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Monday, December 23, 2013

EVAPORATION

1.1) INTRODUCTION

          Evaporation is the process by which water is transferred from the land and water masses of the earth to the atmosphere. Transpiration is the evaporation counterpart for plants. It is the process by which soil moisture taken up by vegetation is eventually evaporated as it exits at plant pores. Evaporation and transpiration combined (evapotranspiration) generally, constitute the largest component  of losses in rainfall-runoff sequences. Accordingly, good estimates of evapotranspiration are a requisite for hydrologic modeling.
          Infiltration is that process by which precipitation moves downward through the surface of the earth and replenishes soil moisture, recharges aquifier, and ultimately supports streamflows during dry periods. Along with interception, depression storage, and storm period evaporation, it determines the availability, if any of the precipitation input for generating overland flows. Furthermore, infiltration rates influences the timing of overland flow inputs to channelized system. Accordingly, infiltration is an important component of any hydrologic model.

1.2) EVAPORATION

           Because there is a continuous exchange of water molecules between an evaporating surface and its overlying atmosphere, it is common in hydrologic practice to define an evaporation as the net rate of vapor transfer.
          Evaporation from a particular surface is directly related to the opportunity for evaporation (availability of water) provided by that surface. For open bodies of water, evaporation opportunities is 100% while for soils it varies from a high 100% to essentially 0% at stages of very low moisture content.
          
         Direct measurement of evaporation are not easily obtained for large bodies of water because of the extensive surface involved. In fact, of all variables included in the general hydrologic equation, surface runoff is the only one that readily permits direct evaluation. The choice of method used to determine evaporation depends on  the required accuracy of results and the type of instrumentation available.

1.3) ESTIMATING EVAPORATION

          The most common method used in estimating evaporation are water budget method and use of evaporation pans. The pan method is less expensive method and will frequently provide good estimates of annual evaporation.

         1.1.1) WATER BUDGET CALCULATIONS
            
        The water budget method for determining evaporation is very simple procedure, but it seldom produces reliable result. In this method, reservoir (lake, pool, etc) evaporation, E can be computed by the formula :

                                           Es = P + R1 – R2 + Rg –Ts – I - ∆Ss

                   
All the terms are in volume units for a time period of interest, and ∆t should be at least a week. In general, however, the method would more likely to be used to estimate monthly or annual evaporation from a particular reservoir. Precipitation, runoff and changes in storage can often be determined within reasonable limits of accucary, but evaluation of net seepage, Os is frequently subject to appreciable errors. Seepage estimates usually come from measurement of groundwater levels ands and/or soil permeability. The water budget is usable on continuous basis if a stage-seepage relation for the lake can be established.

          1.1.2) MASS TRANSFER TECHNIQUE

                    Mass transfer equation are based primarily on the concept of the turbulent transfer of water vapor (by eddy motion) from an evaporating surface to the atmosphere. Many equations, both theoretical and empirical, have been developed. Most are similar in form to a relation between evaporation and vapor pressure first recognized by Dalton :

                                        E = k(eo – ea)

where    E = direct evaporation
              k = coefficient dependent on the wind velocity, atmospheric pressure, and other factor.
       eo , ea = the saturation vapor pressure at water surface temperature and air, respectively.
       
             Theoretical mass transfer equations are based on the concepts of discontinuous and continuous mixing air at the air-liquid interface.
    
                Empirical approaches often require exacting and costly instrumentation and observations, so their general utility is limited. The complexity of the equation varies from simple equation like E = k(eo – ea) to complex relations like Sutton’s equation for a circular lake of radius r :

                                     



Where E = evaporation (cm/day)
           Ρ = atmospheris pressure
           U = average wind velocity (cm/s)
           ρ = mass density of air (g/cm3)
           r = radiusof lake (cm)
           n = empirical constant
          G’ = a complex function

           1.1.3) USE OF EVAPORATION PANS
                  
An evaporation pan is used to hold water during observations for the determination of the quantity of evaporation at a given location. Such pans are of varying sizes and shapes, the most commonly used being circular or square. The best known of the pans are the "Class A" evaporation pan and the "Sunken Colorado Pan". In Europe, India and South Africa, a Symon's Pan (or sometimes Symon's Tank) is used. Often the evaporation pans are automated with water level sensors and a small weather station is located nearby.

   A)    CLASS A EVAPORATION PAN

          In the United States, the National Weather Service has standardized its measurements on the Class A evaporation pan, a cylinder with a diameter of 47.5 in (120.7 cm) that has a depth of 10 in (25 cm). The pan rests on a carefully leveled, wooden base and is often enclosed by a chain link fence to prevent animals drinking from it. Evaporation is measured daily as the depth of water (in inches) evaporates from the pan. The measurement day begins with the pan filled to exactly two inches (5 cm) from the pan top. At the end of 24 hours, the amount of water to refill the pan to exactly two inches from its top is measured.

          If precipitation occurs in the 24-hour period, it is taken into account in calculating the evaporation. Sometimes precipitation is greater than evaporation, and measured increments of water must be dipped from the pan. Evaporation cannot be measured in a Class A pan when the pan's water surface is frozen.
The Class A Evaporation Pan is of limited use on days with rainfall events of >30mm (203mm rain gauge) unless it is emptied more than once per 24hours. Analysis of the daily rainfall and evaporation readings in areas with regular heavy rainfall events shows that almost without fail, on days with rainfall in excess of 30mm (203mm Rain Gauge) the daily evaporation is spuriously higher than other days in the same month where conditions more receptive to evaporation prevailed.
The most common and obvious error is in daily rainfall events of >55mm (203mm rain gauge) where the Class A Evaporation pan will likely overflow.
The less obvious, and therefore more concerning, is the influence of heavy or intense rainfall causing spuriously high daily evaporation totals without obvious overflow.

   B)    SUNKEN COLORADO PAN
         
          The sunken Colorado pan is square, 1 m (3 ft) on a side and 0.5 m (18 in.) deep and made of unpainted galvanized iron. As the name suggests, it is buried in the ground to within about 5 cm (2 in.) of its rim. Evaporation from a Sunken Colorado Pan can be compared with a Class A pan using conversion constants. The pan coefficient, on an annual basis, is about 0.8.





EXERCISE
1)      Consider a lake has a surface area of 1.5mi2. If the average annual evaporation rate s estimated to be 2/3 of the average daily rate calculated in example above, what volume in million gallons per day and cubic meter per day would be lost to evaporation?

            SOLUTION
1)      The surface layer of the lake in square feet,
                         
               52800 x 5280 x 1.5 = 41817760 ft2
               41817760 x 0.21 x 2/3 = 731811 ft3/day

Converting to mgd,

                731181 x 7.48 = 5473946 mgd

For 1 year, the total would be,

               5473946 x 365 = 1997990 mgy

Convert to cubic meter,

              731811 x 0.0283 = 20710 m3 / day
              20710 x 365 = 7559242 m3 / year



         

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