USING TRANSIENT RESPONSE OF SYSTEMS TO PUMPING
Benefits/Disadvantages
•can estimate storativity values
•when estimating aquifer properties, we get results @ early time
•easier to detect extraneous effects
•analysis is more complex
Assumptions
1. homogeneous, isotropic, infinite areal extent
2. pumping well fully penetrates and receives water from
the entire thickness of the aquifer
3. Transmissivity is constant in space and time
4. well has infinitesimal diameter
5. water removed from storage is discharged instantaneously
with decline in head
Q
pumping well
observation well 1
observation well 2
s2
s1
r1
r2
1
Theis Equation
Drawdown given Q T S r t
s = drawdown [L]
ho = initial head @ r [L]
h = head at r at time t [L]
t = time since pumping began [T]
r = distance from pumping well [L]
Q = discharge rate [L3/T]
T = transmissivity [L2/T]
S = Storativity [ ]
2
0.01
0.1
u, Curve A
1
10
10
A
W(u), Curve A
u, Curve B 0.1
1
W(u), Curve B
1
1
B
0.1
0.1
0.01
0.001
0.01
Q
log s drawdown
at the red observation well …..
log time
3
Predict Drawdown Using Theis Equation
Class
Are these
picks:
reasonable
T
and
S how dor we know?
t
Q
Take 3 minutes to calculate: s
What value did you get for s?
Theis Formula cannot be solved directly for T and S
from observations of drawdown (why?)
You cannot rearrange
to solve for T and S
Consequently, we use curve matching techniques
Type Curve is W(u) vs u
OR W(u) vs 1/u
2
s vs t
Plot s vs 1/t or s vs r /t for field data OR
Type curve & field data must be plotted on same log-log paper
Field curve is overlaid on Type Curve
Axes must be kept parallel
Best match of curves is found
Pick any convenient point
read corresponding W(u), u, s and r2/t
Solve Theis Equation for T & S
4
0.01
0.1
u, Curve A
1
10
10
A
W(u), Curve A
W(u), Curve B
1
1
u, Curve B 0.1
1
B
0.1
0.1
0.01
0.001
0.01
log s drawdown
to match the W(u) curve in this format, plot log s vs log 1/t
log 1/time
or sometimes for convenience
log r2/t
5
An example test from Ohio
Pumping well discharges
@ 500 GPM
from a
100ft thick aquifer
Observation well is
200 ft away
Plot as s vs r2/t
on same scale of log paper
as W(u) vs u
Jacob simplified the Theis equation for large t and small r
using only the first 2 terms of the W(u) function
s = drawdown [L]
ho = initial head @ r [L]
h = head at r at time t [L]
t = time since pumping began [T]
r = distance from pumping well [L]
Q = discharge rate [L3/T]
T = transmissivity [L2/T]
S = Storativity [ ]
uSmall
Why
What
is small
are
is
is small?
later
uwhen
< 0.03
terms
r is small
not
needed?
and
Which
t is yields
large, <thus
1%the
error
latter
terms contribute little
Simplified expression
(note constant due to
switch from ln to log)
6
Jacob’s simplified Theis equation
Can you get to
Q
(− 0.5772 − ln u )
note : ln u = 2.3 log u
4πT
Q
(− 2.3 log 1.7822 − 2.3 log u )
s=
4πT
2.3Q
(− log 1.7822 − log u ) note : − log u = log 1
s=
4πT
u
2.3Q ⎛
1⎞
a
s=
⎜ − log 1.7822 + log ⎟ note : log (a − b) = log
4πT ⎝
u⎠
b
⎛ 1 ⎞
⎜
⎟
2.3Q ⎛
1
⎞ 2.3Q
⎜ u ⎟
s=
log
log
1
.
7822
log
−
=
⎜
⎟
u
4πT ⎝
⎜ 1.7822 ⎟
⎠ 4πT
⎜
⎟
⎝
⎠
1
⎛
⎞
⎜ 2 ⎟
r S ⎟
⎜
r 2S
2.3Q
4Tt
⎛
⎞
⎜
⎟ 2.3Q
note : u =
s=
log⎜ 4Tt ⎟ =
log⎜
⎟
r 2S ⎠
4Tt
4πT
1.7822
4πT
1
.
7822
⎝
⎜
⎟
⎜⎜
⎟⎟
⎝
⎠
2.3Q
⎛ 2.25Tt ⎞
s=
log⎜ 2 ⎟
4πT
⎝ r S ⎠
s=
Since Q r T & S are constant, s vs log t is a straight line:
a graph of s (drawdown)vs log(t) (time)
can be described by a straight line
y=s
x = log(t)
y = mx + b
2.25T
2.3Q
2.3Q
log 2
b = intercept =
r S
4πT
4πT
2.3Q
2.3Q
2.25T
s=
log(t) +
log 2
4πT
4πT
r S
m = slope =
7
For Jacob’s simplification of the Theis equation we find an easy
way to estimate T:
s=
2.3Q
2.3Q
2.25T
log(t) +
log 2
4πT
4πT
r S
if we consider the slope in terms of
Δs (changein drawdownover one log cycle of t)
2.3Q
Δs
given log(t) for one log cycle of t is 1
=
log(t) 4πT
2.3Q
and rearranging to solve for T :
Δs =
4πT
T=
2.3Q
4πΔs
at the time intercept t o the straight line approximation of s = 0
2.3Q
2.3Q
2.25T
2.3Q
log(t o ) +
log 2
pulling out
:
4πT
4πT
r S
4πT
2.3Q ⎛
2.25T ⎞
0=
⎜ log(t o ) + log 2 ⎟
4πT ⎝
r S ⎠
with zero on the left hand side the constant is irrelevant
0=
2.25T ⎞
2.25T
⎛
0 = ⎜ log(t o ) + log 2 ⎟
subtract log 2 from both sides
r S ⎠
r S
⎝
2.25T
- log 2 = log(t o )
exponentiate both sides
r S
r2S
rearrange to solve for S
= to
2.25T
2.25Tt o
S=
r2
8
In short:
Jacob’s straight line method for u<0.03
where:
h=
s = drawdown over 1 log cycle of time
to = time intercept for zero drawdown
For the Ohio example Q=500GPM, b=100ft, r=200ft
Plot s vs t
s(ft)
4 ft22 minutes
see next slide
With
a partner,
take
to
T = 13552
~ 1.4 x 10
S = 1.98 x 10
approx.
calculate
T -4and
S 2x10-4 see next slide
to = 2.6 x 10-4 day
one log cycle
h=
s = 1.3 ft
9