Geotechnical Engineering

DHARM

154 GEOTECHNICAL ENGINEERING

k 1 = 8 × 10–4 cm/s h 1 = 7 m

k 2 = 52 × 10–4 cm/s h 2 = 3 m

k 3 = 6 × 10–4 cm/s h 3 = 10 m

kh (or kx) =

()

()

kh kh kh

hh h

11 22 33

123

++

++

=

()8 7 52 3 6 10

20

×+ ×+×

× 10–4

= 13.6 × 10–4 cm/s

∴ Effective average permeability in the horizontal direction

= 13.6 × 10–3 mm/s

kv(or kz) =

h

h

k

h

k

h

k

1

1

2

2

3

3

++

F

HG

I

KJ

=

20

1

10

−^4 [/78 352 106++/ /]

= 7.7 × 10–4 cm/s

∴ Effective average permeability in the vertical direction

= 7.7 × 10–3 mm/s.

Example 5.12. An unconfined aquifer is known to be 32 m thick below the water table. A
constant discharge of 2 cubic metres per minute is pumped out of the aquifer through a tubewell
till the water level in the tubewell becomes steady. Two observation wells at distances of 15 m
and 70 m from the tubewell show falls of 3 m and 0.7 m respectively from their static water
levels. Find the permeability of the aquifer. (S.V.U.—B.Tech., (Part-time)—April, 1982)

The conditions given are shown in Fig. 5.30 :

q=2m^3 /min

Original WT

Drawdown

curve

32 32 mm

29 29 mm

3 3mm

31.331.3 mm

0.7 m

70 70 mm

15 m

Fig. 5.30 Unconfined aquifer (Example 5.12)