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CE 491 / 591 · Groundwater Hydrology · Well Hydraulics, Part 4

Transient Well Hydraulics: Theis and Cooper–Jacob

Self-paced module · about 2.5 hours · prerequisite: steady flow to a well (Thiem) and superposition

Overview

So far the cone of depression has been treated as fully developed (steady state). This module covers what happens from the moment a pump is switched on, and how a pumping test uses that time history to measure an aquifer's transmissivity and storativity.

By the end of this module you should be able to:
  1. Explain physically why the cone of depression grows with time and how \(T\) and \(S\) control its shape.
  2. Compute \(u\) and the Theis drawdown \(s(r,t)\) for a confined aquifer.
  3. Decide when the Cooper–Jacob approximation is valid (\(u < 0.01\)) and use it.
  4. Estimate \(T\) and \(S\) from pumping-test data using the straight-line method.

How to work through it: watch the two short videos, read the notes, spend 15 minutes with the simulator answering the prediction questions, then do the practice problems until you can get them right without opening the solution. Only then take the graded quiz in Brightspace.

1Watch

Two animations, about 2 minutes each. Pause whenever a caption introduces a new idea.

Video 1: the cone of depression, the Theis solution, and the effect of \(T\) and \(S\).
Video 2: the well function \(W(u)\), the Cooper–Jacob approximation, and pumping-test analysis.

2Read

Governing equation

For radial flow to a well in a homogeneous, isotropic confined aquifer, conservation of mass combined with Darcy's law gives

\[ \frac{S}{T}\frac{\partial h}{\partial t}=\frac{\partial^2 h}{\partial r^2}+\frac{1}{r}\frac{\partial h}{\partial r} \]

The left side is new compared with the steady case: water released from storage as the head drops. \(S\) (storativity, dimensionless) is the volume of water released per unit aquifer area per unit decline in head. The ratio \(T/S\) is the hydraulic diffusivity: it sets how fast a pressure disturbance spreads outward.

The Theis solution (1935)

Assumptions: the aquifer is homogeneous, isotropic, laterally infinite and confined; the initial head is uniform, \(h(r,0)=h_0\); the well fully penetrates, has negligible diameter, and pumps at a constant rate \(Q\) from \(t=0\). Under these conditions

\[ s(r,t)=h_0-h(r,t)=\frac{Q}{4\pi T}\,W(u), \qquad u=\frac{r^2 S}{4Tt}, \qquad W(u)=\int_u^{\infty}\frac{e^{-w}}{w}\,dw \]

\(W(u)\) is the well function (mathematically, the exponential integral \(E_1\)). It has no closed form; use a table, the spreadsheet from class, the calculator below, or the series

\[ W(u) = -0.5772-\ln u+u-\frac{u^2}{2\cdot2!}+\frac{u^3}{3\cdot3!}-\frac{u^4}{4\cdot4!}+\cdots \]
Reading \(u\). \(u\) compares the squared distance \(r^2\) with \(4Tt/S\), the squared distance the pressure signal has reached. Large \(u\) (far away, early, high \(S\), low \(T\)) means the cone has not arrived and \(W\approx0\). Small \(u\) means the cone is well developed there.

Cooper–Jacob approximation

When \(u\) is small, every term after \(-\ln u\) in the series is negligible. Keeping two terms gives

\[ s \approx \frac{Q}{4\pi T}\ln\frac{2.25\,T\,t}{r^2 S} = \frac{2.3\,Q}{4\pi T}\log_{10}\frac{2.25\,T\,t}{r^2 S} \qquad (u < 0.01) \]

The error is about 0.25% at \(u = 0.01\), 2% at \(u = 0.05\) and 5% at \(u = 0.1\). Always compute \(u\) before using it.

Worked example

\(Q = 500\) m³/d, \(T = 200\) m²/d, \(S = 1\times10^{-4}\). Find the drawdown at \(r = 200\) m after 1 day.

\(u = \dfrac{(200)^2(10^{-4})}{4(200)(1)} = 0.005\), so \(W(0.005) = 4.726\), and \(s = \dfrac{500}{4\pi(200)}(4.726) = 0.199 \times 4.726 = 0.940\) m.
Cooper–Jacob gives \(0.199\,\ln\!\dfrac{2.25(200)(1)}{(200)^2(10^{-4})} = 0.940\) m, practically identical because \(u = 0.005 < 0.01\).

Now repeat for \(t = 10\) minutes \(= 0.00694\) d: \(u = 0.72\), \(W = 0.360\), \(s = 0.072\) m. Cooper–Jacob would give negative drawdown (−0.049 m), which is physically meaningless. That's why you check \(u\) first.

Pumping-test analysis: the straight-line method

Because Cooper–Jacob is linear in \(\log t\), late-time drawdown data at one observation well plot as a straight line on semi-log paper (\(s\) linear, \(t\) logarithmic).

  1. Plot \(s\) against \(\log_{10} t\). Fit a line through the late points only (early points curve away because \(u\) is still large).
  2. Read \(\Delta s\), the change in drawdown over one log cycle of time. Then \(\displaystyle T = \frac{2.3\,Q}{4\pi\,\Delta s}\).
  3. Extend the line to \(s = 0\) and read the intercept time \(t_0\). Then \(\displaystyle S = \frac{2.25\,T\,t_0}{r^2}\).
  4. Check \(u = r^2S/(4Tt) < 0.01\) for every point used in the fit, using your fitted \(T\) and \(S\).

Keep units consistent: if \(t_0\) is in minutes and \(T\) in m²/day, convert one of them.

3Explore

Before moving a slider, write down what you predict will happen. Then check.

u
W(u)
s (Theis)
s (Cooper–Jacob)
Jacob error

Cone of depression at the current time

Drawdown history at the observation well (shaded: u > 0.01)

Predict, then check

1. Double Q. What happens to the drawdown everywhere?
It doubles at every point and every time. The flow equation is linear and \(s \propto Q\). This is the same linearity that makes superposition work.
2. Increase S by ×10 and watch the drawdown-history plot. Does the late-time slope change?
No. The slope per log cycle is \(2.3Q/(4\pi T)\), which has no \(S\) in it. The line shifts to later times (\(t_0\) grows ×10), so at any fixed time the drawdown is smaller by \(2.3Q/(4\pi T)\). That's why the slope gives \(T\) and the intercept gives \(S\).
3. Increase T. Why does the cone get shallower but wider?
The prefactor \(Q/(4\pi T)\) shrinks, so drawdown near the well drops. But diffusivity \(T/S\) rises, so the pressure signal reaches farther in the same time.
4. Set r = 1000 m and t = 1 hour. Which equation should you use?
Theis. \(u\) is large (much more than 0.01), and Cooper–Jacob underestimates the drawdown, or even gives a negative value.
5. Does the cone ever stop growing in this model?
No. With an infinite aquifer and no recharge, drawdown increases forever, logarithmically in time. Real systems stabilize only when the cone captures a source such as recharge, a river, or leakage. You can model a river with an image well (Part 3).

4Practice (ungraded)

Every problem generates new numbers each time. Aim for two correct attempts in a row on each type before taking the graded quiz. You can enter scientific notation as 2.5e-4.

A · Theis drawdown

B · Cooper–Jacob drawdown

C · Analyze a pumping test

D · Working backwards: time to reach a drawdown

Concept checks

1. In a Cooper–Jacob plot, the late-time slope \(\Delta s\) per log cycle depends on:

2. In a pumping test, the early-time drawdown points curve away from the Cooper–Jacob straight line, plotting above its extension toward \(t_0\). The most likely reason is:

3. Two aquifers have the same \(T\), but aquifer B has 100 times larger \(S\). After the same pumping time, aquifer B's cone of depression is:

5Graded quiz