aquifers is not the same, the specific capacities of wells in different aquifers are not al-
ways comparable.
It is possible, with the use of the equation for Q above, to solve some problems in
gravity wells by measuring two or more rates of flow and corresponding drawdowns in
the well to be studied. Observations in nearby test holes or boreholes are unnecessary.
The steps are outlined in this procedure.
This procedure is the work of Harold B. Babbitt, James J. Doland, and John L. Cleas-
by, as reported in their book, Water Supply Engineering, McGraw-Hill. SI values were
added by the handbook editor.
ANALYZING DRAWDOWN AND RECOVERY
FOR WELL PUMPED FOR EXTENDED PERIOD
Construct the drawdown-recovery curve for a gravity well pumped for two days at 450
gal/min (28.4 L/s). The following observations have been made during a test of the well
under equilibrium conditions: diameter, 2 ft (0.61 m); he = 50 ft (15.2 m); when Q = 450
gal/min (28.4 L/s), drawdown = 8.5 ft (2.6 m); and when rx = 60 ft (18.3 m), (he -hx) = 3
ft (0.91 m). The specific yield of the well is 0.25.
Calculation Procedure:- Determine the value of the constant k
Use the equation
k(he-hx)he QCx IQg 10 W(UAJ
^ QlOg 10 (VO-IW ^ (he-hx)(he)Determine the value of Cx when rw, is equal to the radius of the well, in this case 1.0. The
value of k can be determined by trial. Further, the same value of k must be given when rx
= re as when rx = 60 ft (18.3 m). In this procedure, only the correct assumed value of re is
shown—to save space.
Assume that re = 350 ft (106.7 m). Then, 1/350 = 0.00286 and, from Fig. 6, Cx = 0.60.
Then k = (l)(0.60)(log 350/5)/(8)(50) = (1)(0.6)( 1.843)7400 = 0.00276, rjre = 60/350 =
0.172, and Cx = 0.225. Hence, checking the computed value of k, we have k =
(1)(0.22)(1.843)/150 = 0.0027, which checks with the earlier computed value.
- Compute the head values using k from step 1
Compute he - (h^2 e - 1.7 Q/k)°-^5 = 50- (2500 - 1.7/0.0027)^0 -^5 = 6.8. - Find the values of T to develop the assumed values of re
For example, assume that re = 100. Then T= (0.184)(100)^2 (0.25)(6.8)/1 = 3230 sec = 0.9
h, using the equation
T-(h FTT^i °-
184r|f
r
l*«"V« */ Q- Calculate the radii ratio and d 0
These computations are: rjrw = 100/1 = 100. Then, d 0 = (6.8)(log 10 100)/2.3 = 5.9 ft (1.8
m), using the equation