Lecture No. 4 1

Evapotranspiration

Dr. Mahesh Kumar Jat

Lecture No. 4 2

Evaporation

E E

E

E

P

Standing Water

Soil Moisture

Moisture on PlantsRain

Lecture No. 4 3

Evaporation

• Process by which the phase of water is changed from liquid to a vapor.

• It occurs at the evaporating surface, the contact between water body and overlaying air.

Lecture No. 4 4

Evaporation• Evaporation rate is a function of several

meteorological and environmental factors

The two main factors from an engineering standpoint are:

– Solar energy: it provides latent heat of vapor

– Advective energy: it is the ability to transport

Lecture No. 4 5

Evaporation Measures

– Pan evaporation

– Water budget

– Correlations to climate data (empirical)– Weight and depth

EE

E

E

Lecture No. 4 6

Standard 4 foot diameter pan

Lecture No. 4 7

• Pan evaporation method An evaporation pan is a device designed

to measure evaporation by monitoring the loss of water in the pan during a given time period, usually one (1) day.

Pan coefficient = 0.60 to 0.85 on an annual basis

L c pE p E (4.1)

Evaporation

Lecture No. 4 8

Problem for Practice – A river regulatory person must release water from a reservoir to satisfy a downstream need during the month of July. The average daily class A pan evaporation is 5 mm and the pan coefficient is 0.70. Estimate how much water must be released from the reservoir to satisfy 48,000 m3/day need if the average river width is 61 m and the distance down the center of the river from the reservoir to the point of need is 78 km. Express your answer in terms of both m3 and acre-ft. Neglect or assume that the net infiltration into and out of the river from groundwater sources is negligible and there is no transpiration.

• Input from reservoir to the river = SI = to be sought

• Output from the river release = R = 48,000 m3/day

• Output from river evaporation = E = can be calculated

• Assume S = 0

Lecture No. 4 9

0.7(5 / ) 3.5 / 0.0035 /L c pE p E mm day mm day m day

0I IS S E R S E R

• Evaporation rate from the river

• Evaporation from the river

6 261 78 4.758 10SA m km m

• Surface area of the river

6 2

3 3

0.0035 / (4.758 10 )

16.7 10 /

LE E SA m day m

m day

• Total amount needed3 3 3 3

3 3

16.7 10 / 48 10 /

64.7 10 / 52.45 /

IS E R m day m day

m day ac ft day

Lecture No. 4 10

• Water Budget Method

Evaporation

F

ER

S=?

SI SO

P

Initial Abstract IA

BI

BO

O

E

TE

E

Flowdirection

River Network

Lakes/Reservoirs

Lakes/Reservoirs

SO

SI

Lecture No. 4 11

1. Precipitation (P)

2. Initial Abstraction (IA)3. Infiltration (F)4. Rainfall Excess (R)

5. Surface water inflow (SI)

6. Surface water outflow (SO)

7. Groundwater inflow (BI)

8. Groundwater outflow (BO)9. Evaporation (E)10. Transpiration (T)11. Surface water release (O)

Hydrologic Processes in a Watershed

Lecture No. 4 12

S P S B R B S O Eo oI I

E P S B R B S O So oI I

(4.2)

(4.3)

Evaporation from Water Surface• Water Budget Method over Surface Water Bodies

ER

S=?

SISO

P

BI

BO

O

Lecture No. 4 13

AS P I F R ET

Evaporation from Overland Surface• Water Budget Method over Land Surfaces

R

P

Initial Abstraction, IA

F

E

ET

Lecture No. 4 14

Problem for Problems – Estimate the mean monthly evaporation rate for the O’Hare International Airport watershed. Local climatological data supplied by NOAA for the airport are shown for the months in year 1974. The water leaving the watershed is estimated from a USGS Geological Survey gage station and is also shown. Comment on your answers by explaining why the results are low relative to the annual average values, what assumptions you have made? Assume storage are minimal and no transpiration or infiltration.

Lecture No. 4 15

Month Precipitation (in) Outlet Water (in)

March 2.40 1.42

April 4.27 3.05

May 5.09 3.60

June 4.69 3.50

July 2.96 1.65

AS P I F R ET

• Data

Lecture No. 4 16

0 0 0P R E

Month P-R (in) E (in)

March 2.4-1.42 0.98

April 4.27-3.05 1.22

May 5.09-3.60 1.49

June 4.69-3.50 1.19

July 2.96-1.65 1.31

Average 1.24

• Adopt E = 76 cm/year = 2.49 in./month

• E based on data is lower: precipitation for these five months is less than annual average or outlet flow is too much because impervious natural of the watershed.

Lecture No. 4 17

• Correlations to Climate Data– General Empirical Formula

( , )E f e U

Evaporation

– General Theoretical Formula

(4.4)

Theoretical study of evaporation from large water bodies based on (1) solar energy and (2) advective energy was made in the Ph.D. Dissertation by G. T. Yeh, 1969.

Lecture No. 4 18

– Empirical Formula for Lake Hefner

8 80.00241( )L o aE e e U 1. EL = evaporation rate in inches per day

2. eo = saturation vapor pressure at the water surface in inches of mercury

3. eo8 = vapor pressure in air over the lake at an elevation of 8 m, in inches of mercury

4. U8 = wind speed over the lake at an elevation of 8 m, in miles per day

• As an engineer, you have to find an empirical formula for surface waters in your area of interest

(4.5)

Lecture No. 4 19

– Empirical Formula for Class A pan

( ) ( )np o aE e e m bU

1. Ep = daily pan evaporation, (in./day)

2. eo = saturation vapor pressure at the water surface, (in. of mercury)

3. eo = atmospheric vapor pressure at air temperature, (in. of mercury)

4. U = wind speed at 6 inches above pan rim, (mpd)

5. n, m, and b = 0.88, 0.37, 0.0041, respectively.

(4.6)

• Note: saturated vapor pressure is a function of temperature given in Table 3, Appendix B.

Lecture No. 4 20

Problem for Problems

• This is one of your homework• Hint: Use available Table 3 to find out the

vapor pressures!!

Lecture No. 4 21

Transpiration

F

ER

S=?

SISO

P

Initial Abstraction, IA

BI BO

BO

ET

O

Lecture No. 4 22

Transpiration• Transpiration is the process by which plants

transfer water from the root zone to the leaf surface, where it eventually evaporates into atmosphere.

Lecture No. 4 23

• The process by which transpiration takes place can be described as follows:

– Water is extracted by a plants roots, transported upward through its stem and diffused into the atmosphere through stoma.

Transpiration

Lecture No. 4 24

• Contributing factors:a. Moisture available

b. Vegetation type

c. Vegetation density

d. Vegetation health

Transpiration

Lecture No. 4 25

Transpiration

• Measured with phytometer (plant used as a measuring device)

• Based on monthly consumptive use (if available) and monthly evaporation

1. T = transpiration rate (mm/time)2. ET = evapotransipiration rate (mm/time)3. E = Evaporation rate (mm/time)

T ET E (4.7)

Lecture No. 4 26

Evapotranspiration

• It is the process by which water in the land surface, soil, and vegetation is converted into vapor state and returned to the atmosphere.

• It consists of evaporation from water, soil, vegetative, and other surfaces and includes transpiration by vegetation.

Lecture No. 4 27

F

R

P

E

TE

E

Watershed Evapotranspiration

E E

Lecture No. 4 28

Evapotranspiration • Mass balance

S P R F ET

pET kE

• k = 0.35 to 0.85 = f(soil/plant condition, location of the pan, wind speed, upwind fetch, and humidity)

• Based on Pan Evaporation

• For example, k = 0.7 if wind speed = 170-425 km/day, upwind fetch of green crop = 1,000 m, and low relative humidity = 20-40 percent.

• For a well-watered grass, the function is given in Table 4.8 on page 128 for two cases: in one case the pan is surrounded by green crop and in the other the pan is surrounded by dry-surface ground.

Lecture No. 4 29

Example for practice – Assume the following situations for a small watershed in northern Indiana. The six-month seasonal precipitation is 70 cm, runoff is 20 cm, and the change in groundwater storage is 15 cm. What are the monthly evapotransipiration rates?

S P R ET 15 70 20 ET

70 20 15 35 /6 5.83 /

ET cm monthcm month

Lecture No. 4 30

Evapotranspiration• Irrigation needs based on

evapotranspiration

S P I R F ET

I ET P

Known Known000

Lecture No. 4 31

Evapotranspiration

• Potential Evapotranspiration (PET) is the amount of evapotranspiration that would take place under the assumption of an ample supply of moisture at all times.

• PET is an indication of optimum crop water requirements.

Lecture No. 4 32

Calculation of Potential Evapotranspiration

1. Assume unlimited moisture supply

» Thornwaite’s Equation

» Blaney and Criddle Equation

Lecture No. 4 33

• Thornthwaite’s Equation

101.6

atET

TE

1. ET = Monthly evapotranspiration, (cm)2. t = mean monthly temperature, (oC)3. a = 0.49239 + 0.01792 TE4. TE = Thornthwaite’s temperature efficiency

index, which is given by

1.51412

1

/ 5i

i

TE t

Evapotranspiration

(4.8)

Lecture No. 4 34

• Blaney and Criddle/100ET kpt

1. ET = Monthly evapotranspiration, (in.)2. t = mean monthly temperature, (oF)3. k = consumptive use coefficient4. p = percent of daytime hours per year in the

study month• k = f(crop) are given for different crops e.g., k = 0.7

for tomato• p = f(latitude and month) given in Tables and

depends upon location, e.g., p = 8.08 at North 60o in March

Evapotranspiration

(4.9)

Lecture No. 4 35

Problem for Problems – A conversion of land from tomato farming to citrus is planned. Is it true that less water will be used to supplement rainfall based on a growing season of one year? Explain and support your answers with calculations and referenced assumptions.

• From Eq. (4.9) /100ET kpt

• For tomatoes 0.7 (4 )tomatoesk months

0.5 0.65 (7 )citrusk months

• Assume frost free zone in Florida 25 oN, from Tables

100p

• Mean annual temperature in Florida 65 ot F

Lecture No. 4 36

• From Eq. (4.9) /100ET kpt

0.7 (4 )tomatoesk months

0.5 0.65 (7 )citrusk months

Citrus requires less water than tomatoes based on annual growing season of one year

4(42.79 ) /12 14.25 .tomatoesET in in

7(39.74 ) /12 23.18 .citrusET in in Tomatoes requires less water than citrus based on real growing seasons

42.79 .tomatoesET from next slide in

39.74 .citrusET from next slide in

Lecture No. 4 37

Example 9 of Hand Problems                      

                       

  March April May June July Aug Sep Oct Nov Dec Total

Temp 63.00 65.00 74.00 79.00 81.00 81.00 78.00 69.00 66.00 55.00  

Daytime% 8.39 8.61 9.33 9.23 9.45 9.09 8.32 8.09 7.40 7.42  

ktomoto 0.70 0.70 0.70 0.70 0.70 0.70 0.70 0.70 0.70 0.70  

kcitrus 0.65 0.65 0.65 0.65 0.65 0.65 0.65 0.65 0.65 0.65  

                       

ETtomoto 3.70 3.92 4.83 5.10 5.36 5.15 4.54 3.91 3.42 2.86 42.79

ETCitrus 3.44 3.64 4.49 4.74 4.98 4.79 4.22 3.63 3.17 2.65 39.74

Monthly ET for Tomatoes and Citrus based on Florida Monthly Temperature

/100ET kpt