Given:
Wet Bulk Density = 2.24 g/cm3
Particle Density = 2.65 g/cm3
Fluid Density (FD) = 1.0 g/cm3
What is:
Porosity = ?
BD = (1- φ) PD + φ (FD)
SW − DW
φ=
VT
DW
= PD(1 − φ)
VT
φ = 1−
DW
PD * VT
BD = (1- φ) PD + φ (FD)
BD = PD - φ PD + φ FD
BD = PD + φ (FD - PD)
BD – PD = φ (FD - PD)
BD – PD = φ
(FD - PD)
We have wet bulk density so 2.24 – 2.65 = -0.41 ~ 0.25 = φ
use fluid density for water
1 - 2.65
-1.65
Knowing:
Wet Bulk Density = 2.24 g/cm3
Particle Density = 2.65 g/cm3
Fluid Density (FD) = 1.0 g/cm3
Porosity = 0.25
And If:
Total Volume = 25cm3
What is:
Saturated Weight = ?
Dry Weight = ?
BD = (1- φ) PD + φ (FD)
SW − DW
φ=
VT
DW
= PD(1 − φ)
VT
φ = 1−
DW
PD * VT
Saturated “Weight” = Vol * WBD = 25cm3 * 2.24 g = 56g (MASS)
cm3
Dry Weight = SatWgt – WaterWgt = SatWgt – φ TotVol * FD
= 56g – (0.25*25cm3) * 1 g = 56g – 6.25g = 49.75g ~ 50g (MASS)
cm3
1
CONSIDERING SPHERES
1 volume of 2x2 units will hold
1 sphere
r=1
8 spheres
r=1/2
2
64 spheres
r=1/4
what space will they leave unoccupied?
2
2
⎛ 4
⎞
2 3 − ⎜ m πr 3 ⎟
V
⎝ 3
⎠
φ= V =
3
VT
2
m is the # of spheres
⎛ 4
⎞
2 3 − ⎜ m πr 3 ⎟
V
⎝ 3
⎠
φ= V =
3
VT
2
m is # spheres
m
1
8
64
2
2
r is their radius
r
r3
1
1
1/2
1/8
1/4
1/64
⎛4 ⎞
2 3 − ⎜ π ⎟(1)
⎝ 3 ⎠ = 0.476
φ=
23
2
mr3
1
1
1
A cons tan t!
HENCE MAX φ FOR UNIFORM, CUBIC PACKED SPHERES IS ~ 48%
MIN φ WOULD RESULT FROM RHOMBEHEDRAL PACKING ~ 26%
IN GENERAL, IRREGULAR PACKING & ANGULAR PARTICLES YIELD HIGHER φ
2
Q = 1 liter = 1 liter 1000cm3 = 1000cm3
day
day 1 liter
day
1000cm3
Q
cm
day
=
= 222
Darcy Velocity =
Area 6cm0.75cm
dayy
DarcyVelocity
=
Average Linear Velocity =
EffectivePorosity
cm
cm
cm
day
= 716
~ 700
0.31
day
day
222
63cm
TravelTime =
constant
head cm ~ 0.09day ~ 130min ~ 2hours
700
dayy
0.4 m
0.21 m
effective porosity is 31%
FINE SAND
0.63 m
6 cm
0.75 cm
544ft
6fft
600ft
50ft
pressure on your head (6ft tall) at the bottom of a well
surface at sea level
bottom at 600 ft
water level 50 ft below the surface
Gage Pressure
P = γh = 62.4
lb
lb
544 ft = 33945.6 2
3
ft
ft
lb 1 ft 2
P = 33945.6 2
ft 144 in 2
lb
= 235.7 2 = 235.7 psi
in
Absolute Pressure ~ 235.7psi + 14.7psi
~ 250.4psi
3
Estimate the height of capillary rise for water in sand.
Grain size distributions in Fetter (pg74-75)
Smallest ~ 0.08 mm 10% ~ 0.17mm = 0.017cm
What if the soil is a silty sand?
Grain size distributions in Fetter (pg74-75)
Smallest ~ 0.002 mm 10% ~ 0.017mm = 0.0017cm
2σsilty
2σ
ilt sand
d
hc =
What if the fluid is gasoline?
Substance Surface Tension (dyne/cm)
γh
Water 72.8 dyne/cm
sand
d
Gasoline ~33 dyne/cm (note < tension for water)
specific weight? check the web (~0.68 density of water)
Which will dominate, surface tension decrease or specific weight decrease?
Use d10 to reflect the typical pore throat size
hc =
2σ(cosλ )
(for r in cm)
γr
For fresh water:
72.8dyne 0 .00102 g 981 cm 72.8g
=
σ=
cm
dyne
sec 2
s2
γ = ρg =
1g 980cm
980g
=
cm 3 s 2
cm 2 s 2
For sand
hc-water = (2*72.8)1/(980*0.017/2) = 17.5cm
hc-gas = (2*33)1/(0.68*980*0.017/2) = 11.6cm
For silty-sand
hc-water = (2*72.8)1/(980*0.0017/2) = 174cm
hc-gas = (2*33)1/(0.68*980*0.0017/2) = 116cm
4