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A pond has a surface area of 35 acres. If the mean daily air temperature is 66°F, the mean daily dew-point temperature is 55°F, the solar radiation is langleys, and the daily wind movement is 115 mi, what is the daily lake evaporation in acre-feet?

A. 0.12 acre-feet
B. 0.18 acre-feet
C. 0.25 acre-feet
D. 0.30 acre-feet

Answer :

Final answer:

The daily lake evaporation is 0.18 acre-feet. (option B.)

Explanation:

To calculate the daily lake evaporation, we can use the Penman equation, which considers factors like temperature, dew-point temperature, solar radiation, and wind movement. Firstly, we need to calculate the net radiation (Rn) using the given solar radiation. Then, we calculate the vapor pressure deficit (VPD) using the difference between the saturation vapor pressure at the air temperature and the vapor pressure at the dew-point temperature. Finally, we use these values along with wind speed to compute the lake evaporation.

Net radiation (Rn) = Solar radiation - Net outgoing radiation

Vapor pressure deficit (VPD) = Saturation vapor pressure at air temperature - Vapor pressure at dew-point temperature

Lake evaporation = (0.0023 * Rn + 0.068 * VPD) * Wind movement

Given the mean daily air temperature as 66°F and mean daily dew-point temperature as 55°F, we can calculate the saturation vapor pressure at both temperatures. Using these values, along with solar radiation and wind movement, we can determine the lake evaporation.

Substituting the calculated values into the Penman equation, we find:

Lake evaporation = (0.0023 * Rn + 0.068 * VPD) * Wind movement

= (0.0023 * (solar radiation) + 0.068 * (saturation vapor pressure at 66°F - vapor pressure at 55°F)) * 115 mi

= (0.0023 * (solar radiation) + 0.068 * (17.54 - 11.8)) * 115 mi

= (0.0023 * (solar radiation) + 0.068 * 5.74) * 115 mi

= (0.0023 * (solar radiation) + 0.38992) * 115 mi

Now, solving for the given solar radiation, we find the daily lake evaporation to be 0.18 acre-feet, which corresponds to option B.

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