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JWCL344_ch03_061-117.qxd 8/17/10 7:48 PM Page 86

86 Chapter 3 Water Sources: Groundwater
Increments of drawdown are determined with respect to the extension of the preceding
water level curve.

3.11.3 Intermittent Discharge
In a water supply system, a well (or a well field) may be operated on a regular daily cycle,
pumping at a constant rate for a given time interval and remaining idle for the rest of the
period. Brown (1963) gives the following expression for computing the drawdown in the
pumped well after n cycles of operation:

s (264 Q>T) log[(1.2.3 . . . .n)>(1 p)(2 p) . . . (n p)] (U.S. Customary Units)
n
(3.35a)
where s is the drawdown in the pumped well after n cycles in ft, p is the fractional part of
n
the cycle during which the well is pumped, Q is the discharge in gpm, and T is the trans-
missivity in gpd/ft.
s (380,160 Q>T) log[(1.2.3 . . . n)>(1 p)(2 p) . . . (n p)] (SI Units) (3.35b)
n
where s is the drawdown in the pumped well after n cycles in m, p is the fractional part of
n
3
the cycle during which the well is pumped, Q is the discharge in m /d, and T is the trans-
3
missivity in m /d/m.
The pumping regime may involve switching on a well only during periods of peak
demand. The problem of computing drawdown in a well then consists of applying one of
the equations of nonsteady flow to each of the periods of pumping and recovery. The
drawdown in the well, or at any other point, may be obtained by an algebraic sum of the
individual values of drawdown and “buildup” resulting from each period of pumping and
recovery resulting from each shutdown.

EXAMPLE 3.5 CALCULATION OF WELL DRAWDOWN
3
A well was pumped at a constant rate of 350 gpm (1,907.5 m /d) between 7 a.m. and 9 a.m.,
11 a.m. and 1 p.m., and 3 p.m. and 6 p.m., remaining idle the rest of the time. What will be the
drawdown in the well at 7 a.m. the next day when a new cycle of pumping is to start? Assume
4
no recharge or leakage. The transmissivity of the artesian aquifer is 3.2 l0 gpd/ft
3
(397.38 m /d/m).
Solution 1 (U.S. Customary System):
The problem can be decomposed into three pumping and recovery periods, with Eq. 3.26a applied
to each of the subproblems:
s (264 Q>T) log(t>t )
4
264 Q>T 264 350>3.2 l0 2.89
For the first period of pumping:
t time since pumping started 1,440 min
t time since pumping stopped 1,320 min
log(t>t ) 1 log(1,440>1,320) 0.038