Aquifer and Storage Properties
Contents
import numpy as np
import pandas as pd
import panel as pn
pn.extension("katex", "mathjax")
3. Aquifer and Storage Properties#
(The contents presented in this section were re-developed principally by Dr. P. K. Yadav. The original contents are from Prof. Rudolf Liedl)
3.1. Tutorial Problem 1#
# The given information of the problem are
L = 2 # km as length of garten
W = 1 # km as Width of garten
DP = 3 # m change in pressure head
n = 0.3 # porosity that we assume to be 30%
A = L * W *1e6 # m^2, Area unit convered.
DV = A * DP # m^3, increase in total volume due to change in pressure head
AW = n * DV # m^3, additional Water volumne
print("Park Area is {0:1.1f}".format(A),"m\u00b2")
print("Increase in total volume: {0:1.1E}".format(DV),"m\u00b3")
print("Additional water volume: {0:1.1E}".format(AW),"m\u00b3")
Park Area is 2000000.0 m²
Increase in total volume: 6.0E+06 m³
Additional water volume: 1.8E+06 m³
3.2. Tutorial Problem 2#
3.3. Tutotorial Problem 3#
# Given data
d = 230 # m, diameter of the aquifer
m = 61 # m, thickness of the aquifer
Ss = 1.2 * 10e-5 # 1/m, specific storage
DP_h = 0.75 # m, pressure head difference
# Calculations
A = np.pi *(d/2)**2 # m^2, area of the aquifer
Vt = A*m # m^3 Total volume of the aquifer
DV_w = Ss*Vt*DP_h # m^3, additional water volume
print("The Aquifer Area is {0:1.2E}".format(A),"m\u00b2")
print("The Total Volume is {0:1.2E}".format(Vt),"m\u00b3")
print("The Additional Water is {0:1.2f}".format(DV_w),"m\u00b3")
The Aquifer Area is 4.15E+04 m²
The Total Volume is 2.53E+06 m³
The Additional Water is 228.10 m³
3.4. Tutorial Problem 4:#
#
Head = ["Sub-Area", "Size, (Km2)", "drawdown (m)", "Change in water volume (m3)"]
Sub_Area = ["A", "B", "C", "D"] #name
Size = [36, 18, 72, 85] # km^2, area
Drawdown = [0.85, 1.09, 1.65, 2.37] # m, equivalent to change in pressure head
Q4t = np.transpose([Sub_Area, Size, Drawdown])
data = {"Sub-Area": Sub_Area, "Size, (Km2)": Size, "drawdown (m)": Drawdown, }
df1= pd.DataFrame(data)
df1.set_index('Sub-Area')
| Size, (Km2) | drawdown (m) | |
|---|---|---|
| Sub-Area | ||
| A | 36 | 0.85 |
| B | 18 | 1.09 |
| C | 72 | 1.65 |
| D | 85 | 2.37 |
3.5. Tutorial Problem 4 – Solution#
For details check lecture slides L03 - 28, 29 and 31
In unconfined aquifer Stortavity (\(S\)) is used instead of storage coefficient \(S_s\). They both are related with thickness (\(m\)) as:
and so in unconfined aquifer, we get:
with
\(\Delta V_w\) = change in water volume [L\(^3\)³]
\(A\) = Domain area [L\(^2\)]
\(\Delta \psi\) = change in pressure head [L]
# Given information
s = 0.19
Size = [50, 18, 72, 85] # km^2, area
Drawdown = [0.85, 1.09, 1.65, 2.37] # m, equivalent to change in pressure head
# Solution part
Vol_cha = s*np.multiply(Size, Drawdown)*10**6 # m^3, vol change = s*area size * drawdawn
# output printing
data2 = {"Sub-Area": Sub_Area, "Size, (Km2)": Size, "drawdown (m)": Drawdown, "Change in volume (Km3)":Vol_cha/1e9 }
df2= pd.DataFrame(data2)
df3= df2.set_index('Sub-Area')
print(df3, "\n")
print("The total change in the volume of water is: {0:0.3f} Km\u00b3".format(sum(Vol_cha/1e9)))
Size, (Km2) drawdown (m) Change in volume (Km3)
Sub-Area
A 50 0.85 0.008075
B 18 1.09 0.003728
C 72 1.65 0.022572
D 85 2.37 0.038275
The total change in the volume of water is: 0.073 Km³
3.6. Tutorial Problem 5:#
A confined aquifer is considered in this problem. Specific storage and total porosity equal \(7.5\times 10^{-6}\) 1/m and 30%, respectively. What is the compressibility of the porous medium? (compressibility of water: \(4.6\times 10^{-10}\) m\(^2\)/N, density of water: 998 kg/m\(^3\))
3.7. Tutorial Problem 5 – Solution#
For details check slide nr. L03-28
Specific Storage,
with: \(n\) = Total porosity [-]
\(g\) = acceleration due to gravity [L/T\(^2\)]
\(\alpha_w\) = compressibility of water [LT\(^2\)/M]
\(\alpha_{pm}\) = compressibility of porous medium [LT\(^2\)/M]
\(\rho_w\) = density of water [M/L\(^3\)]
Solve for \(\alpha_{pm}\): \(\frac{S_s}{\rho_w g} -n\alpha_w = \alpha_{pm}\)
# Given data
n = 0.3 # unitless, total porosity
rho_w = 998 # kg/m3, density of water
g = 9.81 # m/s2, accl. due to gravity
alpha_w = 4.6*1e-10 # m2/N, compressibility of water
S_s = 7.5*1e-6 # 1/m, specific storage
# calculated land subsidence (LS)
alpha_pm5 = S_s/(rho_w*g) - n*alpha_w
print("The Compressibility of Porous mdeid is {0:0.2E}".format(alpha_pm5), "m\u00b2/N")
The Compressibility of Porous mdeid is 6.28E-10 m²/N
3.8. Tutorial Problem 6:#
Due to water extraction from a confined aquifer the pressure head is lowered by 183 m. The following aquifer parameters are available: storage coefficient = \(5\times10^{-4}\), total porosity = 0.33, thickness (before water extraction) = 80 m, compressibility of the porous medium = \(6.9\times 10^{-8}\) m²/N and the density of water is 998 kg/m\(^3\)
What is the amount of land subsidence resulting from the water extraction?
3.9. Tutorial Problem 6 – Solution#
For details see slide nr. L03-25 and 26
Change in total volume due to \(\Delta p_{pm}\) :
with:
\(\alpha_{pm}\) = compressibility of porous medium [LT\(^2\)/M]
\(V_T\) = total volume [L\(^3\)]
\(\rho_w\) = density of water [M/L\(^3\)]
\(g\) = acceleration of gravity [L/T\(^2\)]
\(\Delta \psi\) = change in pressure head [L]
\(\Delta V_T = A\times\Delta m\) and
\(V_T = A\times m\) and
with \(A\) = area of the aquifer [L/T\(^2\)] and
\(m\) = Thickness of the aquifer [L].
Substituting these relation in the above equation we get:
# Given data
alpha_pm = 6.9 * 1e-8 # m2/N, compressibility of porous medium
m = 80 # m, thickness
rho_w = 998 # kg/m3, density of water
g = 9.81 # m/s2, accl. due to gravity
DP_h = 30 # m, change in pressure head
# calculated land subsidence (LS)
LS = alpha_pm*m*rho_w*g*DP_h
print("The land subsidence is {0:0.2f}".format(LS), "m")
The land subsidence is 1.62 m
3.10. HOME WORK PROBLEMS#
3.10.1. Homework Problem 1#
The pressure head in an aquifer extending over 200 km\(^2\) is decreased by 1.60 m. Determine the loss of groundwater in the aquifer for two scenarios: The aquifer is unconfined (storage coefficient 0.13). The aquifer is confined (storage coefficient 0.0005).
3.10.2. Homework Problem 2#
Conduct a sieve analysis for a dried soil sample (see data in the table below)
Draw the granulometric curve (cumulative mass distribution) and briefly characterise the sediment with regard to its major constituent(s).
What is the coefficient of uniformity?
#
Head = ["mesh size [mm]", "residue in the sieve [g] ", "∑ total", "∑ / ∑total"]
Size = [6.3, 2, 0.63, 0.2, 0.063, "< 0.063 /cup"]
residue = [11, 62, 288, 189, 42, 8]
data3 = {"mesh size [mm]": Size, "residue in the sieve [g] ": residue}
df4= pd.DataFrame(data3)
df4.set_index("mesh size [mm]")
| residue in the sieve [g] | |
|---|---|
| mesh size [mm] | |
| 6.3 | 11 |
| 2 | 62 |
| 0.63 | 288 |
| 0.2 | 189 |
| 0.063 | 42 |
| < 0.063 /cup | 8 |