GW Engineering EXAM I FALL 2010
NAME _
PROBLEM #1 - 25 points USE UNITS of METERS SECONDS and GRAMS
Prepare a water budget for the year 2009 for the unconfined sedimentary aquifer that
constitutes the basin illustrated below. The basin is surrounded by a

uA = 2.5 x 10-2 (1/4uA = 10)
t = 6 min
= 0.06
W (u, ) = 1
s = 0.55 ft
SLIDE LATERALLY DO NOT SHIFT UP AND DOWN
T DOES NOT CHANGE WITH TIME BUT S DOES
= 0.06
W (u, ) = 1
s = 0.55 ft
t = 53 min
uB = 0.25 (1/4uB =1)
1
Match
early time
Calculate T S Kv Kh Sy

UPPER BOUNDARY IS
CONSTANT HEAD
EQUAL TO WATER TABLE ELEVATION
LIKE MANY RESERVOIRS PROVIDING AND RECEIVING WATER
no exaggeration - head distribution
with aquitard
NOTE THAT
POROSITY DOES NOT EEFECT FLOW FIELD
IT ONLY CHANGES TRAVEL TIME
no exaggeration -

If we only consider advection and
start with a "point" of material with Co=1000mg/l
A point has no volume so can it have a concentration? So why do we say a point?
K = 0.1 cm/sec
dh = 10 cm
dl = 100 cm
= 0.2
How long will it take for the material to move

HOW MUCH WATER DO WE NEED?
Take a moment to consider and estimate
volume per person per day
One Person?
One Person ~ 3 Liters/Day To Maintain Essential Body Fluids (0.8 Gal)
In some arid locations people exist with this as their total consumption
Flushing

GW Engineering FINAL EXAM FALL 2010
_
NAME
NOTE: Supplemental Materials pages 9-10
PROBLEM #1 - 25 points USE UNITS of METERS and DAYS
Write your answer on the following page. SHOW YOUR WORK
The hydrograph shown below is from a basin that is 5 kilometers

NAME
GW Engineering FINAL EXAM 2010
NOTE: Supplemental Materials pages 9'12
PROBLEM #1 -25 points USE UNITS of METERS and DAYS
Write vour answer on the followinq paqe. SHOW YOUR WORK
The hydrograph shown below is from a basin that is 5 kilometers by 5 kil

GW Engineering EXAM II FALL 2010
NAME _
NOTE: Supplemental Materials pages 9-12
Page 9 Piper Graph Paper Page 10-11 Chemical Constants
Page 12 Well Functions
PROBLEM #1 - 25 points USE UNITS of FEET and DAYS
WRITE YOUR ANSWER TO THE FOLLOWING QUESTION ON

http:/inside.mines.edu/~epoeter/_GW/03Budget2/TCB-StreamGage-GWclass.xls
The volume of water
discharged over a
period of time is the
area under the curve
_ft3_ sec = ft3
sec
If you estimate the
average flow rate
and multiply by the #
secs in year you
secs