Problem 1: In an average year, a small area (3.0 km^2) of agricultural catchment
ID: 476563 • Letter: P
Question
Problem 1: In an average year, a small area (3.0 km^2) of agricultural catchment receives 950 mm of precipitation. The catchment is drained by a stream, and a continuous record of stream discharge is available. The total amount of surface-water runoff for the year, determined from a stream discharge, is 1.1*10^6 m^3.
a) What is the volume of water (in m^3) evapotranspired for the year (assume no change in water stored in the catchment)?
b) What is the depth of water (in millimeters (mm)) evapotranspired for the year (again, assume no change in water stored in the catchment)?
c) What is the runoff ratio (rs/p) for the catchment? (p stands for average precipitation rate, and rs stands for average surface run off rate)?
Explanation / Answer
a]
first calculate the volume of total yearly precipitation in m^3.
convert units to meters. 3.0 km^2 x 1000 m/km x 1000 m/km = 3000000 m^2
950 mm x 1m / 1000 mm = 0.950 m
find the volume of total precipitation. volume = area x height
v = 3000000 m^2 x .950m
v = 2,850,000 m^3
subtract the volume of water lost to streamflow. this gives the volume of water lost to evapotranspiration.
v(evapotranspiration) = 2,850,000 m^3 - 1.1 x 10^6 m^3
v(evapotranspiration) = 1,750,000 m^3
b]
find the depth of water lost to evaporation.
v(evapotranspiration) = 1,750,000 m^3
volume = area x height
1,750,000 m^3 = 3000000 m^2 x height
height = 0.583 m x 1000 mm / 1 m = 583 mm
c]
Run off ratio = Average surface run off rate / Average precipitaion rate = 1.1 x 10^6 m^3 m^3 / 2,850,000 m^3
= 0.3859