Transpiration / EvapotranspirationAn Important Phase of Hydrologic CycleJeaneth Sena

TRANSPIRATION - is the process of water movement through a plant and its evaporation from aerial parts such as from leaves but also from stems and flowers.-This process constitutes an important phase of the hydrologic cycle since it is the principal mechanism by which the precipitation falling on land areas is returned to the atmosphere.- Transpiration is transfer of water from plants to the atmosphere.-Transpiration also includes a process called guttation, which is the loss of water in liquid form from the uninjured leaf or stem of the plant, principally through water stomata.

Studies have revealed that about 10 percent of the moisture found in the atmosphere is released by plants through transpiration. The remaining 90 percent is mainly supplied by evaporation from oceans, seas, and other bodies of water (lakes, rivers, streams).

Measurement of Transpiration Phytometer Ceramic and Piche atmometers

1.Phytometer a large vessel filled with soil in which one or more plants are rooted

2.Ceramic and Piche atmometers Such instruments automatically feed water from a reservoir to an exposed, wetted surface. The change in content of the reservoir serves as an index to transpiration. With proper care and exposure, atmometers are useful in experimental work for estimating temporal and spatial variations in potential transpiration.

EVAPOTRANSPIRATION is the sum ofevaporation andplant transpiration from the Earth's land and ocean surface to theatmosphere. total evaporation (or evapotranspiration) is the evaporation from all water, soil, snow, ice, vegetation, and other surfaces plus transpiration. An element (such as a tree) that contributes to evapotranspiration can be called anevapotranspirator. Consumptive use is the total evaporation from an area plus the water used directly in building plant tissue.

Potential Evapotranspiration 1. Thornthwaite defined the term potential evapotranspiration as the water loss which will occur if at no time there is a deficiency of water in the soil for the use of vegetation. 2. Penmann suggested that the original definition be modified to include the stipulation that the surface be fully covered by green vegetation.3. Priestly and Taylor have proposed that potential evapotranspiration be considered equivalent to the first term of the Penmann equation multiplied by the factor 1.26 and taking into account the heat flux into the soil.

Methods of Estimating Evapotranspiration 1. Tanks and Lysimeter Experiments2. Field Experimental Plots3. Installation of Sunken (Colorado) tanks4. Evapotranspiration Equations

1. Tanks and Lysimeter Determination of Evapotranspiration- Many observations of evapotranspiration are made in soil containers, variously known as tanks, evapotranspirometers, and lysimeters. - Alysimeteris a measuring device which can be used to measure the amount of actualevapotranspirationwhich is released by plants, usually crops or trees.

Types of Lysimeter Weighing Non-weighing

Weighing Lysimeter A weighing lysimeter, for example, reveals the amount of water crops use by constantly weighing a huge block of soil in a field to detect losses of soil moisture. Non-weighing Lysimeter Non-weighing types are used to determine ET as a residual by measuring all other components of the soil water balance, including water inputs (rain and irrigation), outputs (drainage and runoff), and change in soil water storage. Non-weighing lysimeters, also called percolation or drainage lysimeters, are also used in percolation and solute transport studies..

2. Field Experimental Plots - The different elements of the water budget (other than ET) in a known interval of time are measured in special experimental plots established in the field. ET is then estimated as:

ET = Precipitation + Irrigation Input Runoff Increase in Soil Moisture Storage Groundwater Loss Since groundwater loss due to deep percolation is difficult to measure, it is minimised by maintaining the soil moisture condition in the plot at field capacity. This method provides fairly reliable results.

3. Installation of Sunken (Colorado) Tanks Sunken Colorado pan or tank Thesunken Colorado panis square, 1m (3ft) 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 5cm (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. Main advantage of this pan its aerodynamic and radiation characteristics are similar to that of a lake Disadvantages difficult to detect leaks, expensive to install, extra care is needed to keep the surrounding area free from tall grass, dust etc

Jomelle Sulpa

HYDROMETEOROLOGICAL EQUATIONS Blaney Criddle Equation Penman Equation Evaporation Index Method1.Blaney Criddle Equation A method for estimating reference crop evapotranspiration. A relatively simplistic method for calculating evapotranspiration. Given the coarse accuracy of this equation, it is recommended that it is used to calculate evapotranspiration for periods of one month or greater.

(1 ENGLISH)(2 SI)where:U = seasonal consumptive use (inches in English and cm in SI units)t = mean monthly temperature (F in English units and C in SI units)p = monthly percentage of hours of bright sunshine (of the year)k = monthly consumptive use coefficient determined from experimental data

where:f = monthly consumptive use factorK = seasonal values of consumptive use coefficient F = seasonal values of consumptive use factor ( refers for the summation for all the months of the growing season.)

Example. Determine the evapotranspiration and irrigation requirement for wheat, if the water application efficiency is 65% and the consumptive use coefficient for the growing season is 0.8 from the following data :

Month Mean Monthly Temp (C) Monthly Percentage of Sunshine (hours) Effective Rainfall (cm)

November 18 7.20 2.6

December 15 7.15 2.8

January 13.5 7.30 3.5

February 14.5 7.10 2.0

Month Mean Monthly Temp (C)Monthly Percent of Sunshine (Hours)Effective Rainfall (cm)Monthly Consumptive Use Factor

t pPe

November 18 7.20 2.6 11.82

December 15 7.15 2.8 10.74

January 13.5 7.30 3.5 10.48

February 14.5 7.10 2.0 10.50

Blaney Criddle Equation (monthly)ETo=p(0.46Tmean+ 8)where: ETo = the reference evapotranspiration [mm day1] (monthly)Tmean = is the mean daily temperature [C] given as Tmean= (Tmax+ Tmin)/ 2 p = is the mean daily percentage of annual daytime hours. Given the limited data input to this equation, the calculatedevapotranspirationshould be regarded only as broadly accurate. The output of this equation only provides an order of magnitude. The inaccuracy of this equation is also exacerbated by extreme variants of weather. In particular, evapotranspirationis known to be exaggerated by up to 40% in calm, humid, clouded areas and depreciated by 60% in windy, dry, sunny areas.

2.Penman Equation It is the most general and widely used equation for calculating reference ETo. It also describes evaporation(E) from an open water surface, and was developed byHoward Penmanin 1948. It requires daily mean temperature,wind speed, air pressure andsolar radiation.

3.Evaporation Index Method It is the traditional method of estimating Evapotranspiration. It uses evaporation from an open water surface that provides an index of the integrated effect of radiation, air temperature, air humidity and wind on evapotranspiration. Analysis of data on consumptive use indicate a high degree of correlation between pan evaporation values and consumptive use. Et= kpEp where: Et= Consumptive usekp = Pan coefficient (depends on the climate, the environment, and the type of the pan) Ep = Water evaporated in the pan.

Other Hydrometeorological Equations used. PriestleyTaylor - only radiation (irradiance) observations are required. Shuttleworth - modified the Penman Equation to use SI units. JensenHaise - based on extensive field data on alfalfa reference. Makkink - calibrated to a specific location

Factors that affect Evapotranspiration Evapotranspiration is affected by a variety of environmental factors, management conditions, plant characteristics, and geographical factors.

The rate of evapotranspiration at any location on the Earth's surface is controlled by several factors: Energy availability. The more energy available, the greater the rate of evapotranspiration. It takes about 600 calories of heat energy to change 1 gram of liquid water into a gas. Humidity gradient away from the surface. The rate and quantity of water vapor entering into the atmosphere both become higher in drier air. Wind Speed. The process of evapotranspiration moves water vapor from ground or water surfaces to an adjacent shallow layer that is only a few centimeters thick. When this layer becomes saturated evapotranspiration stops. However, wind can remove this layer replacing it with drier air which increases the potential for evapotranspiration. Winds also affect evapotranspiration by bringing heat energy into an area. Water Availability. Evapotranspiration cannot occur if water is not available. Physical attributes of the vegetation. Such factors as vegetative cover, plant height, leaf area index and leaf shape and the reflectivity of plant surfaces can affect rates of evapotranspiration. Stomatal Resistance. Plants regulate transpiration through adjustment of small openings in the leaves called stomata. As stomata close, the resistance of the leaf to loss of water vapor increases, decreasing to the diffusion of water vapor from plant to the atmosphere. Soil characteristics. Soil characteristics that can affect evapotranspiration include its heat capacity, and soil chemistry and albedo. Seasonal Trends. Minimum evapotranspiration rates generally occur during the coldest months of the year. Maximum rates generally coincide with the summer season. The seasonal maximum evapotranspiration actually may precede or follow the seasonal maximum solar radiation and air temperature by several weeks.

DEPRESSION STORAGECarla Dacillo

Depression In terms of depressions area, it represents the loss or "initial abstraction" caused by such phenomena as surface ponding, surface wetting, interception and evaporation. Depression storage may be treated as a calibration parameter, particularly to adjust runoff volumes.

Areas of Depression Storage:1.Impervious Depression Storage- Impervious surfaces are mainly artificial structures such as pavements that are covered by impenetrable materials such as asphalt, concrete, brick, and stone and rooftops. Soils compacted by urban development are also highly impervious.-Impervious area depression storage, in. [mm]. Water stored as depression storage on impervious areas is depleted by evaporation. A relationship for depression storage versus catchment slope has been developed as follows (Kidd, 1978).

Dp = (0.0303)( S -0.49 )Where:Dp= depression storage, inch.S= catchment slope, percent.

2.Pervious Depression Storage Porous surface with spaces in the material, such as landscaping, gravel, and alternative pavers. Pervious surfaces allow rainwater or snowmelt to pass through into the ground, thereby reducing runoff and filtering pollutants. Pervious area depression storage, in. [mm]. Water stored as depression storage is subject to both infiltration and evaporation. This parameter is best represented as an interception loss, based on the type of surface vegetation. For grassed urban surfaces, a value of 0.10 in. (2.5 mm) is typical.

Factors affecting the magnitude of depression storage: Surface Type Slope Rainfall Return Period (recurrence interval)

Depression storage capacity insoil science, is the ability of a particular area of land to retain water in its pits and depressions, thus preventing it from flowing. Depression storage capacity, along withinfiltration capacity, is one of the main factors involved inHorton Overland Flow.

DEPRESSION Focused Recharge Recharge: Implies replenishing a supply of water held within ageological formationunderground. If water falls uniformly over a field such thatfield capacityof the soil is not exceeded, then negligible water percolates togroundwater. If instead water puddles in low lying areas, the same water volume concentrated over a smaller area may exceed field capacity resulting in water that percolates down to recharge groundwater. The recurring process of water that falls relatively uniformly over an area, flowing to groundwater selectively under surface depressions is Depression Focused Recharge.

GROUNDWATER RECHARGE Groundwater rechargeordeep drainageordeep percolationis ahydrological process wherewatermoves downward from surface water togroundwater. This process usually occurs in thevadose zonebelow plantrootsand is often expressed as afluxto thewater tablesurface According to Linsley et al. the volume of water stored by surface depression at any given be approximated using

Where:V= volume actually in storage at same time of interestSd= maximum storage capacity of the depressionsPe= rainfall excess (gross rainfall minus evaporation, interception, and infiltration)

Example:Assume that a rainfall event of intensity 2.25 cm/h falls over a uniformly forested watershed of area 20 km2. If there is no infiltration and no surface depression storage, compute the volume of water that leaves the basin as storm runoff for a 30-min and a 105-min rainfall. But assuming that in addition to interception there are also losses due to depression storage.Assume that K is 1.5, S is 0.2 cm and that the evaporation rate E is 0.05 cm/h.

Solution: Assume that K is 1.5, S is 0.2 cm and that the evaporation rate E is 0.05 cm/h.Compute total precipitation volume: P = (2.25 cm/h) (0.5 h) = 1.125 cmP = (2.25 cm/h) (1.75 h) = 3.9375 cm Use equation 3 to obtain:Li = (0.2 cm) (1 - exp(-1.125/0.2)) + (1.5) (0.05 cm/h) (0.5 h) = 0.2368 cmLi = (0.2 cm) (1 - exp(-3.9375/0.2)) + (1.5) (0.05 cm/h) (1.75 h)= 0.33125 cm but assuming that in addition to interception there are also losses due to depression storage.Assume that Sd is 0.2 cm.0.5-hour:Pe = P - Li = (1.125 - 0.2368) cm = 0.8882 cm.1.75-hour:Pe = P - Li = (3.9375 - 0.33125) cm = 3.60625 cm.Volume of surface depression (use eq.):0.5-hour:V = (0.2 cm) (1 - exp(-0.8882/0.2)) = 0.1976 cm1.75-hour:V = (0.2 cm) (1 - exp(-3.60625/0.2)) = 0.2 cm

Again, assuming that there is no change in basin storage, then the output of the basin is equal to:0.5-hour:Volume of Output = (Pe - V)*Abasin = ((0.8882 cm - 0.1976 cm)/100 cm/m ) (20 *10^6 m2)= 138,120. m3 1.75-hour:Volume of Output = (Pe - V)*Abasin = ((3.60625 cm - 0.2 cm)/100 cm/m ) (20 *10^6 m2)= 681,250. m3