Civil Engineering Reference
In-Depth Information
and, integrating between test limits:
r
1
h
2
2
∫
∫
q
=
∂ =
r
k
2
π
h h
∂
r
r
h
1
1
h
2
−
h
2
2
1
=
k
2
π
2
i.e.
r
r
2
q
ln
=
k h
π
(
2
−
h
2
)
2
1
1
or
q r /r
h
ln
2 3
.
q
log
r /r
2
1
10 2
1
k
=
=
π
(
2
−
h
2
)
π
(
h
2
−
h
2
)
Pumping tests can be expensive as they require the installation of both the pumping and the observa-
tion wells as well as suitable pumping and support equipment. Care must be taken in the design of a
suitable test programme and, before attempting to carry out any pumping test, reliable data should be
obtained about the subsoil profile, if necessary by means of boreholes specially sunk for the purpose.
Suction pumps can be used where the groundwater does not have to be lowered by more than about
5 m below the intake chamber of the pump but for greater depths submersible pumps are generally
necessary.
Where a pumping test has been completed in which there are no observation well data, it is still pos-
sible to obtain a very rough estimate of k with the formula proposed by Logan (
1964
), k
≈
1.22q/(s
w
/h
o
),
where h
o
is the thickness of confined ground and is
w
is the corrected drawdown in the pumping well.
(The observed value of drawdown in the well has to be corrected for head loss through the well screens
before being used in calculation of permeability. It is usual to reduce the observed value by 25% to give
s
w
, unless the head losses from water entering the well are observed to be obviously much greater.)
Example 2.3:
Pumping out test
A 9.15 m thick layer of sandy soil overlies an impermeable rock. Groundwater level is at
a depth of 1.22 m below the top of the soil. Water was pumped out of the soil from a
central well at the rate of 5680 kg/min and the drawdown of the water table was noted
in two observation wells. These two wells were on a radial line from the centre of the
main well at distances of 3.05 and 30.5 m.
During pumping the water level in the well nearest to the pump was 4.57 m below
ground level and in the furthest well was 2.13 m below ground level.
Determine an average value for the permeability of the soil in m/s.
Solution:
3
3
q
=
5680
kg min
/
=
5 68
.
m min
/
=
0 0947
.
m s
/
h
=
9 15 4 57
.
−
.
=
4 58
.
m h
=
9 15 2 13
.
−
.
=
7 02
.
m
1
2
q r /r
h
ln
0 0947 2 3026
28 3
.
×
×
.
2
1
k
=
=
=
2 45 10
.
×
−
3
m s
/
(
2
−
h
2
)
π
.
π
2
1
