CHAPTER 9 — NUMERICALS
Groundwater Resources of Nepal

2 Problems — Confined Aquifer Storage & Well-Field Planning

KEY FORMULAE — CHAPTER 9

A. Annual Groundwater Storage Volume:

$$ \boxed{V = S \times A \times \Delta h} $$

B. Number of Wells:

$$ \boxed{n = \frac{V}{q \times t}} $$

Physical meaning: The storage coefficient \(S\) for a confined aquifer represents the volume of water released per unit area per unit decline in head. It is very small (10−4 to 10−3) because water is released only by elastic compression of the aquifer skeleton and expansion of water.

SymbolTypical RangeSI Unit
\(S\) (confined)10−4–10−3dimensionless
\(S_y\) (unconfined)0.01–0.30dimensionless
\(A\) (plan area)10–10,000 km²m²
\(\Delta h\) (annual fluctuation)2–15 mm
\(q\) (well yield)10–50 m³/hrm³/hr

WHY IS CONFINED AQUIFER STORAGE SO SMALL?

When piezometric pressure drops in a confined aquifer, water is released by two mechanisms:

Both mechanisms are elastic and produce very small volumes. In contrast, an unconfined aquifer releases water by actual draining of pore spaces (gravity drainage), which is 100-1000× more efficient. Hence \(S_y \gg S\).

SOLUTION METHODOLOGY — Storage & Well-Field Planning

  1. Convert area to m²: \(A_{m^2} = A_{km^2} \times 10^6\)
  2. Compute storage volume: \(V = S \times A \times \Delta h\)
  3. Compute annual yield per well: \(q_{annual} = q \times 24 \times \text{pumping days}\)
  4. Number of wells: \(n = V / q_{annual}\)
CONFINED AQUIFER STORAGE CONCEPT Confined Aquifer (S = storage coeff.) Wet season Dry season Δh Plan Area = A (km²) V = S × A × Δh
The piezometric surface fluctuates annually between wet and dry seasons. The volume V = S × A × Δh represents the recoverable groundwater storage.
Problem 20: Groundwater Storage — Pokhara 2072 Ashwin Q3 — 8 marks

Question: In Pokhara, a confined aquifer of 30 m average thickness extends over 800 km². The piezometric surface fluctuates annually from 19 m to 9 m above the top of the aquifer. S = 0.0008. What groundwater storage can be expected annually? With average well yield 30 m³/hr and 200 days pumping, how many wells can be drilled?

GIVEN DATA

Aquifer thickness30 m
Plan area, \(A\)800 km² = \(800 \times 10^6\) m²
Piezometric fluctuation19 m to 9 m ⇒ \(\Delta h = 10\) m
Storage coefficient, \(S\)0.0008
Well yield, \(q\)30 m³/hr
Pumping days200 days/year
POKHARA A = 800 km² S = 0.0008, Δh = 10 m RESULTS V = 6.4 × 10&sup6; m³ n = 44 wells (each @ 30 m³/hr)

SOLUTION

Step 1 — Annual groundwater storage. The recoverable volume is the product of storage coefficient, area, and the annual range of piezometric surface fluctuation:

$$ V = S \times A \times \Delta h = 0.0008 \times 800 \times 10^6 \times 10 $$

Breakdown: \(0.0008 \times 8 \times 10^8 = 6.4 \times 10^5\) per m of Δh; multiplied by 10 m:

$$ = 0.0008 \times 8 \times 10^9 = 6.4 \times 10^6 \text{ m}^3 $$
\( V = 6.4 \times 10^6 \text{ m}^3 = 6.4 \text{ million m}^3/\text{year} \)

Step 2 — Annual yield per well. Each well operates continuously (24 hr/day) for 200 days:

$$ q_{annual} = 30 \text{ m}^3/\text{hr} \times 24 \text{ hr/day} \times 200 \text{ days} = 30 \times 4800 = 144{,}000 \text{ m}^3/\text{year per well} $$

Step 3 — Number of wells.

$$ n = \frac{V}{q_{annual}} = \frac{6{,}400{,}000}{144{,}000} $$
\( n = 44.4 \approx \mathbf{44 \text{ wells}} \)
Physical insight: 800 km² is a vast area (roughly 28 km × 28 km), yet only 6.4 million m³ of recoverable storage per year — this is because S = 0.0008 is extremely small (confined artesian storage). In contrast:

WHAT IF THE AQUIFER WERE UNCONFINED?

If the same aquifer were unconfined with specific yield \(S_y = 0.15\):

$$ V_{unconfined} = 0.15 \times 800 \times 10^6 \times 10 = 1.2 \times 10^9 \text{ m}^3 $$

That is 187 times more water! This illustrates why confined aquifers, while capable of artesian flow, are not volumetrically rich. Their advantage lies in pressure-driven flow (artesian wells that flow without pumping) and protection from surface contamination.

Problem 21: Groundwater Storage — Janakpur 2072 Magh Q3 — 8 marks

Question: In Janakpur, a confined aquifer of 20 m average thickness extends over 850 km². Water table fluctuates from 15 m to 10 m above the top. S = 0.0009. What is the annual groundwater storage? With well yield of 30 m³/hr and 210 pumping days, how many wells can be drilled?

GIVEN DATA

Aquifer thickness20 m
Plan area, \(A\)850 km² = \(850 \times 10^6\) m²
Piezometric fluctuation15 m to 10 m ⇒ \(\Delta h = 5\) m
Storage coefficient, \(S\)0.0009
Well yield, \(q\)30 m³/hr
Pumping days210 days/year
JANAKPUR A = 850 km² S = 0.0009, Δh = 5 m RESULTS V = 3.825 × 10&sup6; m³ n = 25 wells (each @ 30 m³/hr)

SOLUTION

Step 1 — Annual groundwater storage.

$$ V = S \times A \times \Delta h = 0.0009 \times 850 \times 10^6 \times 5 $$

Breakdown: \(A \times \Delta h = 850 \times 10^6 \times 5 = 4.25 \times 10^9\) m³ of aquifer volume affected

$$ V = 0.0009 \times 4.25 \times 10^9 $$
\( V = 3.825 \times 10^6 \text{ m}^3 = 3.825 \text{ million m}^3/\text{year} \)

Step 2 — Annual yield per well. Each well operates 24 hr/day for 210 days:

$$ q_{annual} = 30 \times 24 \times 210 = 30 \times 5040 = 151{,}200 \text{ m}^3/\text{year per well} $$

Step 3 — Number of wells.

$$ n = \frac{3{,}825{,}000}{151{,}200} $$
\( n = 25.3 \approx \mathbf{25 \text{ wells}} \)
Comparison with Problem 20 (Pokhara): Despite Janakpur having a slightly larger area (850 vs 800 km²) and higher S (0.0009 vs 0.0008), its smaller Δh (5 vs 10 m) halves the recoverable storage. The dominant factor is piezometric fluctuation Δh, not area or S.
COMPARISON: Pokhara vs Janakpur Groundwater Storage Pokhara V = 6.4 M m³ Janakpur V = 3.825 M m³ Wells 44 wells 25 wells Pokhara: A=800km², S=0.0008, Δh=10 m Janakpur: A=850km², S=0.0009, Δh=5 m Δh is the dominant factor — double Δh ≈ double the storage
Pokhara has twice the recoverable storage primarily because Δh is twice as large. Area and S have minor effects.
ParameterPokharaJanakpurRatio
Area800 km²850 km²0.94
Δh10 m5 m2.0
S0.00080.00090.89
Storage V6.4 M m³3.825 M m³1.67
Pumping days2002100.95
Wells44251.76

Despite similar area and S, Pokhara has 76% more wells because Δh is twice as large — the dominant factor.

These problems appeared in 2072 Ashwin Q3 and 2072 Magh Q3. They are the simplest numericals in the course — easy marks if you remember V = S × A × Δh and n = V/(q × t).