Q4 (a) A confined stratified aquifer has a total thickness of 22 m and is made up of three layers. The bottom layer has a coefficient of permeability of 12 m/day and a thickness of 6.0 m. The middle and top layers have permeability of 14 m/day and 20 m/day respectively and are of equal thickness. Calculate the following: (i) Equivalent hydraulic conductivity (kp(eg), if the flow is perpendicular to the stratification. (ii) Equivalent hydraulic conductivity (knceg), if the flow is parallel to the stratification. (iii) Ratio of equivalent hydraulic conductivities.
Q4 (a) A confined stratified aquifer has a total thickness of 22 m and is made up of three layers. The bottom layer has a coefficient of permeability of 12 m/day and a thickness of 6.0 m. The middle and top layers have permeability of 14 m/day and 20 m/day respectively and are of equal thickness. Calculate the following: (i) Equivalent hydraulic conductivity (kp(eg), if the flow is perpendicular to the stratification. (ii) Equivalent hydraulic conductivity (knceg), if the flow is parallel to the stratification. (iii) Ratio of equivalent hydraulic conductivities.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Please use FORMULAE SHEET
Transcribed Image Text:Q4 (a) A confined stratified aquifer has a total thickness of 22 m and is made
up of three layers. The bottom layer has a coefficient of permeability of
12 m/day and a thickness of 6.0 m. The middle and top layers have
permeability of 14 m/day and 20 m/day respectively and are of equal
thickness. Calculate the following:
(i)
Equivalent hydraulic conductivity (kp(eq), if the flow is
perpendicular to the stratification.
(ii)
Equivalent hydraulic conductivity (kn(eg)), if the flow is
parallel to the stratification.
(ii)
Ratio of equivalent hydraulic conductivities.
Transcribed Image Text:* l.
formula sheet.pdf +968 9326 0937 >
FORMULAE SHEET
10.67 × L x Q1.85
hf
hf
= KQ?
C1.85 x d4.87
Q = 0.00278 × C × i × A
30 x 25.4
i =
40 × 25.4
i =
or
Q =
= 0.00278 × Ap ×i
T + 10
Т+ 20
2nKb(h2 – h1)
2nT(s1 – $2)
Q :
In (2)
In (2
T×K × (h; – h})
In
3/8
QFs
= 23.97 × D°%3 × VĪ
Dest
Qтах
[23.97×11/2
1
Keq.H
(K,H, + K2H2 + K3H3)
H
H
Ky(eq)
Н
H2 H3
K1
K2 " K3
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