Chapter 3. Evaporation and Transpiration – Elementary Engineering Hydrology

3

Evaporation and Transpiration

Chapter Outline

3.1 DEFINITION

Evaporation is an important process in the hydrologic cycle preceding precipitation. It is the process by which water in the liquid form transforms into vapour through the transfer of energy. The exact laws governing this phenomenon are not completely known.

The transformation from solid to the vapour state without passing through the usual intermediate liquid state is known as sublimation. In the atmosphere, evaporation occurs from free water surfaces like (1) seas, (2) lakes and rivers, (3) moisture in solid surfaces of (4) land and soil, (5) vegetation, (6) snowfields, (7) glaciers and (8) even from the falling rain drops.

3.1.1 Process of Evaporation

Any water body, regardless of its size, is made up of a number of molecules. Each one of them is constantly in motion at varying velocities and in different directions. This velocity of movement is dependent on temperature. So also, in a given mass, every molecule is attracted by another by a force inversely proportional to the square of the distance between them and directly proportional to the product of their masses. The molecules underneath attract the molecules which are near the top surface. Thus, some molecules near the surface, which have more kinetic energy than the attracting force—quite enough to overcome the attraction of the molecules in the lower part of the water body, escape into the atmosphere. This process is known as evaporation.

The molecules that escape into the atmosphere carry with them some heat energy. Thus evaporation is a cooling process. Some molecules are continuously moving into the atmosphere and some are returning from the atmosphere to the water surface. There may be a collision of molecules moving from and to the water surface. Thus, the rate of evaporation is the net rate of movement of molecule to and from the water surface.

When the rate of molecules coming from the atmosphere to the water surface is more than the rate of molecules going out from the water surface, then the process is known as condensation.

Evaporation is expressed in terms of the depth of water per unit area per unit time (e.g. mm/m2/h).

3.1.2 Dalton’s Law of Evaporation

John Dalton was the first scientist to describe this process, scientifically.

Immediately adjacent to the water surface, there is a thin layer of air that is saturated with vapour and its temperature is the same as that of water.

Evaporation depends upon the difference of vapour pressure in the atmosphere and the saturated vapour pressure of this air film.

where, E = Evaporation rate

          C = A constant

         Vapour pressure gradient in the vertical direction.

If the air above the film has vapour pressure same as that of the film, i.e. when de = 0, then there will be no evaporation.

However, if the air above the film has vapour pressure less than that of the film, i.e. de ≠ 0, then there will be evaporation.

Dalton suggested the formula

 

E = C (esea)

 

where, E = Evaporation rate

         C = A constant

         es = Saturated vapour pressure at the water surface temperature

         ea = Vapour pressure in the air above

Dalton’s Law of evaporation is fundamental and all further work is based on this law.

3.2 NECESSITY OF ESTIMATION OF EVAPORATION

Large amounts are spent in constructing a dam and creating a reservoir to store water. Canals are also constructed to carry water. Some water stored in these reservoirs and flowing in canals is lost due to evaporation. The amount of the loss of water due to evaporation can be substantial (about 30%).

It is, therefore, necessary to estimate the loss due to evaporation to assess the water available for use in different seasons.

3.3 FACTORS AFFECTING EVAPORATION

The factors affecting evaporation are (1) temperature, (2) solar radiation, (3) wind, (4) barometric pressure and altitude, (5) dissolved solids, (6) turbidity, depth of water, (7) shape of surface, (8) extent, i.e., total area of water surface, (9) colour of water, (10) velocity of water, and (11) waves at water surface. These are discussed below in detail.

3.3.1 Temperature

Evaporation is directly affected by the temperature of water as well as that of the air above. The kinetic energy of the molecules of water increases with temperature. Similarly, the saturated vapour pressure of the air film near the water surface and the vapour pressure of the air above increases. Since evaporation depends upon the difference of these two vapour pressures, change in temperature may not affect evaporation in the same proportion and, as such, effect of temperature on evaporation is somewhat complicated.

Evaporation observed in a lake for one complete year indicates that for the same mean temperature observed in different months, the evaporation figures differ. However, evaporation increases with temperature, even though a close relation between evaporation and temperature cannot be established.

3.3.2 Solar Radiation

About 590 calories are required for converting one gram of water into water vapour. This heat energy is directly or indirectly derived from solar radiation. It is quite evident that evaporation rate changes during day, night and during different months in a year, since incidence of the Sun’s radiation on the earth is different. Air and water temperatures depend upon solar radiation, and one can expect a close relation between the Sun’s radiation and evaporation.

3.3.3 Wind

The air above the water surface receives the evaporated water molecules, and the rate of evaporation depends upon the moisture content in the air. If the water molecules in the air above are removed quickly, then the rate of evaporation increases. Removal of water molecules in the air above water surface is done by the wind and naturally depends on the velocity of wind. The more the velocity of wind, the faster will be the process of removal of water vapour particles and more will be the evaporation rate.

However, there is a limiting velocity of wind up to which the rate of evaporation increases with the increase in the wind velocity. Further increase has no effect on the evaporation rate. This higher limit of wind velocity depends on other factors also.

The velocity of wind referred above is naturally near the water surface, because the variation in the velocity of wind in the vertical direction is not a straight-line relation. Wind speed observed by an anemometer at a higher level will have to be modified considering the variation in the vertical direction.

So also, the velocity of wind will affect the evaporation rate if the air movement is turbulent. The water vapour will be removed quickly by turbulent diffusion. If the airflow is laminar, it will have no significant effect on the removal of water vapour and indirectly on the evaporation rate.

3.3.4 Barometric Pressure and Altitude

It is evident that the barometric pressure has an influence on the rate of evaporation. As the barometric pressure reduces, the rate of evaporation increases. However, the change in barometric pressure brings about changes in other variables that influence evaporation, and, as such, it is very difficult to establish an exact relation between the change in barometric pressure and evaporation rate.

Similarly, as altitude increases, the barometric pressure reduces and so the temperature. However, evaporation increases with the reduction of barometric pressure, but reduces with the reduction in temperature. The result will be the combined effect.

However, field observations indicate that evaporation goes on reducing with altitude.

3.3.5 Quality of Water

When water contains some dissolved solids, the vapour pressure reduces as compared to the pure water. Thus, the evaporation rate reduces with an increase in soluble solids. Evaporation is about 2–3% less from sea water as compared to fresh water. It is also noticed that the evaporation rate reduces by 1% with 1% increase in the specific gravity due to the dissolved solids.

Turbidity of water has no direct effect on the evaporation rate. The reflection of the solar radiation may increase and thus turbidity may indirectly affect the evaporation rate.

3.3.6 Depth of Water

The depth of water in the lake or reservoir has definitely an effect on the evaporation rate. The heat energy received by water from the Sun’s radiation is utilized in three different ways:

  1. Reflecting it back into atmosphere
  2. Increasing the temperature of water
  3. For evaporation

In the case of water of high depth during summer season, heat is utilized to increase the temperature of water at lower depths and thus evaporation reduces slightly. In cold season for higher depths, heat is available for evaporation from water and hence the evaporation rate increases. In the case of shallow waters, it is exactly the opposite way. The evaporation rate observed in two different lakes having different depths is shown in Fig. 3.1.

3.3.7 Other Factors

The other factors that have an effect on evaporation are as follows:

  • Shape of the water surface
  • Size, i.e. total area of the water surface

    Fig. 3.1 Evaporation in lakes with different depths

  • Colour of water surface
  • Flow velocity of water
  • Surface waves

However, efforts are still on to evaluate the effect of these factors on the rate of evaporation, though it is quite certain that these factors affect the evaporation rate. No exact relation has been concluded so far.

3.4 ESTIMATION OF EVAPORATION

The following methods are generally adopted to evaluate the rate of evaporation from a reservoir.

  • Water budget method
  • Energy budget method
  • Mass transfer method
  • Actual observations
  • Evaporation formulae

3.4.1 Water Budget Method

This method is the application of continuity equation to the water content in a reservoir which is expressed as

 

E = P + I−(UiUo) − O ± dS

where, E = Evaporation

         P = Precipitation

          I = Surface inflow

       Ui = Underground inflow into the reservoir

      Uo = Underground outflow from the reservoir

       O = Outflow from the reservoir, if any

     dS = Change in storage

The total evaporation during the observed time (may be a week or so) can be evaluated provided all terms on the RHS of the above equation are known. The precipitation, if any, during the period of observation has to be taken into account. The inflow in the form of surface flow from the various streams discharging into the river/reservoir, if any, can be measured. Underground inflow or underground outflow is very difficult to calculate.

Outflow from the reservoir may be in the form of irrigation canal flow or hydroelectric power generation outflow which can be evaluated. During the observation period, there might be some change in the reservoir storage. This may be positive or negative.

Thus, knowing the variables on the RHS, evaporation loss during the specified period can be estimated.

This method is not accurate but may give a rough idea about the loss due to evaporation.

Example 3.1

The catchment area of an irrigation tank is 70 km2. The constant water spread during October 2006 was 2 km2. During that month, the uniform precipitation over the catchment was recorded to be 100 mm. 50% of the precipitation reaches the tank. The irrigation canal discharges at a uniform rate of 1.00 m3/s in the month of October.

Assuming seepage losses to be 50% of the evaporation losses, find out the daily rate of evaporation for October 2006.

Solution:

Total inflow

Outflow from canal = 1 × 3600 × 24 × 31

                                  = 2.68 × 106m3

Loss of water = 3.50 × 106 − 2.68 × 106

                       = 0.82 × 106 m3

                       = Seepage loss + evaporation loss

Since seepage loss is 50% that of evaporation loss,

 

Loss of water = 1.5 × evaporation loss

Therefore, evaporation loss

Rate of evaporation

                                 = 8.8 mm/day/m2

Example 3.2

The surface area of a reservoir in m2 given by A = 100 y2, where y is the depth of water in metres in the reservoir. In one week, the water depth in the reservoir has reduced from 10 to 9 m. Find the average hourly rate of evaporation. Assume seepage loss to be 40% of the evaporation loss.

Solution:

Example 3.3

The catchment area of a reservoir is 10.0 km2 . A uniform precipitation of 0.5 cm/h for 2.0 h was observed on 7th of July. 50% of the runoff reached the reservoir. A canal carrying a discharge of 1.25 m3/s is taken from the reservoir. The rate of evaporation observed was 0.7 mm/m2/h. The seepage loss was observed to be 50% of the evaporation loss. Find the change in the reservoir level on 4th July from 8:00 a.m. to 6:00 p.m. if the water spread of the reservoir was 0.476 km2.

Solution:

Inflow into the reservoir

Canal outflow = 1.25 × 10 × 3600 = 4.5 × 104 m3

Evaporation and seepage losses

Total outflow from the reservoir = 4.5 × 104 + 0.5 × 104 = 5.0 × 104 m3

                    = inflow.

Hence, there will not be any change in the reservoir water level.

Example 3.4

A trapezoidal channel of bed width 4.0 m and side slopes 1:1 carries water at a depth of 2.0 m. The rate of evaporation observed was 0.35 mm/m2/h. Find the daily loss due to evaporation from the canal in a length of 10 km in ha m.

Solution:

Top width of water level = 4.0 + 2.0 + 2.0 = 8.0 m

Total water spread in the canal in 10.0 km = 8.0 × 10.0 × 1000 = 8 × 104 m2

Rate of evaporation from the canal = 0.35 mm/m2/h

3.4.2 Energy Budget Method

In this method, in a specific time, the energy received by water from the Sun or from the atmosphere is accounted and then the energy utilized for evaporation is calculated; thus, the actual evaporation is worked out.

Thus

 

QI = QR ± Qs + QE

 

where, QI = Total energy received from Sun’s radiation

        QR = Energy reflected back into the atmosphere by water

        Qs = Change in the energy of the stored water

        Qe = Energy required for evaporation

Qs may be positive or negative depending on energy utilized by the water or energy received from the water. This will depend upon the season. QE can be calculated if all other factors in the equation above are evaluated. However, this requires a lot of instrumentation. Once QE is calculated, the actual evaporation can be calculated knowing the following:

  • Latent heat of evaporation
  • Temperature of water
  • Barometric pressure

This method too is not accurate, as it requires very accurate instrumentation. However, it may give a rough idea.

Example 3.5

A reservoir having a water spread of 0.5 km2 and water storage of 1 million m3 receives the Sun’s radiation at the average rate of 0.025 Langley per second. 5% of this radiation is reflected back into the atmosphere. The rise in the average temperature of the water is 2 °C during 9:00 a.m. to 5:00 p.m.

Assuming no exchange of heat between water and ground as well as the atmosphere, find the average rate of evaporation.

Solution:

1 Langley = 1 cal/cm2

Latent heat of evaporation: 539.55 cal/g

Specific heat of water: 1 cal/g/°C

  1. Energy received from the Sun = 0.5 × 106 ×104 × 0.025 × 8 × 3600

                                             = 3.6 × 1012 cal

  2. Energy received by water = 0.95 × 3.6 × 1012

                                             = 3.42 × 1012 cal

  3. Energy utilized by water = 2 × 1 × 106 × 106 cal

                                             = 2.0 × 1012 cal

  4. Energy utilized by water for evaporation = (3.42 − 2.0) × 1012cal

                                             = 1.42 × 1012 cal

  5. Volume of water evaporated

3.4.3 Mass Transfer Method

This method is based on the determination of the mass of water vapour transferred from the water surface to the atmosphere. It is also known as vapour flow approach or aerodynamic approach. The concept of boundary layer theory and continuous mixing, are applied. It is assumed that wind velocity in the vertical is logarithmic and atmosphere is adiabatic.

The evaporation is expressed as:

where, E = Evaporation in mm/h

       zI, z2 = Arbitrary levels above water surface levels in metres

       el, e2 = Vapour pressures at zl z2 in mm Hg

       vl, v2 = Wind velocity at z1, z2 in km/h

            T = Average temperature in °C between z1 and z2

3.4.4 Methods of Actual Observations

Evaporation can be measured by an instrument known as atmometer.

In an atmometer, there is a continuous supply of water to a surface. This surface is kept constantly wet and evaporation from this surface is measured. Atmometers are not common because of their small size. They do not have sufficient exposure. The rate of evaporation observed from an atmometer is on higher side than that from any other method.

The following are normal atmometers: (1) Livingstone atmometer, and (2) Piche atmometer

  1. Livingstone atmometer It consists of a 50-mm diameter spherical surface of 2.5-mm thick porous material as shown in Fig. 3.2.

    The bottle is filled with distilled water that is supplied continuously to the porous bulb. The loss of water from the bottle is due to evaporation.

  2. Piche atmometer It consists of a graduated glass tube of 15-mm diameter and 300-mm in length. It is filled with water and covered with a filter paper. The tube is kept in an inverted position so that there is continuous supply of water to the filter paper. The loss of water from the glass tube is the loss due to evaporation. The Piche atmometer is shown in Fig. 3.3.

3.4.5 Pan Observations

The estimation of evaporation can be done by taking actual observations from a pan and correlating these results to a reservoir.

Fig. 3.2 Livingstone atmometer

Fig. 3.3 Piche atmometer

3.4.5.1 Pan

A pan is a metal container, square or circular, and of uniform cross-section. Normally it is circular. The diameter may range from 300 to 1500 mm.

This pan is filled with water and the loss of water from this pan in a specified period is measured. The rate of evaporation observed here is correlated to the evaporation from a reservoir.

3.4.5.2 Observations on a Pan

The observations over a specified time are taken by two methods.

  1. In the first method, the pan is filled up to a specific level and the loss of water is calculated by observing the level of water over a specified time. Knowing the cross-section of the pan and the reduction in the water level, the loss due to evaporation over the specified time can be calculated. The water levels in the pan are taken accurately by a hook gauge.
  2. In the second method, the pan is filled up to a specific level. This level is maintained constant by adding water to the pan periodically to meet the evaporation loss. The loss due to evaporation over that period is the quantity of water added to the pan to maintain a constant level.

Example 3.6

The drop in water level in the case of a 1.5-m diameter evaporation pan during 24 h is 10 mm. The precipitation recorded during this period was 15 mm. Find the rate of evaporation from the pan.

Solution:

Area of pan

Evaporation in 24 h

Therefore, Rate of evaporation

Example 3.7

The amount of water added to a 1.0-m diameter circular land pan over a period of 10 h to maintain the water level constant is 9.56 lit. Find the hourly rate of evaporation from the pan.

Solution:

Area of pan

Rate of evaporation per hour

Rate of evaporation = 1.21 mm/h/m2

Example 3.8

Observations taken on a 1.0-m diameter circular land pan on 7 July 2006 from 8:00 a.m. to 6:00 p.m. were as follows:

  1. Quantity of water taken out of the pan = 5.0 lit
  2. Precipitation during this time period = 20 mm

If there was no change in the water level in the pan, find the rate of evaporation.

Solution:

  1. Area of pan
  2. Amount of water received due to precipitation = 0.785 × 103 × 20 ×10−3 = 15.706 lit
  3. Evaporation from the pan: 15.706 − 5.0 = 10.706 lit

Therefore, the rate of evaporation

Example 3.9

In the case of a 1.5-m diameter circular pan, following observations were taken from 8:00 a.m. to 6:00 p.m.

  1. Quantity of water added to keep the water level in the pan constant is 5.0 lit
  2. Precipitation during the observation period is 10 mm
  3. Leakage from the pan is 2.0 lit

Find the rate of evaporation from the pan.

Solution:

Area of the pan

Quantity of water received by the pan = 1.767 × 103 × 10 × 10−3

                                                                      = 17.67 lit by way of precipitation

Evaporation from the pan = 17.67 + 5.0 − 2.0 = 20.67 lit

3.4.5.3 Pan Coefficient

The observations from a pan are extrapolated to a reservoir by using a pan coefficient. A pan coefficient may be defined as follows:

The value of pan coefficient is always less than 1.0 and is dimensionless because, in the case of a pan, a very small area is exposed to atmosphere and, secondly, the metal container absorbs more energy and in turn is utilized for evaporation.

Example 3.10

The rate of evaporation from a 1.25-m pan having a pan coefficient of 0.8, was 1 mm/m2/h. Find the total evaporation from a reservoir in a week, having a water spread of 2.0 ha.

Solution:

Example 3.11

The monthly evaporation (in mm) observed on a pan from January to December 2006 is as follows:

 

111, 126, 127, 132, 141, 146, 148, 143, 138, 126, 118, 114

 

The water spreads of reservoir in January 2006 and December 2006 were 2.56 and 2.69 km2, respectively. Assuming a pan coefficient of 0.8, find the total loss of water due to evaporation in 2006 from the reservoir. Neglect all other losses.

Solution:

3.4.5.4 Factors Affecting Pan Coefficient

The following factors affect the pan coefficient.

  • Diameter of pan
  • Depth of water in the pan
  • Height of rim of pan above the water level
  • Location of pan
  • Colour of pan
  • Material used for pan

The pan coefficient can be evaluated by taking observations on the different sizes of pans and extrapolating the observations as shown in Fig. 3.4.

The different types of pans are: land pan, floating pan and sunken pan

3.4.5.5 Land pan

This type of pan is located on land above the ground normally on the bank of the lake or the reservoir. There are different types and sizes. The standard one is shown in Fig. 3.5.

It is made of ungalvanized iron sheet 1200 mm in diameter and 250 mm in depth. The bottom is supported on a wooden frame placed 150 mm above ground level, and the water level is kept 50–75 mm below the rim. The pan coefficient varies between 0.6 and 0.7.

3.4.5.6 Floating pan

In this type, the pan is kept floating in the reservoir. Thus with this arrangement, similar surrounding conditions are maintained. Initially, the water level in the pan is kept the same as that of reservoir and observations are taken regularly by approaching it from the bank.

Fig. 3.4 Variation of pan coefficient w.r.t. pan diameter

The standard Floating Pan of 900-mm2 area and 450-mm depth is as shown in Fig. 3.6. The pan coefficient varies from 0.7 to 0.82.

3.4.5.7 Sunken pan

In this type, the pan is sunk in the ground on the banks of the reservoir. This arrangement avoids splash of rain, drifting of dust or trash and obstruction to the wind. It also eliminates boundary effects such as the radiation of the side walls and heat exchange between the pan and the atmosphere.

Fig. 3.5 Land pan

Fig. 3.6 Floating pan

The standard type is made of ungalvanized unpainted iron sheet with an area of 900 mm2 and 450 mm in depth. The rim is kept 100 mm above water level and the water level same as that of ground, as shown in Fig. 3.7. The pan coefficient varies between 0.75 and 0.86.

Fig. 3.7 Sunken pan

3.4.6 Formulae to Estimate the Evaporation from a Reservoir

There are several formulae to estimate evaporation from a reservoir. All of them are empirical.

The formulae normally used are: Fitzgerald’s formula, Rohwer’s formula, Meyer’s formula and Lake Mead formula.

The various variables used in the formulae in estimating evaporation from a reservoir are as follows:

E = Evaporation in mm/day

es = Saturated vapour pressure in mm Hg

ea = Actual vapour pressure in mm Hg

V = Wind velocity at the water surface in km/h

Pa = Mean barometric pressure in mm Hg

Ta = Average air temperature in °C

Tw = Average water temperature in °C

The formulae are as follows:

  1. Fitzgerald formula

     

    E = (0.4 + 0.124 V )(esea)

     

    where, E is evaporation in mm/day.

     

  2. Rohwer’s formula

     

    E = 0.771 (1.465 − 0.000732 Pa) (0.44 + 0.07334 V) (esea)

     

    where, E is evaporation in mm/day.

     

  3. Meyer’s formula

     

    E = C (es − ea) (1 + 0.06215 V10)

     

    where, E is evaporation in mm/month

    C a constant having a value of 11 for deep water bodies and 15 for shallow water bodies.

            V 10 = Velocity at 10 m above water surface in km/h

    where, E is evaporation in mm/day.

     

  4. Lake Mead formula

     

    E = 0.0331 V(esea) [1 − 0.03 (TaTw)]

     

    where, E is evaporation in mm/day.

Example 3.12

For a reservoir the following observations were taken:

  1. Saturation vapour pressure of air = 30 mm Hg
  2. Relative humidity = 0.5
  3. Velocity of wind at the rate of 0.5 m above ground = 25 km/h
  4. Water spread of reservoir = 10 km2
  5. Atmospheric pressure = 10.4 m of water
  6. Average air temperature = 29 °C
  7. Average water temperature = 27 °C

Evaluate monthly evaporation by using different formulae.

Solution:

The monthly evaporation rate was calculated by four empirical formulae.

  1. Fitzgerald’ s equation

    Therefore, ea = 15 mm Hg

    Assuming average velocity of wind

    V = Velocity of wind at the rate of 0.5 m above ground

    E = (0.4 + 0.124 V) (esea)

       = (0.4 + 0.124 × 25) (30 − 15)

       = (0.4 + 3.1) (15)

    E = 52.5 mm/day

  2. Meyer’s equation

     

    E = C(esea)(1 + 0.0621 V10)

     

    where, E = Evaporation in mm per month

                   V = Velocity at 10 m above ground

                   C =15

    Velocity at the rate of 10 m above ground = V10

    Velocity at the rate of 0.5 above ground = V0.5 = 25 km/h

  3. Lake Mead equation

     

    E = 0.0331 V (esea) [1 − 0.03 (TaTw)]

     

    where, Ta = Average air temperature = 29 °C

             Tw = Average water temperature = 27 °C

               E = 0.0331 × 25 (30 − 15) [1 − 0.03 (29 − 27)]

               E = 11.66 mm/day

  4. Rower’s equation

     

    E = 0.771 (1.465 − 0.000732 Pa) × (0.44 + 0.07334 V) (esea)

     

    where, Pa = Pressure in mm Hg

           

             E = 0.771 (1.465 − 0.000732 × 764.70) × (0.44 + 0.07334 × 25) (30 − 15)

                = 0.771 (1.465 − 0.559) (0.44 + 1.833) 15

                = 23.81 mm/day

3.5 control of evaporation from a reservoir

Evaporation from a reservoir can be reduced by as much as 30%. The different methods to control evaporation from a reservoir are as follows.

3.5.1 Surface Area Reduction

Evaporation depends mostly on the exposed surface area. If by any means the surface area exposed is reduced, the evaporation from a reservoir can be reduced.

This can be achieved by the following ways:

  • Reducing the meandering length of the streams
  • Selecting the dam site such that the ratio of surface area to storage is the minimum
  • Storing water underground

This method has a limited scope, but evaporation can be controlled by this method.

3.5.2 Mechanical Covers

If a reservoir is covered, completely or even partially, the movement of water vapour can be controlled and thus evaporation will reduce.

This is a very costly method and can be practised for small reservoirs only.

3.5.3 Wind Shields

Wind velocity near the water surface affects the evaporation. Less the velocity, less will be the evaporation. The wind velocity near the water surface can be reduced by providing obstruction to wind. This obstruction can be achieved by providing plants and bushes all around the reservoir. However, the exact effect due to this method cannot be estimated.

This method can be adopted for small lakes and reservoirs only.

3.5.4 Surface Films

Reduction in evaporation from a reservoir is achieved by spreading a film over the water surface. This can be done by means of some chemicals. This film will retard the movement of water molecules from the water surface to the air above.

This film is one-molecule thick and hence is termed as monomolecularfilm (one millionth of a millimetre). The reduction in evaporation due to such films is observed to be 40–70% in the laboratory and 25% in the lakes and the reservoirs.

The chemical to be used should have the following properties:

  • It should prevent movement of water molecules into the atmosphere.
  • It should allow movement of oxygen and carbon dioxide.
  • It should not get disturbed due to wind or rain.
  • It should not be harmful to plants and animals.
  • It should be cheap and available in the required quantity.

3.5.4.1 Chemicals normally used

Cetyl alcohol (C16H33OH) known as hexadecanol or stearyl alcohol (C18H37OH) known as octa- decanol are the chemicals normally used. These are available in powder form and a solution is prepared by using turpentine, petrol, kerosene, or so.

3.5.4.2 Application of chemicals

The chemical can be applied by one of the following methods:

  • Hand spreading from the banks
  • Spraying from a boat
  • Aerial application using aeroplanes
  • By automatic dispensing units mounted on well-spaced anchored barges

3.5.5 Floating Covers

Evaporation from a reservoir can be controlled by providing floating covers over the water surface. This retards the movement of water molecules from the water surface to the air above.

These floating covers may be any of the following:

  • Wooden planks or sheets
  • Plastic balls
  • Polyethylene sheets

This method is very costly. However, it has definitely some positive result.

This method can be adopted for very small lakes and reservoirs.

3.6 TRANSPIRATION

Plants absorb water from the soil through their roots. Mineral salts are also absorbed in dilute solution using water as its vehicle. This solution is transported through roots and stems to the leaves where plant food is produced from this sap, CO2 from atmosphere and using energy from the Sun through chlorophyll. The plant food thus produced is again distributed using water as a vehicle for the cell growth and tissue building.

Most of the water thus absorbed by the plants is discharged back into the atmosphere in the form of vapour. This process is known as transpiration. It is observed that only 1% of water sucked in by the roots is retained by the plants.

3.6.1 Factors Affecting Transpiration

Practically, all the meteorological factors that affect evaporation, influence transpiration also. In addition, the following factors affect transpiration:

 

Sunlight: The growth of plants depends on sunlight. About 95% of the transpiration occurs during the day time.

Moisture available: Transpiration is limited by the rate of moisture that becomes available to the plants. The plants can extract water between field capacity and the wilting point. If the water available is less than the wilting point, then the plants may not be able to suck up water from the soil. Thus, it may affect transpiration.

Stage of plant development: Transpiration depends upon the plant growth. The growth of plant varies (1) diurnal, (2) seasonal and (3) annual. Also, the growth of the plants is more during its early stage. Transpiration varies accordingly.

3.6.2 Measurement of Transpiration

Transpiration can be measured in terms of the depth of water transmitted daily or annually.

A phytometer provides a practical method for measuring transpiration. This is a large vessel filled with soil in which one or two plants are rooted. The soil surface is sealed to prevent evaporation from the soil. The initial weight of the container along with the soil and the plants is recorded. The transpiration loss can be worked out by observing the weight of the container over a known period. A small phytometer containing water only is known as potometer.

3.6.3 Transpiration Reduction

Water conservation can be achieved to some extent through transpiration reduction. The following are the different methods to conserve water:

  • Use of chemicals to inhibit water consumption (similar to mono molecular films in the case of evaporation)
  • Harvesting of plants
  • Improved irrigation methods
  • Removing unwanted, unproductive and useless trees, bushes and grasses

Transpiration ratio (TR) is the ratio of the total weight of water transpired by a plant during its complete development to the weight of the dry matter produced. It is dimensionless.

The TR for rice is between 300 and 600 and that for wheat is between 600 and 800.

3.6.4 Evapotranspiration

In studying water balance from an area, it is difficult to separate evaporation and transpiration and hence these two processes are combined and treated as a single and termed as evapotranspiration.

3.6.4.1 Potential Evapotranspiration

When adequate quantity of water is available for evaporation and transpiration, the combined process is known as potential evapotranspiration. It is normally known as PET. Actual evapotranspiration, normally denoted by AET is, therefore, always less than potential evapotranspiration (AET < PET).

3.6.4.2 Measurement of Evapotranspiration

Evapotranspiration can be measured by the following methods:

  • Lysimeter
  • Inflow-outflow measurements

Lysimeter: It consists of a circular tank, its diameter ranging from 600 mm to 3000 mm. It is buried in the ground so that its top is flush with the ground. It is filled with the soil similar to the field conditions, and the crop or the tree of which AET is to be found is grown in the lysimeter. A typical lysimeter is shown in Fig. 3.8.

Water is added to the lysimeter and an account of water added, etc. is kept. Then AET is calculated as follows:

 

P + W = O + AET + ΔS

 

where, P = Precipitation during the period of observations

       W = Water supplied

       O = Water drained

  AET = Actual evapotranspiration

     ΔS = Change in soil moisture in lysimeter

Lysimeters are expensive and the process of observation is time-consuming.

Fig. 3.8 Lysimeter

Inflow—outflow measurements: This method is applicable to large areas, such as up to 100 ha. An account of the inflow of water is maintained for a specific period. The inflow may be precipitation or surface inflow or even groundwater inflow. During the same period, an account of outflow is maintained. The outflow may be the surface outflow or the groundwater outflow. Naturally, the difference between these two will be the loss due to evapotranspiration.

If the observation period is small, then it is assumed that the groundwater inflow and outflow are equal. In this method, it is very difficult to measure inflow, outflow and groundwater storage to the desired accuracy.

3.7 SOIL EVAPORATION

Immediately after the rain stops, evaporation from the soil starts. Evaporation from a saturated soil surface is approximately the same as that from a water surface at that temperature. As the soil begins to dry, the evaporation goes on reducing. Finally it virtually stops as there is no supply of water. If the groundwater table is within 1.0 m from the ground, evaporation from soil is noticed since there is some supply of water.

Evaporation opportunity is defined as ‘the ratio of evaporation from soil and evaporation from equivalent water surface’. It is dimensionless and is generally expressed in terms of percentage. Thus,

Evaporation from soil is measured by using a lysimeter of size 1 m × 1 m × 1 m. The soil is filled flush with the tank edge and the groundwater table is maintained at a specific level. The evaporation from the soil can be determined by weighing the tank at known intervals and from the quantity of water added to maintain the groundwater table.

REVIEW QUESTIONS
  1. Define evaporation. Explain the process of evaporation.
  2. Explain Dalton’s law of evaporation.
  3. Discuss the factors that affect evaporation.
  4. How is evaporation from a reservoir estimated?
  5. Explain the water budget method of estimation of evaporation from a reservoir.
  6. Explain the energy budget method of estimation of evaporation from a reservoir.
  7. Explain the mass transfer method of estimation of evaporation from a reservoir.
  8. What is an atmometer? Discuss the ones normally used.
  9. What is a pan? How are observations taken on a pan?
  10. What is a pan coefficient? Discuss the factors that affect pan coefficient.
  11. Explain land pan with the help of a neat sketch.
  12. Explain sunken pan with the help of a neat sketch.
  13. Explain floating pan with the help of a neat sketch.
  14. Discuss the merits and demerits of the different types of pans.
  15. Discuss the various parameters involved in the different formulae used to estimate evaporation from a reservoir.
  16. State and explain the different formulae normally used to estimate evaporation from a reservoir.
  17. Can evaporation from a reservoir be reduced? Explain the different methods.
  18. Write a detailed note on the use of surface films to reduce evaporation from a reservoir.
  19. Explain the process of transpiration. Discuss the different methods to reduce transpiration.
  20. Discuss the different methods to measure transpiration.
  21. What is evapotranspiration? Explain the different methods to estimate it from a catchment.
  22. Discuss soil evaporation.
  23. Write a note on the pattern of evaporation over India.
  24. Write short notes on
    1. Potential evapotranspiration.
    2. Factors affecting transpiration.
    3. Livingstone atmometer.
    4. Pitch atmometer.
    5. Transpiration ratio.
    6. Phytometer.
    7. Potometer.
    8. Lysimeter.
    9. Pan coefficient.
    10. Factors affecting transpiration.
    11. Annual evaporation in India.
    12. Atmometer.
    13. Evaporation opportunity.
  25. Differentiate between
    1. Evaporation, sublimation and transpiration.
    2. Phytometer and potometer.
    3. Land pan, sunken pan and floating pan
    4. Evapotranspiration and potential evapotranspiration.
NUMERICAL QUESTIONS
  1. The uniform precipitation over a catchment area covering 80 km2 received a total precipitation of 80 mm in a month. 60% of the precipitation reached the reservoir. The spread of reservoir during that month was 2.5 km2. The irrigation canal discharges at a constant rate of 1.1 m3/s. If the seepage losses were 60% of the evaporation losses, find the daily rate of evaporation.

     

    Ans: 8.24 mm/h/m2

  2. In the case of a reservoir, the surface area in m2 is given by A = 90 y2, y is the average depth of water in the reservoir in metres. In a week, the depth of water reduced from 9.0 m to 8.5 m. If the evaporation loss is double the seepage loss, find the rate of hourly evaporation loss.

     

    Ans: 2 mm/h/m2

  3. The bed width of a trapezoidal channel is 5.0 m and the side slopes are 1:1. It carries water at a depth of 1.8 m. The rate of evaporation observed on a 1.5-m diameter floating pan was 0.25 mm/h/m2. Find the loss of evaporation in a day from the canal in a length of 5 km. Assume the pan coefficient to be 0.8.

     

    Ans: 0.0206ham

  4. During a test, the drop in water level in the case of a land pan of diameter 1.0 m in 24 h was observed to be 15 mm. The precipitation recorded during this period was 10 mm. Find the rate of evaporation from the pan.

     

    Ans: 1.32mm/h/m2

  5. A test was conducted on a 0.5-diameter sunken pan. The quantity of water added to maintain the water level constant over a period of 10 h was 950 cc. Find the rate of evaporation.

     

    Ans: 0.5 mm/h/m2

  6. The following observations were taken on a 1.0-m diameter land pan from 8 a.m. to 5 p.m.
    1. Precipitation during the observation period = 15 mm
    2. Leakage from the pan = 1.51
    3. Quantity of water added to keep the level constant = 4.01

       

      Find the rate of evaporation.

     

    Ans: 2.02 mm/h/m2

  7. The following observations were taken from a reservoir:
    1. Velocity of wind at 0.5 m above ground = 24 m/s
    2. Atmospheric pressure = 10.4 m of water
    3. Average air temperature = 30 ° C
    4. Average water temperature = 27 ° C
    5. Relative humidity = 0.5
    6. Water spread of reservoir = 15 km2
    7. Saturation vapour pressure of air = 30 mm Hg

       

      Find the monthly evaporation using empirical formulae.

     

    Ans: (i) 22.50 × 106m3
    (ii) 11.25 × 106m3
    (iii) 4.87 × 106m3
    (iv)10.36 × 106 m3

MULTIPLE CHOICE QUESTIONS
  1. The pan coefficient is given by
    1. Rate of evaporation from reservoir — Rate of evaporation from pan
    2. Rate of evaporation from reservoir × Rate of evaporation from pan
  2. The dissolved salts in water
    1. Reduce the rate of evaporation
    2. Increase the rate of evaporation
    3. Have no effect on the rate of evaporation
    4. Double the rate of evaporation
  3. Lysimeter is an instrument used to measure
    1. Evaporation
    2. Transpiration
    3. Evapotranspiration
    4. Sublimation
  4. The unit of evaporation is
    1. mm/h
    2. mm/m2/h
    3. mm/m/h
    4. mm/m3
  5. The pan coefficient is always
    1. More than 1
    2. Equal to 1
    3. Less than 1
    4. Less than 0.5
  6. Dalton’s law of evaporation is given by
    1. E = C (eaes)
    2. E =C (esea)
  7. Transpiration ratio is given by
    1. Weight of dry matter produced + Weight of water transpired
    2. Weight of dry matter produced − Weight of water transpired
  8. Transpiration is confined to
    1. Day time hours
    2. Night hours
    3. All the day
    4. Any of the three
  9. The reduction of evaporation by using monomolecular surface is approximately
    1. 2%
    2. 10%
    3. 25%.
    4. 50%
  10. Out of the water sucked by the roots, the percentage retained by plants is
    1. 1
    2. 5
    3. 10
    4. 15
  11. The transpiration ratio for rice is between
    1. 50 to 300
    2. 300 to 600
    3. 600 to 900
    4. 900 to 1200
  12. 11. PET means
    1. Practical evapotranspiration
    2. Progressive evapotranspiration
    3. Perfect evapotranspiration
    4. Potential evapotranspiration
  13. AEP means
    1. Additional evapotranspiration
    2. Accumulated evapotranspiration
    3. Actual evapotranspiration
    4. Allowable evapotranspiration
  14. Evapotranspiration means
    1. Evaporation − transpiration
    2. Evaporation + transpiration
    3. Evaporation ÷ transpiration
    4. Evaporation × transpiration
  15. Evaporation is a
    1. Cooling process
    2. Heating process
    3. Combined process
    4. None of the three
  16. Annual evaporation loss from a reservoir is about
    1. 5%
    2. 10%
    3. 30%
    4. 60%
  17. From a reservoir evaporation loss can be evaluated from
    1. Land pan
    2. Floating pan
    3. Sunken pan
    4. Any of the three
  18. The chemical notation of cetyl alcohol is
    1. C16H32OH
    2. C10H20OH
    3. C20H40OH
    4. C16H33OH
  19. The chemical symbol for stearyl alcohol is
    1. C18H38OH
    2. C18H40OH
    3. C18H37OH
    4. C19H38OH
  20. The thickness of a monomolecular film is
    1. One tenth of a millimetre
    2. One hundredth of a millimetre
    3. One thousandth of a millimetre
    4. One millionth of a millimetre
  21. In the case of a crop TR means
    1. True ratio
    2. Transit ratio
    3. Transpiration ratio
    4. Technical ratio
  22. The dimensions of evaporation opportunity are
    1. mm/m2/h
    2. Dimensionless
    3. mm/h
    4. mm/m2
ANSWERS TO MULTIPLE CHOICE QUESTIONS

1. a      2. a      3. c      4. b      5. c      6. b      7. b      8. a      9. c      10. a      11. b      12. d      13. c      14. b      15. a      16.      c      17. d      18. d      19. c      20. d      21. c      22. b