02 10 evaporation-and_infiltration_jrs-rzl

INCI 4138 Hydrologic Processes 1

Hydrologic Processes

INCI 4138

Introduction to Water Resources Engineering

Evaporation and Infiltration Processes

Dr. Jorge Rivera-Santos & Raúl E. Zapata-López


Evaporation

The process of water changing from its liquid phase to the vapor phase.

Water vapor is the state of water in the atmosphere.

It is a two-phase process. Water molecules escape

from the water surface to the atmosphere (heat?).

Transport of water vapor molecules away from the evaporating surface (wind?).


Evaporation Rate of evaporation depends on…

* Solar Radiation==================================================================================================================================================================

* Atmospheric Pressure * Air Temperature* Relative Humidity * Wind Speed==================================================================================================================================================================

* Water Temperature * Quality of Water* Geometry of the evaporating surface===========================================================================================================================================


Evaporation: Radiation effect


Evaporation Latent Heat is the heat that is given up or absorbed

when a phase changes.

Latent heat of vaporization is the heat given up during vaporization of liquid water to water vapor.

where T is the temperature in °C and lv is in joules per kilogram (J/Kg).

Tlv 237010501.2 6


Evaporation Process

Water molecules are always moving around.

This is caused by solar radiation.

When there is enough energy, water molecules are knocked away from the surface. Wind moves them out of the upper layer.

Water molecule transferred to the atmosphere become water vapor.


Evaporation from Water Bodies

Four methods to determine evaporation. Comparative method (evaporation pan)

Energy balance method (energy budget)

Aerodynamic method (mass transfer)

Combined balance method (Penman Method)


pLpL EKE

EvaporationEvaporation (yr or monthly) from water bodies are estimated by measuring daily evaporation from a pan and then corrected using a given factor based on local climatic conditions and measured daily rainfall (if present).

pzsp

zsLL E

ee

eeKE

Lp

'

KLp= 0.7

Klp = 1.5 at z=4m


Evaporation using Pan near a Lake

Consider the ability to transport water vapor away from the water surface.

The transport is generated by the humidity gradient in the air near the surface and wind speed across the surface.

Saturation vapor pressure – pressure at which saturated vapor exists.

Saturated water vapor - the maximum moisture content the air can hold for a given temperature.


Evaporation using Pan near a Lake

EL and Ep has units of mm/day or in./day.

esL and esp are vapor pressure at the water surface of lake and pan, which is the saturation vapor pressure at maximum water temperature.

ez is the air vapor pressure at height z2 above the water surface which is taken as the ambient vapor pressure in air considering its relative humidity, Rh.

Vapor pressure data in Appendix C (Gupta)

pzsp

zsLL E

ee

eeKE

Lp

'asha eRe Klp = 1.5 at z=4m

(See Ex.2.7, Problems 2.19 & 2.20 -Gupta).

Example 2.7 (Gupta)

Pan constant = 0.7

Pan Evaporation = 9 mm

a) E L = K*EP = 0.7*9 = 6.3 mm

b) Lake Temperature Max. = 20°C & Min = 18 ° C

Pan Temperature Max. = 28 °C & Min = 25°C

Air Temperature @ 4m = 30°C

With Relative Humidity = 25%

Wind speed @ 4m =6 m/s


E L = K’*(esL – ez)/(esp-ez)*EP = 1.5*(2.337-1.061)/(3.781-1.061)*9 = 6.33 mm

From Table C.2: vapor pressure values are.esL = 2.337 KPa esp = 3.781 KPa by interpolation esz = 4.243 KPa But with Rh = 25% ez = 0.25*4.243 = 1.061 KPa

pLpL EKE

pzsp

zsLL E

ee

eeKE

Lp

'


Evapotranspiration Considers evaporation from natural surfaces whether

the water source is in the soil, plants or combination of both.

Plants extract water from soil and then is liberated to the air by the transpiration process.

Consumptive use is the amount of water required to support the optimum growth of a particular crop under field conditions. It considers the water loss from soil and plants of the particular crop.

Thornthwaite (1948) introduced the term potential evapotranspiration to define evapotranspiration that will occur when the soil contains an adequate moisture supply at all times.


Evapotranspiration Reference crop evapotranspiration, Eto, is based on

an idealized crop of a uniform height, completely covering the ground, growing actively, and not experiencing any shortage of water.

Evapotranspirometers and lysimeters are used for these measurements but they are rare. The water balance budget is used to estimate its value.

Penman-Monteith Method is one of the equations used to estimate evapotranspiration.

Actual evapotranspiration, Et, from a surface depends on correction factors to account for the specified crop growth stage (Kc) and available soil water (Ka). Et = Kc Ka Eto (Eq 2.34 –Gupta)


Infiltration process

Figure 7.4.1 (p. 234)Subsurface water zones and processes (from Chow et al. (1988)).

Figure 7.4.1 (p. 234)Subsurface water zones and processes (from Chow et al. (1988)).



Process of water penetrating into the soil.

Rate of infiltration is influenced by:

soil surface condition

vegetative cover condition

soil properties- Porosity [n=Vv/Vt =e/(1+e)] or void ratio [e =Vv/Vs =n/(1-n)]

- Hydraulic conductivity (K)

- Moisture content (θ)



Moisture zones during infiltration. Moisture profile as a function of time for water added to the soil surface.


Infiltration process: EquationsΦ = Phi Index, SCS approach

Horton Equation

I

∑(Pi-Φ) ∆t = R


Infiltration - Horton’s equation

fp is the infiltration rate at time t, in./hr or mm/hr.

f0 is initial infiltration rate, in./hr or mm/hr.

fc is ultimate infiltration rate, in./hr or mm/hr.

k is the decay constant, (time units)-1.

ktccp effff 0


Infiltration - Horton’s equation

Ft = cumulative infiltration capacity in inches

(or mm) at time t in hours . k = decay constant in (1/time units = 1/hr) See example in file “Example-Horton-RZL”

available in INCI4138 group at uprm.edu site.

ktcct e

k

fftfF

10


Rainfall Infiltration rate and Cumulative Infiltration

The rainfall hyetograph illustrates the rainfall pattern as a function of time. The cumulative infiltration at time t is Ft or F(t) and at time t + Δt is Ft + Δt or F(t + Δt) is computed using equation 7.4.15. The increase in cumulative infiltration from time t to t + Δt is Ft + Δt – Ft or F(t + Δt) – F(t) as shown in the figure. Rainfall excess is defined in Chapter 8 as that rainfall that is neither retained on the land surface nor infiltrated into the soil.


Infiltration - Ponding time.

This figure illustrates the concept of ponding time for a constant intensity rainfall.

Ponding time is the elapsed time between the time rainfall begins and the time water begins to pond on the soil surface.

Review Example 2.14 (pgs 85-88, Gupta) & Problems 2.34 & 2.35 (Pg 119)Also review Excel file “Example Horton-RZL”

Infiltration – Ponding Process


ktccp effff 0

ktcct e

k

fftfF

10

fo’ = 5.24 in./hr

Example Horton-RZL

The standard f curve can be given by the equation f = 1.2 + (9 - 1.2) e-4.56t where f in in in./hr and t is in hr. (a & b) Develop and plot the standard infiltration and the cumulative infiltration curve. Rainfall intensity is 1.5 in./hr for the first 40 minutes and then 6.0 in./hr thereafter.

P40 =1.5in./hr(40min)(hr/60min)= 1 in.

f’ = 1.2 + (5.24 - 1.2) e-4.56t’ where t’ = t-40

Infiltration – Ponding Process

( c) Rainfall intensity is 1.5 in./hr for the first 40 minutes and then 6.0 in./hr thereafter. For that event, draw the corresponding rainfall intensity curve as well as the actual infiltration and cumulative infiltration curve.


ktccp effff 0

ktcct e

k

fftfF

10

After the 40 minutes the time value is adjusted to have t’ = t – 40 and the fo’ = 5.24 in./hr.

f = 1.2 + (5.24 - 1.2) e-4.56t’

and

F=1.0+1.2t’+(5.24-1.2)*(1- e-4.56t’)/4.56

For the first 40 minutes (f = I) and not

f = 1.2 + (9 - 1.2) e-4.56t.

Therefore, F=Σ(f*Δt) =Σ(I*Δt) up to F=1in.


Infiltration: Φ-Index (SCS)

Ptotal =∑(i∆t) =∑(Pi∆t)

R=∑Ri =∑(Qi∆t)

∑(Pi-Φ) ∆t =∑Ri=R

Φ = (∑(Pi∆t) – R)/(m∆t)

See Examples 2.19 & 2.20 (pg. 106-107, Gupta) & Problems 2.43, 2.44 & 2.45 (pg. 120, Gupta). Also review file “Example Phi-Index-RZL”

Φ-Index=Constant RATE of abstraction (in./hr or cm/hr).

Calculated by finding the loss difference between gross precipitation ( Ptotal) and observed surface runoff measured as a hydrograph (R).

Assumes uniform loss across the rainfall pattern.

Infiltration: Φ-Index (SCS)EXAMPLE 2.19 (Gupta) Φ-INDEX EVALUATION

5

4.5 RunoffExcess=2.52in Infiltration

4

P 3.5 R E 3

C I 2.5

P I 2 Φ valueT

A 1.5 I O 1

N 0.5

(in./hr) 0

0 0.5 1 1.5 2 2.5 3 3.5

TIME (hr)

DATA:

Area (acre) = 500

Direct Runoff (acre-ft) 105 ‘=105/500*12 in. = 2.52 in.Assuming

Φ <2.0 in./hrTime Time Time Precipitation Amount of

Increment Intensity Precipitation (hr) (min) (hr) (in./hr) (in.) (in.)

δt P (P*δt) (P-Φ)δt0 0

0.5 4.5 2.25 (4.5-Φ)*0.50.5 30

0.5 3.0 1.5 (3-Φ)*0.51 60

0.5 1.0 0.5 (0)*0.51.5 90

0.5 3.5 1.75 (3.5-Φ)*0.52 120

0.5 2.0 1 (2.0-Φ)*0.52.5 150

0.5 0.0 0 (0)*0.53 180

Sumation 14 7 (13-4Φ)0.5

Total Precipitation (in.) = 7 Runoff (in.) = 105/500/12 = 2.52

Infiltration = P*t-R (in.) = 4.48

Therefore: (13-4Φ)*0.5 = 2.52Φ = (13 - 2.52*2) / 4 = 7.96/ 4 = 1.99 in./hr

1.99 OK.

INCI 4138 Hydrologic Processes25

Ptotal =∑(i∆t) =∑(Pi∆t) R=∑Ri =∑(Qi∆t)

∑(Pi-Φ) ∆t =∑Ri=R or solving for Φ

Φ = (∑(Pi∆t) – R)/(m∆t)

Infiltration: Φ-Index (SCS)EXAMPLE 2.19 (Gupta) Φ-INDEX EVALUATION

5

4.5 RunoffExcess=2.52in Infiltration

4

P 3.5 R E 3

C I 2.5

P I 2 Φ valueT

A 1.5 I O 1

N 0.5

(in./hr) 0

0 0.5 1 1.5 2 2.5 3 3.5

TIME (hr)

DATA:

Area (acre) = 500

Direct Runoff (acre-ft) 105 ‘=105/500*12 in. = 2.52 in.Assuming

Φ <2.0 in./hrTime Time Time Precipitation Amount of

Increment Intensity Precipitation (hr) (min) (hr) (in./hr) (in.) (in.)

δt P (P*δt) (P-Φ)δt0 0

0.5 4.5 2.25 (4.5-Φ)*0.50.5 30

0.5 3.0 1.5 (3-Φ)*0.51 60

0.5 1.0 0.5 (0)*0.51.5 90

0.5 3.5 1.75 (3.5-Φ)*0.52 120

0.5 2.0 1 (2.0-Φ)*0.52.5 150

0.5 0.0 0 (0)*0.53 180

Sumation 14 7 (13-4Φ)0.5

Total Precipitation (in.) = 7 Runoff (in.) = 105/500/12 = 2.52

Infiltration = P*t-R (in.) = 4.48

Therefore: (13-4Φ)*0.5 = 2.52Φ = (13 - 2.52*2) / 4 = 7.96/ 4 = 1.99 in./hr

1.99 OK.

INCI 4138 Hydrologic Processes26

Ptotal =∑(i∆t) =∑(Pi∆t) R=∑Ri =∑(Qi∆t)

∑(Pi-Φ) ∆t =∑Ri=R or solving for Φ

Φ = (∑(Pi∆t) – R)/(m∆t)

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