Chapter 5 – Pumping Test & Aquifer Parameter Estimation

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Chapter 5 – Pumping Test Methods

Fig: Pumping test setup showing cone of depression in a confined aquifer

a) Theis Method (Distance-Drawdown)

Confined aquifer

Steps:

Plot r vs s in semi-log graph (r → log, s → linear)
Draw best fit straight line
Extend to s = 0 → Radius of Influence (R₀)
Take 1 log cycle: Δs = s₁ − s₂
T:
\( T = \frac{2.303\,Q}{2\pi\,\Delta s} = \frac{Q}{2.72\,\Delta s} \)

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b) Cooper and Jacob Method (Time-Drawdown)

Steps:

Plot t vs s in semi-log graph (t → log, s → linear)
Draw best fit straight line
Extend to s = 0 → find t₀
Take 1 log cycle: Δs = s₂ − s₁

\( T = \frac{2.303\,Q}{4\pi\,\Delta s} \)

\( S_y = \frac{2.246\,T\,t_0}{r^2} \)

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Unconfined Aquifer Correction

If s²/(2H) > 0.003, correction required:

\( s' = s - \frac{s^2}{2H} \)

Then use corrected s′ in same semi-log method:

\( T = \frac{2.303\,Q}{4\pi\,\Delta s'} \)

\( S_y = \frac{2.246\,T\,t_0}{R^2} \)

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Chapter 5 – Pumping Test Numericals

Numerical 1 – Distance-Drawdown (Confined)

Q. Q = 0.025 m³/s, confined aquifer:

r (m)	7	30	60	100	160	200
s (m)	1.3	0.95	0.71	0.535	0.36	0.29

Fig: Semi-log r vs s plot — distance-drawdown method

Solution:

Step 1 — Plot r (log scale) vs s (linear) on semi-log paper.

Step 2 — Draw best-fit straight line. Extend to s=0 to get R₀ ≈ 460 m.

Step 3 — Read Δs for 1 log cycle (r=10 to r=100): Δs = 1.33 − 0.530 = 0.800 m

Step 4 — Compute T:

\[ T = \frac{Q}{2.72\,\Delta s} = \frac{0.025}{2.72 \times 0.800} = \boxed{0.0115\;\text{m}^2\text{/s}} \]

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[Semi-log graph paper — Q1 Distance-Drawdown Plot]

x-axis (log): r (m), 4 cycles 1–1000
y-axis (linear): s (m), 0–1.5
6 data points plotted, best-fit line
Extrapolated to s=0 at r ≈ 460m
1 log cycle: r=10 (s=1.33) to r=100 (s=0.53), Δs=0.80

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Numerical 2 – Time-Drawdown (Cooper-Jacob, Confined)

Q. r = 100 m, Q = 1000 m³/day:

t (days)	0.001	0.005	0.01	0.05	0.1	0.5	1	5	10
s (m)	0.083	0.196	0.249	0.346	0.431	0.539	0.614	0.742	0.797

Solution:

Step 1 — Plot t (log scale) vs s (linear) on semi-log paper.

Step 2 — Draw best-fit straight line through the data points.

Step 3 — Extend line to s=0 to find t₀ = 0.00045 days.

Step 4 — Read Δs per log cycle: Δs = 0.43 − 0.24 = 0.19 m

Step 5 — Compute T:

\[ T = \frac{2.303 \times Q}{4\pi \times \Delta s} = \frac{2.303 \times 1000}{4\pi \times 0.19} = \frac{2303}{2.388} = \boxed{964.6\;\text{m}^2\text{/day}} \]

Step 6 — Compute S:

\[ S = \frac{2.246\,T\,t_0}{r^2} = \frac{2.246 \times 964.6 \times 0.00045}{100^2} = \frac{0.975}{10000} = \boxed{9.75 \times 10^{-5}} \]

Check: S in range 10⁻⁵ to 10⁻³ for confined aquifer. ✓

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[Semi-log graph paper — Q2 Time-Drawdown Plot]

x-axis (log): t (days), 4 cycles
y-axis (linear): s (m), 0–0.8
Best-fit line extrapolated to s=0 at t₀ ≈ 0.00045 days
Δs per log cycle ≈ 0.19 m

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Numerical 3 – Unconfined, Distance-Drawdown (with correction)

Q. Q = 4000 m³/day, H = 45 m. Observation wells:

r (m)	15	20	40	60	80	100	150
s (m)	1.8	1.5	1.0	0.8	0.5	0.3	0.1
s′ (m)	1.764	1.475	0.989	0.793	0.497	0.299	0.0999

Step 1 — Apply Jacob correction: s′ = s − s²/(2H), where H = 45 m

Sample calculation for r=15: s′ = 1.8 − 1.8²/(2×45) = 1.8 − 3.24/90 = 1.8 − 0.036 = 1.764

Correction is small here because s/H < 5%, but becomes significant for larger drawdowns.

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From graph (s′ vs r, semi-log):

Step 2 — Plot corrected s′ vs r on semi-log paper.

Step 3 — Read R₀ = 210 m (where line meets s′=0), Δs′ = 0.64 m per log cycle.

Step 4 — Compute T:

\[ T = \frac{2.303 \times Q}{2\pi \times \Delta s'} = \frac{2.303 \times 4000}{2\pi \times 0.64} = \frac{9212}{4.021} = \boxed{2290.6\;\text{m}^2\text{/day}} \]

Step 5 — Compute K:

K = T/H = 2290.6/45 = 50.9 m/day

Step 6 — S_y (requires t₀ from time data):

\( S_y = \frac{2.25\,T\,t_0}{r_0^2} \)

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[Semi-log graph paper — Q3 Corrected Drawdown Plot]

x-axis (log): r (m), 4 cycles 1–10000
y-axis (linear): s′ (m), 0–2.0
7 data points, best-fit line
Extrapolated to s′=0 at r₀ ≈ 210 m
Δs′ per log cycle ≈ 0.64 m