Oblique view
Hydrostatic before
tunneling
1500ft
Cross section view
40ft
K=50ft/d
ht=1500ft
length=90ft
200ft
ht=150ft
ht=130ft
length=90ft
K=0.00005ft/d
Use average
ht=140ft and
multiply flow by 2
for the two sides
ht=1500ft
40ft
K=50ft/d
Q = KiA = 5x10-5 ft 1500-140 ft 1ft1ft 2sides = 1.5x10-3 ft3
day
90ft
day ft2
If the repository was 2000ftx1000ft then
Q = 1.5x10-3 ft3 2000ft 1000ft ~ 3000ft3 ~ 2 ft3
= 16 gallons
day ft2
day
minute
minute
What about the local flow to each tunnel? It will make no difference.
Random distribution of K
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*
*
*
6
1x10-3cm/s
4
1x10-5cm/s
2
1x10-6cm/s
⎛1
⎞
⎜ (logK1 +logK2 +...+logKN ) ⎟
⎝N
⎠
10
(
* sample locations
assumed to be
representative of the
proportions
ti
)
⎛1
−3
−5
−6 ⎞
⎜ 6log(1x10 )+4log(1x10 )+...+2log(1x10 ) ⎟
⎝ 12
⎠
= 10
= 6.8x10−5
This would be treated as isotropic (same K in every direction)
1
Realize you would only know the values shown
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You might have clues based on lithology
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*
Calculate Flow and Heads between boundaries
L1
H1
H3
D K1
K2
H4
L2
H2
H1 = 20cm
H2 = 10cm
K1 = 1cm/sec
K2 = 0.2cm/sec
L1 = 30cm
L2 = 30cm
D = 2cm
Q @ H2 = ??
H3 = ??, H4 = ??
2
Calculate Flow and Heads between boundaries
L1
H1
H3
D K1
K2
H4
L2
H2
cm
cm
+ 1cm0.2
1cm1
H1 = 20cm
sec
sec = 0.6 cm
K eq = Weighted Aritmetic Average =
H2 = 10cm
2cm
sec
K1 = 1cm/sec Q = KiA, no width is given so calculate per unit width
K2 = 0.2cm/sec
cm 20cm − 10cm
cm
cm 2
2cm = 0.1
2cm = 0.2
per unit width
Q = 0.6
L1 = 30cm
sec
60cm
sec
sec
L2 = 30cm
b inspection
by
i
ti gradient
di t is
i linear,
li
D = 2cm
and H 3 = H 4
Q @ H2 = ??
and they are at the midpoint
H3 = ??, H4 = ??
20cm − 10cm
H3 = H4 =
2
+ 10cm = 15cm
Calculate Flow and Heads between boundaries
L1
H1
H3
D K1
K2
H4
L2
H2
Δh
VΔl
H1 = 20cm
OR : head loss from H1 to H3 V = Ki = K
so Δh =
Δl
K
H2 = 10cm
Use proper combination:
K1 = 1cm/sec
All equivalent V and K
K2 = 0.2cm/sec
Or all layer 1 V and K
L1 = 30cm
Or all layer 2 V and K
L2 = 30cm
cm
0.1
30cm
VΔl
D = 2cm
sec
Δh =
=
= 5cm
cm
K
Q @ H2 = ??
0.6
sec
H3 = ??, H4 = ??
H 3 = H 1 − Δh = 20cm − 5cm = 15cm
3
Calculate Flow and Heads between boundaries
L1
H1
H3
K1
L2
D
H2
K2
H1 = 20cm
H2 = 10cm
K1 = 1cm/sec
K2 = 0.2cm/sec
L1 = 30cm
L2 = 30cm
D = 2
2cm
Q @ H2 = ??
H3 = ??
Calculate Flow and Heads between boundaries
L1
H1
K1
H3
D
L2
H2
K2
60cm
cm
K eq =
= 0.33
H1 = 20cm
30cm 30cm
sec
+
cm
cm
H2 = 10cm
1
0.2
sec
sec
Velocity
K1 = 1cm/sec
K2 = 0.2cm/sec Q = VA = KiA, no width is given so calculate per unit width
L1 = 30cm
cm
cm 2
cm 20cm − 10cm ⎞
⎛
Q = ⎜ 0.33
2cm = 0.11
per unit width
⎟2cm= 0.05555
L2 = 30cm
sec
60cm
sec
sec
⎝
⎠
D = 2cm
Δh
VΔl
head loss from H1 to H3 V = Ki = K
so Δh =
Q @ H2 = ??
Δl
K
cm
H3 = ??
0.0555
30cm
K for path
VΔl
=
K
sec
= 1.66cm
cm
1
sec
H 3 = H 1 − Δh = 20cm − 1.66cm = 18.33cm
Δh =
length of
interest
4
20
Calculate Flow and Heads between boundaries
1.66
1.66
10
15
18.33
L1
H1
K1
H1 = 20cm
H2 = 10cm
K1 = 1cm/sec
K2 = 0.2cm/sec
L1 = 30cm
L2 = 30cm
D = 2cm
Q @ H2 = ??
H3 = ??
H3
D
8.33
L2
K2
H2
Confirm by calculating from
the other side
head loss from H3 to H2 V = Ki = K
VΔl
Δh =
=
K
Δh
VΔl
so Δh =
Δl
K
Velocity
cm
30cm
K for path
sec
= 8.33cm
length of
cm
0.2
interest
sec
0.0555
H 3 = H 2 + Δh = 10cm + 8.33cm = 18.33cm
Review keys for homework from September
13 exercises
i
6
6c 6d 6
6e
Remember to continually work on your
cheat sheets
And
Work the sample exam problems
5