Theim Equation

Hydrology and Water Resources · FE Reference Handbook section

Hydrology and Water Resources

7 formulas

10 exam-style examples

~59 min

All Hydrology and Water Resources lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Theim Equation within Hydrology and Water Resources. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what theim equation describes physically and when it applies.
State every one of the 7 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: acre-in/hr ≈ cfs makes the rational formula work in US units.

Lecture

Why this section exists. Theim Equation is the part of Hydrology and Water Resources that lets you connect a watershed, aquifer or detention facility to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a rainfall-runoff or well-drawdown calculation with one lookup. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. acre-in/hr ≈ cfs makes the rational formula work in US units. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Row of centrifugal pumps and valved steel piping inside a water pumping station. — Photo 1. Where this shows up in practice: theim equation.
Capstone Studio instructional photograph

Hydrology and Water Resources — Theim Equation: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a watershed, aquifer or detention facility. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 7 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

T	Quantity produced by "T = Kb = transmissivity (ft2/sec)" — read its definition and unit from the handbook line directly above the equation.
b	Quantity produced by "b = thickness of confined aquifer (ft)" — read its definition and unit from the handbook line directly above the equation.
h1, h2	Quantity produced by "h1, h2 = heights of piezometric surface above bottom of aquifer (ft)" — read its definition and unit from the handbook line directly above the equation.
r1, r2	Quantity produced by "r1, r2 = radii from pumping well (ft)" — read its definition and unit from the handbook line directly above the equation.
ln	Quantity produced by "ln = natural logarithm" — read its definition and unit from the handbook line directly above the equation.
H	Quantity produced by "H = height of peizometric surface prior to pumping (ft)" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

2r T `h 2 − h1 j
ln d r2 n
where
Sewage Flow Ratio Curves
Ratio of Minimum or of Peak-to-Average Daily Sewage Flow
P 0.2
Curve A 2: 5
14 +
Curve B: 1
4+ P
18 + P
Curve G:
4+ P
Population in Thousands (P)
Design and Construction of Sanitary and Storm Sewers, Water Pollution Control Federation and American Society of Civil Engineers, 1970.
Reprinted with permission from ASCE.
This material may be downloaded from ncees.org for personal use only. Any other use requires prior permission of ASCE.
Hydraulic-Elements Graph for Circular Sewers
VALUE OF: f and n
ff nf
1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6
n, f VARIABLE WITH DEPTH
n, f CONSTANT
INDEPENDENT OF n, f

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Thiem equation for steady radial flow to a well in a confined aquifer — Theim Equation

A fully penetrating well pumps a confined aquifer of thickness 29 m and hydraulic conductivity 48 m/day. Steady observation heads are 19.5 m at 20 m and 22.0 m at 250.0 m from the well. Compute the pumping rate and the transmissivity.

Given

b = 29 m, K = 48 m/day
r₁ = 20 m, h₁ = 19.5 m
r₂ = 250.0 m, h₂ = 22.0 m

Find

Q and T

Start with the thinking

The Thiem equation applies only to steady state; the Theis solution is needed while drawdown still grows.
For a confined aquifer the heads appear linearly; an unconfined aquifer uses h² terms.

Step-by-step solution

Formula
$Q = \frac{2 π K b ( h _{2} - h _{1} )}{ln ( r _{2} / r _{1} )}$
Head difference
$h_{2} - h_{1} = 22.0 - 19.5 = 2.50 m$
Log term
$ln (250.0/20) = 2.526$
Substituting
$Q = 2 π (48) (29) (2.50) /2.526 = 8, 657 m^{3} / d a y$
Formula
$T = K b$
Substituting
$T = 48 (29) = 1, 392 m^{2} / d a y$
Rate check
$8, 657 m^{3} / d a y = 100.2 L / s$

Answer: Q = 8,657 m³/day (100.2 L/s), T = 1,392 m²/day

Why the other options are there

19,934 m³/day (log₁₀ used)
400.0 m³/day (transmissivity reported as discharge)

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

Example 2

Thiem equation for steady radial flow to a well in a confined aquifer — Theim Equation (2)

A fully penetrating well pumps a confined aquifer of thickness 14 m and hydraulic conductivity 17 m/day. Steady observation heads are 19.5 m at 10 m and 22.1 m at 90 m from the well. Compute the pumping rate and the transmissivity.

Given

b = 14 m, K = 17 m/day
r₁ = 10 m, h₁ = 19.5 m
r₂ = 90 m, h₂ = 22.1 m

Find

Q and T

Start with the thinking

The Thiem equation applies only to steady state; the Theis solution is needed while drawdown still grows.
For a confined aquifer the heads appear linearly; an unconfined aquifer uses h² terms.

Step-by-step solution

Formula
$Q = \frac{2 π K b ( h _{2} - h _{1} )}{ln ( r _{2} / r _{1} )}$
Head difference
$h_{2} - h_{1} = 22.1 - 19.5 = 2.60 m$
Log term
$ln (90/10) = 2.197$
Substituting
$Q = 2 π (17) (14) (2.60) /2.197 = 1, 770 m^{3} / d a y$
Formula
$T = K b$
Substituting
$T = 17 (14) = 238.0 m^{2} / d a y$
Rate check
$1, 770 m^{3} / d a y = 20.48 L / s$

Answer: Q = 1,770 m³/day (20.48 L/s), T = 238.0 m²/day

Why the other options are there

4,074 m³/day (log₁₀ used)
600.0 m³/day (transmissivity reported as discharge)

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

Example 3

Thiem equation for steady radial flow to a well in a confined aquifer — Theim Equation (3)

A fully penetrating well pumps a confined aquifer of thickness 12 m and hydraulic conductivity 37 m/day. Steady observation heads are 24.5 m at 25 m and 27.0 m at 160.0 m from the well. Compute the pumping rate and the transmissivity.

Given

b = 12 m, K = 37 m/day
r₁ = 25 m, h₁ = 24.5 m
r₂ = 160.0 m, h₂ = 27.0 m

Find

Q and T

Start with the thinking

The Thiem equation applies only to steady state; the Theis solution is needed while drawdown still grows.
For a confined aquifer the heads appear linearly; an unconfined aquifer uses h² terms.

Step-by-step solution

Formula
$Q = \frac{2 π K b ( h _{2} - h _{1} )}{ln ( r _{2} / r _{1} )}$
Head difference
$h_{2} - h_{1} = 27.0 - 24.5 = 2.50 m$
Log term
$ln (160.0/25) = 1.856$
Substituting
$Q = 2 π (37) (12) (2.50) /1.856 = 3, 757 m^{3} / d a y$
Formula
$T = K b$
Substituting
$T = 37 (12) = 444.0 m^{2} / d a y$
Rate check
$3, 757 m^{3} / d a y = 43.49 L / s$

Answer: Q = 3,757 m³/day (43.49 L/s), T = 444.0 m²/day

Why the other options are there

8,651 m³/day (log₁₀ used)
410.0 m³/day (transmissivity reported as discharge)

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

Example 4

Thiem equation for steady radial flow to a well in a confined aquifer — Theim Equation (4)

A fully penetrating well pumps a confined aquifer of thickness 25 m and hydraulic conductivity 34 m/day. Steady observation heads are 27.0 m at 40 m and 28.9 m at 90 m from the well. Compute the pumping rate and the transmissivity.

Given

b = 25 m, K = 34 m/day
r₁ = 40 m, h₁ = 27.0 m
r₂ = 90 m, h₂ = 28.9 m

Find

Q and T

Start with the thinking

The Thiem equation applies only to steady state; the Theis solution is needed while drawdown still grows.
For a confined aquifer the heads appear linearly; an unconfined aquifer uses h² terms.

Step-by-step solution

Formula
$Q = \frac{2 π K b ( h _{2} - h _{1} )}{ln ( r _{2} / r _{1} )}$
Head difference
$h_{2} - h_{1} = 28.9 - 27.0 = 1.90 m$
Log term
$ln (90/40) = 0.811$
Substituting
$Q = 2 π (34) (25) (1.90) /0.811 = 12, 513 m^{3} / d a y$
Formula
$T = K b$
Substituting
$T = 34 (25) = 850.0 m^{2} / d a y$
Rate check
$12, 513 m^{3} / d a y = 144.8 L / s$

Answer: Q = 12,513 m³/day (144.8 L/s), T = 850.0 m²/day

Why the other options are there

28,813 m³/day (log₁₀ used)
810.0 m³/day (transmissivity reported as discharge)

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

Example 5

Thiem equation for steady radial flow to a well in a confined aquifer — Theim Equation (5)

A fully penetrating well pumps a confined aquifer of thickness 17 m and hydraulic conductivity 54 m/day. Steady observation heads are 19.5 m at 20 m and 22.3 m at 100.0 m from the well. Compute the pumping rate and the transmissivity.

Given

b = 17 m, K = 54 m/day
r₁ = 20 m, h₁ = 19.5 m
r₂ = 100.0 m, h₂ = 22.3 m

Find

Q and T

Start with the thinking

The Thiem equation applies only to steady state; the Theis solution is needed while drawdown still grows.
For a confined aquifer the heads appear linearly; an unconfined aquifer uses h² terms.

Step-by-step solution

Formula
$Q = \frac{2 π K b ( h _{2} - h _{1} )}{ln ( r _{2} / r _{1} )}$
Head difference
$h_{2} - h_{1} = 22.3 - 19.5 = 2.80 m$
Log term
$ln (100.0/20) = 1.609$
Substituting
$Q = 2 π (54) (17) (2.80) /1.609 = 10, 035 m^{3} / d a y$
Formula
$T = K b$
Substituting
$T = 54 (17) = 918.0 m^{2} / d a y$
Rate check
$10, 035 m^{3} / d a y = 116.1 L / s$

Answer: Q = 10,035 m³/day (116.1 L/s), T = 918.0 m²/day

Why the other options are there

23,106 m³/day (log₁₀ used)
870.0 m³/day (transmissivity reported as discharge)

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

Example 6

Thiem equation for steady radial flow to a well in a confined aquifer — Theim Equation (6)

A fully penetrating well pumps a confined aquifer of thickness 16 m and hydraulic conductivity 51 m/day. Steady observation heads are 25.0 m at 35 m and 27.8 m at 120.0 m from the well. Compute the pumping rate and the transmissivity.

Given

b = 16 m, K = 51 m/day
r₁ = 35 m, h₁ = 25.0 m
r₂ = 120.0 m, h₂ = 27.8 m

Find

Q and T

Start with the thinking

The Thiem equation applies only to steady state; the Theis solution is needed while drawdown still grows.
For a confined aquifer the heads appear linearly; an unconfined aquifer uses h² terms.

Step-by-step solution

Formula
$Q = \frac{2 π K b ( h _{2} - h _{1} )}{ln ( r _{2} / r _{1} )}$
Head difference
$h_{2} - h_{1} = 27.8 - 25.0 = 2.80 m$
Log term
$ln (120.0/35) = 1.232$
Substituting
$Q = 2 π (51) (16) (2.80) /1.232 = 11, 651 m^{3} / d a y$
Formula
$T = K b$
Substituting
$T = 51 (16) = 816.0 m^{2} / d a y$
Rate check
$11, 651 m^{3} / d a y = 134.9 L / s$

Answer: Q = 11,651 m³/day (134.9 L/s), T = 816.0 m²/day

Why the other options are there

26,828 m³/day (log₁₀ used)
590.0 m³/day (transmissivity reported as discharge)

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

Example 7

Thiem equation for steady radial flow to a well in a confined aquifer — Theim Equation (7)

A fully penetrating well pumps a confined aquifer of thickness 22 m and hydraulic conductivity 11 m/day. Steady observation heads are 26.0 m at 30 m and 27.7 m at 250.0 m from the well. Compute the pumping rate and the transmissivity.

Given

b = 22 m, K = 11 m/day
r₁ = 30 m, h₁ = 26.0 m
r₂ = 250.0 m, h₂ = 27.7 m

Find

Q and T

Start with the thinking

The Thiem equation applies only to steady state; the Theis solution is needed while drawdown still grows.
For a confined aquifer the heads appear linearly; an unconfined aquifer uses h² terms.

Step-by-step solution

Formula
$Q = \frac{2 π K b ( h _{2} - h _{1} )}{ln ( r _{2} / r _{1} )}$
Head difference
$h_{2} - h_{1} = 27.7 - 26.0 = 1.70 m$
Log term
$ln (250.0/30) = 2.120$
Substituting
$Q = 2 π (11) (22) (1.70) /2.120 = 1, 219 m^{3} / d a y$
Formula
$T = K b$
Substituting
$T = 11 (22) = 242.0 m^{2} / d a y$
Rate check
$1, 219 m^{3} / d a y = 14.11 L / s$

Answer: Q = 1,219 m³/day (14.11 L/s), T = 242.0 m²/day

Why the other options are there

2,807 m³/day (log₁₀ used)
550.0 m³/day (transmissivity reported as discharge)

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

Example 8

Thiem equation for steady radial flow to a well in a confined aquifer — Theim Equation (8)

A fully penetrating well pumps a confined aquifer of thickness 18 m and hydraulic conductivity 48 m/day. Steady observation heads are 23.5 m at 20 m and 24.5 m at 230.0 m from the well. Compute the pumping rate and the transmissivity.

Given

b = 18 m, K = 48 m/day
r₁ = 20 m, h₁ = 23.5 m
r₂ = 230.0 m, h₂ = 24.5 m

Find

Q and T

Start with the thinking

The Thiem equation applies only to steady state; the Theis solution is needed while drawdown still grows.
For a confined aquifer the heads appear linearly; an unconfined aquifer uses h² terms.

Step-by-step solution

Formula
$Q = \frac{2 π K b ( h _{2} - h _{1} )}{ln ( r _{2} / r _{1} )}$
Head difference
$h_{2} - h_{1} = 24.5 - 23.5 = 1.00 m$
Log term
$ln (230.0/20) = 2.442$
Substituting
$Q = 2 π (48) (18) (1.00) /2.442 = 2, 223 m^{3} / d a y$
Formula
$T = K b$
Substituting
$T = 48 (18) = 864.0 m^{2} / d a y$
Rate check
$2, 223 m^{3} / d a y = 25.73 L / s$

Answer: Q = 2,223 m³/day (25.73 L/s), T = 864.0 m²/day

Why the other options are there

5,118 m³/day (log₁₀ used)
590.0 m³/day (transmissivity reported as discharge)

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

Example 9

Thiem equation for steady radial flow to a well in a confined aquifer — Theim Equation (9)

A fully penetrating well pumps a confined aquifer of thickness 25 m and hydraulic conductivity 32 m/day. Steady observation heads are 23.0 m at 30 m and 25.9 m at 190.0 m from the well. Compute the pumping rate and the transmissivity.

Given

b = 25 m, K = 32 m/day
r₁ = 30 m, h₁ = 23.0 m
r₂ = 190.0 m, h₂ = 25.9 m

Find

Q and T

Start with the thinking

The Thiem equation applies only to steady state; the Theis solution is needed while drawdown still grows.
For a confined aquifer the heads appear linearly; an unconfined aquifer uses h² terms.

Step-by-step solution

Formula
$Q = \frac{2 π K b ( h _{2} - h _{1} )}{ln ( r _{2} / r _{1} )}$
Head difference
$h_{2} - h_{1} = 25.9 - 23.0 = 2.90 m$
Log term
$ln (190.0/30) = 1.846$
Substituting
$Q = 2 π (32) (25) (2.90) /1.846 = 7, 897 m^{3} / d a y$
Formula
$T = K b$
Substituting
$T = 32 (25) = 800.0 m^{2} / d a y$
Rate check
$7, 897 m^{3} / d a y = 91.40 L / s$

Answer: Q = 7,897 m³/day (91.40 L/s), T = 800.0 m²/day

Why the other options are there

18,184 m³/day (log₁₀ used)
510.0 m³/day (transmissivity reported as discharge)

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

Example 10

Thiem equation for steady radial flow to a well in a confined aquifer — Theim Equation (10)

A fully penetrating well pumps a confined aquifer of thickness 11 m and hydraulic conductivity 11 m/day. Steady observation heads are 21.0 m at 35 m and 23.5 m at 240.0 m from the well. Compute the pumping rate and the transmissivity.

Given

b = 11 m, K = 11 m/day
r₁ = 35 m, h₁ = 21.0 m
r₂ = 240.0 m, h₂ = 23.5 m

Find

Q and T

Start with the thinking

The Thiem equation applies only to steady state; the Theis solution is needed while drawdown still grows.
For a confined aquifer the heads appear linearly; an unconfined aquifer uses h² terms.

Step-by-step solution

Formula
$Q = \frac{2 π K b ( h _{2} - h _{1} )}{ln ( r _{2} / r _{1} )}$
Head difference
$h_{2} - h_{1} = 23.5 - 21.0 = 2.50 m$
Log term
$ln (240.0/35) = 1.925$
Substituting
$Q = 2 π (11) (11) (2.50) /1.925 = 987.2 m^{3} / d a y$
Formula
$T = K b$
Substituting
$T = 11 (11) = 121.0 m^{2} / d a y$
Rate check
$987.2 m^{3} / d a y = 11.43 L / s$

Answer: Q = 987.2 m³/day (11.43 L/s), T = 121.0 m²/day

Why the other options are there

2,273 m³/day (log₁₀ used)
700.0 m³/day (transmissivity reported as discharge)

Reference: FE Reference Handbook — Hydrology and Water Resources → Theim Equation

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a watershed, aquifer or detention facility, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Theim Equation contains 7 relations; you must be able to find this page in under 15 seconds.
Exam style: a rainfall-runoff or well-drawdown calculation with one lookup.
Unit rule: acre-in/hr ≈ cfs makes the rational formula work in US units.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

acre-in/hr ≈ cfs makes the rational formula work in US units
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Theim Equation

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

Q=Needs technical review

T=Kb=transmissivity(ft2​/sec)

b=thicknessofconfinedaquifer(ft)

h1​,h2​=heightsofpiezometricsurfaceabovebottomofaquifer(ft)

r1​,r2​=radiifrompumpingwell(ft)

ln=naturallogarithm

H=heightofpeizometricsurfacepriortopumping(ft)

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section