

### Example 11.1 page no : 401¶

In [2]:
#Calculate the volumetric flow rate
import math

# variables
g=32.2                                             #ft/s**2
dz=1.25                                            #ft
Dp=0.03                                            #in (Diameter of particle)
eta=0.33                                           #dimentionless
rho=62.3                                           #lbm/ft**3
mew=1.002                                          #cP
dx=1                                               #ft
#1 cP.ft.s = 6.72*10**(-4)#lbm
#1 ft = 12 in

# calculation
Vs=g*dz*(Dp/12.0)**2*eta**3*rho/(150*mew*(1-eta)**2*dx*6.72*10**(-4))                  #ft/s
d=2                                                 #in (diameter of pipe)
A=(math.pi)/4.0*(d/12.0)**2                         #ft**2
Q=Vs*A                                              #ft**3/s
R=(Dp/12)*Vs*rho/(mew*6.72*10**(-4)*(1-eta))        #dimentionless (Reynold's number)

# results
print "The velocity of water is %f ft/s\n"%Vs
print "The volumetric flow rate is %f ft**3/s\n"%Q
print "Reynolds number is %f"%R

The velocity of water is 0.012422 ft/s

The volumetric flow rate is 0.000271 ft**3/s

Reynolds number is 4.288586


### Example 11.2 page no : 403¶

In [1]:
#Calculate the pressure gradient

# variables
Vs=2.                          #ft/s
dp=0.03                        #in (diameter of particle)
rho=62.3                       #lbm/ft**3
eta=0.33                       #dimentionless

# calculation
#let DP denote pressure gradient
#1 ft = 12 in
#1 lbf.s**2 = 32.2lbm.ft
DP=1.75*rho*Vs**2*(1-eta)/((dp/12)*eta**3*32.2*144)                 #psi/ft

# result
print "The pressure gradient is %d psi/ft"%DP

The pressure gradient is 701 psi/ft


### Example 11.3 page no : 405¶

In [3]:
#Calculate the permeability

# variables
Q=1.                                   #ft**3/min
mew=0.018                              #cP
dx=0.5                                 #in
A=1.                                   #ft**2
dP=2.                                  #lbf/in**2

# calculations
#1 ft = 12 in
#1 min = 60 sec
#1 ft**2.cP = 2.09*10**(-5) lbf.s
#1 darcy = 1.06*10**(-11) ft**2
k=(Q*mew*(dx/12.0)/A/dP)*(1/144.0)*2.09*10**(-5)*(1/60.0)*(1.0/(1.06*10**(-11)))            #darcy

# result
print "The permeability is %.3f darcy"%k

The permeability is 0.086 darcy