Drainage area = 400 mi2
very rough average annual baseflow? ~ 40 cfs?
average annual ground water discharge ~ 1.26x109 ft3
Drainage area = 400 mi2
What is the baseflow discharge in 1987?
1
Drainage area = 400 mi2
1987
Qo = 150 cfs
tlog = time for Q to drop 1 log cycle = 0.6yr
t = time for recession = 0.7yr
1 yr
1 log
cycle Q
2
TOTAL GW THAT COULD DISCHARGE AT START OF RECESSION, Vtp:
Vtp is evaluated ∫
∞
0
Vtp =
Qotlog
2.3
TOTAL GW THAT COULD DISCHARGE AT END OF RECESSION, VR:
VR is evaluated
∫
∞
t @ end
VR =
1987, 1988, 1989, 1993
Qo ~ 150 cfs
tlog = time for Q to drop 1 log cycle ~ 0.6 yr
t = time for recssion
recession~~0.7
0.7yryr
Vtp ~ 1.2x109 ft3
VR ~ 8.4x107 ft3
Vdischarged ~ 1.2x109 ft3 ~26,000 AF
Given:
Wet Bulk Density = 2.24 g/cm3
Particle Density = 2.65 g/cm3
Fluid Density (FD) = 1.0 g/cm3
What is:
Porosity = ?
Qotlog
⎛ tt
2.3⎜10 log
⎜
⎝
⎞
⎟
⎟
⎠
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
3
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)
DW
= PD(1 − φ)
VT
SW − DW
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
⎛ 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
4
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
5