# 	**Problem 3.1**
# What is the weight in newtons of an object with a mass of 14.5 kg?
mass = 14.5 kg
weight = mass*grav(); N

# 	**Problem 3.3**
# An object has a mass of 78.5 kg and a volume of 0.45 m^3.

# Part A:  What is its density?
mass = 78.5 kg
volume = 0.45 m^3
density = mass / volume

# Part B:  What is its specific weight?
Spec_weight = density * grav(); N/m^3

# Part C: Is the object more or less dense than water?
# density less than density of water (1000 kg/m^3) so less dense

# 	**Problem 3.5**
# The hydraulic conductivity of a silty sand was measured in a laboratory
# permeameter and found to be 3.75e-5 cm/s at 25 C. What is the 
# intrinsic permeability in cm^2? Refer to Appendix 14 for values of 
# density and viscosity.
Ko = 3.75e-5 cm/s
density_water = 0.997044 g/cm^3
viscosity_water = 0.008937 g / cm s
ki = Ko * viscosity_water / (density_water * grav()); cm^2

# 	**Problem 3.7**
# A constant-head permeameter has a cross-sectional area of 78.5 cm^2. 
# The sample is 23 cm long. At a head of 3.4 cm, the permeameter 
# discharges 50 cm^3 in 38 s.

# Part A: What is the hydraulic conductivity in centimeters per second 
# and feet per day?
area = 78.5 cm^2
Lo = 23 cm
delH = 3.5 cm
to = 38 sec
Vol_water = 50 cm^3
Ko = Vol_water * Lo / (area * to * delH); cm/s; ft/day

# Part B:  What is the intrinsic permeability if the hydraulic conductivity 
# was measured at 15 degrees C?
density_water = 0.99909  g/cm^3
viscosity_water = 0.011404 g / cm s
ki = Ko * viscosity_water / ( density_water * grav() ); cm^2


# Part C: From the hydraulic conductivity value, name the type of soil.
# well sorted sand or gravel

# 	**Problem 3.9**
# An aquifer has a specific yield of 0.19. During a drought period, the following
# average declines in the water table were noted:
Aa = 15 km^2
Ab = 7.5 km^2
Ac = 18.3 km^2
Ad = 22.5 km^2
Ae = 9.44 km^2
Af = 22.7 km^2
dHa = 2.34 m
dHb = 1.22 m
dHc = 0.76 m
dHd = 3.44 m
dHe = 1.89 m
dHf = 0.35 m
Sy = 0.19
Vw = Sy*dHa*Aa + Sy*dHb*Ab + Sy*dHc*Ac + Sy*dHd*Ad + Sy*dHe*Ae + Sy*dHf*Af

# 	**Problem 3.11**
# A confined aquifer has a specific storage of 1.022e-6 ft and a thickness 
# of 23 ft. How much water would it yield if the water declined an average of 1.75
# ft over a circular area with a radius of 418 ft?
Ss = 1.022e-6 /ft
b = 23 ft
dH = 1.75 ft
ro = 418 ft
Vw = Ss*b*dH*pi()*ro^2; ft^3

# 	**Problem 3.13**
# A confined aquifer has a specific storage of 4.033e-3 /m and a porosity of 
# 0.274. The compressibility of water is 4.6e-10 m^2/N. What is the
# compressibility of the aquifer skeleton? 
Ss = 4.033e-3 /m
n = 0.274
beta = 4.6e-10 m^2/N
density_water = 1000 kg/m^3
alpha_aquifer = (Ss - density_water*grav()*n*beta) / (density_water*grav()); m^2/N

# 	**Problem 3.15**
# An aquifer has three different formations. Formation A has a thickness of 22 ft
# and a hydraulic conductivity of 17.0 ft/d. Formation B has a thickness of 3.5 ft 
# and a conductivity of 99 ft/d. Formation C has a thickness of 26 ft and a 
# conductivity of 22 ft/d. Assume that each formation is isotropic and 
# homogeneous. Compute both the overall horizontal and vertical conductivities.
Ka = 17 ft/day
Kb = 99 ft/day
Kc = 22 ft/day
ba = 22 ft
bb = 3.5 ft
bc = 26 ft
K_horizontal = (Ka*ba + Kb*bb + Kc*bc)/(ba + bb + bc); ft/day
K_vertical = (ba + bb + bc)/(ba/Ka + bb/Kb + bc/Kc); ft/day

# 	**Problem 3.17**
# Use the Hazen method to estimate the hydraulic conductivity of the sediments 
# graphed in Figure 3.33.
Ko = 100 /cm s
d10 = 0.16 mm
Conductivity = Ko*d10^2; cm/s