Estimation of Evaporation Evaporation Pans – placed near a water body from which evaporation is to be estimated In the event that precipitation occurs during observation and water level goes above the indicated water level to be maintained, pour/remove the excess water during the periodic measurement to bring the water back to the indicated level. πππ πΆππππππππππ‘ = πΏπππ πΈπ£ππππππ‘πππ πππ πΈπ£ππππππ‘πππ πππ πΈπ£ππππππ‘πππ = π·πππ‘β ππ ππππππππ‘ππ‘πππ + π·πππ‘β ππ πππ‘ππ π΄ππππ or πππ πΈπ£ππππππ‘πππ = π·πππ‘β ππ ππππππππ‘ππ‘πππ − π·πππ‘β ππ πππ‘ππ π ππππ£π If no precipitation occurs during the observation, simply disregard the precipitation parameter on the above formula. Examples of Evaporation Pans: a.) United States “Class A” Evaporation Pan - mounted on a raised and level wooden base to allow free circulation of air b.) United Kingdom’s Symon’s Pan or S-Pan - square prism sunk in the ground - one of its drawback is that the heat of the ground may contribute to the evaporation of water inside the pan leading to overestimation of pan evaporation - another drawback would be the difficulty in detecting probable leakage from the pan which could again lead to overestimation of pan evaporation c.) Indian Standard Evaporation Pan - similar in orientation with United States Class A Evaporation Pan but includes a protective mesh to prevent birds from drinking water from the pan - its drawback is that the mesh could absorb the surrounding heat and radiate it within the pan, contributing to the evaporation, and could lead to overestimation of pan evaporation Sample Problem: An Indian Standard evaporation pan with internal diameter of 1200mm is 255mm deep and is filled with water up to a height of 190mm. After 24 hrs., it was observed that 2.93 liters of water has to be added to the pan to bring the water level back to its original level of 190mm. A nearby rain gauge recorded 2.1mm of rainfall during this 24-hr period. If the pan coefficient is 0.8, what is the lake evaporation during this period? Given: Diameter of pan = 1200mm Depth of pan = 255mm Vadded = 2.93 L Pan Coefficient = 0.8 Solution: Convert volume added from liters to cubic meters: π = 2.93πΏ π₯ 1π = 2.93π₯10 1000 πΏ π Compute the depth of water added: π· = π 2.93π₯10 1000ππ = = 2.50π₯10 π π₯ = 2.51ππ π΄πππ ππ πππ 0.25π(1.22) 1π Compute for the Pan Evaporation: πππ πΈπ£ππππππ‘πππ = π·πππ‘β ππ ππππππππ‘ππ‘πππ + π·πππ‘β ππ π€ππ‘ππ πππππ πππ πΈπ£ππππππ‘πππ = 2.1 + 2.51 πππ πΈπ£ππππππ‘πππ = 4.61 ππ Compute for the Lake Evaporation: πππ πΆππππππππππ‘ = 0.8 = πΏπππ πΈπ£ππππππ‘πππ πππ πΈπ£ππππππ‘πππ πΏπππ πΈπ£ππππππ‘πππ 4.61 πΏπππ πΈπ£ππππππ‘πππ = 0.8 π₯ 4.61 π³πππ π¬ππππππππππ = π. ππ ππ Study well! ο Prepared by: Engr. Lea Diane P. Lacbao, RMP
