CHAPTER 7 — NUMERICALS
Water Well Design (Recuperation Test)

3 Problems — Open Well Specific Yield & Design

KEY FORMULAE — CHAPTER 7 (RECUPERATION TEST)

Concept: An open well is depressed (pumped down) to create a depression head, then allowed to recover. The rate of recovery indicates how fast the aquifer can supply water.

A. During recovery, the depression head decays exponentially:

$$ s(t) = s_1\,e^{-Ct} $$

where \(C\) is the specific yield of the open well (rate of inflow per unit area per unit depression head).

B. Specific yield from test data:

$$ \boxed{C = \frac{2.303}{T}\,\log_{10}\!\left(\frac{s_1}{s_2}\right)} \quad \text{[per hour or per second]} $$

C. Yield from any well under any depression:

$$ \boxed{Q = C \times A \times H} $$

D. Required well diameter for a target yield:

$$ \boxed{D = \sqrt{\frac{4Q}{\pi\,C\,H}}} $$
SymbolTypical RangeSI Unit
\(C\) (specific yield)0.2–2.0 per hourm³/(hr·m²·m)
\(s_1\) (initial depression)2–8 mm
Recovery time \(T\)30–120 minhours (convert!)
\(D\) (open well diameter)2–8 mm
\(Q\) (open well yield)5–50 m³/hrm³/hr

DERIVATION HINT — Specific Yield Formula

The recuperation test assumes that the rate of water inflow into the well is proportional to the depression head: \(dV/dt = C \cdot A \cdot s\), where \(V\) = volume of water in the well, \(A\) = cross-sectional area, and \(s\) = depression head. Since \(dV = A \cdot ds\):

$$ \frac{ds}{dt} = -C \cdot s \quad \Rightarrow \quad s(t) = s_1 e^{-Ct} $$

At time T: \(s_2 = s_1 e^{-CT}\). Taking natural log: \(CT = \ln(s_1/s_2)\). Converting to log10: \(C = (2.303/T)\log_{10}(s_1/s_2)\).

OPEN WELL vs TUBE WELL

SOLUTION METHODOLOGY — Recuperation Test Problems

  1. Convert time to hours (minutes ÷ 60)
  2. Compute s2 = s1 − rise
  3. Find C using \(C = (2.303/T)\log_{10}(s_1/s_2)\)
  4. For yield: \(Q = C \times (\pi/4)D^2 \times H\)
  5. For diameter: \(D = \sqrt{4Q/(\pi CH)}\)
RECUPERATION TEST SETUP Static WL Start (depressed) After time T s1 s2 Rise D = well diameter D RECOVERY CURVE Time → Depression head (s) s1 s2 T Rise s(t) = s₁ · e-Ct (exponential decay)
Left: Open well recuperation test — the well is pumped down (s1) and allowed to recover (to s2) over time T. Right: The depression head decays exponentially.
Problem 17: Recuperation Test — Specific Yield & Well Yield 2078 Chaitra Q7 — 6 marks

Question: In Dhangadi, a recuperation test was conducted on an open well. The water level was depressed by 4 m and rose 2.2 m in 90 minutes. Calculate specific yield. Also determine yield from a 5 m diameter well under 2.5 m depression.

GIVEN DATA

Initial depression, \(s_1\)4 m
Rise in 90 min2.2 m ⇒ \(s_2 = 4 - 2.2 = 1.8\) m
Recovery time, \(T\)90 min = 1.5 hr
Well diameter (for yield)5 m
Depression head (for yield)2.5 m

SOLUTION

Convert time! Recovery time = 90 min = 1.5 hr. Also compute \(s_2 = s_1 - \text{rise} = 4 - 2.2 = 1.8\) m.

Part (a) — Specific yield C.

Apply the recuperation formula with T = 1.5 hr, s1 = 4 m, s2 = 1.8 m:

$$ C = \frac{2.303}{T}\,\log_{10}\!\left(\frac{s_1}{s_2}\right) = \frac{2.303}{1.5}\,\log_{10}\!\left(\frac{4}{1.8}\right) $$

Ratio: \(s_1/s_2 = 4/1.8 = 2.222\)

Logarithm: \(\log_{10}(2.222) = 0.3468\)

Product: \(1.535 \times 0.3468 = 0.532\)

\( C = 0.532 \text{ per hour (m}^3\text{/hr per m}^2\text{ per m of depression)} \)

Interpretation: C = 0.532/hr means the aquifer can supply 0.532 m³ of water per hour for every 1 m² of well cross-section per 1 m of depression head. This is a moderate-to-good aquifer for an open well in the Dhangadi region (Terai alluvium).

RECOVERY DIAGRAM — Problem 17 (Dhangadi) Time (min) Depression (m) 4.0 1.8 90 min Rise = 2.2 m s2 = 1.8 m
Depression head decays exponentially from 4.0 m to 1.8 m in 90 minutes during the recuperation test.

Part (b) — Yield from a 5 m diameter well under 2.5 m depression.

Well cross-sectional area:

$$ A = \frac{\pi}{4}(5)^2 = \frac{\pi \times 25}{4} = 19.635 \text{ m}^2 $$

Yield:

$$ Q = C \times A \times H = 0.532 \times 19.635 \times 2.5 $$
\( Q = 26.1 \text{ m}^3/\text{hr} \approx 7.26 \text{ L/s} \)
Physical insight: The specific yield of 0.532/hr means the aquifer can replenish the well at about half the well volume per hour per metre of depression. This is a moderate-to-good aquifer for an open well. The Dhangadi region (Terai) has alluvial deposits with reasonable permeability. Note that yield depends linearly on both well area and depression head — doubling either doubles the yield.
Problem 18: Recuperation Test — Yield & Required Diameter 2073 Bhadra Q8(b) — 2+2+3 marks

Question: During a recuperation test, water level depressed by 3 m rose 1.75 m in 75 minutes. (i) Specific yield? (ii) Yield from 5 m diameter well at 2.5 m depression? (iii) Required diameter for 12 L/s at 2 m depression?

GIVEN DATA

Initial depression, \(s_1\)3 m
Rise in 75 min1.75 m ⇒ \(s_2 = 3 - 1.75 = 1.25\) m
Recovery time, \(T\)75 min = 1.25 hr

SOLUTION

Convert time! 75 min = 75/60 = 1.25 hr. And \(s_2 = 3 - 1.75 = 1.25\) m.

Part (i) — Specific yield.

$$ C = \frac{2.303}{1.25}\,\log_{10}\!\left(\frac{3}{1.25}\right) = 1.842 \times \log_{10}(2.4) $$

\(\log_{10}(2.4) = 0.3802\)

$$ C = 1.842 \times 0.3802 $$
\( C = 0.700 \text{ per hour} \)

Interpretation: This C = 0.700/hr is higher than Problem 17’s 0.532/hr, indicating a more permeable aquifer that can replenish faster.

Part (ii) — Yield from 5 m diameter well at H = 2.5 m.

Area: \(A = (\pi/4)(5)^2 = 19.635\) m²

$$ Q = 0.700 \times 19.635 \times 2.5 = 0.700 \times 49.09 $$
\( Q = 34.4 \text{ m}^3/\text{hr} \approx 9.54 \text{ L/s} \)

Part (iii) — Required diameter for Q = 12 L/s = 43.2 m³/hr at H = 2 m.

First convert target yield: \(Q = 12 \times 3.6 = 43.2\) m³/hr

Rearranging \(Q = C \times (\pi/4)D^2 \times H\) for D:

$$ D^2 = \frac{4Q}{\pi \cdot C \cdot H} = \frac{4 \times 43.2}{\pi \times 0.700 \times 2} = \frac{172.8}{4.398} = 39.29 \text{ m}^2 $$
\( D = \sqrt{39.29} = 6.27 \text{ m} \approx 6.3 \text{ m} \)
Physical insight: Achieving 12 L/s from an open well requires a large diameter (6.3 m) because open wells depend on the cross-sectional area for intake. In contrast, a tube well with a screen can achieve the same yield with a 20–30 cm diameter because it draws water through a much larger contact surface along the depth.
COMPARISON: Open Well vs Tube Well for Same Yield D = 6.3 m Open Well (~12 L/s) D = 0.25 m Tube Well (~12 L/s) Same Q
Open wells require large diameters because intake area = well cross-section. Tube wells draw from the entire screened length.
Problem 19: Recuperation Test (Identical to Problem 18) 2071 Bhadra Q10 — 3+3+4 marks

Note: This problem has identical data to Problem 18 (s1 = 3 m, rise = 1.75 m, time = 75 min). The answers are the same:

(a) \(C = 0.700\) per hour
(b) \(Q = 34.4\) m³/hr = 9.54 L/s (for D = 5 m, H = 2.5 m)
(c) \(D = 6.27\) m (for Q = 12 L/s, H = 2 m)
EXAM FREQUENCY — Recuperation Test Problem 2071 Q10, 10 marks 2072 Not asked 2073 Q8b, 7 marks 2078 Q7, 6 marks 3 out of 4 years — MUST PREPARE 2071 & 2073 have identical data. 2078 has different data but same method.

This problem appeared in 2 exam years (2071, 2073) with identical data, and a similar problem in 2078. It is a high-priority problem for exam preparation.

Exam tip: Always draw a neat labelled sketch of the open well showing s1, s2, the static water level, and the recovery. Write the formula clearly before substitution. Convert minutes to hours (students often forget this). Show units at every step. Include the exponential decay diagram for full marks.