Civil Engineering Dept.

CE 331 Engineering Statics

Dr. Rashid Allayla

 

 

Part 1 
"Surface Hydrology"

 

 

 

 

 

 


Hydrology:

 

According to the US Federal Council for Science and Technology (ad hoc Panel on Hydrology , 1962) hydrology is a “science that treats all the waters of the earth, their occurrence, circulation and distribution, their chemical and physical properties, and their reaction with their environment including their relations to living things”

 

 Hydrologic Cycle:

 

                Hydrologic cycle is the water circulatory system on earth. The cycle has no beginning or end as the evaporated water rises to atmosphere due to solar energy. The evaporated water can be carried hundreds of miles before it is condensed and returned to earth in a form of precipitation. Part of the precipitated water is intercepted by plants and eventually returned to the atmosphere by evapotranspiration from plants and upper layers of soil, runs overland eventually reaching open water bodies such as streams, oceans or natural lakes or infiltrates through the ground forming deep or shallow groundwater aquifers. A good portion of the precipitated water evaporates back to the atmosphere thereby completing the hydrologic cycle.

 

 

 

hydrologic cycle image (From Purdue University)

          

                    

                            From EPA

 

Elements of Hydrologic Cycle:

               

-          Evaporation, E

-          Transpiration, T

-          Precipitation, P

-          Surface runoff, R

-          Groundwater flow, G, and,

-          Infiltration, I

 

 

 

 

Text Box: Falls into ground forming surface runoff, lakes, rivers or reservoirs

 

 

 

 

System Concept in Hydrologic Cycle:

        Oceans

 

                        Aquifers

 
 

 

 

 


                       

 

Illustration courtesy of: Colorado Division of Water Resources, Office of the State Engineer

Hydrologic Budget

 

            The hydrologist must be able to estimate components of hydrologic cycle in order to design projects and, more importantly protect the public from excessive floods and draughts. This can be accomplished by careful accounting technique that is not unlike keeping track of money in the bank.

               

                Precipitation                       Watershed                 outflow           Inflow           Change in Storage                                                                                                  

          

In engineering hydrology, the hydrologic budget is a quantitative accounting technique linking the components of hydrologic cycle. It is a form of a continuity equation that balances the gains and losses of water with the amount stored in a region. The components of water budget are inflow, outflow and storage.

 

S  INFLOWS - S OUTFLOWS   =  D STORAGES,   Or in mathematical term,

 

I – O = ∂S / ∂t

 

Breaking system into individual component as shown in schematic figure:

 

P – E + ([(Rin + Gin)] – (Rout + Gout)] – T = dS/dt

Where:

P: Areal mean rate of precipitation (L/T)

E: Evaporation (L/T)

Rin, Gin: Inflow from surface and groundwater (L/T)

             Rout, Gout:  Outflow from surface and groundwater (L/T)

             S: Storage (L) and,

             T: Transpiration (evaporation from plants, L/T)

 

 

All the above is measured in volume per unit area of watershed.

 

Breaking the above into surface and subsurface balance equations:

 


Surface:        P + Rin – Rout + Gup – I - E - T = ∆ Ss

 


Subsurface:  I +  Gin -  Gout – Gup  = ∆ Sg

 

 

The water budget formula is often used to estimate the amount of evaporation and evapotranspiration. Combining & dropping subscripts to represent net flows we get hydrologic budget formula,

 

P  -  R  -  G  -  E  -  T  =  D S / ∆t

 

Balance Equation for Open Bodies with Short Duration:

 

And, for an open water bodies, short duration,

 

I  -  O  =  D S/D t 

Where,

 

I    =  inflow volume per unit time

O  =  outflow per unit time

 

Balance Equation for Urban Drainage:

 

For urban drainage system, ET (evapotranspiration) is often neglected,

 

P  -  I  -  R  -  D  =  0

 

P = precipitation

I = infiltration

R = direct runoff

D = Combination of interception and depression storage

 

 

 

Illustrative Example:

 

            In a given year, a 10,000 Km2 watershed received 30 cm of precipitation. The annual rate of flow measured in the river draining the area is 60 m3/sec. Estimate the Evapotranspiration. Assume negligible change of storage and net groundwater flow.

Solution: 

 

Combining E and T, then ET = P – R

 

In the above, the precipitation term P is given in cm and the runoff term R is given in discharge unit. Since units in the equation must be consistent, and since the area of the watershed is constant, the volume of flow into the watershed is converted to equivalent depth.

 

Volume due to runoff = 60 m3/s x 86400 sec/day x 365 day/yr = 1.89216 x 109 m3

 

                                 = 1.89216 x 109 m3 x (100 cm/m)3 = 1.89216 x 1015 cm3

 

Equivalent depth = volume of water / area of watershed

                                                                   = 1.89216 x 1015 / [(10,000) (100,000 cm/km)2 ] = 18.92 cm

 

Amount of Evapotranspiration ET = P – R = 30 – 18.92 = 11.08 cm / yr

 

 

 

 

 

                                                      

 

               

 

 

 

 

                                                   

 

 

 

 

Illustrative Example: (From Viessman 2003)

           

            The drainage area of a river in a city 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. Assume negligible changes in groundwater flow and storage (i.e.  G and ΔS = 0) .

 

Solution:

 

Then:  ET = P – R, converting runoff from m3/yr to m/yr, then,

 

R = [144.4 m3/s x 86400 s/day x 365 day/yr] / [11,839 km2 x 106 m2/km2] = 0.38 m

 

ET = P – R = 1.08 – 0.38 = 0.7 m

 

 

 

 

 

 

 
 

 

 

 

 

 

 

 

 

 

 

 


Precipitation:


 

 

Illustrative Example:

 

At a particular area, the storage in a river reach is 40 acre – ft. The inflow at that time was measured to be 200 cfs and the outflow is 300 . The inflow after 4 hours was measured to be 260 and the outflow was 270. Determine a) the change in storage during the elapsed time and b) The final storage volume.

 

a) Average inflow rate  =  ( 200 + 260 ) / 2 = 230 cfs

Average outflow rate = ( 300 + 270 ) / 2 =  285 cfs   and Since :  I  -  O  =  D S/D t 

 

D S/D t  = 230 – 285 = - 55 cfs

 

D S = (- 55) (4 hr) = - 220 cfs-hr = (- 220 cfs-hr) (60 x 60 Sec / 1-hr ) ( 1 acre – ft / 43,560 ft3 ) =

       = - 18.182 acre-ft

 

b) S2 = 40 – 18.182 = 21.82 AF

 

 

Illustrative Example:

 

At a particular area, the storage in a river reach is 40 acre – ft. The inflow at that time was measured to be 200 cfs and the outflow is 300 . The inflow after 4 hours was measured to be 260 and the outflow was 270. Determine a) the change in storage during the elapsed time and b) The final storage volume.

 

a) Average inflow rate  =  ( 200 + 260 ) / 2 = 230 cfs

Average outflow rate = ( 300 + 270 ) / 2 =  285 cfs   and Since :  I  -  O  =  D S/D t 

 

D S/D t  = 230 – 285 = - 55

 

D S = (- 55) (4 hr) = - 220 cfs-hr = (- 220 cfs-hr) (60 x 60 Sec / 1-hr ) ( 1 acre – ft / 43,560 ft3 ) =

       = - 18.182 acre-ft

 

b) S2 = 40 – 18.182 = 21.82 AF

 

 

 

 

 

 

 

 

Precipitation:

 

            Precipitation is the discharge of water out of the atmosphere. The principal form of precipitation is rain and snow and to a lesser extent is hail, sleet. The physical factor producing precipitation is the condensation of water droplets due to atmosphere cooling.

The chief source of moisture producing precipitation is evaporation from oceans, seas. Only one tenth of the precipitation comes from continental sources in the form of soil evaporation and transpiration. Some of the precipitated water returns back to oceans and seas but the major portion  is retained as replenishment to surface water bodies, groundwater recharge and soil and plant moisture, The steps required to form precipitation include:

 

a)       Cooling which condensate moist air to near saturation,

 

b)      Phase change of water vapor to liquid or solid and,

 

c)       Growth of water droplet to perceptible size. This mechanism requires cloud elements to be large enough so that their falling speed can exceeds the upward movement of the air.  If the last step does not occur, cloud will eventually dissipate. 

 

 

 

           

 

Distribution of Precipitation:   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Measurement of Precipitation:

 

Errors:

·         Splashing

·         Moisture deficiency of gage

·         Wind blowing

 

 

2. Tipping Bucket Gage:

 

 

Counter

(Will tip over when filled by 0.01 inch of rain) to record intensity of rain

 
                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                       

 

 

 

 

 

 


-         

To measuring tube

 
Interception: The amount of precipitation that is intercepted by vegetation.

-          Depression Storage: This is the part of precipitation that is intercepted by holes in the ground and uneven surfaces. This part of storage will eventually be evaporated or slowly seeps underground.

-          Infiltration: This portion of precipitation makes its way to replenish groundwater through seepage.

-          Overland Flow: This portion of precipitation is the access water after the local rate of infiltration reaches its maximum. It develops as a film of water that moves overland eventually reaching streams, reservoirs or lakes.

 

Methods of Estimating Areal Precipitation in the Ground:

 

a)       Arithmetic Mean: Equal weights are assigned to all gage stations.

 

 

b) Thiessen Polygon: Each gage is assigned an area bounded by a perpendicular bisect between the station and those surrounding it. The polygon represent their respective areas of influence (see figure). The average precipitation is calculated as:

 

            Average PPT = ∑ ai pi / AT

 

Illustrative Method:

 

Observed Precipitation

(L)

Area of Polygon

(L2)

Precipitation x Area

(L3)

 

P1

 

A1

 

P1 A1

 

P2

 

A2

 

P2 A2

 

P3

 

A3

 

P3 A3

 

Pn

 

An

 

Pn An

Average P =( P1 A1  + P2 A2 + P3 A3 + …. + Pn An ) / (A1 + A2 + A3 + …. + An )

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


c) Isohyetal Method: The area between two successive isohyets is measured using planimeter or simply by counting sub grids.  The average precipitation is computed by multiplying the average precipitation between two successive isohyets by the inter-Isohyetal area, adding them and  dividing by the total area of the watershed.

 

Average PPT = ∑ ai pi / AT

 

 

 

The method of calculation is similar to that of Thiessen method except the area is the one bounded by two isohyets.

 

Illustrative Example:

 

 

 

 

Ave PPT = 65610/ 666

                     = 98.51 inch

 

 

 

 

                                                    Adjusted values

 

 

 

 

Alternative Method:

 

Inverse Distance Method: The method is based on the assumption that the precipitation at a given point is influenced by all stations. The method of solution is to subdivide the watershed area into m rectangular areas. The mean precipitation is calculated using the following formula:

                           m             n                       n

            P = 1/A ∑ Aj ( ∑ dij –b )-1 ∑ d ij –b P i

                          j = 1         i = 1                  i = 1

Where: Aj is the area of the jth sub area, A is the total area, dij is the distance from the center of the jth area to the ith precipitation gage, n is the number of gages and b is a constant and in most applications, it is taken as equal to 2. Note that if b is 0, the equation is reduced to the following:  P = 1/A ∑ Aj Pi

                         

Which is simply the arithmetic mean.

 

1

 

2

   O     d2,18

3

4

5

       O d5,18

6

7

8

 

9

10

          

11

      12

 

13

14

15

16

17

18

19

20

21

 

22

 O   d22,18

23

24

25

26

 O d26,18

27

28

 

O: Precipitation Gage

 

For example, the precipitation over sub area 18 is determined as:

                           4                                    4

            P18 =  [ ∑ d-2 i,18   Pi  / ( ∑ d-2 i,18 )

                          i =1                               i =1

 

Illustrative Example:

 

The sub areas shown below are 4.5 km2 each, find the precipitation x in subarea shown below:

                                                                                                                                                                        Y = 1.5 km

                        X = 3 km                                             X = 3 km                                                      X = 3 km

                             1               d= 3 km

                                                                                 

               

                        P = 0.2 “

                              2  

                              X

                   Unknown PPT

            d = 4.5 km   3

 


                        

                                            P = 0.15”

 

 

P2 =   1 →  3  d -2 i , 2 Pi / (d -2 i , 2)

 

P2 = [d-21,2] P1 / [d -2 1 , 2 + d -2 3,2] + [d-23,2] P3 / [d -2 1 , 2 + d -2 3,2]

 

P2 = [ (3) -2 ] (0.2) / (0.111+ 0.0494) + (4.5) -2 (0.15) / (0.111+ 0.0494 = 0.13854 + 0.04619 = 0.1847


Methods of Estimating Missing Precipitation:

 

a) Normal Ratio Method:

 

The missing precipitation at station x is calculated by using weights for precipitation at individual stations. The precipitation at station x is,

                       n

            Px = ∑ wi Pi

                     i=1

where  n is the number of stations and wi designates the weight for station i and computed as

            wi = Ax / n Ai

 

Where Ai is the average annual rainfall at gage i, Ax is the average annual rainfall at station x in question.

 

Combining the above two equations,

 

                                   n

            Px = (AX / n) ∑ Pi / Ai 

                                i =1

 

 

Illustrative Example:

 

The following data was taken from 5 gage stations including stations:

 

 

Gage

Average Annual Rainfall (cm)

Total Annual Rainfall (cm)

A

32

2.2

B

28

2.0

C

25

2.0

D

35

2.4

X

26

?

 

Solution:

 

Px = wA PA + wB PB  + WC PC + WD PD

 

   = (Ax  / n AA) PA  + (Ax  / n AB) PB + (Ax  / n AC) PC + (Ax  / n AD) PD

 

    = (26 / 4 x 32) (2.2) + (26 / 4 x 28) (2.0) + (26 / 4 x 25) (2.0) + (26 / 4 x 35) (2.4) = 1.877 cm

 

Then: Px = 1.877 cm

 

 

 

 

 

 

b) Quadrant Method:

 

To account for the closeness of gage stations to the missing data gage, quadrant method is employed. The position of the station of the missing data is made to be the origin of the four quadrants containing the rest of stations. The weight for station i is computed as:

 

                       4

            wi =  ∑ ( 1 / d2i )

                      i=1

 

and the missing data is calculated as,

                           n                 n

                Px = ∑wi . Pi / ∑ wi

                      i=1              i=1

 

 

 

Gauge Consistency:

 

Double Mass Curve

 

This is a method used to check inconsistency in gage reading. The inconsistency could be attributed to  environmental changes such as sudden weather changes that adversely effect  gage reading, vandalism, instrument malfunction, etc. Double Mass Curve is a plot of accumulated annual or seasonal precipitation at the effected station versus the mean values of annual or seasonal accumulated precipitation for a number of stations surrounding the station that have been subjected to similar hydrological environment and known to be consistent. The double mass curve produced is then examined for trends and inconsistencies which is reflected by the change of slope. A typical Double Mass Curve of infected station (station A) versus mean values of similar stations is shown below:

 

 

 

Station A (mm)

            PA

                                PB

                    12 Station Mean PB (mm)

 

Method of Correction:

 

As shown in the figure, the slope of the Double Mass Curve changed abruptly from m1 prior to 1990 to m2 after 1990. The record can then be adjusted using the following ratio:

            Pa = (ma/mo) po       where:  ma  (adjusted)     & mo  (observed)

In the above, subscript a denotes adjusted and o denotes observed. If the initial part of the record need to be adjusted then m2 is the correct slope and PA2 and PB2 are correct. So, if m2 is the correct slope, the slope m1 should be removed from PA1 and replaced by m2 by using the formula:  p A1 = (m2 / m1) P A1 where pA1 is the adjusted data.

 

 

 

 

 

 

 

 

 

 

 

 

 

 


                          p1 = (m2 / m1) P1                                                                                     p1 = (m1 / m2) P2

                              

 

 

 

 

 

 

Illustrative Example (McCuen 1998)

 

Gage H was permanently relocated after a period of 3 years. Adjust the double mass curve and find the values of h79, h80 and h81.

 

 

year

E

F

G

H

Total

∑E+F+G

Cumulative

E+F+G

Cumulative

H

h

1979

22

26

23

28

71

71

28

24.7

80

21

26

25

33

72

143

61

29.1

81

27

31

28

38

86

229

99

33.5

82

25

29

29

31

83

312

130

 

83

19

22

23

24

64

376

154

 

84

24

25

26

28

75

451

182

 

85

17

19

20

22

56

507

204

 

86

21

22

23

26

66

573

230

 

 

 

                    

 

 

 

 

 

 

 

 

 

Evaporation:

 

Evaporation is the process by which water is transferred from liquid state to gaseous state through transfer of energy. When a sufficient kinetic energy exists on the surface, water molecules can escape to the atmosphere by turbulent air.

 

Factors Affecting Evaporation:

 

Temperature: It is a measure of combined potential and kinetic energy of the body’s atom.

 

Humidity and Vapor Pressure: The amount of water vapor in the atmosphere is very small compared to the other elements that exist (like Oxygen, Nitrogen, CO2, etc.). However, Water vapor in the air plays an important part in controlling weather patterns and evaporation processes. When the air is dry, evaporation takes place, causing an increase in the quantity of vapor in the air and an increase in vapor pressure. This process will continue until vapor pressure in the air is equal to vapor pressure at the surface. At this point, saturation will occur and further evaporation ceases.

                         

Denoting e as the vapor pressure and es as the saturated vapor pressure, the relative humidity is defined as,

 

R = e / es

 

Vapor pressure is commonly expressed in bars, where,

 

1 bar = 105 Newtons / square meters  (106 dynes / cm2)

1 mb = 1000 dynes / cm2

 

Radiation: (From AMS Glossary): It is the process by which electromagnetic radiation is propagated through free space. The propagation takes place at the speed of light (3.00 x 108 m s−1 in vacuum) by way of joint (orthogonal) oscillations in the electric and magnetic fields. This process is to be distinguished from other forms of energy transfer such as 1. conduction and convection. 2. Propagation of energy by any physical quantity governed by a wave equation.

 

Wind Speed: Wind speed varies with the height above water surface. It can be calculated using the empirical formula,

 

V/VO = (Z/ZO)0.15

 

Where V is the wind speed in mi/hr at Z height, VO is the wind speed at height ZO measured in ft of the anemometer (instruments designed to measure total wind speed)

 

 

Measurement  of Evaporation:

 

Evaporation Pans: The most popular method of estimating evaporation. The best known is the US Weather Bureau Class A Pan. (complete description of Class A Pan can be found in AMS Glossary link): The U.S. Weather Bureau evaporation pan (Class-A pan) is a cylindrical container fabricated of galvanized iron or other rust-resistant metal with a depth of 25.4 cm (10 in.) and a diameter of 121.9 cm (48 in.). The pan is accurately leveled at a site that is nearly flat, well sodded, and free from obstructions. The water level is maintained at between 5 and 7.5 cm (2 and 3 in.) below the top of the rim, and periodic measurements are made of the changes of the water level with the aid of a hook gauge set in the still well. When the water level drops to 17.8 cm (7 in.), the pan is refilled. Its average pan coefficient is about 0.7.

 

 

                               

                                    

        

 From: The Earth & Geographic Sciences Department / University of Mass, Boston

 

Empirical Formulas:

 

Using the general mass transfer relation is

 

E = K f(u)(e0 –ea)

 

Where K is constant, f(u) is the wind speed at a given height, ea is the actual vapor pressure at a given height, and e0 is the saturated vapor pressure at water surface temperature.

 

The rate of evaporation from a lake can be calculated using empirical laws, one of which has been developed by Meyer (1944),

 

E = C (es – e) (1 + W/10)

 

Where,  E:        Lake evaporation (inches / day)

              es – e: Water vapor deficit (difference between saturated vapor   pressur and actual vapor pressure of   

                          atmosphere in-Hg)

              C:        Constant (0.36 for open water, 0.5 for wet soil)       

              W:       Wind Speed 25 ft above water level (mph

 

Another equation developed by Dunne ((1978) is

 

E = (0.013 + 0.00016 u2) e [(100 – Rh) / 100]

 

Where, Rn:       The relative humidity in %

             e:          Vapor pressure of air (mill bars

             u2:        Wind speed 2 m above water in km/day

 

      

 

 

Illustrative Example:

 

Find the lake evaporation for a lake with mean value of air temperature is 87 F, and for water temperature is 63 F, average wind speed is 10 mph and relative humidity is 20%.

 

Solution:

 

-Using Meyer formula: From table, the saturated vapor pressure, es (@63oF) = 0.58 in. Hg

                                                                                                                          es (@87oF) = 1.29 in. Hg                             

 e = 1.29 x 0.20 = 0.26 in Hg = 8.75 mb

 

For C = 0.36 then E = 0.36 (0.58 – 0.26) [1+10/10    = 0.23 in/day

 

-Using Dunne’s formula: Converting wind speed to km/day,

 

E = [0.013 + (0.00016 x 386)] (8.75) [(100-20)/100]

 

   = 0.527 cm/day or 0.21 in/day which is comparable to the previous value.

 

 

 

 

 

 

 
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


Some Empirical Equations:

 

Author

Equation

Explanation

 

Dalton

 

E (i/mo) = C(eO-ea)

 

C=15 for small, shallow water and 11 for large deep water

 

Meyer

 

E (i/mo) = 11 (1+0.1 ug) (eO-ea)

 

ea measured 30 above ground surface

 

Horton

 

E (i/mo) = 0.4 [2 – exp(-2u)]) (eO-ea)

 

u = speed of wind

 

Penman

 

E (i/day) = 0.35 (1+0.24 u2) (eO-ea)

 

u2 = wind speed 2 meters above surface

 

Harbeck

 

E(i/day)=0.001813 u (eO-ea)[1-0.03(Ta-Tw)]

 

Ta = Average Temp OC + 1.9OC

TW = Average water surface temperature

 

 

Coaxial Chart: Penman (1948)

 

Penman developed an equation based on aerodynamic and energy balance equations for daily evaporation E and later Kohler (1955) developed an expression for lake evaporation in inches per day that is based on Penman’s theory, If EL designates average daily lake evaporation (in/day), then,

 

EL = 0.7 [EP + 0.00051 P αP (0.37 + 0.0041 uP) (T0 –Ta)0.88

       

                                                                                                                To Outer face Temp of pan O F

                   Lake Evap. in/day                                                                                     Ta Air Temp. in o F

  Windspeed in mi/day

 Advected Energy (For given Temp & wind speed ,use  chart to obtain αP)

 Atmospheric pressure in in-Hg

 

αP = 0.13 + 0.0065T0 – (6.0 x 10-8 T03) + 0.016 uP0.36   (Use Chart below to obtain αP)

 

Steps of Using the Coaxial Chart:

 

-           From pan water temperature (measured in oF) and pan wind speed (measured mi/day), use chart A to obtain αP .

-           From wind speed uP start at the upper left hand of the coaxial chart (chart B) to the elevation (in feet) above mean sea level.

-           Turn right to the value of  αP in the lower left chart.

-           Turn left to the value of (T0 – Ta) in the lower right chart.

-           Turn left to pan evaporation in the upper right chart.

-          Turn right to read lake evaporation.

                                                                               

 

 

 

 

 

 

 

 NO SCALE

                   

 

Infiltration:

 

Infiltration: is a process of water entry into a soil through surface. Percolation: movement of water within soil profile, a process that follows infiltration. Infiltration Rate: Rate at which water enters soil measured in L/T. Cumulative Infiltration: Volume of infiltrated water at given time, measured in L. Infiltration Capacity: Is the maximum rate at which a given soil can absorb, a potential value that may or may not be satisfied following rain. It is measured in L/T.

 Elements of Infiltration

 

 

 

 

 

Infiltration Capacity:

 

                Infiltration capacity is an aspect of infiltration that is associated with soil.  It is defined as the maximum amount of water per unit time that can be absorbed under given conditions. The greater the infiltration capacity of soil, the greater amount of water that can be infiltrated.

 

Empirical Models of Infiltration:

 

Horton Model:

 

Horton theorized that the process of infiltration is analogues  to exhaustion process where the rate of performing work is proportional to the amount of work remaining to be performed. The rate of performing work is analogous to df/dt  is related to the amount of work remaining is analogous to (f – fc) in infiltration model.

 

 

 

 

 

 

 

 

 

 

 

 

 

 


Horton Infiltration Formula:

 

                                                              K: Constant depending on soil & surface & cover conditions, t is time

 

f = f+ (f0  –  f) e-Kt                                  

 


                                                                                                                                                   Initial infiltration capacity (L/T)

 

                                                                                                                                                Final infiltration capacity (equilibrium capacity)

 

                                                                                                                          Infiltration capacity at a given time t

 

The assumption inherent is that water is “ponded” and, therefore, it is a potential infiltration curve

Horton equation requires evaluation of f0, fand K .which can be derived from infiltration tests

 

 

 

 

Observations:

 

-          If rainfall intensity exceeds the infiltration capacity, the infiltration capacity decreases exponentially.

 

-          The area under the curve is the volume of infiltration. However, the actual infiltration rate is equal the infiltration capacity f only when rain intensity, less the rate of retention, equals or exceeds f (see figure below).

 

                                       

-          The value of f is maximum ( = f0 ) at the beginning of the storm. This becomes constant ( =f) as the soil becomes saturated.

-          When rain intensity at any given time is less than the infiltration capacity, adjustments to the infiltration capacity curve must be made.

-          Infiltration capacity depends on soil type (porosity and pore-size distribution are being the most important parameters), moisture content, organic matter, vegetative cover, season and soil content of organic matter (organic content enhances infiltration because it increases porosity).The values of f0 and k can be calculated by observing the variation of infiltration with time, plotting f vs. t and selecting two points from the graph. The two values can then be determined from Horton equation.

 

Illustrative Example

 

Given initial infiltration capacity of 60 cm/day and time constant, k of 0.4 hr-1. Derive infiltration capacity vs. time curve if the equilibrium capacity is 10 cm/day. Estimate the total infiltrated water in m3 for the first 10 hours for a 100 km2 watershed.

 

 Solution:

 

Horton curve: f = f + (f0 – f) e-Kt substituting,

 

                         f = 10 + (50) e-0.4t

Integrating,

 

V = ∫[10 + 50 e-0.4t] dt 

 

V = [10t -(50)e-0.4t0 to 10 = [(10)(10) –(50 / 0.4)e-4] + 50 = 147.7cm

 

Volume in  = (147.7 / 100 cm/m)100 km) (1000 m/km)2  = 1.471 x 108 m3

 
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


Correction of Horton Curve:

 

It is often the case that the intensity of  rain is  much  smaller  than the  values of initial infiltration capacity f0 and the equilibrium capacity f of the soil. Since the Horton’s formula assumes that the intensity of rain is always larger than the infiltration capacities of the soil, solving the equation for f as function of time only, would show continuous decrease in f even if the rain intensities are very low and  much less than the soil capacity (see figure)

Text Box: Infiltration capacity decreases
even though rain intensity is
low. 
                         

 

 

 

 

 


                       Hand 7 icon. 

 

                       Problem!!

                                      

 

                                                    0                         t1  

 


  Equivalent time τ for the actual

  Accumulated infiltration

 

 

 


 

 


                                                          τ

                                                  Volume of water infiltration that has been over- estimated by Horton curve!

                                                                                                       

What to do?  It is clear that if the total intensity of rain at the first time increment is less than the infiltration capacity, it is more reasonable to assume that the reduction in f is dependent on the infiltrated volume of water rather than the elapsed time. Therefore, it becomes necessary to shift the curve to t = τ, which would produce an infiltrated volume equal to the volume of the actual rainfall.          

     

In order to accommodate for the possible infiltration deficiency, the following procedure is employed:

 

Let f(t) be the maximum infiltration capacity of the soil at given time (as prescribed by infiltration capacity curve), then

                 

∆F = ∫t to ∆t  f (t) dt

 

Where  ∆F is the amount of infiltrated water between t and ∆t. Depending on the initial intensity of rain compared to the maximum infiltration capacity f, the equation for f(t) can take one of  two forms:

 

 

If           1)    i . dt    ∆F,   then τ = ∆t       where i is the rain intensity between t to ∆t

 

Then    f0 (new) = f+ (f0  –  f) e - K t      which is simply the Horton formula

 

But, if    2)   i . dt < ∆F,   then τ < ∆t

 

Then    f0 (new) = f+ (f0  –  f) e - K τ  

 

Application of the above equations for every time step results in actual infiltration.

 

Steps:

1)       Draw the infiltration Capacity curve (Horton Curve, fig. A).

2)       Establish Mass curve (f vs. volume) by finding  the area under Horton curve at Δt. This is calculated by multiplying the average value of f by Δt

 


 Horton Curve

 

                 ( fi + fi+1 ) / 2

                                                                              Δt

 

 

                                                   fi                   fi+1

 

3)       After establishing the mass curve from Horton curve (Fig. B), compute the volume of  rain that is actually infiltrated during the interval when i < f and find the corresponding infiltration capacity (f1) from the established mass curve (f fig. B)

 

4)       The new infiltration capacity curve is now drawn where f0 = f1  at  t = t (fig D)

 

                Horton Curve                                                        Mass Curve = ∫ f(t)

        

                      t                                                                      t

                              Actual infiltrated water (total rain @ t

 

Illustrative Example:

 

a) Curve Shifting Method:

 

                Given the following rain pattern and the infiltration curve (based on Horton formula). Determine the surface runoff neglecting all losses.

                                                                                          

 

 

4

 

 

3

 

 

2

 

1

 

R

A

I

N

 

I

N    cm

T    per

E    hr

N

S

I

T

Y

 

I

N

F

I

L

 

C

A

P

A

C

I

T

Y

 

 

                              4.5

 

5

 

3

                                         2.5

                                                                                 1.8

2                 

                       1.5                                                

   0.8

1

 
    

Average height = 4 cm/h

Infiltration = 4 x 0.5 =2 cm

 
                                                                                                                                                                         

 


                          

 

 

 

 

 

 

 


         

             0                0.5               1.0            1.5               2.0             2.5             0                 0.5             1.0                1.5                2.0              2.5     Time hr

 

 

Procedure:

 

1) Calculate the mass curve (F vs. f) at the end of each time interval where F is the area under Horton curve:

 

F = f t + [(f0 – f) / k] (1 - e-K t).                                     

                                                                                  

 

 

                                                                                                                       Mass Curve

Text Box: Time Increment     Ave. height cm/hr     Cumulative infiltration Cm

       0.0 → 0.5                      4                            4 x 0.5 = 2
       0.5 → 1.0                      2.5                         2 + 2.5 x 0.5 = 3.25
       1.0 → 1.5                      1.7                         3.25 + 1.7 x 0.5 = 4.1
       1.5 → 2.0                      1.5                         4.1 + 1.5 x 0.5 = 4.85
       2.0 → 2.5                      1.5                         4.85 + 1.5 x 0.5 = 5.6
              

                                                                                             Note that:    In the first 1 hr all rainfall has infiltrated

The values of f vs. F are:

Text Box: End of Time Increment (hr)    f  (cm / hr)          Cumulative infiltration  F (Cm)

        0.5                                            3                                         2
        1.0                                            2                                         3.25
        1.5                                            1.7                                      4.1
        2.0                                            1.5                                      4.85
        2.5                                            1.5                                      5.6
 


                               

 

 

 

               

 

 

 

 

The resulting mass curve is:

 

                             

 

1)       Calculate the actual amount of infiltrated water

 

      Time (hr)    Rain Intensity (Cm/hr)           Cumulative precipitation( Cm)

 

       0.5                                0.8                                                 0.8 x 0.5 = 0.4

       1.0                                1.5                                                 0.4 + 1.5 x 0.5 = 1.15

       1.5                                4.5                                                 1.15 + 4.5 x 0.5 = 3.4

       2.0                                2.5                                                 3.4 + 2.5 x 0.5 = 4.65

       2.5                                2.0                                                 4.65 + 2.0 x 0.5 = 5.65

      

 
                                                               

 

 


                                                                                                                                                                                                                                  

 

 

 

 

2)       Calculate the amount of rainfall that is totally infiltrated  and shift the original capacity curve to the right and obtain the storm infiltration curve (storm infiltration curve is shown as solid curve below) and calculate the cumulative infiltration & direct runoff.

 

                             Shift Horizontally

                             To the point where

                             Intensity of rain >  f

 

 

2.3

 

 

1.8

 

 

   5

 

   4

    3.6

   3

     f

  Cm/h

   2

 

   1

 

   0

    

 
  Original Horton Curve                                          

                                                                                                                                Note that the shift the

                                                                                                                                curve would not have

                                                                                                                                been necessary if

                                                                                                                                intensity from t = 0 t1

                                                                                                                                were ≥ f

 

 


                                                                                                                                                           Shifted Horton

                                                                                                                                                           curve

                                                                                                                             

 

                                                    0                0.5              1.0               1.5                2.0             2.5   Time  (hr)   

 

Actual rain

        

 

Time (hr)      Cumulative Precipitation (cm)           f (After shifting in cm/hr)       Cumulative infil.  F (cm )     R.O.= P - F

                                      (From Step 3)

  0 -  0.5                           0.4                                                         0.8                           0.8 x 0.5 = 0.4                                0

  0.5 -  1.0                           1.15                                                       1.5                           0.4 + 1.5 x 0.5 = 1.1                       0

  1.0 -  1.5                           3.4                                                         2.3                           1.1 + 2.3 x 0.5 = 2.25            3.4- 2.3=1.1

  1.5 - 2.0                            4.65                                                       1.8                           2.25 + 1.8 x 0.5 =  3.15         4.65-3.15= 1.5

   2.0 - 2.5                           5.65                                                       1.6                           3.15 + 1.6 x 0.5 = 3.95          5.65-3.95= 1.7

                                        

                                                         Read from the new curve @ middle beween t = 1 & t =1.5

Alternative Methedology:

 

b) Integrated Horton Method:

 

Illustrative Example:

 

Given the Horton parameters: f = 0.2 inches per hr, f0 = 3 inches per hr and k = 1.4 per hr

 

The rainfal hyetograph:

 

Time (hrs)

Intensity (in/hr)

Depth (in)

0.0-0.5

0.8

0.40

0.5-1.0

2.1

1.05

1.0-1.5

0.9

0.45

1.5-2.0

0.4

0.20

2.0-2.5

0.8

0.40

2.5-3.0

1.1

0.55

 

Horton equation:              f = 0.2 + (2.8) e-1.4t  ……………………………………… (1)

Accumulated Infiltration: F = 0.2 t + 2 (1 - e-1.4 t) ………………………… (2)

 

1) Time step 1 : 0 – 0.5 hrs

 

tP + ∆t = 0.5 hrs

initial condition: At tP = 0, F(tP) = 0

Potential infiltration F @ tP+dt equation 2 yields

F (tP+∆t) = (0.2) (0.5) + (2)(1-e-1.4 x 0.5) = 1.107 inches

Potential ∆F = F (tP+∆t) - F (tP) = 1.107 in

Compare the rainfall depth with the potential infiltration ∆F (which represents the area under Horton curve at ∆t = 0.5 hrs):

If rainfall depth ≥ the potential ∆F, set the actual infiltration = ∆F

If rainfall depth < the potential ∆F, the actual infiltration = the volume rain.

Since the volume of rainfall is 0.4 < 1.107 the actual infiltration = the volume of rain = 0.4 in

Now compute the accumulative infiltration (column 7) = The  accumulative infiltration during the previous time step plus the actual infiltration after ∆t:

F = 0 + 0.4 = 0.4 in

Access rainfall during the first time step = Rainfall minus infiltrated water:

= 0.4 – 0.4 =  0 inches   ………………………………………………………………………………..  (Column 9)

 

 

 

2) Time step 2 : 0.5 – 1.0 hrs

 

If the rainfall quantity is ≥ potential infiltration as prescribed by Horton curve, set the time to tP + ∆t. However, if the rainfall quantity is < the potential infiltration, Find the equivelent time by setting F = accumulative infiltration and solve for t that corresponds to F = actual accumulative infiltration using the integral form of Horton formula eq. 2). Since the infiltrated water (=0.4) < Potential infiltration (=1.107), then:

 

F = 0.4  = 0.2 t + 2 (1 - e-1.4 t) → Using successive substitution or Newton-Raphson iteration Method (1)

 

t = 0.146 hrs  …………………………………………………………………………………………….. (Column 8)

 

2) set tP = 0.146 hrs and go to the next time step.

 

The potential infiltration at time t = tP = 0.146

F(0.146) = 0.2 (0.146) + (2) ( 1 – e-(1.4)(0.146) ) = 0.4 inches …. (Column 2) .. We already know that!

The potential infiltration at time t = tP + ∆t = 0.146 + 0.5 = 0.646 hrs is:

F(0.646) = 0.2 (0.646) + (2) ( 1 – e-(1.4)(0.646) ) = 1.320 inches  … (Column 3)

∆F = 1.32 – 0.4 = 0.920 inches … (Column 4)  (This is how much water available for future infiltration)

Compare potential ∆F with the actual volume of rainfall:

Potential infiltration (=0.92) < Rainfall volume (=1.05), the actual infiltration is 0.92” only ...(Column 6)

Accumulative infiltration = 0.4 + 0.92 = 1.32 inches ……………………………………………… (Column 7)

Since the actual rainfall volume (=1.05) >The potential infiltration (=0.92”), Excess rain = 0.13 inches

 

t = 0.646 hrs ……………………………………………………………………………………………… (Column 8)

 

3) set tP = 0.646 hrs and go to the next time step.

 

The potential infiltration at time t = tP = 0.646

F(0.646) = 0.2 (0.646) + (2) ( 1 – e-(1.4)(0.646) ) = 1.32 inches ……………………………………….. (Column 2)

The potential infiltration at time t = tP + ∆t = 0.0.646 + 0.5 = 1.146 hrs is:

F(1.146) = 0.2 (1.146) + (2) ( 1 – e-(1.4)(1.146) ) = 1.827 inches ……………………………………… (Column 3)

∆F = 1.827 – 1.32 = 0.507 inches…(Column 4) (This is how much water is available for future infiltration). Compare potential ∆F with the actual volume of rainfall:

Potential infiltration (=0.507) > Rainfall volume (=0.45), the actual infiltration is 0.45”only .(Column 6)

Accumulative infiltration = 1.32 + 0.45 = 1.77 inches ……………………………………………. (Column 7)

F = 1.77  = 0.2 t + 2 (1 - e-1.4 t) → Using successive substitution or Newton-Raphson iteration Method

the  equivelent time = 1.073 hr

Since the actual rainfall volume (=0.45) < The potential infiltration (=0.507), Excess rain = 0.0 inches

 

t = 1.073 hrs …. Column 8

 

3) set tP = 1.073 hrs and go to the next time step.

 

The potential infiltration at t = tP + 0.5 = 1.573 hrs is:

F(1.073) = 0.2 (1.073) + (2) ( 1 – e-(1.4)(1.073) ) = 1.77 inches ………………………………………. (Column 2)

F(1.573) = 0.2 (1.573) + (2) ( 1 – e-(1.4)(1.573) ) = 2.0935 inches ……………………………………. (Column 3)

∆F= 2.0935–1.769 = 0.3245 inches (Column 4)(This is how much water available for future infiltration)

Compare potential ∆F with the actual volume of rainfall:

Potential infiltration (=0.3245) >Rainfall volume (=0.2), the actual infiltration is 0.2 in only.(Column 6)

Accumulative infiltration = 1.77 + 0.2 = 1.97 inches …………………………………………….. (Column 7)

F = 1.97  = 0.2 t + 2 (1 - e-1.4 t) → Using successive substitution or Newton-Raphson iteration Method

the  equivelent time = 1.353 hr

Compare potential ∆F with the actual volume of rainfall:

Since the actual rainfall volume (=0.2) < The potential infiltration (=0.3235), Excess rain = 0.0 inches

t = 1.353 hrs ……………………………………………………………………………………………. (Column 8)

 

3) set tP = 1.353 hrs and go to the next time step.

 

The potential infiltration at t = tP + 0.5 = 1.573 hrs is:

F(1.353) = 0.2 (1.353) + (2) ( 1 – e-(1.4)(1.353) ) = 1.970 inches …. Column 2

The potential infiltration at t = tP + 0.5 = 1.853 hrs is:

F(1.853) = 0.2 (1.853) + (2) ( 1 – e-(1.4)(1.853) ) = 2.221 inches …. Column 3

∆F= 2.221–1.970 = 0.251 inches…Column 4(This is how much water available for future infiltration)

Compare potential ∆F with the actual volume of rainfall:

Potential infiltration (=0.251) <Rainfall volume (=0.4), the actual infiltration is 0.251 in only.(Column 6)

Accumulative infiltration = 1.97 + 0.251 = 2.221 inches …. Column 7

F = 2.221 = 0.2 t + 2 (1 - e-1.4 t) → Using successive substitution or Newton-Raphson iteration Method

the  equivelent time = 1.853 hr

Compare potential ∆F with the actual volume of rainfall:

Since the actual rainfall volume (=0.4) >The potential infiltration (=0.251), Excess rain = 1.49 inches

 

 

 

Interval t (hrs)

 

1

F(tP)

(in)

 

2

F(tP + ∆t)

(in)

Potential

3

∆F

(in)

 

4

Rain Depth (in)

5

Actual ∆F

(in)

 

6

Accumulation F (in)

5+7

7

Equivalent Tme tP

(hrs)

8

Rainfall Excess

(in)

9

0-0.5

0

1.107

1.107

0.4

0.4

0.4

0.146

0

0.5-1.0

0.4

1.320

0.920

1.050

0.920

1.320

0.646

0.13

1.0-1.5

1.320

1.827

0.507

0.450

0.450

1.770

1.073

0.0

1.5-2.0

1.769

2.0935

0.3245

0.200

0.200

1.970

1.353

0.0

2.0-2.5

1.970

2.221

0.252

0.400

0.251

2.221

1.853

0.149

2.5-3.0

2.221

2.396

0.175

0.550

0.175

2.396

2.353

0.375

 

 

 

 

 

 

 

 

 

 

 

 

 

______________________________________________________________________________________________

 

(1) Newton-Raphson iteration Method:

the Newton-Raphson method is an iterative numerical method used to approximate the roots of an equation: Given one approximation.  The sequence of approximate roots x0, x1, x2, x3, ... is created by the rule:

xn+1 = xn - f(xn)/f '(xn)

 

where f is the function whose roots we want, and f ' is its derivative

 

Example: Let f (x) = x2 – 4, we know the solution is + or – 2

 

f(x) = x2 – 4

f ’(x) = 2 x

 

Let initial x0 = 6 then: f(x0) = 32, f ’(x0) = 12,   then: xn+1 = 3.33

Let x1 = 3.33, Then: xn+2 = 2.27

    .

    .

x3 = 2.01, then: x4 = 2.00

 

 

To Determine derivative of Horton Curve W/R to tP:

0 = f + (f0 – f) / k (1 – e – k tp) – F

t (i+1) = t (i) – f [t(i)] / f ‘(t(i))

 

Initial t (i) is set = tP + dt/2 and the calculated value t(i+1) is used for the next value of t (i)

 

i

 

t(i)

(hrs)

f [t(i)]

(in/hr)

f ‘ [t(i)]

(in/hr)

f/f ’

t (i+1)

(hrs)

1

0.25

0.240624

2.173127

0.110727

0.139273

2

0.139273

-0.01784

2.503978

-0.00713

0.146399

3

0.146399

-8.2x10-5

2.481106

-3.3x10-5

0.146432

______________________________________________________________________________________________

 

Summary (Integrated Horton Method):

 

1.        Set ti = tP + ∆t

2.        Calculate the average maximum infiltration capacity over time step f i * = [F(ti) – F(ti-1)] / ∆t

3.        The actual infiltration is: f * = Min (i , fi *)

4.        Calculate the accumulative infiltration F(t+∆t) = F(t) + f * = F(t) + (f *) (∆t)

5.        Calculate new tP using one of the functions:

 

If:  ∆F = f i * ∆t     then: f i  = f i *

If:  ∆F = i ∆t         then: f I  = i

Solve for tP using iterative methods