# 	**Problem 2.1**
# Standard U .S. Class A evaporation land pan has an inside 
# diameter of 47.5 in. and a depth of 10.0 in.

# Part A: Surface area of water in the pan in square meters.
diameter = 47.5 inch
depth = 10 inch
area = pi()*(diameter/2)^2; m^2

# Part B: Volume of the pan in cubic meters.
volume = area * depth; m^3

# Part C: initial volume of water in the pan is 11.5 U.S. gallons, 
# what is the depth of the water in millimeters?
init_volume = 11.5 gal
init_depth = init_volume/area; mm # Note: Todd Rayne is incorrect

# Part D: After a 24-h period with no precipitation the volume of 
# water in the pan is 10.2 U.S. gallons, what is the evaporation rate
# in millimeters/ day?
evap_rate = (init_volume -10.2 gal)/ (24 hr) / area ; mm/day

# Part E: What is new depth of water in millimeters?
new_depth = init_depth - evap_rate * 1 day; mm

# Part F: During the succeeding day there was a 3-h period of 
# precipitation at a constant rate of 5 mm/h. Assuming that the 24-h
# evaporation rate calculated in step D also occurs during this 24-h 
# period, what would be the depth of water in the pan?
rain_rate = 5 mm/hr
rain_time = 3 hr
final_depth = new_depth + rain_rate * rain_time - evap_rate * 1 day; mm

# Part G: If there is no further rain, and no water is added to the pan, 
# how long would it take for the water in the pan to totally evaporate, 
# assuming the net 24-h evaporation rate of step D?
time_evap = final_depth / evap_rate; day

# 	**Problem 2.5**
# A pond has a surface area of 35 ac. If the mean daily air temperature
# is 66°F, the mean daily dew-point temperature is 55°F, the solar 
# radiation is 480 langleys, and the daily wind movement is 115 mi, 
# what is the daily lake evaporation in acre-feet?
# Using the nomograph in figure 2.1 the evaporation rate is 0.15 in/day
evap_rate = 0.15 inch/day
area = 35 acre
Evap_volume = evap_rate * area; acre feet/day

# 	**Problem 2.9**
# The flow of a river at the start of a base flow recession was 
# 712 m^3 / s; after 60 d the flow declined to 523 m3 / s. 
# Part A: What is the recession constant? 
Qo = 712 m^3/s
time = 60 days
Qt = 523 m^3/s
a = (-1/time)*ln(Qt/Qo); /day

# Part B: What would be the flow after 112 d?
time = 112 day
Q = Qo*exp(-a*time)

# 	**Problem 2.11**
# A 90 degree V-notch weir is placed in a road culvert to measure the 
# flow of a stream passing through the culvert. The value of Ho is 
# 2.72 ft. Compute the discharge of the stream.
# For dimensional consistency with Q = ko * Ho^(5/2)
# ko = 2.5 ft^(1/2) / s
ko = 2.5 ft^(1/2) / s
Ho = 2.72 ft
Q = ko * Ho^(5/2); ft^3/s

# 	**Problem 2.13**
# An industrial park with flat-roofed buildings, large parking lots, and 
# little open area has a drainage basin area of 398 ac. The 25-year 
# rainfall even has a precipitation intensity of 2.382 in./h. 
# If the C factor is 0.75, what is the maximum rate that overland flow 
# will drain from the industrial park?
Area = 398 acre
Io = 2.382 inch/hr
Co = 0.75
Q = Co * Io * Area; ft^3/s

# 	**Problem 2.15**
# Figure 2.28 is the hydrograph of a river with a long summer baseflow 
# recession. Compute the volume of annual recharge that occurs between 
# runoff year 1 and runoff year 2.
Qo1 = 280 ft^3/s
t1 = 5.5 * 30 days
Vtp1 = Qo1 * t1 / 2.3026
time1 = 6.4 * 30 days
Vt1 = Vtp1 / 10^(time1/t1)
Qo2 = 320 ft^3/s
t2 = 5.9 * 30 days
Vtp2 = Qo2 * t2 / 2.3026
recharge = Vtp2 - Vt1; ft^3

# 	**Problem 2.17**
# An aqueduct has smooth earthen sides and bottom. The slope of the water surface 
# is 1.7 ft/mi. The channel is trapezoidal in shape with a 45° angle to the sides 
# of the trapezoid and a bottom segment that is 8.5 ft wide. The water in the aqueduct 
# is 3.6 ft deep in the center.
# Part A: What is the average velocity of water in the aqueduct?
depth = 3.6 ft
width_top = 15.7 ft
width_bottom = 8.5 ft
area = depth * 0.5*(width_bottom + width_top); ft^2
side_length = sqrt( (0.5*(width_bottom - width_top))^2 + depth^2); ft
wetted_perimeter = width_bottom + 2*side_length; ft
slope = 1.7 ft / mile
hydraulic_radius = area / wetted_perimeter
Manning_n = 0.02
Manning_const = 1.0 m^(1/3) / s
velocity = (1/Manning_n) * Manning_const * hydraulic_radius^(2/3) * slope^(1/2); ft/s

# Part B: What is the volume of flow in the aqueduct?
Q = velocity * area; ft^3/s