Refer to the stratigraphic column of the cross section of the North Atlantic Coa
ID: 802159 • Letter: R
Question
Refer to the stratigraphic column of the cross section of the North Atlantic Coastal Plain aquifer system shown on the attached page. To simplify, lump the Chesapeake and surficial aquifers into a single hydro- stratigraphic unit (a single hydrologic "layer'). Assume that the Castle-Hayne aquifer is continuous; although it is not shown to be in the attached cross section, the aquifer is indeed continuous in 3-D. The hydraulic conductivities of the aquifer and confining units have been measured by pump tests, and average values are reported in the table below. 2. Unit K (ft day ) 1.0 10.0 3.5 0.5 0.002 Chesapeake&Surficial; Aquifer Castle-Hayne Aquifer Severn-Magothy Aquifer Potomac Aquifer All confining units (A) What is the equivalent vertical hydraulic conductivity of the entire layered sequence at the Eastern edge of the cross section [3 pts]? (B) What is the equivalent horizontal hydraulic conductivity at this location [3 pts] (C) What is the hydraulic anisotropy (KxK, given in a ratio such as 2:1) of the layered sequence at this location [1 pt]? (D) If the hydraulic conductivity of the confining units increased to 2.0 ft/day, what would the equivalent horizontal and vertical conductivities be [4 pts? What is the anisotropy [1 pt? (E) If the hydraulic conductivity of the confining units decreased to 10 ft/day, what would the equivalent horizontal and vertical conductivities [4 pts)? What is the anisotropy [1 pt]?
Explanation / Answer
A)Equivalent vertical hydraulic conductivity = [ k1z1 + k2z2 +k3z3....]/[z1+z2+z3....]
= [0.002*2800+0.5*900+0.002*500+3.5*500+1*200+10*100 ]/5000
= 0.681 ft/day
B) Equivalent horizontal hydraulic conductivity = [x1+x2+x3....]/[x1/k1 +x2/k2+x3/k3...]
= 50mile / [10mile/0.002+ 35mile/1 +5mile/3.5]
=0.01 ft/day
c) Anisotropy = kx/kz = 0.01/0.681 = 0.0147
d)Equivalent vertical hydraulic conductivity = 0.61+2= 2.61 ft/day
Equivalent horizontal hydraulic conductivity = 50mile / [10mile/2.002+ 35mile/3 +5mile/5.5]
= 2.85 ft/day
anisotropy = 2.85/2.61 = 1.092
e) Equivalent vertical hydraulic conductivity = [ k1z1 + k2z2 +k3z3....]/[z1+z2+z3....]
= [10^-6*2800+0.5*900+0.002*500+3.5*500+1*200+10*100 ]/5000
= 0.680 ft/day
Equivalent horizontal hydraulic conductivity = [x1+x2+x3....]/[x1/k1 +x2/k2+x3/k3...]
= 50mile / [10mile/10^-6+ 35mile/1 +5mile/3.5]
=5*10^-6 ft/day
anisotropy = kx/kz = 5e-6/0.68 = 7.35*10^-6 ft/day
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