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8/11/2019 Chapt13 Unsteady Well Hydraulics Theis Cooper Jacob
1/4
cmutsvangwa irrigation systems design: dept. of civil and water eng., nust, 16/02/2006 15:23:35
UNSTEADY WELL HYDRAULICS
Cooper-Jacob solution (Modified Theis method)
Determining the aquifer constants, Scand T=kb from a pumping test of onesingle well and one observation well.
From Theis equations:
==u
u
o duu
e
kb
Qhhs
4 (1)
Where: ho-h = drawdown (s) at an observational well at distance rfrom the pumping well
( )uWkb
Qhhs o
4== (2)
( ) ++++= .....!4.4!3.3!2.2
ln577216.0432
uuuuuuW (3)
And u is the argument:
t
r
kb
Su c
2
4
= (4)
The argument u decreases as the pumping time increases. For large values of tand small values r, u becomes small enough so that the terms after the first twobecome negligible. Thus the drawdown (s) can be expresses as:
=
Tt
Sr
T
Qs c
4ln577216.0
4
2
(5)
Rewriting the above equation to base 10 and simplifying:
cSr
Tt
T
Qs
2
25.2log
4
3.2
= u
8/11/2019 Chapt13 Unsteady Well Hydraulics Theis Cooper Jacob
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cmutsvangwa irrigation systems design: dept. of civil and water eng., nust, 16/02/2006 15:23:35
Where rwis the radius of the pumping well.
A plot of equation 6 on a semi-log paper gives a best fit straight line (Fig. 1) andthe drawdown difference per log cycle, s is:
TQs
43.2=
Therefore the constant T=kb can be computed.
Fig. 1
From equation 6, s=0 (at t0) when 125.22
0 =cSr
Tt
The time tois found by extending the best fit line to meet the zero drawdown line.
From this equation, 125.22
=c
Sr
Tt the storage coefficient can be determined:
2
25.2
r
TtSc =
Determining the aquifer constants, Scand T=kb from a pumping test of onesingle well and several observation wells
The pumping will done for a fixed time t and draw downs are recorded in severalobservation wells.
From
One log-cycleDrawdown,s,m
10
s
100t
t0
(0, 0)
2
8/11/2019 Chapt13 Unsteady Well Hydraulics Theis Cooper Jacob
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cmutsvangwa irrigation systems design: dept. of civil and water eng., nust, 16/02/2006 15:23:35
c
Sr
Tt
T
Qs
2
25.2log
4
3.2
=
=
cSrTt
TQ
225.2log
43.2
+=
2
1log
25.2log
4
3.2
rS
Tt
T
Q
c
+= rS
Tt
T
Q
c
log21log25.2
log4
3.2
=
= r
S
Tt
T
Q
c
log225.2
log4
3.2
rTS
Tt
T
Q
C
log24
3.225.2log4
3.2
=
=r
TS
Tt
T
Q
C
log22
3.225.22log
4
3.2=
constant i.e t=const
Therefore above a plot of the above equation a semi-log paper gives a
straight line andTQ
23.2 is the gradient and also equal to s , which is also
equal to one log cycle.
T
Qs
2
3.2=
Therefore T can be computed as:
s
QkbT
==2
3.2
3
8/11/2019 Chapt13 Unsteady Well Hydraulics Theis Cooper Jacob
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cmutsvangwa irrigation systems design: dept. of civil and water eng., nust, 16/02/2006 15:23:35
At s=0, r=ro (Fig. 2) and fromc
Sr
Tt
T
Qs
2
25.2log
4
3.2
=
co Sr
Tt
T
Q2
25.2log
4
3.20
= , and therefore the term 125.22
0
=cSr
Tt
Thus the storage coefficient can be computed from:
2
0
25.2
r
TtSC=
Fig.2
References
1. Raghunath H. M., (1991), Hydrology, Wiley Eastern, Bombay.2. Shaw E. M., (1988), Hydrology in Practice, Van Nostrand Reinhold,
Wokingham3. Viessman J.R., Lewis G. L., and Knapp J.W., (1989), Introduction to
hydrology, Harper Collins, USA4. Wanielista M., (1990), Hydrology and Water Quality Control, John Wiley,
Canada5. Wilson E.M., (1990), Engineering Hydrology, Macmillan Education, UK
ro at s=0cycle1-log
r m
s, m
4
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