(b) A well penetrates an unconfined aquifer. The water level in the well prior to pumping was 14 m. After long period of pumping at a constant rate of 50 l/s, the drawdown in the two observation wells located at 10 m and 15 m were observed to be 3 and 1.4 m, respectively. (i) Calculate the hydraulic conductivity of the aquifer; (ii) Determine the transmissivity of the aquifer material; (ii) Draw a neat sketch to show the labelled diagram of the well.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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(b) A well penetrates an unconfined aquifer. The water level in the well
prior to pumping was 14 m. After long period of pumping at a constant
rate of 50 l/s, the drawdown in the two observation wells located at 10
m and 15 m were observed to be 3 and 1.4 m, respectively.
(i)
Calculate the hydraulic conductivity of the aquifer;
(ii)
(iii)
Determine the transmissivity of the aquifer material;
Draw a neat sketch to show the labelled diagram of the well.
Transcribed Image Text:(b) A well penetrates an unconfined aquifer. The water level in the well prior to pumping was 14 m. After long period of pumping at a constant rate of 50 l/s, the drawdown in the two observation wells located at 10 m and 15 m were observed to be 3 and 1.4 m, respectively. (i) Calculate the hydraulic conductivity of the aquifer; (ii) (iii) Determine the transmissivity of the aquifer material; Draw a neat sketch to show the labelled diagram of the well.
* 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
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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