# 	**Problem 4.1**
# A fluid in an aquifer is 6.5 m above a reference datum, the fluid 
# pressure is 1800 N/m2, and the flow velocity is 3.4e-5 m/s. 
# The fluid density is 1.01e3 kg/m3.

# Part A:  What is the total energy per unit mass?
Zo = 6.5 m
Po = 1800 N/m^2
Vo = 3.4e-5 m/s
Density = 1.01e3 kg/m^3
Energy_per_mass = Vo^2/2 + grav()*Zo + Po/Density

# Part B:  What is the total energy per unit weight?
Energy_per_weight = Energy_per_mass / grav()

# 	**Problem 4.3**
# A piezometer is screened 723.4 m above mean sea level. 
# The point-water pressure head in the piezometer is 17.9 m 
# and the water in the aquifer is fresh at a temperature of 20 degrees C.

# Part A:  What is the total head in the aquifer at the point 
# where the piezometer is screened?
Zo = 723.4 m
Hp = 17.9 m
Htotal = Hp + Zo

# Part B:  What is the fluid pressure in the aquifer at the point 
# where the piezometer is screened?
Density = 998.203 kg/m^3  # at 20 degrees C
Po = Density * grav() * Hp

# 	**Problem 4.5**
# A piezometer in a saline water aquifer has a point-water pressure 
# head of 18.73 m. If the water has a density of 1022 kg/m3 and is 
# at a field temperature of 18 degrees C, what is the equivalent 
# fresh-water pressure head?
Hp_salt = 18.73
Density_salt_water = 1022 kg/m^3
Density_fresh_water = 998.595 kg/m^3   # at 18 degrees C
Po = Density_salt_water * grav() * Hp_salt
Hp_fresh = Po / (Density_fresh_water * grav() )

# 	**Problem 4.7**
# A sand aquifer has a median pore diameter of 0.232 mm. The fluid density
# is 1.003e3 kg/m^3 and the fluid viscosity is 1.15e-3 N-s/m^2. If 
# the flow rate is 0.0067 m/s, is Darcy's law valid? What is the reason 
# for your answer? 
diameter = 0.232 mm
density = 1.003e3 kg/m^3
viscosity = 1.15e-3 N s / m^2
velocity = 0.0067 m/s
Re = density * velocity * diameter / viscosity
# Because Re less than 10, Darcy's law is valid

# 	**Problem 4.9**
# A confined aquifer is 8 ft thick. The potentiometric surface drops 1.33 
# ft between two wells that are 685 ft apart. The hydraulic conductivity 
# is 251 ft/day and the effective porosity is 0.27.

# Part A: How many cubic feet per day are moving through a strip of 
# the aquifer that is 10 ft wide?
b = 8 ft
w = 10 ft
dH = 1.33 ft
dL = 685 ft
Ko = 251 ft/day
n = 0.27
Q = w * b * Ko * dH/dL; ft^3/day

# Part B: What is the average linear velocity?
velocity = Q / (w*b*n); ft/day

# 	**Problem 4.11**
# An unconfined aquifer has a hydraulic conductivity of 8.7 x 10-2 cm/s. 
# There are two observation wells 597 ft apart. Both penetrate the aquifer
# to the bottom. In one observation well the water stands 28.9 ft above
# the bottom, and in the other it is 26.2 ft above the bottom.

# Part A: What is the discharge per 100-ft-wide strip of the aquifer in 
# cubic feet per day?
h1 = 28.9 ft
h2 = 26.2 ft
dL = 597 ft
Ko = 8.7e-2 cm/s
w = 100 ft
Q = w*Ko*(h1^2 - h2^2) / (2*dL); ft^3/day

# Part B: What is the water-table elevation at a point midway between the 
# two observation wells?
x = dL/2
head = sqrt( h1^2 - (h1^2 - h2^2)*x/dL); ft

# 	**Problem 4.13**
# Refer to Figure 4.19. The hydraulic conductivity of the aquifer is 14.5 
# m/day. The value of h1 is 17.6 m and the value of h2 is 15.3 m. The 
# distance from h1 to h2 is 525 m. There is an average rate of recharge of 
# 0.007 m/d.
Ko = 14.5 m/day
h1 = 17.6 m
h2 = 15.3 m
dL = 525 m
W_ave = 0.007 m/day

# Part A: What is the average discharge per unit width at x = 0?
x = 0 m
qx = ( Ko*(h1^2 - h2^2)/(2*dL) ) - W_ave*( (dL/2) - x); m^2/day

# Part B: What is the average discharge per unit width at x = 525 m?
x = 525 m
qx = ( Ko*(h1^2 - h2^2)/(2*dL) ) - W_ave*( (dL/2) - x); m^2/day

# Part C: Is there a water-table divide? If so, where is it located?
divide_location = (dL/2) - (Ko * (h1^2 - h2^2)/(W_ave * 2 * dL))

# Part D: What is the maximum height of the water table?
h_max = sqrt(h1^2 - (h1^2 - h2^2) * divide_location/dL + (W_ave/Ko)*(dL-divide_location)*divide_location)

# 	**Problem 4.15**
# An earthen dam is constructed on an impermeable bedrock layer. It is 550
# ft across (i.e., the distance from the water in the reservoir to the 
# tailwaters below the dam is 550 ft). The average hydraulic conductivity
# of the material used in the dam construction is 0.77 ft/day. The water in
# the reservoir behind the dam is 35 ft deep and the tailwaters below the 
# dam are 20 ft deep. Compute the volume of water that seeps from the 
# reservoir, through the dam, and into the tailwaters per a 100-ft-wide 
# strip of the dam in cubic feet per day.
Ko = 0.77 ft/day
dL = 550 ft
h1 = 35 ft
h2 = 20 ft
width = 100 ft
q = width*0.5*Ko*(h1^2 - h2^2)/dL; ft^3/day