# 	**Problem 1.1**
# A vertical water tank is 15 ft in diameter and 60 ft. high. 
# What is the volume of the tank in cubic feet?
d = 15 ft
h = 60 ft
Vol = pi()*h*(d/2)^2; ft^3

# 	**Problem 1.3**
# If the above tank were measured and found to have an inside 
# diameter of exactly 15.00 ft and a height of 60.00 ft, 
# what would be the volume in cubic feet?
d = 15 ft
h = 60 ft
Vol = pi()*h*(d/2)^2; ft^3

# 	**Problem 1.5**
# If a well pumps at a rate of 8.4 gal per minute, how long 
# would it take to fill the tank described above?
d = 15 ft
h = 60 ft
Vol = pi()*h*(d/2)^2
Pump = 8.4 gal/min
t = Vol/Pump; hr

# 	**Problem 1.7**
# If during the next week the pool still loses 2.35 in. of water
# to evaporation, even with 29 mm of rainfall, how many liters of 
# water must be added?
Area = 60 m^2
Evap = 2.35 inch/week
Pcpt = 29 mm/week
H2O = (Evap - Pcpt)*Area; l/week

# 	**Problem 1.9**
# The parking lot of the Spendmore Megamall has an area of 128 ac. 
# It is partially landscaped to provide some areas of grass. 
# Assume that an average of 63% of the water that falls on the 
# parking lot will flow into a nearby drainage ditch, and the 
# rest either evaporates or soaks into unpaved areas. 
# If as summer thunderstorm drops 3.23 cm of ran, how many cubic 
# feet of water will flow into the drainage ditch?
Area = 128 acre
Pcpt = 3.23 cm
Rain_Vol = Area*Pcpt
Runoff = 0.63*Rain_Vol; ft^3

# 	**Problem 1.11**
# What mass of water at 15o C can be cooled 1 degree C by the amount 
# of heat needed to sublime (go from a solid to a vapor state) 
# 18 g of ice at 0 degrees C?
mass_ice = 18 g
sublime_constant = 677 cal/g
sublime_energy = mass_ice * sublime_constant
delT = 1 degCdiff
specific_heat = 1 cal/degC/g
mass_water = sublime_energy/(specific_heat*delT); g

# 	**Problem 1.13**
# At a water elevation of 6391 ft, Mono Lake has a volume of 
# 2,939,000 ac-ft, and a surface area of 48,100 ac. Annual 
# inputs to the lake include 8.0 in. of direct precipitation, 
# runoff from gauged streams of 150,000 ac-ft per year, and 
# ungauged runoff and groundwater inflow of 37,000 ac-ft per year. 
# Evaporation is 45 in. per year.

# Part A: Input vs. Output
lake_area = 48100 acre
pcpt_rate = 8 inch/yr 
evap_rate = 45 inch/yr 
pcpt_volume = pcpt_rate * lake_area
evap_volume = evap_rate * lake_area
stream_volume = 150000 acre * 1 ft/yr
groundwater_volume = 37000 acre * 1 ft/yr
input = pcpt_volume + stream_volume + groundwater_volume
outflow = evap_volume

# Part B
# The lake level will rise over the long term because input is 
# greater than outflow

# Part C
# What would be the lake surface area when the inputs 
# balance the outputs? 
excess_volume = input - outflow
NetEvap_rate = evap_rate - pcpt_rate
increase_area = excess_volume/NetEvap_rate
NewLakeArea = lake_area + increase_area; acre

# Part D 
# What is the residence time for water in Mono Lake when the 
# water surface is at 6391 ft?
lake_volume = 2939000 acre * 1 ft
residence_time = lake_volume/outflow; yr