Page 94 - Water Engineering Hydraulics, Distribution and Treatment

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Water Sources: Groundwater
Chapter 3
should be such as to avoid drawing this poor quality water
into the pumping well.
The following procedure is carried out to estimate the maxi-
mum yield of a well:
3.15.2 Specific Capacity–Drawdown Curve
1. Calculate the specific capacity of the fully penetrating
A graph of specific capacity versus drawdown is prepared
well having the proposed diameter from Eq. (3.15)
from the data on existing wells in the formation if such data
or Eq. (3.18).
are available. Specific capacities should be adjusted for well
2. Reduce the specific capacity obtained above for par-
losses and partial penetration and should be reduced to a
tial penetration. This can be done by using Eq. (3.41).
common well radius and duration of pumping. If no data
3. Adjust the specific capacity for the desired duration
are available, a step-drawdown test is conducted on the pro-
of pumping from Eq. (3.40).
duction well. The curve is extended to cover the maximum 3.15.3 Maximum Yield
available drawdown. For a well receiving water from more 4. Calculate the maximum available drawdown.
than one aquifer, the resultant specific capacity is the sum 5. Compute the maximum yield of the well by multiply-
of the specific capacities of the aquifer penetrated, reduced ing the specific capacity in step 3 by the maximum
appropriately for partial penetration. available drawdown.
EXAMPLE 3.9 COMPUTATION OF THE SPECIFIC CAPACITY OF A WELL AND ITS MAXIMUM YIELD
A well having an effective diameter of 12 in. (305 mm) is to be located in a relatively homogeneous artesian aquifer with a
−4
3
transmissivity of 10,000 gpd/ft (124.18 m /d/m) and a storage coefficient of 4 × 10 . The initial piezometric surface level is 20 ft
(6.1 m) below the land surface. The depth to the top of the aquifer is 150 ft (45.72 m) and the thickness of the aquifer is 50 ft
(15.24 m). The well is to be finished with a screen length of 20 ft (6.1 m). Compute the specific capacity of the well and its maximum
yield after 10 days of pumping. Neglect well losses.
Solution 1 (US Customary System):
The specific capacity of a l2 in., fully penetrating well after 40 days of pumping can be calculated from Eqs. (3.15) and (3.16):
s = 114.6 QW (u) ∕T (3.15)
2
u = 1.87(S∕T)(r ∕t) (3.16)
−4 4 2 −9
u = 1.87(4 × 10 ∕10 )(0.5 ∕10) = 1.87 × 10 .
From Table 3.2: W (u) = 19.52.
(Q∕s) = T∕114.6 W(u) = 10, 000∕(114.6 × 19.52)
3
= 4.47 gpm∕ft (80 m ∕d∕m).
The percentage of aquifer screened is K = 20∕50 = 40%.
p
The slenderness of the well factor is b∕r = 50∕0.5 = 100.
w
The value of F from Fig. 3.12 is 0.65.
p
3
The expected specific capacity of the well = 0.65 × 4.47 = 2.9 gpm∕ft (51.87 m ∕d∕m).
The maximum available drawdown = 150 − 20 = 130 ft (39.62 m).
3
The maximum yield of the well = 130 × 2.9 = 377 gpm (2,055 m ∕d).
Solution 2 (SI System):
The specific capacity of a 305 mm, fully penetrating well after 40 days of pumping can be calculated from Eqs. (3.12) and (3.13):
s = (Q∕4 T)W(u).
2
u = (r s)∕(4 Tt).
Given data are
t = 10 days
3
T = 124.18 m ∕d∕m,
S = 4 × 10 −4
r = (305 mm∕2) = 0.1525 m.