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
TOTAL GW THAT COULD DISCHARGE AT START OF RECESSION, Vtp:
Vtp is evaluated ∫
∞
0
Vtp =
NEED CONSISTENT UNITS
Qotlog
2.3
TOTAL GW THAT COULD DISCHARGE AT END OF RECESSION, VR:
VR is evaluated
∫
∞
VR =
Qotlog
⎛ tt ⎞
2.3⎜10 log ⎟
⎟
⎜
⎠
⎝
1987, 1988, 1989, 1993 Q at end of recession is
not used in equations
Qo ~ 150 cfs
tlog = time for Q to drop 1 log cycle ~ 0.6 yr SAME EVERY YEAR
t = time for recssion
recession~~0.7
0.7yryr
CONNECT LOW FLOWS
9 3
t @ end
Vtp ~ 1.2x10 ft
VR ~ 8.4x107 ft3
Vdischarged ~ 1.2x109 ft3 ~26,000 AF
FROM Qo TO NEXT UP
TURN
1
What are the elevation and pressure heads?
Total head
25 inches
Total head
17 5 inches
17.5
Pressure head
21.2 inches
Total head
10.8 inches
Pressure head
15.5 inches
Elevation
3.8 inches
Pressure head
15.8 inches
d t
datum
Q = KiA
K=Q
iA
K=
⎡
⎢
⎢
⎢
⎣
Elevation
2 inches
Elevation
-5 inches
⎡
1cm 3
⎢ Volume ml
1ml
⎢
⎢
time sec
⎢⎣
( h2 − h1 )in
distance in
⎤
⎥
⎥
⎥
⎥⎦
⎤
⎥⎡
⎤
2 54cm) 2⎥
⎥ ⎢π ( radius in 2.54cm
in
⎦
⎥⎣
⎦
=
XX cm
sec
Reasonable for sand?
yielded
145ml discharge in 13min48sec
2
VDarcy = Ki = K
VInterstitial =
φ=
dh
dl
Ki K dh
=
φ
φ dl
Kii
VInterstitial
Reasonable for sand?
Q = KiA
K=Q
iA
K=
Q = 1.5cm3
sec
⎡
⎢
⎢
⎢
⎢⎣
1cm 3 ⎤
⎥
1ml ⎥
⎥
60 sec
⎥⎦
90
ml
=
0.2 cm
sec
⎡
⎤
⎢ ( 17.5 – 10.8 )in (2.54cm/in) ⎥ ⎡
⎤
2 54cm ) 2⎥
⎢
⎥ ⎢π ( 1 in 2.54cm
in
⎦
⎥⎣
⎢
(45cm)
⎣
⎦
A = 20cm2
Reasonable for sand?
yieldedi = 0.37
Yes see:
145ml discharge in 13min48sec
http://en.wikipedia.org/wiki/Hydraulic_conductivity#Ranges_of_values_for_natural_materials
3
Calculate Effective Porosity
using
data from
Calculate
Effective
Porosity
Darcy
using Apparatus
data from
Darcy Apparatus
What
Wh t measurements
t will
ill you need?
d?
Travel time of a tracer through the sand = 30inch in 7 min
What equation will you solve?
What equation will you solve?
Q= 1.5cm3
sec
A= 20cm 2
Vdarcy
=
0 075 cm/sec
0.075
=
=
= 0.41
Dye
moves through
the tube
30in 2.54cm/in
0.18 cm/sec
It moves
~30 cm in 7 min
7min 60sec/min
Vtracer
With a different gradient:
Q = KiA
K=Q
iA
Q = 2.3cm3
sec
⎡
⎢
⎢
⎢
⎢⎣
3
140 ml 1cm
1ml
60 sec
⎤
⎥
⎥
⎥
⎥⎦
Recall the sand has
been removed and
replaced since the
prior
i measurements,
t
so we do not expect
the same value of K
=
K=
0.8 cm
sec
⎡
⎤
( 2.5
)in
⎢
⎥⎡
⎤
2 54cm ) 2⎥
⎢
⎥ ⎢π ( 1 in 2.54cm
in
⎦
⎢ (45cm) (1in/2.54cm) ⎥ ⎣
⎣
⎦
A = 20cm2
Reasonable for sand?
yieldedi = 0.14
Yes see:
145ml discharge in 13min48sec
http://en.wikipedia.org/wiki/Hydraulic_conductivity#Ranges_of_values_for_natural_materials
4
Calculate Effective Porosity
using data from
Darcy Apparatus
What
Wh t measurements
t will
ill you need?
d?
Travel time of a tracer through the 30 inch tube = 300 sec
What equation will you solve?
Q= 2.3cm3
sec
A= 20cm 2
Vdarcy
=
=
0.1 cm/sec
=
= 0.39
0 39
Dye moves through the tube
Vtracer
30in 2.54cm/in
0.254 cm/sec
It moves
~30
cm
in
7 min
300 sec
With a our class gradient and Q:
Q = KiA
K=Q
iA
Q = 2cm3
sec
⎡
⎢
⎢
⎢
⎢⎣
3
120 ml 1cm
1ml
60 sec
⎤
⎥
⎥
⎥
⎥⎦
=
K=
0.7 cm
sec
⎡
⎤
( 1
)in
⎢
⎥⎡
⎤
2 54cm ) 2⎥
⎢
⎥ ⎢π ( 1 in 2.54cm
in
⎦
⎢ (15cm) (1in/2.54cm) ⎥ ⎣
⎣
⎦
A = 20cm2
Reasonable for sand?
yieldedi = 0.17
Yes see:
145ml discharge in 13min48sec
http://en.wikipedia.org/wiki/Hydraulic_conductivity#Ranges_of_values_for_natural_materials
5
Calculate Effective Porosity
using data from
Darcy Apparatus
What measurements will you need?
What
Wh t measurements
will
ill cm3/sec
you need?
d?
For the previous
example witht Q=2.3
Travel time of a tracer through the 30 inch tube = 300 sec
What equation will you solve?
Vdarcy
=
Q= 2.3cm3
sec
A= 20cm 2
0.1 cm/sec
= moves through
= the tube
= 0.45
Dye
Vtracer
30in 2.54cm/in
cm/sec
It moves
~30 cm in0.254
7 min
300 sec
We must have used the wrong numbers in
class because we did not get this result
by construction
draw the
perpendicular
from the 1125
contour to the
1132 point
assume linear head drop so
1125 occurs 7/9 of the way
between 1132 and 1123
(7/9)*390 = 303’
230
i = (1132-1125) / 230 ~ 0.03
6