CHAPTER 5 — NUMERICALS
Pumping Test & Aquifer Properties

4 Problems — Cooper-Jacob Time-Drawdown & Distance-Drawdown Methods

KEY FORMULAE — CHAPTER 5

A. Theis Equation (general transient well flow):

$$ s = \frac{Q}{4\pi T}\,W(u), \qquad u = \frac{r^2 S}{4Tt} $$

B. Cooper-Jacob Approximation (valid for small u, i.e., large t or small r):

$$ s = \frac{2.303\,Q}{4\pi\,T}\,\log_{10}\!\left(\frac{2.25\,T\,t}{r^2\,S}\right) $$

C. Cooper-Jacob Time-Drawdown Method (plot s vs log10 t):

$$ \boxed{T = \frac{2.303\,Q}{4\pi\,\Delta s}}, \qquad \boxed{S = \frac{2.25\,T\,t_0}{r^2}} $$

D. Thiem Distance-Drawdown Method (steady state, plot s vs log10 r):

$$ \boxed{T = \frac{2.303\,Q}{2\pi\,\Delta s}}, \qquad R_0 = r \text{ where } s = 0 $$
SymbolTypical RangeSI Unit
\(T\)50–5000 m²/daym²/day or m²/s
\(S\) (confined)10−5–10−3dimensionless
\(S\) (unconfined)0.01–0.30dimensionless
\(\Delta s\)0.1–2 m per log cyclem
\(t_0\)10−5–10−2 daysdays
\(R_0\)100–1000 mm

DERIVATION HINT — Cooper-Jacob Approximation

The Theis well function \(W(u)\) is an infinite series: \(W(u) = -0.5772 - \ln u + u - u^2/4 + \ldots\). When \(u < 0.01\) (which happens for large \(t\) or small \(r\)), terms beyond \(-\ln u\) become negligible. Substituting \(u = r^2S/(4Tt)\):

$$ s \approx \frac{Q}{4\pi T}\left(-0.5772 - \ln\frac{r^2 S}{4Tt}\right) = \frac{2.303\,Q}{4\pi T}\,\log_{10}\!\left(\frac{2.25Tt}{r^2 S}\right) $$

This is a semi-log straight line when plotting s vs log(t). The slope = \(\Delta s\) per log cycle = \(2.303Q/(4\pi T)\), from which T can be extracted.

COOPER-JACOB VALIDITY CHECK

The approximation requires \(u = r^2S/(4Tt) < 0.01\). Check after finding T and S:

STEP-BY-STEP METHOD — Cooper-Jacob Time-Drawdown

  1. Plot s (y-axis, linear) vs t (x-axis, log scale) on semi-log paper
  2. Fit a best-fit straight line through the late-time data points
  3. Read Δs = drawdown difference over one log cycle of time
  4. Compute \(T = 2.303Q/(4\pi\Delta s)\)
  5. Extend line to s = 0, read the time intercept t0
  6. Compute \(S = 2.25Tt_0/r^2\)
TIME-DRAWDOWN (Cooper-Jacob) log10(t) → s (m) ↓ Δs 1 log cycle t0 DISTANCE-DRAWDOWN (Thiem) log10(r) → s (m) ↑ Δs 1 log cycle R0 Both methods use the slope of the straight line (Δs per log cycle) to find T
Left: Cooper-Jacob time-drawdown semi-log plot. Right: Thiem distance-drawdown semi-log plot. The slope gives T; the zero intercept gives S or R0.
Problem 13: Cooper-Jacob Time-Drawdown Method 2078 Chaitra Q5(b) — 6 marks

Question: A pumping test in a confined aquifer (Q = 1000 m³/day) has an observation well at r = 100 m. Use the Cooper-Jacob method to find T and S from the following data.

GIVEN DATA

Pumping rate, \(Q\)1000 m³/day
Observation well distance, \(r\)100 m
t (days)0.0010.0050.010.050.10.51510
s (m)0.0830.1960.2490.3760.4310.5590.6140.7420.797
SEMI-LOG PLOT: Drawdown s vs log₁₀(t) 10-3 10-2 10-1 100 101 t (days) → 0 0.1 s (m) ↓ Δs = 0.183 1 log cycle s=0 → t0 ≈ 4.4×10-4 d
Semi-log plot of drawdown vs time. The data forms a straight line whose slope and intercept yield T and S.

SOLUTION

Step 1 — Plot s vs log10(t) and fit a straight line. The data is remarkably linear from the start, confirming the Cooper-Jacob approximation is valid.

Step 2 — Determine Δs per log cycle. Reading from the data:

Cross-check: t = 0.1 → 0.431, t = 1 → 0.614: \(\Delta s = 0.183\) m. ✓

Average \(\Delta s \approx 0.183\) m per log cycle.

Step 3 — Compute Transmissivity T.

$$ T = \frac{2.303\,Q}{4\pi\,\Delta s} = \frac{2.303 \times 1000}{4\pi \times 0.183} = \frac{2303}{2.300} $$
\( T \approx 1001 \text{ m}^2/\text{day} \approx 1000 \text{ m}^2/\text{day} \)

Step 4 — Find t0 (where s = 0 on the line).

Using the line equation from point (t = 1, s = 0.614):

$$ s = \Delta s \times [\log_{10}(t) - \log_{10}(t_0)] $$
$$ 0.614 = 0.183 \times [0 - \log_{10}(t_0)] \quad \Rightarrow \quad \log_{10}(t_0) = -\frac{0.614}{0.183} = -3.355 $$
$$ t_0 = 10^{-3.355} = 4.42 \times 10^{-4} \text{ days} $$

Step 5 — Compute Storage Coefficient S.

$$ S = \frac{2.25\,T\,t_0}{r^2} = \frac{2.25 \times 1000 \times 4.42 \times 10^{-4}}{100^2} = \frac{0.9945}{10000} $$
\( S \approx 1.0 \times 10^{-4} \)
Physical insight: S = 10−4 is typical of a confined aquifer (artesian storage). It means that a 1 m drop in piezometric head over 1 m² releases only 0.1 litre of water — all from elastic compression of the aquifer skeleton and water expansion. T = 1000 m²/day indicates a very transmissive aquifer (good for production wells).
Cooper-Jacob validity check: For the earliest data point (t = 0.001 d, r = 100 m): \(u = r^2S/(4Tt) = 100^2 \times 10^{-4}/(4 \times 1000 \times 0.001) = 1/4 = 0.25\). This is > 0.01, so strictly the earliest point should not be fitted. However, the data still forms a good line because the deviation is small for practical purposes.
This exact problem has appeared in 3 out of 4 exam years (2072, 2073, 2078). It is the MOST frequently asked numerical in the entire course.
Problems 14 & 15: Cooper-Jacob (Identical Data) 2073 Bhadra Q5(b) 2072 Ashwin Q6 8 marks each

Note: These two examinations present the exact same data as Problem 13 (Q = 1000 m³/day, r = 100 m, identical time-drawdown table). The solution method and answers are identical:

\( T \approx 1000 \text{ m}^2/\text{day}, \qquad S \approx 1.0 \times 10^{-4} \)

This problem has appeared in 3 out of 4 exam years (2072, 2073, 2078). It is the most frequently asked numerical in the entire course. Master the Cooper-Jacob procedure above and you are guaranteed marks.

EXAM FREQUENCY — Cooper-Jacob Time-Drawdown Problem 2071 Not asked 2072 Q6, 10 marks 2073 Q5b, 8 marks 2078 Q5b, 6 marks 3 out of 4 years — MUST PREPARE Same data in all three years. Memorise the method and you will score full marks.
Exam tip: Always include a sketch of the semi-log plot showing the data points, the straight line, the slope (Δs per log cycle), and the t0 intercept. This earns presentation marks even if your arithmetic has minor errors. Also state the Cooper-Jacob validity condition \(u < 0.01\) to show conceptual understanding.
Problem 16: Distance-Drawdown Method (Multiple Wells) 2071 Bhadra Q8 — 10 marks

Question: Steady-state discharge Q = 0.025 m³/s. Data from multiple observation wells in a confined aquifer. Calculate radius of influence and transmissivity.

GIVEN DATA

Discharge, \(Q\)0.025 m³/s
r (m)73060100160200
s (m)1.300.950.710.5350.360.29
SEMI-LOG PLOT: Drawdown s vs log₁₀(r) 1 10 100 1000 r (m) → 1.5 1.0 0.5 0 s (m) ↑ r=7 r=30 r=60 r=100 r=160 r=200 R0 ≈ 613 m Δs ≈ 0.68 m 1 log cycle
Distance-drawdown semi-log plot. The straight line extends to s = 0 at R0 (radius of influence).

SOLUTION

Step 1 — Plot s vs log10(r). The drawdown decreases with distance (as expected — water further from the well is less affected). On a semi-log plot with linear s-axis and log r-axis, the data forms a straight line.

Step 2 — Determine Δs per log cycle of r.

Reading the best-fit line between r = 10 (interpolated s ≈ 1.21) and r = 100 (s = 0.535):

$$ \Delta s \approx 1.21 - 0.535 = 0.68 \text{ m per log cycle} $$

Cross-check with another log cycle: from r = 100 (s = 0.535) to r = 1000 (extrapolated s ≈ 0, where the line reaches zero). This gives Δs ≈ 0.535/0.787 cycles ≈ 0.68 m/cycle. ✓

Important difference from time-drawdown: In the distance-drawdown method, the formula for T has \(2\pi\) in the denominator (not \(4\pi\)) because we use the Thiem steady-state equation, not the Cooper-Jacob transient approximation.

Step 3 — Compute Transmissivity.

$$ T = \frac{2.303\,Q}{2\pi\,\Delta s} = \frac{2.303 \times 0.025}{2\pi \times 0.68} = \frac{0.05758}{4.273} $$
\( T = 0.01348 \text{ m}^2/\text{s} = 1164 \text{ m}^2/\text{day} \approx 1.35 \times 10^{-2} \text{ m}^2/\text{s} \)

Step 4 — Find Radius of Influence R0.

Extend the straight line to s = 0. Using the line equation from (r = 100, s = 0.535):

$$ s = \Delta s \times [\log_{10}(R_0) - \log_{10}(r)] $$

At \(s = 0\), using the known point (r = 100, s = 0.535):

$$ 0 = 0.535 - 0.68 \times [\log_{10}(R_0) - 2] $$
$$ \log_{10}(R_0) = 2 + \frac{0.535}{0.68} = 2 + 0.787 = 2.787 $$
\( R_0 = 10^{2.787} = 613 \text{ m} \approx 600 \text{ m} \)
Physical insight: The radius of influence (~600 m) tells us that beyond this distance, the aquifer is essentially unaffected by pumping. This is critical for:
  • Well spacing: Another well should be placed >600 m away to avoid interference
  • Contamination assessment: A contaminant source within 600 m could potentially reach the pumping well
  • Aquifer yield: T ≈ 1165 m²/day indicates a very transmissive aquifer (consistent with coarse alluvium or karstified limestone)

TIME-DRAWDOWN vs DISTANCE-DRAWDOWN: COMPARISON

FeatureTime-Drawdown (Cooper-Jacob)Distance-Drawdown (Thiem)
Data sourceOne observation well, multiple timesMultiple observation wells, one time
Plots vs log(t)s vs log(r)
T formula\(T = 2.303Q/(4\pi\Delta s)\)\(T = 2.303Q/(2\pi\Delta s)\)
Second parameterS from t0R0 from s = 0 intercept
Flow conditionTransient (time-varying)Steady-state
Equipment needed1 observation well + time readingsMultiple observation wells at different r