Chapter 3-evaporation+transpiration+infiltration.pdf

HYDROLOGY (BFC 32002) Prepared by:-

WAN AFNIZAN BIN WAN MOHAMED DEPT. OF WATER & ENVIRONMENTAL ENGINEERING FAC. OF CIVIL & ENVIRONMENTAL ENGINEERING e-mail: [email protected]

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

1

CHAPTER 3 EVAPORATION, TRANSPIRATION & INFILTRATION

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

2

Learning Objectives Define evaporation & transpiration. Describe methods for estimating the evaporated & transpired quantity of water. Indicate the importance of evapotranspiration in hydrologic modelling. Explain the infiltration process and rate. Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

3

CONTENT  EVAPORATION (E) ☻ AFFECTS OF METEOROLOGY FACTORS

☻ ESTIMATING EVAPORATION  TRANSPIRATION (T) ☻ ESTIMATING TRANSPIRATION

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

4

.... Cont ‘

CONTENT

 EVAPOTRANSPIRATION (ET) ☻ ESTIMATING EVAPOTRANSPIRATION  INFILTRATION (I) ☻ INFILTRATION CAPACITY ☻ FACTORS AFFECTING INFILTRATION ☻ INFILTRATION MEASUREMENT ☻ INFILTRATION METHODS

☻ INFILTRATION INDEX

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

5

EVAPORATION (E)

What is Evaporation

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

6

EVAPORATION (E) 

Process  water is transformed from the liquid phase to vapor phase.



Occurs on water surface, water flow, highway, roof, land surface, land subsurface and plants.



Important factor  study of water resources  mainly affects the flow rate of river, irrigation, water requirement and reservoir water.

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

7

EVAPORATION (E) METEOROLOGY FACTORS

4 factors affecting :1. SUN RADIATION  Process will continue if there is energy.  Cloud  barrier for evaporation

2. WIND  If wind blows the component of vapor  changed dry air  evaporation occurs. Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

8

EVAPORATION (E) METEOROLOGY FACTORS

3. HUMIDITY RELATIVE  If humidity relative  potential of air to absorb water   evaporation 

4. SUN RADIATION  If air & soil temperature   evaporation is faster.

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

9

.... Cont ‘

EVAPORATION (E)

How to ESTIMATE EVAPORATION

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

10

.... Cont ‘

EVAPORATION (E) ESTIMATING EVAPORATION

 FOUR (4) methods used to estimate EVAPORATION :-



Water Budjet



Energy Budjet



Mass Transfer Techniques



Pans Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

11

.... Cont ‘

EVAPORATION (E) 1st METHOD

WATER BUDJET METHOD

P

E

I

 Common method used. O

 Formula to calculate :G

E  P  I  G  O  ΔS OR

ET  P  R Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

12

.... Cont ‘

EVAPORATION (E)

1st METHOD WHERE :ET E R G P I O ΔS

= Evapotranspiration = Evaporation = Surface runoff = Groundwater = Rainfall depth = Surface runoff that ENTER catchment area = Surface runoff OUT from catchment area = Change in storage, above and below land surface Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

13

EXAMPLE 3.1 The drainage area of the Sembrong River at Johore Malaysia, is 11,839 km2. If the mean annual runoff is determined to be 144.4 m3/s and the average annual rainfall is 1.08 m, estimate the ET losses for the area. How does this compare with the lake evapotranspiration of 1 m/year measured at Muar River, Negeri Sembilan . Given :A = 11,839 km2 R = 144.4 m3/s per year P = 1.08 m

Find :ET ???

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

14

SOLUTION Step !! 1.

Calculate ET using the following formula :-

ET  P  R Given in (meter)

2.

But first, need to calculate R IN METER Conversion of R :-

m3 1 1 km 1 km 86400 sec R  144.4     365 days  s 11839 km2 1000 m 1000 m 1 day R  0.38 m per year Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

15

SOLUTION Therefore :ET  1.08  0.38 ET  0.7 m per year < 1 m/year (drainage)

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

(Muar River)

16

.... Cont ‘

EVAPORATION (E) 2nd METHOD

ENERGY BUDJET METHOD

 Enter all resources & thermal energy losses  evaporation as wanted variable.  Apply continuity equation.

 Accuracy depend on the reliability & preciseness of data.

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

17

.... Cont ‘

EVAPORATION (E) 2nd METHOD

ENERGY BUDJET METHOD

 Formula to calculate evaporation :-

Qs  Qr  Qa  Qar  Qbs  Qo  Qv E   L(1  B)  c p (Te  Tb )  whereby :p (To  Ta ) B  0.61 . 1000 (eo  ea ) Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

18

.... Cont ‘

EVAPORATION (E) 2nd METHOD

ENERGY BUDJET METHOD

which :E Qo Qs Qr Qa Qv Qar Qbs B Cp To Ta p ρ eo ea

= evaporation (cm3/cm2 – day) = increase in stored energy by the water = solar radiation incident at the water surface = reflected solar radiation = incoming long wave radiation from atmosphere = net energy advected into the water body = reflected long wave radiation = long wave radiation emitted by the water = Bowen’s ratio = the specific heat of water (cal/ g – oC) = the water surface temperature (oC) = the air temperature (oC) = the atmospheric pressure (mb) = the mass density of evaporated water (g/cm3) = the saturation vapor pressure at the water surface temperature (mb) = the vapor pressure of the air (mb) Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

19

.... Cont ‘

EVAPORATION (E)

3rd METHOD MASS TRANSFER TECHNIQUES

 Based on the concept of the turbulent transfer of water vapor to atmosphere.  Three (3) techniques applied :-



Dalton’s equation



Meyer’s equation



Dunne’s equation Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

20

.... Cont ‘

EVAPORATION (E)

3rd METHOD MASS TRANSFER TECHNIQUES

i. DALTON’S EQUATION  Formula used :-

E  κ e o  e a 

where :E  Refer Table 3.1

e o , ea

= Direct evaporation (in/day) = a coefficient dependent on the wind velocity, atmospheric pressure,and other factors (GIVEN) = the saturation vapor pressure at the water surface temperature and the vapor pressure of air,

respectively (

in.Hg)

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

21

.... Cont ‘

EVAPORATION (E)

3rd METHOD MASS TRANSFER TECHNIQUES Table 3.1 a: Values of vapor pressure ea & eo (Traditional U.S Unit)

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

22

.... Cont ‘

EVAPORATION (E)

3rd METHOD MASS TRANSFER TECHNIQUES Table 3.1 b: Values of vapor pressure ea & eo (SI

Unit)

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

23

.... Cont ‘

EVAPORATION (E)

3rd METHOD MASS TRANSFER TECHNIQUES

ii. MEYER’S EQUATION

 Formula used :-

where :-

W  E  Ce o  e a 1   10   Mile per hour

E

= Daily evaporation (in/day)

W

= Wind velocity in mph measured about 25ft above the water surface = Pan empirical coefficient (GIVEN)

C

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

24

.... Cont ‘

EVAPORATION (E)

3rd METHOD MASS TRANSFER TECHNIQUES

iii. DUNNE’S EQUATION In percentage

 Formula used : 100  R h  E   0.013  0.00016 U2  e a   100   where :E

= Evaporation rate (cm/day)

U2 eo, ea

= Wind velocity measured at 2 m above the surface (km/day) = Saturation vapor pressure at the water surface temperature and the vapor pressure of air, (milibars)

Rh

= Relative humidity (%) Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

25

EXAMPLE 3.2 Using the Meyer and Dunne equations, find the daily evaporation rate for a lake given that the mean value for air temperature was 87o F, the mean value for water temperature was 63oF, the average wind speed was 10 mph, and the relative humidity was 20 %. Refer to Table 3.1 for vapor pressure values. Assume pan empirical coefficient is 0.36 Given :Ta = 87 F  ea To = 63 F  eo W = 10 mph Rh = 20 %

Find :E ??? Using :i) Mayer ii) Dunne Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

26

SOLUTION Step !! 1. Using Meyer’s equations:W  E  Ce o  e a  1  10  

 Need to find eo & ea :At To = 63 F  have to interpolate eo

eo

0.74

0.52

60

63

70 (°F)

0.74  0.52 e o  0.52  70  60 63  60 e o  0.59 inHg

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

27

SOLUTION At Ta = 87 F

ea

 have to interpolate ea

1.42

1.42  1.03 e a  1.03  90  80 87  80

1.03

80

87

90 (°F)

e a  1.30 inHg

But Rh = 20 % therefore ; 20 e a  1.30   0.26 in.Hg 100 Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

28

SOLUTION Substitute all values into Mayer’s eq. :-

 10  E  0.360.59  1.301    10  E  0.23 in/day

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

29

SOLUTION

2. Using Dunne’s equations: 100  R h  E   0.013  0.00016 U2  e a    100 

 Need to convert ea in milibars & U2 in km/day:25.4 mm 1 mb e a  0.26 in.Hg    8.75 mb 1in 0.75 mm.Hg

mile 1.6093 km 24 hrs km U2  10    386 hr 1 mile 1 day day Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

30

SOLUTION Substitute all values into Dunne’s eq. : 100  20  E  0.013  0.00016 386 8.75   100  E  0.527 cm/day

OR

0.21 in/day

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

31

.... Cont ‘

EVAPORATION (E)

4th METHOD USE OF EVAPORATION PANS

 The least expensive.

 Provide good estimate of annual evaporation.  Position of the pan :-



On the land surface



In the land surface



Floating on the water Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

32

.... Cont ‘

EVAPORATION (E)

4th METHOD

Evaporation pan Class A Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

33

.... Cont ‘

EVAPORATION (E)

4th METHOD If pan is located as in situation 3 (floating in water) :-

Evaporation in lake is calculated using :-

Where for both evaporation situation aboved :espan eslake e μ c Epan Elake

= saturation vapor pressure at certain temperature in the pan = saturation vapor pressure at certain temperature in the lake = air vapor pressure on the water = wind velocity = constant = pan evaporation = lake evaporation Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

34

.... Cont ‘

EVAPORATION (E)

4th METHOD If daily evaporation – Class A pan is used, the following formula is referred :-

The difference betweem eo and ea is calculated using this formula :-

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

35

.... Cont ‘

EVAPORATION (E)

4th METHOD Where :-

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

36

.... Cont ‘

EVAPORATION (E)

4th METHOD Annual lake evaporation can be calculated using Penman formula :Pan coefficient (0.70 – 0.75)

Where :-

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

37

TRANSPIRATION (T)

What is Transpiration

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

38

.... Cont ‘ 

TRANSPIRATION (T)

Process  water moves through plants & evaporates through leaf stomata.

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

39

.... Cont ‘

TRANSPIRATION (T)

How to ESTIMATE TRANSPIRATION

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

40

.... Cont ‘

TRANSPIRATION (T)

BLANNEY-CRIDDLE METHOD 

Consumptive use of water during GROWING SEASON by the following formula :-

whereby :-

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

41

.... Cont ‘

TRANSPIRATION (T)

BLANNEY-CRIDDLE METHOD U

= Consumptive use of water during the growing season (inch).

Ks

= Seasonal consumptive use coefficient applicable to a particular crop,

B

empirically derived (Table 3.2). = the summation of monthly consumptive use factor for a given season.

t

= the mean monthly temperature (in

p k

= the monthly day time hour given = the consumptive use coefficient

F).

as % of the year (Table 3.3).

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

42

.... Cont ‘

TRANSPIRATION (T)

BLANNEY-CRIDDLE METHOD

 Monthly consumptive use is determined by the following formula :-

where :u k

= monthly consumptive use (inch) = consumptive use coefficient Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

43

.... Cont ‘

TRANSPIRATION (T)

BLANNEY-CRIDDLE METHOD

Table 3.2: Seasonal consumptive use crop coefficients (Ks) for irrigated crops

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

44

.... Cont ‘

TRANSPIRATION (T)

BLANNEY-CRIDDLE METHOD

Table 3.3: Daytime hours coefficient (p)

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

45

EXAMPLE 3.3 Determine the monthly consumptive use of an alfalfa crop grown in southern California for the month of July if the average monthly temperature is 72oF, the average value of daytime hours in percentage of the year is 9.88, and the mean montly consumptive use coefficient for alfalfa is 0.85 Given :Crop = Alfalfa Location = South Time = July T = 72F

p = 9.88 k = 0.85

Find :u ???

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

46

SOLUTION Step !! From equation:-

0.85  72  9.88 u 100 u  6.05 inch of water

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

47

EXAMPLE 3.4 Determine the seasonal consumptive use of a tomato crop grown in New Jersey if the mean monthly temperature for May, June, July and August are 61.6, 70.3, 75.1 and 73.4o F, respectively and the percent daylight hours for the given months are 10.02, 10.8,10.22 and 9.54 as percent of the year, respectively Given :T1 = 61.6F, T2 = 70.3F, T3 = 75.1F, T4 = 73.4F p1 = 10.02, p2 = 10.08, p3 = 10.22, p4 = 9.54 Find :U ??? Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

48

SOLUTION Step !! From equation:-

Find B value first :Growing season is 4 months

 61.6  10.2   70.3  10.08   75.1 10.22   73.4  9.54   B        100 100 100 100         B  27.9

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

49

SOLUTION

Ks value is obtained from Table 3.2 :Because New Jersey is humid area  take lower values of Ks

 Ks = 0.65 Therefore; U  0.65 27.9 U  18.1 in of water (4 months growing season) Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

50

EVAPOTRANSPIRATION (ET)

What is Evapotranspiration

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

51

.... Cont ‘ 

EVAPOTRANSPIRATION (ET)

Process sum of evaporation from the

earth's land surface & transpiration from plant to atmosphere. 

Used for hydrologic modelling & 2 concepts applied :-



Evaporation potential (ETp)



Evaporation Actual (ETa) Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

52

.... Cont ‘

EVAPOTRANSPIRATION (ET)

How to ESTIMATE EVAPOTRANSPIRATION

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

53

.... Cont ‘

EVAPOTRANSPIRATION (ET)

ESTIMATING EVAPOTRANSPIRATION

 TWO (2) methods used to estimate EVAPOTRANSPIRATION :-



Thornhwaite method



Penman method

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

54

.... Cont ‘

EVAPOTRANSPIRATION (ET) 1st METHOD

THORNHWAITE METHOD

 Method used to calculate ETp.  Formula to calculate :ET p  GI  k  E p 

SS a    AWC

   

x

whereby :x

= AWC/G  REFER Table 3.4 Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

55

.... Cont ‘

EVAPOTRANSPIRATION (ET) 1st METHOD

THORNHWAITE METHOD ETp GI k

= evapotranspiration potential (in/day) = growth index of crop in percentage of maturity = ratio of GI to pan evaporation, usually 1.0 – 1.2 for short grasses, 1.2 – 1.6 for crops up to shoulder height, and 1.6 – 2.0 for forest

Ep S Sa AWC

= pan evaporation ( in/day) = total porosity  REFER Table 3.4 = available porosity (unfilled by water) = porosity drainable only by evapotranspiration  REFER Table 3.4 = moisture freely drained by gravity

G

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

56

.... Cont ‘

EVAPOTRANSPIRATION (ET) 1st METHOD

THORNHWAITE METHOD

Table 3.4: Hydrologic capacities of soil texture classes

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

57

.... Cont ‘

EVAPOTRANSPIRATION (ET) 2ndMETHOD

PENMAN METHOD

 Combine mass transport & energy budget theories  Widely used method  more reliable approaches  Good result if no observation by pan evaporation & water balance study.

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

58

.... Cont ‘

EVAPOTRANSPIRATION (ET) 2ndMETHOD

PENMAN METHOD

 Formula used to calculate ETp :-

Δ H  0 . 27 E o ET p  Δ  0.27

ΔH  γEo OR Et  Δ  γ

whereby :-

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

59

.... Cont ‘

EVAPOTRANSPIRATION (ET) 2ndMETHOD

PENMAN METHOD n  H  R A (1  r )  0.18  0.55 D 





 0.5  0.10  0.9 n   B 0 . 56  0 . 092 e   a  D  

OR

H  R l  RB Which

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

60

.... Cont ‘

EVAPOTRANSPIRATION (ET) 2ndMETHOD

PENMAN METHOD es

= saturation vapor pressure at mean air temperature (mmHg)  (Figure 3.7 or Table 3.1) ea = actual vapor pressure in air (mm/day)  h = ea/es   ea = h  es h = relative humidity (in decimal) RI = net of short wave radiation (mm/day) RB = long wave radiation accepted by earth (mm/day) R @ RA = mean monthly extraterrestrial radiation (mmH2O evaporated/day)  (Table 3.6) Δ = curve slope of vapor pressure at t (mmHg)  (Figure 3.8) n/D = cloud ratio n = actual duration of bright sunshine D = maximum possible duration of bright sunshine r = reflection coefficient of surface (in decimal) Eo = evaporation (mm/day) Et @ ETp = potential evapotranspiration (mm/day) u2 = mean wind speed at 2m above the ground (mi/day) H = daily heat budget at surface (mm/day) B = variable given (Table 3.5) Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS  = physchometer constant (  = 0.27)

61

.... Cont ‘

EVAPOTRANSPIRATION (ET) 2ndMETHOD

PENMAN METHOD Figure 3.7: Relation between temperature, T and saturated vapor pressure, es

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

62

.... Cont ‘

EVAPOTRANSPIRATION (ET) 2ndMETHOD

PENMAN METHOD

Table 3.6: Getting values of R or RA

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

63

.... Cont ‘

EVAPOTRANSPIRATION (ET) 2ndMETHOD

PENMAN METHOD



Figure 3.8: Getting values of

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS



64

.... Cont ‘

EVAPOTRANSPIRATION (ET) 2ndMETHOD

PENMAN METHOD Table 3.5: Getting variables of

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

B

65

EXAMPLE 3.5 Using the Penman method, estimate ETp, given the following data : temperature at water surface = 22oC, temperature of air = 33oC, relative humidity = 45%, wind velocity = 1.5 mph (36 mi/day). The month is June at latitude 33o north, r is given as 0.07 and n/D is to be found 0.70 Given :Tw = 22C, Ta = 33C h = 0.45, u = 36 mi/day Month = June Latitude = 33

r = 0.07 n/D = 0.7 Find :ETp ???

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

66

SOLUTION Step !! From equation:-

ET p  ΔH  0.27 Eo Δ  0.27 1. Find  value first from figure 3.8:-

At Ta = 33C   = 1.2 mm Hg 2. Calculate Eo value from the following formula :-

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

67

SOLUTION where:-

es  From Table 3.1, (at Ta = 33C)  have to interpolate es 42.18

es

42.18  31.83 e s  31.83  35  30 33  30

31.83

30

35 (°C)

33

ea  h = ea/es

e s  38.04 inHg

ea = h  es ea = 0.45  38.04 = 17.12 mmHg

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

68

SOLUTION therefore:Eo  0.3538.04  17.12 1  0.0098 (36) Eo  9.91 mm/day

3. Calculate H value from the following formula :n  H  R A (1  r )  0.18  0.55 D 





n   0 . 5  0.10  0.9    B 0.56  0.092e a D  

RA  From Table 3.6, (at North, Latitude = 33, June) Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

69

SOLUTION where:-

RA  have to interpolate

RA

16.7

16.5

16.7  16.5 R A  16.5  40  30 33  30 R A  16.56

30

33

40

B  From Table 3.5, B  at T = 33C (air) (C need to convert to K) Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

70

SOLUTION

K = C + 273 K = 33C + 273 = 306K Again B  have to interpolate

B

18.60

17.46

18.60  17.46 B  17.46  310  305 310  305 B  17.69

305

306

310 (°K)

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

71

SOLUTION therefore:H  (16.56) (1  0.07)  0.18  (0.55  0.7) 



 (17.69) 0.56  (0.092 17.12)0.5

 0.10  (0.9  0.7)

H  6.38 mm / day

Thus:(1.2  6.38)  (0.27  9.91) ETp  1.2  0.27 ETp  7.02 mm/day Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

72

EXAMPLE 3.6 Estimate the monthly potential evapotranspiration for June. The mean monthly temperatures are shown in the Table below. The average relative humidity is 50%. The wind speed is 130 mi/day. Assume that n/D = 70%, γ = 0.27, and r = 25% at 50 latitude north. Given :Ta = 24.2C or 75.5F h = 0.5, u = 130 mi/day Month = June Latitude = 50 north

r = 0.07 n/D = 0.7

 = 0.27

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

Find :Et ???

73

EXAMPLE 3.6 Month

Jan

Feb

Mar

Apr

May

June

Tm(0F)

-1.5

5.2

30.2

40.2

58.1

75.5

Tm(0C)

-18.6

-14.9

-1.0

4.6

14.5

24.2

Month

July

Aug

Sep

Oct

Nov

Dec

Tm(0F)

70.3

67.5

51.0

40.2

31.2

15.2

Tm(0C)

21.3

19.7

10.6

4.6

-0.4

-9.3

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

74

SOLUTION Step !! From equation:-

This is 

Et  ΔH  0.27 Eo Δ  0.27 1. Find  value first from figure 3.8:-

At Ta = 24.2C (June)   = 0.75 mm Hg 2. Calculate Eo value from the following formula :-

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

75

SOLUTION where:-

es  From Table 3.1, (at Ta = 24.2C)  have to interpolate es 23.76

es

23.76  17.53 es  17.53  25  20 24.2  20

17.53

20

25 (°C)

24.2

ea  h = ea/es

es  22.76 inHg

ea = h  es ea = 0.5  22.76 = 11.38 mmHg

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

76

SOLUTION therefore:Eo  0.3522.76  11.38 1  0.0098 (130) Eo  9.06 mm/day

3. Calculate H value from the following formula :-

H  Rl  RB where:-

?

n  R l  R A  (1  r )  0.18  0.55  D  n  R B  B 0.56  0.092 e a  0.10  0.90  D 

? 



Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

77

SOLUTION Calculate Rl first : Need to find RA :RA  at North, Latitude 50, June RA  From Table 3.6, RA = 16.7 Therefore :R l  16.7  (1  0.25) 0.18  0.55  0.7  R l  7.08 mm/day

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

78

SOLUTION

Next, compute RB : Need to find B :B  at T = 75.5F

B

 have do interpolation 16.25  15.65 B  15.65  80  75 75.5  75

16.25

15.65

B  15.71 75

75.5

80

(°F)

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

79

SOLUTION Therefore :-



R B  15.71 0.56  0.092 11.38

 0.10  (0.90  0.7)

R B  2.86 mm/day

Thus :H = 7.08 – 2.86 H = 4.22 mm/day

Hence :-

(0.75  4.22)  (0.27  9.06) Et  0.75  0.27 Et  5.5 mm/day Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

80

INFILTRATION (I)

What is Infiltration

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

81

.... Cont ‘ 

INFILTRATION (I)

Process  precipitation moves

downwards through earth surface. 

It replenishes soil moisture, recharge aquifer & support stream flows during dry periods.



Terminology :-



Infiltration rate – The max. rate at which ground can absorb water.



Field capacity – The volume of water that ground can hold Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

82

.... Cont ‘

INFILTRATION (I) INFILTRATION CAPACITY, fc



Definition  max. rate in absorb water

at a certain type of soil and in a given time. 

Unit in cm/h or mm/h.



Two condition exists :-

1. 2.

f = fc when i > fc f=i

when i < fc

Note :-

i = Rainfall intensity fc = Constant infiltration rate ( t  )

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

83

.... Cont ‘

INFILTRATION (I) INFILTRATION CAPACITY, fc

Horton model

Figure 3.10 : Infiltration capacity curve

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

84

.... Cont ‘

INFILTRATION (I)

FACTORS AFFECTING INFILTRATION



Three (3) main factors affecting fc :-

1.

Characteristics of soil  Texture  loose soil has larger fc Underneath drainage  good drainage has larger fc  Type of soil  dry soil absorb > water   has larger fc Land use  forest soil has higher fc Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

85

.... Cont ‘

INFILTRATION (I)

FACTORS AFFECTING INFILTRATION

2.

Soil surface  If covered by grass or other vegetation  high infiltration

3.

Fluid characteristics Turbidity  if  will block pores in soil  reduce fc Temperature  if   increase fc Contamination  if   reduce fc Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

86

.... Cont ‘

INFILTRATION (I) INFILTRATION MEASUREMENT



Using two (2) equipments :-



Flooding type infiltrometer



Rainfall simulator

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

87

.... Cont ‘

INFILTRATION (I)

FLOODING TYPE INFILTROMETER

Figure 3.11 : Infiltrometer set

 Consists of metal cylinder (30 cm , L = 60 cm) & open both ends  Driven 50 cm  5 cm depth water is poured

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

88

.... Cont ‘

INFILTRATION (I)

FLOODING TYPE INFILTROMETER

 Water is added from burette  maintain level at the tip of pointer.  Until uniform rate of infiltration is obtained (2 – 3 hours).

 Surface protected by perforated disk  prevent turbidity.  Disadvantage : Infiltered water spread at the outlet of the tube Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

89

.... Cont ‘

INFILTRATION (I)

FLOODING TYPE INFILTROMETER OVERCOME ????

Figure 3.12 : Ring Infiltrometer

 Consists of two concentric rings  Inserted to the ground & water is maintained  Outer ring as water jacket  prevent water spreading out Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

90

.... Cont ‘

INFILTRATION (I)

FLOODING TYPE INFILTROMETER

 Disadvantage : Raindrop effect not simulated  Driving ring disturb the soil  Result depends on size  border effect

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

91

.... Cont ‘

INFILTRATION (I)

RAINFALL SIMULATOR

 Give low values  due to rainfall and turbidity effects  Small plot of land (2m  4m)  Series of nozzles on the longer side.  Produce raindrops at 2m height.  Controlled conditions  various of intensities & durations  surface runoff is measured.  Infiltration rate & time is calculated.  If i > f  fc is obtained. Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

92

.... Cont ‘

INFILTRATION (I) INFILTRATION METHODS



Two (2) methods :-

- focussed



Horton model



Green – Ampt Model

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

93

.... Cont ‘

INFILTRATION (I)

HORTON MODEL

 Represented by the this empirical eq. :

Whereby :-

f = Infiltration rate capacity (depth/time) at a given time (mm/hr or cm/hr) k = Decay constant which is representing the rate of decrease in f capacity fc = Asymptotic/ constant infiltration rate (t  ) fo = Initial infiltration capacity, t = 0 t = time Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

.... Cont ‘

INFILTRATION (I)

HORTON MODEL

Figure 3.13 : Infiltration rate graph

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

.... Cont ‘

INFILTRATION (I)

HORTON MODEL

 Total infiltration :



F(t )   f (t )dt   fc  fo  fc  e kt t

t

0

0

 fo  fc    kt  F  f c t  1e  k  0



Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS



t

 dt

EXAMPLE 3.7 A catchment soil has Horton infiltration parameters: fo = 100 mm/h, fc = 20 mm/h and k = 2 min-1. What rainfall rate would result in ponding from beginning of the storm? If this rainfall rate is maintained for 40 minutes, describe the infiltration as a function of time during the storm . Given :fo = 100 mm/h fc = 20 mm/hr k = 2 min-1

Find :f?

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

97

SOLUTION Step !! From Horton’s equation :-

Substitude fo, fc and k values :

( 0 < t < 40 min ) Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

98

EXAMPLE 3.8 An initial infiltration was recorded as 5.5 cm/hr during 10 hours of rainfall. Given that fc and k is 0.4 cm/hr and 0.32 hr-1 respectively, determine; (i) Infiltration at 5 hours. (ii) Total infiltration within first 8 hours. (iii) Total infiltration between 5 and 10 hours from rainfall begin Given :fo = 5.5 mm/h fc = 0.4 mm/hr k = 0.32 hr-1

Find :f at 5 hours ? F within 8 hours? F (5 < t < 10 hours) ?

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

99

SOLUTION Step !! (i)

f at t = 5 hours

(-kt) f  fc  (fo  fc) e Substitude fo, fc and k and t :

(-0.32  5) f  0.4  (5.5  0.4) e f = 1.43 cm/hr

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

100

SOLUTION Step !! (ii)

F within 8 hours

 fo  fc    kt  F  f c t  1e  k  0





t

 5.5  0.4  ( 0.328 ) F  (0.4)(8)  1 e 0.32 





  -0 

F = 17.91 cm

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

101

SOLUTION Step !! (iii)

F (5 < t < 10 hours)

 fo  fc    kt  F  f c t  1e  k  0





t

 5.5  0.4  ( 0.328 )  F  (0.4)(8)  1 e  0 . 32  5



 5.5  0.4  F  (0.4)(8)  1  e( 0.328) 0.32 





10

  (0.4)(5)  5.50.320.4 1  e

( 0.32 5 )

 



F = 4.56 cm Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS



102

.... Cont ‘

INFILTRATION (I)

GREEN-AMPT MODEL

 Also called the delta function model.  Most realistic models of infiltration.

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

.... Cont ‘

INFILTRATION (I) INFILTRATION INDEX

What is Infiltration Index () Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

104

.... Cont ‘

INFILTRATION (I) INFILTRATION INDEX

 The average infiltration rate.  Average rainfall at which P volume = R volume.

 Derived from rainfall hyetograph  Treat as constant infiltration capacity

 If i <   f = i  If i >   excess rainfall (surface runoff)

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

105

.... Cont ‘

INFILTRATION (I) INFILTRATION INDEX

 If i >   excess rainfall (surface runoff)

Figure 3.14 : Infiltration index

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

106

.... Cont ‘

INFILTRATION (I) INFILTRATION INDEX

 Formulae for  index :

PR Φ index  te P = Total rainfall or precipitation (cm) R = Total runoff (cm) te = Time of rainfall excess

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

107

EXAMPLE 3.9 A storm with 10 cm rainfall produced a direct runoff of 5.8 cm. Table below show the time distribution of the storm, estimate the Φ index. Time (hour) Rainfall Intensity (cm/h) Given :P = 10 cm R = 5.8 cm

1

2

3

4

5

6

7

8

0.4 0.9 1.5 2.3 1.8 1.6 1.0 0.5 Find :? Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

108

SOLUTION Step !! (i)

Sketch first rainfall hyetograph Rainfall hyetograph 2.3

2

1.8 1.6

1.5 1.5

0.9

1

0.5

 (1st assumption) ( 0 <  < 0.4 cm/hr )

1

0.5

0.4



Ranfall Intensity (cm/hr)

2.5

0 1

2

3

4

5

6

7

Time (hour)

t = 8 hours

Wan e Afnizan b Wan Mohamed ; JKAP ; FKAAS

8

SOLUTION Step !! (i)

Calculate the  index :

PR Φ index  te 10  5.8 Φ index   0.525 cm/hr 8 (ii)

Do checking (is  follows the 1st assumption??) Answer  NO !!!!! Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

110

SOLUTION Cheking …. Rainfall hyetograph 2.3

2

1.8 1.6

1.5 1.5

 (calculated)

0.5

1

0.9

1

 = 0.525 cm/hr

0.5

0.4



Ranfall Intensity (cm/hr)

2.5

0 1

2

3

4

5

6

7

Time (hour)

t = 8 hours

Wan e Afnizan b Wan Mohamed ; JKAP ; FKAAS

8

 (1st assumption) ( 0 <  < 0.4 cm/hr ) 111

SOLUTION (iii)

Modify the te value : Rainfall hyetograph 2.3

2

1.8 1.6

1.5

 (2nd assumption)

1.5

( 0.5 <  < 0.9 cm/hr )

0.5

1

0.9

1

0.5

0.4



Ranfall Intensity (cm/hr)

2.5

0 1

2

3

4

5

6

7

8

Time (hour)

te = 6 hours Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

112

SOLUTION (iv)

Calculate again the  index :

PR Φ index  te However P become new :

P = 10 – (0.4  1 + 0.5  1) = 9.1 cm therefore :

9.1  5.8 Φ index   0.55 cm/hr 6 Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

113

SOLUTION Again!!! Do checking (is  follows the 2nd assumption??)

(v)

Rainfall hyetograph 2.3

2

1.8 1.6

1.5

 (2nd assumption) ( 0.5 <  < 0.9 cm/hr )

1.5

0.5

1

0.9

1

 = 0.55 cm/hr 0.5

0.4



Ranfall Intensity (cm/hr)

2.5

0 1

2

3

4

5

6

7

 (calculated)

8

Time (hour)

te = 6 hours

Answer  YES !!!!! GOT IT Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

114

SOLUTION

Thus :  = 0.55 cm/hr Checking Rainfall Excess with Runoff depth given in the question : Rainfall hyetograph

Ranfall Intensity (cm/hr)

2.5

2.3

2

1.8 1.6

1.5 1.5

0.5

1

0.9

1

 = 0.55 cm/hr 0.5

0.4

0 1

2

3

4

5

6

7

8

Time (hour) Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

115

SOLUTION Hence : P excess = (0.9 – 0.55)(1) + (1.5 – 0.55 )(1) + (2.3 – 0.55)(1) + (1.8 – 0.55)(1) + (1.6 – 0.55)(1) + (1 – 0.55)(1)

P excess = 5.8 cm = R (5.8 cm) OK !!!! :

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

116

EXAMPLE 3.10 The rainfall intensity in the 50 hectar of catchment area is given table below. If volume of surface runoff is 30 000 m3, estimate Φ index for the catchment area and sketch the circumstances in form of hyetograph. Time (hour) Rainfall Intensity (cm/h)

1

2

3

4

5

6

7

5

10

38

25

13

5

0

Given :A = 50 ha  need to convert unit (m) R = 30 000 m3 Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

Find :? 117

SOLUTION Step !! (i)

Sketch first rainfall hyetograph Rainfall hyetograph 38

35 30 25 25 20 15

 (1st assumption) ( 0 <  < 0.5 cm/hr )

13 10

10 5

5

5 0 0 1

2

3

4

5

6

Time (hour)

t = 6 hours

Wan e Afnizan b Wan Mohamed ; JKAP ; FKAAS

7



Ranfall Intensity (mm/hr)

40

SOLUTION Step !! (i)

Calculate the  index :

PR Φ index  te

P = Intensity  time =

where :-

mm  hr hr

P = (5 + 10 + 38 + 25 + 13 + 5) (1) = 96 mm

Volume (m 3 ) Runoff  Area (m 2 )

1 1 ha Runoff  30 000 m    60 mm 2 50 ha 10 000 m 3

Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

119

SOLUTION Therefore;

96  60 Φ index   6 mm/hr 6

Do checking (is  follows the 1st assumption??)

(ii)

Rainfall hyetograph

Answer  NO !!!!!

38

35

 (calculated)  = 6 mm/hr

30 25 25 20

 (1st assumption) ( 0 <  < 0.5 cm/hr )

13

15 10 10

5

5 5

0 0 1

2

3

4

5

6



Ranfall Intensity (mm/hr)

40

7

Time (hour)

te = 6Wanhours Afnizan b Wan Mohamed ; JKAP ; FKAAS

120

SOLUTION (iii)

Modify the te value : Rainfall hyetograph 38

35 30 25 25

 (2nd assumption)

20

( 5 <  < 10 cm/hr )

15

13 10

10 5

5

5



Ranfall Intensity (mm/hr)

40

0

0 1

2

3

4

5

6

7

Time (hour)

te = 4 hours Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

121

SOLUTION (iv)

Calculate again the  index :

PR Φ index  te However P become new :

P = 96 – (5 + 5)(1) = 86 mm therefore :

86  60 Φ index   6.5 mm/hr 4 Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

122

SOLUTION Again!!! Do checking (is  follows the 2nd assumption??)

(v)

Rainfall hyetograph 38

 (calculated)

35

 = 6.5 mm/hr

30 25 25

 (2nd assumption)

20 15

( 5 <  < 10 cm/hr )

13 10

10 5

5

5 0



Ranfall Intensity (mm/hr)

40

0 1

2

3

4

5

6

7

Time (hour)

te = 4 hours

Answer  YES !!!!! GOT IT Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

123

TIME’S UP …

THANK YOU Wan Afnizan b Wan Mohamed ; JKAP ; FKAAS

124