Recall that potential evapotranspiration (PET) is the rate of evapotranspiration that would occur if it was only limited

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Recall That Potential Evapotranspiration Pet Is The Rate Of Evapotranspiration That Would Occur If It Was Only Limited 1 (134 Bytes) Viewed 48 times

B) Calculate the daily evaporation from a forest in
Virginia on a sunny day, as a rate in units of cm/day. Assume the
net solar radiation is 200.0 Wm -2 , the density of
water is 1000.0 kg m-3 and the latent heat of
vaporization is 2.45x106 J kg-1 . Simplify
the above equation based on the stated assumptions (use symbols
first, then numbers) and show your work.
C) 3) If we actually accounted for heat lost to the air and
to the ground, would your calculated evaporation rate be higher or
lower? Why?
D) If the forest had a higher albedo (more solar radiation
reflected) would you expect the et to increase or decrease?
Why?

Recall That Potential Evapotranspiration Pet Is The Rate Of Evapotranspiration That Would Occur If It Was Only Limited 2 (70.26 KiB) Viewed 48 times

Recall that potential evapotranspiration (PET) is the rate of evapotranspiration that would occur if it was only limited by the amount of energy available i.e.: not by the amount of water or by biological processes). Below you will practice calculating PET rates using watershed. In class we will look at a method which uses physical parameters to estimate PET. This example uses energy balance. The energy balance at the surface of a watershed is defined as: dQ de = R. -G-H-E (1) R. - net solar radiation input to the watershed (fraction of solar radiation received by the surface and not reflected back to the atmosphere) G= energy output through conduction of heat to the ground (heat transferring from the surface of the catchment down into the soil) H = net output of sensible heat to the atmosphere (heat transferred to the air) (note that air has a lower heat capacity than both land and water - so it takes less heat energy to increase the temperature] E, - output of latent heat to the atmosphere (also called the latent heat flux) due to evaporation (remember latent heat is from changes of state - evaporation requires energy so the process of evaporation, a change of state from liquid to vapor, results in an energy loss for the catchment energy balance - energy the catchment gives to the water during evaporation) Q = heat energy stored in control volume (watershed) per unit surface area R, G, H, and E, are all energy fluxes (energy per unit area per unit time). Units for an energy flux are [Jms] - [W m) and do/dt is the change in energy over the change in time. Thus, all together, the change in energy at the surface of a watershed over time is a balance between energy input from the sun (R), energy lost to heat the air (H), energy lost to heat the ground below the surface (H) and energy used to evaporate surface water (EI).

At the surface, the rate of evaporation is related to the latent heat flux, the density of water, and the latent heat of vaporization as follows: E Pwho (2) et et = evapotranspiration rate [LT] (note this is usually expressed as a length per time, or "height" of the total volume of water evaporated from the watershed - just like precipitation is often expressed as a height) Es = latent heat flux (same as above) Av density of water in units of [ML-*) lv = latent heat of vaporization [LPT4 So how fast the water evaporates is related to the density of water (which varies with temperature), the latent heat of vaporization (how much energy it takes for the phase change to happen) and the energy lost due to that phase change. By solving both equations for E, and setting the results equal to each other, we find: R.-G-H- Pulo et - (3) Make the following (common) assumptions: dQ/dt = 0 Assume heat energy within the watershed remains approximately constant G = 0 Because soil has a low heat capacity relative to water, we can assume this for watersheds where the surface soil does not contain a large amount of water. Note this assumption does not work if your time period is less than one day. H = 0 This neglects sensible heat, because of the low heat capacity of air relative to water. Note this assumption can only be used if there is a very small mass of water evaporating.