# 	**Problem 10.1**
# A saline solution with a concentration of 1823 mg/L is introduced into a 
# 2-m-long sand column in which the pores are initially filled with 
# distilled water. If the solution drains through the column at an average 
# linear velocity of 1.43 m/day and the dynamic dispersivity of the sand 
# column is 15 cm, what would the concentration of the effluent be 0.7 day 
# after flow begins?
Co = 1823 mg/L
Lo = 2 m
v = 1.43 m/day
alpha = 15 cm
t = 0.7 day
DL = alpha*v
Conc = (Co/2)*(erfc( (Lo - v*t)/(2*sqrt(DL*t)))+exp(v*Lo/DL)*erfc( (Lo + v*t)/(2*sqrt(DL*t)))); mg/L

# 	**Problem 10.3**
# A landfill is leaking an effluent with a concentration of sodium of 1250
# mg/L. It seeps into an aquifer with a hydraulic conductivity of 9.8 
# m/day, a gradient of 0.0040, and an effective porosity of 0.15. A down-
# gradient monitoring well is located 25 m from the landfill. What would 
# the sodium concentration be in this monitoring well 300 days after the 
# leak begins?
Co = 1250 mg/L
Ko = 9.8 m/day
gradient = 0.004
ne = 0.15
Lo = 25 m
t = 300 days
v = Ko * gradient/ne; m/day
log_constant = 1 /m
constant = 0.83 m
alpha = constant*(log10(log_constant*Lo))^2.414
DL = alpha*v
Conc = (Co/2)*(erfc( (Lo - v*t)/(2*sqrt(DL*t)))+exp(v*Lo/DL)*erfc( (Lo + v*t)/(2*sqrt(DL*t)))); mg/L

# 	**Problem 10.5**
# What is the relative velocity of a solute front of a solute-soil system 
# with a distribution coefficient of 83 mL/g, a dry bulk density of 2.12 
# g/cm3, and a volumetric water content of 0.26?
Kd = 83 ml/g
density_bulk = 2.12 g/cm^3
theta = 0.26
relative_velocity = 1/(1+(density_bulk/theta)*Kd)   # Eq. 10-15

# 	**Problem 10.7**
# A capture well is pumping at a rate of 37,000 ft3/day from a confined 
# aquifer with a hydraulic conductivity of 920 ft/day, an initial 
# hydraulic gradient of 0.0027, and an initial saturated thickness of 
# 40 ft. 
Q = 37000 ft^3/day
Ko = 920 ft/day
gradient = 0.0027
b = 40 ft

# Part A: What is the maximum width of the capture zone?
Width = Q/(2*Ko*b*gradient); ft    # Eq. 10-18

# Part B: What is the distance from the well to the stagnation point?
Xo = - Q/(2*pi()*Ko*b*gradient); ft   # Eq. 10-17