Question 1) You have made measurements of groundwater heights in late August at
ID: 803404 • Letter: Q
Question
Question 1) You have made measurements of groundwater heights in late August at two locations in Sagehen creek. At 8000 ft elevation the groundwater table is 35 ft below the soil surface. At 7000 ft the groundwater table is 15 ft below the soil surface. The points are 2000 ft apart horizontally (i.e. on a planar map). Use information from the soils table in HW #3.
1a) What is the effective length (L) of travel in ft (hint use the geometry of the surface)?
1b) What is the Q in cubic feet per hour if the groundwater moves perpendicular through an area that is 10 ft deep and 1000 ft wide and made out of sand?
1c) What is the Q in cubic feet per hour if the groundwater moves perpendicular through an area that is 10 ft deep and 1000 ft wide and made out of clay?
1d) What is the Darcy velocity in part b and part c (hint: think about the units!) in feet per day?
1e) Assuming that discharge to the stream in late august is made entirely from groundwater behaving as Darcy flow. What is the hydraulic conductivity in cm/hr if the discharge in the stream is 0.12 cfs? What type soil material is that most likely to be (hint: look at the table in HW#3)?
CLAY=0.03(cm/hr)
SAND= 11.78(cm/hr)
Range of elevation Average elev (ft) 8250 7750 7250 6750 6250 5750 Minimum (ft Maximum (ft Area (km^2) % of total area 8000 7500 7000 6500 6000 5500 8500 8000 7500 7000 6500 6000 1.5 2.3 5.4 5.9 7.8 4.3 5% 5% 20% 22% 29% F 16% FExplanation / Answer
1b. height of aquifer h=10 ft
length of aquifer L=1000ft
Cross section of the aquifer A= Lh
=10 *1000=10,000 ft
Q=-KA h/L
K= hydraulic conductivity for sand=11.78cm/hr=0.386 ft
Q= discharge of aquifer=??
Q=-0.386*10,000*10/1000
=-0.386*10,000*0.01
=38.6ft3/hr
1c. Hydraulic conductivity of clay K=0.03 cm/hr=0.00098 ft/hr
Length of aquifer L=1000 ft
Height of aquifer h=10 ft
Cross section area of aquifer= L * h=1000*10=10,000 ft
Now Q=-kA*h/L
=-0.00098*10,000*10/1000
=0.00098*10,000*0.01
=0.098 ft3/hr
1d. Darrcy velocity v=Q/A
Q1= 38.6 ft3
Q2=0.098ft3/hr=0.098/0.041=2.4 ft3/day
A1=A2=10,000ft
Now Darcy velocity v1=Q1/A1
v1=38.6/10,000=3.86*10^-3ft/hr
Since 1 hour=0.041 day
therefore v1=38.6*10-3/0.041
=941.46*10-3 ft/day
=0.941 ft/day
v2=Q2/A2
=2.4ft3/day/10,000 ft
=2.4*10^-4 ft3/day
1e. Q=discharge of stream=0.12cfs
length of aquifer L=2000 ft
h1=35 ft
h2=15 ft
hydraulic gradient dh/dl= h2-h1=35-15=20 ft
k= hydraulic conductivity??
Cross sectional area of aquifer= length* height of aquifer
=2000*20
=40,000 ft
Since darcy law=Q=-KAdh/dl
Q/(A*dh/dl)=K
=0.12/40,000*20
=0.12/800,000
=1.5*10-7ft/s
Now convert feet into cm/hr
1.5*10-7ft/s
We know 1feet=30.48 cm
1hr=3600 s
=1.58*10-7*30.48/3600
=48.15*10-7/3600
=0.013*10-7cm/hr
=0.013*10_7cm/hr
This value is more close to hydraulic conductivity of clay(0.003 cm/hr) so the aquifer is made up of clay
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