Question 13.13: A non-equilibrium pumping test has been conducted in a confi......

Understanding Hydraulics [985713]

A non-equilibrium pumping test has been conducted in a confined aquifer. The time since pumping started and the corresponding drawdown in an observation hole 91 m from the pumped well are listed below. The pumping rate was 1800 m³/d. Using a Jacob analysis, determine the value of T and S.

Time (t min)	Drawdown (s m)	Time (t min)	Drawdown (s m) (continued)
2	0.16	60	2.39
4	0.44	80	2.72
6	0.60	100	2.87
8	0.85	120	3.13
10	0.91	180	3.54
15	1.15	260	3.78
20	1.30	360	4.20
30	1.68	600	4.71
40	1.98	840	5.05
50	2.25

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The data are plotted as the drawdown–log time graph shown in Fig. 13.27. From the graph, Δs = 2.33 m and $t_{0}$ = 5.6 min = 5.6/(60 × 24) days. Thus:

T = \frac{2.30 Q}{4 \pi \Delta s} = \frac{2.30 \times 1800}{4 \times \pi \times 2.33}=141  m ^2 / d \quad \text { and } S = \frac{2.25 T t_0}{r^2} = \frac{2.25 \times 141 \times 5.6}{91^2 \times 60 \times 24}=0.000149

Figure 13.27

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