Aquifer Heterogeneity/Anisotropy
Contents
import numpy as np
import matplotlib.pyplot as plt
import panel as pn
pn.extension('katex', 'mathjax')
6. Aquifer Heterogeneity/Anisotropy#
6.1. Tutorial Problem 12: Hydrologic Triangle#
6.2. Solution of Tutotrial Problem 12#
The following 4 steps are to be performed:
Step I : Connects all the points
Step II : Divide the connected lines at equal head-level (here = 1 m)
Step III : Join all the equal head lines
Step IV : Mark the flow direction from higher head towards lower head
6.3. Tutorial Problem 13: Flow Nets#
6.4. Solution of Tutorial Problem#
This is to be sketched and demonstrated.
6.5. Tutorial Problem 14 (special topic)#
From the laboratory test the degree of saturation(\(\theta\)) of the unsaturated core (temperature = 9\(^\circ C\)) sample was found to be 30% and relative permeability (\(k_r\)) is assumed to be 0.1. From the grain analysis the sample was determined to be predominantly medium sand (intrinsic permeability, \(k = 1.61 \times 10^{-7}\) cm\(^2\)). Provided that density (\(\rho\)) and dynamic viscosity of water (\(\mu\)) at 9\(^\circ C\) is 999.73 kg/m\(^3\) and 0.0013465 N\(\cdot\)s/m\(^2\) respectively, find the conductivity of the sample. What will be the conductivity of the same sample when the moisture content is 1% (\(k_r \approx 0.001\)) and 80% (\(k_r \approx 0.4\)). Explain the effect of moisture content on the sample.
6.6. Solution of Tutorial Problem 14#
Lecture contents on the topic in L02- slides 02, 22 & 26
Hydraulic conductivity of the unsaturated sample (\(\theta < 100\%\)) can be obtained from the following expression:
# Given
kr_30 = 0.1 # (-), relative permeability for moist. cont. 30%
i_p = 1.61 * 10**-7 # cm^2, intrinsic permeability
rho = 999.73 # kg/m^3, Sample density
mu = 0.0013467 # N-s/m^2, dynamic visc.
g_c = 9.81 # N/kg, force unit used for gravitational constant
# Solutions 1
i_pm = i_p/10000 # m^2 unit conversion for int. permeab.
K_30 = (i_p*rho*g_c/mu)*kr_30
# Solution 2 when moisture content is 1% and 80%
kr_1 = 0.001 # (-), relative permeability for moist. cont. 1%
kr_80 = 0.4 # (-), relative permeability for moist. cont. 80%
K_1 = (i_p*rho*g_c/mu)*kr_1
K_80 = (i_p*rho*g_c/mu)*kr_80
# output
print("The conductivity of water when moisture content is 30% is: {0:1.1e}".format(K_30),"m/s \n")
print("The conductivity of water when moisture content is 1% is: {0:1.1e}".format(K_1),"m/s \n")
print("The conductivity of water when moisture content is 80% is: {0:1.1e}".format(K_80),"m/s \n")
print("The conductivity of media increases very rapidly with increase of moisture content")
The conductivity of water when moisture content is 30% is: 1.2e-01 m/s
The conductivity of water when moisture content is 1% is: 1.2e-03 m/s
The conductivity of water when moisture content is 80% is: 4.7e-01 m/s
The conductivity of media increases very rapidly with increase of moisture content
6.7. Tutorial Problem 15#
6.8. Solution Tutorial Problem 15#
# Given
K_s = 2 # cm/d # saturated conductivity
al_a = 0.04 # 1/cm, fit constant
Ph_a = -100 # cm, pressure head at A
Ph_b = -90 # cm, pressure head at B
Z_a = 300 # cm, elevation head at A from datum
Z_b = 200 # cm, elevation head at B from datum
# Solution 1
Ph_m = (Ph_a+Ph_b)/2 # mean pressure head
K_psi = K_s*np.exp(al_a*Ph_m) # cm/d, from the given model
#Solution 2
H_A = Ph_a+Z_a # cm, hydraulic head at A
H_B = Ph_b+Z_b # cm, hydraulic head at B
dh_dz = (H_B - H_A)/(Z_b - Z_a) # (-), hydraulic head gradient
q_z = -K_psi*dh_dz # cm/d, Darcy velocity
print("The unsaturated conductiviy of the sample is: {0:1.3f}".format(K_psi), "cm/d\n")
print("The Darcy velocity is: {0:1.3f}".format(q_z), "cm/d\n")
print("The negative sign indicates the direction opposite to increase in z.")
The unsaturated conductiviy of the sample is: 0.045 cm/d
The Darcy velocity is: -0.040 cm/d
The negative sign indicates the direction opposite to increase in z.