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Abstractions from Precipitation

A. Evaporation and Evapotranspiration

Evaporation�Heat energy from the sun causes water in puddles, streams, rivers, seas

or lakes to change from a liquid to a water vapor. This is calledevaporation.The vapor rises into the air and collects in clouds.

� The process by which water isevaporated from wet surfaces andtranspired by plants together is calledevapotranspiration.

� Major factors affecting evaporation areRadiation, Temperature, wind,barometric pressure and altitude,surface area, dryness, depth of water,quality of water etc.

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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

� The amount of water lost through these exposed surfaces via evaporation can be substantial to upto 30%

� The methods adopted to evaluate rate of evaporation from a reservoirare Water budget method, Energy budget method, Mass transfermethod, actual observations and empirical formulae.

Estimation of evaporation

� Water budget method� Apply mass balance equation to the various components of inflow,

outflow and storage

0( )R S d

E P Q Q Q Q Q dS= + + + − − ±

Surface runoff - Qr

Subsurface runoff - Qs

Inflow- Q

Outflow- Q0

Evaporation- E

Subsurface seepage losses- Qd

Precipitation - P

dS = change in storage

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Water budget method

Characteristics:

- Simple

- Difficult to estimate Qd and Qs

- Unreliable, accuracy will increase as Δt increases

Example on Water Budget Method

� A reservoir had an average surface area of 20 km2 during June 2012. In that month the mean rate of inflow = 10 m3/s, outflow = 15 m3/s, monthly rainfall = 10 cm and change in storage = 16 million m3. Assuming the seepage losses to be 1.8 cm, estimate the evaporation (in cm) in that month.

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Solve

� 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 from the tank discharges at a uniform rate of 1.00 m3/s in the month of October.Assume seepage losses to be 50% of the evaporation losses, find out the daily rate of evaporation for October 2006.

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

Estimation of evaporation� Energy-Budget Method

� Apply law of conservation of energy to the various components of incoming energy, outgoing energy and energy stored in the water body over a known time interval

� Calculates evaporation on a daily time interval

energy inflows = energy outflows

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� Energy-Budget Method

n a e g s iH H H H H H= + + + +

Where, Hn = net heat energy received by the water surface

( )1c b

H r H= − −

( )1c

H r− = incoming solar radiation into a surface of reflection coefficient rHb = back radiation (long wave) from water body

Ha = sensible heat transfer from water surface to air

He = heat energy used up in evaporation = ρLEL where, ρ = density of water, L = latent heat of evaporation and EL = evaporation in mm

Hg = heat flux into the ground

Hs = heat stored in water body

Hi = net heat conducted out of the system by water flow (advected energy)

� Energy-Budget Method� All energy terms are in calories per square mm per day

� If time periods are short, the terms Hs and Hi can be neglected as negligibly small

� All terms except Ha can be measured or evaluated indirectly� Ha is estimated using Bowen’s ratio β

46.1 10a a w a

a

e L s a

H H T Tp

H LE e eβ

ρ− −

= = = × ×−

Where, pa = atm. pressure in mm of Hges = saturated vapour pressure in mm of Hgea = actual vapour pressure of air in mm of HgTw = temp. of water surface in °CTa = temp. of air in °C

( )1

n g s i

L

H H H HE

Lρ β

− − −=

+

Gives satisfactory results, with errors of the order of 5% when applied to periods less than a week

Gives satisfactory results, with errors of the order of 5% when applied to periods less than a week

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Measurement of solar energy

� Pyranometer

� Pyrheliometer

� Radiometer

� The fraction of solar energy that is reflected back to space is called the albedo. Different parts of the Earth have different albedos. � For example, ocean surfaces and rain forests have low albedos, which

means that they reflect only a small portion of the sun's energy.

� Deserts, ice, and clouds, however, have high albedos; they reflect a large portion of the sun's energy.

� Over the whole surface of the Earth, about 30 percent of incoming solar energy is reflected back to space.

Example on energy budget method

� Calculate by the energy budget method the evaporation rate per day from an open water surface, if the net radiation is 200 W/m2

and the air temperature is 25°C, assuming no sensible heat or ground heat flux.

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Empirical evaporation equations

� Quite a number of equations available

� Most formulae based on Dalton’s Equation (1802)

� Where, E = evaporation in mm/day or mm/month,� es = saturation vapour pressure at the water surface temperature in mm of

Hg.� ea = actual vapour pressure of overlying air in mm of Hg.� f(u) = wind speed correction function� K = a coefficient depends on atmospheric pressure and other factors,

� f(u) = a + b*u� a , b are constants ;� u : wind speed (m/s) ; ( u2 : wind speed at height 2m.)

( ) ( )s aE K f u e e= ⋅ ⋅ −

Empirical formulae

� Meyer’s Equation

� Where, E = evaporation in mm/month,

� es = saturation vapour pressure at the water surface temperature in mm of Hg corresponding to mean monthly temperature of air.

� ea = actual vapour pressure of overlying air in mm of Hg based on mean monthly temperature and relative humidity.

� u10 = Monthly wind velocity in km/h, at 10m above ground

� K = 15 for small shallow ponds, 11 for large or deep water bodies

10(1 0.06215 ) ( )s aE K u e e= ⋅ + ⋅ −

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Empirical formulae

� Rohwer’s Equation

� Where, E = evaporation in mm/day,

� es = saturation vapour pressure at the water surface temperature in mm of Hg corresponding to mean monthly temperature of air.

� ea = actual vapour pressure of overlying air in mm of Hg based on mean monthly temperature and relative humidity.

� u = Mean wind velocity in km/h, at the surface

� Pa = Mean barometric reading in mm of mercury

( )0.771 1.465 0.000732 (0.44 0.07334 ) ( )a s aE P u e e= ⋅ − ⋅ + ⋅ −

Empirical formulae

� Horton’s Equation

� Where,

� E = evaporation in mm/day,

� es = saturation vapour pressure at the water surface temperature in mm of Hg corresponding to mean monthly temperature of air.

� ea = actual vapour pressure of overlying air in mm of Hg based on mean monthly temperature and relative humidity.

� u = Mean wind velocity in km/h, at the surface

� For large areas the above equation is to be multiplied by the area factor

0.4 ( )s aE e eψ= ⋅ −

0.1242 ueψ −= −

( )( )

1(1 )P P

h

ψ

ψ

−− +

−

Where, P = fraction of time during which the wind is turbulenth = relative humidity expressed as a fraction

Where, P = fraction of time during which the wind is turbulenth = relative humidity expressed as a fraction

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Actual observations

� Direct measurement by evaporation pans:

� The most commonly employed pans are� Class A Land Pan,

� ISI standard pan

� Colorado Sunken Pan,

� Floating pan.

� The evaporation pans are installed in the vicinity of reservoir or lakefor which loss of water through evaporation is to be determined.

Class A evaporation pan

� Used by US Weather Bureau

� Depth of water maintained between 18 cm to 20 cm

� Made of unpainted GI sheet

� Evaporation measurements are made by measuring the depth of water with a hook gauge in a stilling well

Measurements in diagram are in mm

� Advantages� Stable pan coefficient� Easy access for observation� More stability compared to floating pan� Relative freedom from dirt and trash� Reasonable cost of installation

� Advantages� Stable pan coefficient� Easy access for observation� More stability compared to floating pan� Relative freedom from dirt and trash� Reasonable cost of installation

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ISI standard pan (IS: 5973-1970)

� Modified Class A pan� Made of copper sheet of 0.9 mm thickness, tinned inside and

painted white outside� Fixed point gauge indicates the level of water� Top of the pan is covered fully by hexagonal wire net made of iron

to protect water in the pan from birds

Colorado sunken pan� Square in shape

� Radiation and aerodynamic characteristics are similar to that of a lake.� Direct solar radiation on side walls and heat exchange through side

walls are negligible

� However,� Difficult to detect leaks

� Extra care is needed to keep the surrounding area free from tall grass, dust etc.

� Expensive to install

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US geological survey floating pan

� Square in plan (900 mm side and 450 mm depth)

� Pan supported by drum floats in the middle of a raft is set afloatin a lake

� Water level keep at same level as the lake leaving a rim of 75 mm

� However,� High cost of installation and maintenance

� Difficulty in performing measurements� Not easily accessible

� Splashing of water into or out of the pan due to wave action whichis very difficult to quantify

Types of Evaporation Pans

USWB Class A Land Pan

Floating Pan

Sunken Pan

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Problems

� A test was conducted on a 0.5 m dia. sunken pan. The quantity ofwater added to maintain the water level constant over a period of10 h was 950 cc. Find the rate of evaporation.

� Observations taken on a 1.0 m diameter circular land pan on 7th

July 2013 from 8.00 am to 6.00 pm were as follows:� Quantity of water taken out of the pan = 5.0 lit

� Precipitation during this time period = 20 mm

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

Problems

� 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.� Quantity of water added to keep the water level in the pan constant

is 5.0 lts

� Precipitation during the observation period is 10 mm

� Leakage from the pan is 2.0 lt

Find the rate of evaporation from the pan.

� 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 per hour per unit area.

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Pan coefficient� Evaporation pans are not exact models of large reservoirs

� Differ in heat storing capacity and heat transfer from the sides and bottom� Sunken pan and floating pan aims to reduce this deficiency

� As a result of this factor, the evaporation from a pan depends on a certain extent on its size

� A 3 m diameter pan may give evaporation value same as a neighbouringlarge lake while a 1 m diameter pan may give 20% more value.

� Height of rim affects the wind action over the surface� Also casts a shadow of variable magnitude over the water surface

� Heat transfer characteristics of the pan material different from that of the reservoir� Colour of the pan

� Material used for pan

� Location of pan

Pan coefficient

� The observations from the pan are extrapolated to a reservoir byusing a pan coefficient.

� Value of pan coefficient < 1.0� Area smaller

� Metal container absorbs more energy

pan a fromn evaporatio of rate

reservoir thefromn evaporatio of ratetcoefficienPan =

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Control of evaporation from a reservoir� Surface area reduction

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

� Mechanical covers� Wind shields� Surface films

� Chemical films [cetyl alcohol (C16H33OH) known as hexadecanol or stearyl alcohol(C18H37OH) known as octadecanol]� Prevent movement of water molecules in the atmosphere� Allow movement of oxygen and carbon dioxide� Not be disturbed by wind or rain� Not be harmful to plants and animals� Cheap and available in the required quantity

� Available in powdered form; liquified using turpentine, petrol, kerosene or so.� Monomolecular film (one molecule thick); one millionth of a mm� Application: hand spreading from banks, spraying from a boat, aerial application using

airplanes, automatic dispensers on regularly spaced barges.� Evaporation reduction: 40-70% in laboratory and 25% in lakes and reservoirs.

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Control of evaporation from a reservoir

� Floating covers� Wooden planks or sheets

� Plastic balls

� Polyethylene sheets

Solve

� A reservoir with a surface area of 250 ha. had the following average values of climate parameters during a week:

� Water temperature = 20°C� Relative humidity = 40%� Wind velocity at 1.0 m above ground surface = 16 km/h

� Estimate the average daily evaporation from the lake by using Meyer’s formula.

� An ISI standard evaporation pan at the site indicated a pan coefficient of 0.80 on the basis of calibration against controlled water budgeting method. If this pan indicated an evaporation of 72 mm in the week under question, � Estimate the accuracy if Meyer’s method relative to the pan evaporation

measurements� Also estimate the volume of water evaporated from the lake in that week

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� The average watershed areas in the reservoirs and the pan evaporation at the reservoir site for 12 months are given below:

The pan coefficient may be assumed as 0.8. If the water in the reservoir fetched at the rate of Rs. 0.75 per m3, how much additional revenue accrues to the project per year if the evaporation losses are reduced to 25%?

Solve

Month Water spread area in ha Pan evaporation in cmJan 681 11.5Feb 623 17.2Mar 590 28.5Apr 578 34.3May 567 31.5Jun 560 20.0Jul 574 17.2Aug 598 15.7Sep 630 15.7Oct 664 17.2Nov 696 14.3Dec 716 11.5

Transpiration

� Process by which water vapour escapes from the living plant, principally through the leaves and enters the atmosphere.� Water extracted by the plant roots from the soil is transferred

through the plant to the intercellular space within the leaves

� Air enters the leaf through the stomata openings� Chloroplasts within the leaf use CO2 and a small portion of available water to

manufacture carbohydrates required for plant growth (PHOTOSYNTHESIS)

� As air enters the leaf, water escapes through the open stomata (TRANSPIRATION)

� Transpiration is essentially confined to daylight hours and the rate of transpiration depends upon the growth periods of the plant.� Plant growth varies diurnally, seasonally, annually

� Also more growth occurs in the early stage of the plant

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Factors affecting Transpiration� Plant Factors

� Extent and efficiency of the root systems in moisture absorption

� Leaf area

� Leaf arrangement and structure

� Stomatal behaviour

� Soil Factors (governing the water supply to the roots)� Field Capacity: amount of water held in the soil after excess gravitational

water has drained

� Wilting Point: moisture content at which permanent wilting of plants occurs

� Available Water: moisture in the soil between Field Capacity and Wilting Point

� Climate Factors� Solar radiation, light intensity, atmospheric pressure, temperature and

wind

Measurement of Transpiration

� Not possible to measure transpiration loss from an appreciable area under natural conditions

� Measured using phytometer

� Transpiration Ratio (TR) is the ratio of the total weight of water transpired by a plant during its complete development to the weight of dry matter produced� TR for wheat = 300 – 600

� TR for rice = 600 – 800

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Evapotranspiration

� In an area covered with vegetation it is difficult and unnecessary from practical view point to separately evaluate evaporation and transpiration

� Sum of water used by plants in a given area in transpiration and the water evaporated from the adjacent soil in the area in any specified time

� Also known as CONSUMPTIVE USE

� POTENTIAL EVAPOTRANSPIRATION (PET): Evapotranspiration when sufficient moisture is always available to completely meet the needs of vegetation fully covering the area� Depends essentially on climatic factors (for a given soil and plant condition)

� ACTUAL EVAPOTRANSPIRATION (AET): Real Evapotranspirationoccurring in a specific situation

� If water supply to the plant is adequate, soil moisture will be at the field capacity and AET will be equal to PET

� If water supply is less than PET, the soil dries out and the ratio of AET/PET is <1� The decrease of the ratio AET/PET with available moisture depends

on the type of soil and rate of drying.� For Clayey soils, AET/PET = 1 for nearly 50% drop in the available moisture

Evapotranspiration

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Measurement of Evapotranspiration

� Lysimeter� Circular tank 60 – 90 cm in diameter; 180 cm deep

� Filled with soil and crops or natural vegetation, for which ET is required, are grown

� Sides impervious; bottom pervious

Abstractions from Precipitation

B. Infiltration

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Infiltration

� Entry and movement of water through the land surface into the sub-strata below

� Infiltration Capacity: Maximum rate at which the soil is capable of absorbing water (f)

� Abstractions or precipitation loss� Part of precipitation that is not available as surface runoff

� Interception

� Depression storage

� Evaporation

� Infiltration (Dominant Abstraction)

� Excess rainfall� Part of precipitation available in the form of surface flow after meeting

all abstractions

� Infiltration rate� Rate at which water enters the ground surface and then flows downwards� High in the beginning but attains a steady constant lower value with time� Unit: mm/h, cm/h

� Factors affecting infiltration� Rainfall characteristics� Ground surface condition

� Type of surface: bare/vegetated/covered with mulch or litter� Surface slope

� Slope > 16° reduces infiltration� Soil characteristics� Depth of surface detention and thickness of the saturated layer� Soil moisture

� Infiltration rate affected in the initial stage� Human activities� Climatic conditions� Other minor factors

� Entrapped air, quality of water

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Measurement of infiltration

� Infiltrometer� Flooding type

� Single tube flooding infiltrometer

� Double tube flooding infiltrometer

� Sprinkling type infiltrometers or rain simulators� Conducted in lab

� Area selected: 8 m2

� Water sprinkled @ 50 mm/h from a height of 5 m

Drawbacks of infiltrometers

� Tube infiltrometers� Soil is disturbed while driving the tubes� Raindrop impact on soil not taken in account� Slope of ground not taken into account� Cannot be conducted on soil with boulders� Infiltration is affected because of ring size� Lateral flow of infiltrated water may occur� Lateral escape of entrapped air

� Sprinkling infiltrometers� Difficult to have uniform sprinkling of water of uniform intensity over the

entire catchment area for a sufficient duration� Discharge has to be measured very accurately� Area under experiment being small, significant lateral flow

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Infiltrometer with adjustable depth of

flooding

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Solve

� An infiltration ring test conducted on a double ring infiltrometerwith an inner ring of diameter 30 cm yielded the following data

� Determine the infiltration capacity rates for the time intervals in the test and plot the variation of infiltration capacity rate with time.

Time (mins) 0 5 10 30 60 120 180 240 300 360

Cum. Vol of water added (cm3)

0 46 90 246 435 662 842 1000 1154 1300

y = -0.154ln(x) + 1.0958R² = 0.954

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 50 100 150 200 250 300 350 400

Infiltration Rate (cm/h)

Time (mins)

Chart Title

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Example

� The quantity of water added to a double ring infiltrometer of 1.00 m dia. at 30 min interval to maintain a constant water level is as follows:

Find: 1. Rate of infiltration for every 30 min and plot the graph.

2. Average rate of infiltration

Time (min) 0 30 60 90 120 150 180

Quantity of water added (lit) 0 10 9.2 8.6 8.2 8.0 8.0

Expression of infiltration

� Horton’s equation

� Where, f = infiltration rate at time tf0 = initial infiltration ratefc = final infiltration ratek = a constant (depends on type of soil and condition of

vegetable cover)t = time in hours

� k varies between 2 to 5

� Infiltration indices

( )0

kt

c cf f f f e

−= + −

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Infiltration indices

� However,� Problem lies in selecting the appropriate infiltration curve

representative of the conditions existing at the start of the storm� Results satisfactory when i > f� But when i < f, problem becomes complicated

� Infiltration indices express infiltration as an average rate throughout the storm

� However, � Average value assumes too little infiltration during the first part of

the storm and too much near end of it

� Therefore best suited for use � In analysis of major flood producing storms occurring on wet soils,

or storms of such intensity and duration that the infiltration rate might be assumed to have reached a final constant rate prior to or early in the storm

Infiltration indices

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� Infiltration rate is most often expressed as an uniform

average rate for simplicity of calculations:

� φ index

� W index

Infiltration indices

Solve

� A storm with a uniform intensity of 1.6 cm/h for a period of 8 hours occurring over a basin of area 650 km2 produced a runoff estimate to be 57.2 million m3. Find the average infiltration rate during the storm.

� The ordinates of a rainfall mass curve of a storm over a basin of area 850 km2 measured in mm at one hour interval are� 0, 10, 22, 30, 39, 45.5, 50, 55.5, 60, 64 and 68

� If the infiltration during this storm can be represented by Horton’s equation with f0 = 6.5 mm/h, fc = 1.5 mm/h and k = 0.15 /h, estimate the resulting runoff volume.

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Solve

� A seven hour storm produced the following rainfall intensities (in mm/h) at half an hour intervals over a basin of area 1830 km2.� 4, 9, 20, 18, 13, 11, 12, 2, 8, 16, 17, 13, 6 and 1.

� If the corresponding observed runoff is 36.6 million m3, estimate the φ-index for the storm.

� A storm during a dry weather has rainfall intensities of 8, 12, 40, 38, 30, 26, 28, 5, 16, 32, 36, 24, 14 and 4 mm/h at half a hour intervals. � What is the runoff volume from a basin area of 600 km2 if the initial

abstractions are 10 mm and the φ-index for the basin is 10 mm/h?

� What is the percent error in runoff estimate if the initial abstractions are neglected?

Solve

� The average precipitation during a storm over a

catchment area of 10 km2 is as follows:

� 40 mm/h for 1 h

� 60 mm/h for 1 h

� 30 mm/h for 1 h

The resulting hydrograph was plotted on a graph paper with

the following scale.

� 1 cm = 1 h on x-axis

� 1 cm = 10 m3/s on y-axis

If the area of the hydrograph was measured as 30 cm2,find the φ index of infiltration

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� The storm over a catchment area of 50 km2 was having the following intensity:

� 40 mm/h for 1 h� 70 mm/h for 2 h� 30 mm/h for 1 h

The catchment area had infiltration rate as follows:� 20% area φ = 10 mm/h

� 60% area φ = 15 mm/h

� Balance impervious

Find the runoff due to the storm.

� A catchment area having an average rate of infiltration of 15 mm/h experienced the storm of the following intensities� 50 mm/h for 2 h� 30 mm/h for 0.5 h

The resulting runoff was 107 m3.Find the catchment area.

Solve

� The infiltration capacity in a basin is represented by Horton’s equation as

� Where f is in cm/h and t is in hours

� Assuming the infiltration to take place at capacity rates in a storm of 60 minutes duration, estimate the depth of infiltration in � The first 30 minutes and

� The second 30 minutes of the storm

tef

20.3 −+=