Engineering Hydrology

King Fahd University of Petroleum & Minerals

Civil Engineering Dept. CE 331 Engineering Statics Dr. Rashid Allayla

Dr. Rashid Allayla

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

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” Application of hydrology include:      

Determining water balance of a region (see subsequent discussion) Predicting floods and droughts Designing irrigation projects Preventing catastrophic events due to excess flooding Provide means to supply drinking water to communities Helps design dams, reservoirs, sewers, bridges and other structural and hydrologic projects

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. Elements of hydrologic cycle are:      

Evaporation, E Transpiration, T Precipitation, P Surface runoff, R Groundwater flow, G, and, Infiltration, I

A full discussion of each of the elements will be discussed in detail in subsequent chapters.

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Hydrologic Cycle in Visual Form:

(From Purdue University)

Fate of Precipitation

Goes into groundwater Storage or retained as soil moisture

Falls into ground forming surface runoff, lakes, rivers or reservoirs

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King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

System Concept in Hydrologic Cycle: n

I n f i i t r a t i o n

P r e c i p i t a t i o n

Transpiration

Atmosphere

Evaporation Transpiration

Vegetation

Surface RO

Surface

Streams/Lakes/Rivers Rivers Exfiltration

Precipitation Evaporation Surface runoff

Soil Interflow Capillary Rise

Groundwate r Oceans Illustration courtesy of: Colorado Division of Water Resources, Office of theflow State Engineer Percolation Aquifers

Aquifers

Oceans

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. Salary

Bank

Precipitation

Watershed

Activity Cash in & Cash out

outflow

Inflow

Statement of account

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.  INFLOWS -  OUTFLOWS =  STORAGES, Or, in mathematical term, 4

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

I – O = ∂S / ∂t Breaking system into individual component as shown in schematic figure: Where:

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

P Gup

T Ground Surface

Rin Gin

Rout Gout

I Water Table

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 =  S / ∆t Balance Equation for Open Bodies with Short Duration: And, for an open water bodies, short duration,

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Where,

Dr. Rashid Allayla

I - O =  S/ t 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

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Dr. Rashid Allayla

Illustrative Example: (From Viessman 2003) The drainage area of a river in a city is 11,839 km 2. 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 Precipitati ET = P – R = 1.08 – 0.38 = 0.7 m

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Dr. Rashid Allayla

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 =  S/ t  S/ t = 230 – 285 = - 55 cfs  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 =  S/ t  S/ t = 230 – 285 = - 55  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

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King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

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.

Droplets become heavy enough to fall (0.1 mm)

Cloud

Droplet breaks down to smaller size & returns to atmosphere Further Condensation

Droplet forms by condensation On particles of aerosols (0.001-10μm)

Rain drop (0.1-3mm)

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King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Distribution of Precipitation:

Total Precipitation Evap.

Evap.

Vegetation

Land Surface Direct Surface RO

Infiltration

Stream/Lake/river/ Sea Direct Surface RO

Interception & Depression Storage

Soil

i ii t t

Percolation

When filled

Total Rain

If pervious

InfiltRation

Direct RO

Interception & Deprssion Storage

Aquifers

ii t

ii t

Impervious Overland Flow (Direct RO)

Stream/lake/river/sea

ii t

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King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Measurement of Precipitation: 1. National Weather Service Gage:

20.3 cm

Side view

Receiver ( Collector ) One-tenth the area of the collector

Measuring tube (0.1 inch of rain will fill tube to 1 inch depth)

Errors:   

Splashing Moisture deficiency of gage Wind blowing

Measuring stick

2. Tipping Bucket Gage:

-

Counter (Will tip over when filled by 0.01 inch of rain) to record intensity of Interception: The amount of precipitation that is intercepted by vegetation. To measuring tube rain

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-

Dr. Rashid Allayla

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.

Watershed

Precipitation gage

Equivalent depth of total rain at watershed.

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 ) 12

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Theissen method provides weighting factors for each gage by enabling data from adjacent areas to be incorporated in calculating the mean value of precipitation

Watershed

Perpendicular bisector Area a representing gage P1 Computation Procedure: The average precipitation is computed by multiply the precipitation of each station by its assigned area, adding them and dividing the total area of the watershed. Average PPT = ∑ aipi / AT

Polygon

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 interIsohyetal area, adding them and dividing by the total area of the watershed. Average PPT = ∑ ai pi / AT Watershed Rain Gage

o o

o o o o o

Lines of equal rainfall (Isohyets)

a1”

o

a2”

o o o

a3” a4”

a5”

The method of calculation is similar to that of Thiessen method except the area is the one bounded by two isohyets. 13

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Illustrative Example:

80 mm 90 mm

100 mm 110 mm

100 mm 80 mm

Average PPT

Area

Product

87*

120

10440

95

156

14820

105

190

19950

105

140

14700

95*

60

5700

Σ 666

Ave PPT = 65610/ 666 = 98.51 inch

Σ 65610

Adjusted values

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King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

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

j=1

i=1

i=1

P = 1/A ∑ Aj ( ∑ dij –b )-1 ∑ d ij –b P i Where: m is the number of the subareas, 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

3

4

d2,18

5 O d5,18 12

6

7

13

14

8

9

10

11

15

16

17

18

19

20

21

23

24

25

26 O d26,18

27

28

22 O d22,18

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: X = 3 km 1

d= 3 km

X = 3 km 2 X Unknown PPT

P = 0.2 “

P2 =

Y = 1.5 km X = 3 km d = 4.5 km 3 P = 0.15”

∑ 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

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Dr. Rashid Allayla

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 = ∑ w i Pi i=1

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

Gage A B C D X

Average Annual Rainfall (cm) 32 28 25 35 26

Total Annual Rainfall (cm) 2.2 2.0 2.0 2.4 ?

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

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

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

i=1

i=1

Px = ∑wi . Pi / ∑ wi

Illustrative Example: Point PPT

PPT

∆Y

∆X

A

??

0

0

-

++

++

B

1.6

2

4

20-

0.05

0.08

C

1.8

6

1

37

0.027

0.49

D

1.5

2

3

13

0.0769

0.1154

E

2.0

3

3

18

0.056

0.1112

F

1.7

2

2

8

0.125

0.2125

* D2 = ΔX2 +ΔY2

(D2)*

W=1/D2

∑ 0.3345

PxW

0.5677

∆x xC xD

∆y

xB

A xF xE

PA = 0.5677 / 0.3345 = 1.7

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:

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Station A (mm) PA Part 1

Part 2

PA2

2000 1995

PA1

1990 1980 1975

1985 PB1

PB2

PB

12 Station Mean PB (mm)

Method of Correction: (READ ONLY) As shown in the figure, the slope of the Double Mass Curve changed abruptly from m 1 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.

Adjustment of segment 1

Adjustment of segment 2 m2

m2 P1

m1

p1 = (m2 / m1) P1

m1 P2

p1 = (m1 / m2) P2

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King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

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

1979 80 81 82 83 84 85 86

22 21 27 25 19 24 17 21

26 26 31 29 22 25 19 22

23 25 28 29 23 26 20 23

28 33 38 31 24 28 22 26

Total ∑E+F+G 71 72 86 83 64 75 56 66

Cumulative E+F+G 71 143 229 312 376 451 507 573

Cumulative H 28 61 99 130 154 182 204 230

h 24.7 29.1 33.5

∞

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King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

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.

Dry air

Initial condition: Dry air causing evaporation

O

O

O

As evaporation continues, pressure of air increases due to pressure of vapor (note: If pressure of dry air is P1 and pressure of indicator is P2, the vapor pressure is P2 – P1

O O O O O O O

When vapor pressure = surface vapor pressure, saturation occurs.

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 = 0.03 Hg 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

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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: a) 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.

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

b) Empirical Formulas: Using mass transfer, evaporation takes the following general form: E = C f(u)(e –ea) Where K is constant, f(u) is a function of wind speed at a given height, e is the actual vapor pressure at a given height, and es is the saturated vapor pressure at water surface, The rate of evaporation from a lake can be calculated using empirical laws, E = C (es – e) (1 + W/10)

……………………… Meyer (1944)

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

………………. Dunne ((1978)

The relative humidity in % Vapor pressure of air (mill bars) Wind speed 2 m above water in km/day

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Illustrative Example: Use Meyer formula and Dunn formula to 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, e s (@63oF= air temp) = 0.58 in. Hg es (@87oF = water temp) = 1.29 in. Hg

E = C (es – e) (1 + W/10)

e = 1.29 x 0.20 = 0.26 in Hg / (0.03 Hg/mb) = 8.7 mb For open water, C = 0.36 then E = 0.36 (0.58 – 0.26) [1+10/10 = 0.23 in/day -Using Dunne’s formula:

E = (0.013 + 0.00016 u2) e [(100 – Rh) / 100] Some Empirical Equations:

Converting wind speed to km/day = (10 mph) (24hr/d) (1.6 km/mi) = 384 km/d E = [0.013 + (0.00016 x 384)] (8.7) [(100-20)/100] = 0.518 cm/day or = 0.204 in/day which is comparable to the previous value.

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Author

Equation

Dalton

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

Meyer

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

Horton

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

Penman

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

Dr. Rashid Allayla

Explanation C=15 for small, shallow water and 11 for large deep water ea measured 30 above ground surface

Harbeck

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

u = speed of wind u2 = wind speed 2 meters above surface 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 E L 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.

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NO SCALE Elev. above Sea level (ft)

αP Pan Evaporation (in/day) Pan wind movement (m/d)

400 100 50 0

Finish Start EL (in/d)

αP Pan water temperature T0 - Ta in O F

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Infiltration: Infiltration: is a process of water entry into a soil through surface it is influenced by the extent of vegetal cover, conditions of soil, rainfall intensity, physical properties of soil and quality of water. Percolation: is the movement of water within soil profile, a process that follows infiltration. Infiltration Rate: is the rate at which water enters soil measured in L/T. Cumulative Infiltration: is the 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

Storm Conditions: - Rainfall intensity - Rainfall duration - Evaporation Transpiration

Underground Conditions: -Transmission characteristics underlying soil -Volume of storage available below ground

of

Soil Surface Conditions: - Vegetation Covers - Compaction/cracks/ - Degree of urbanization

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. Infiltration capacity is measured using infiltrometers which is a device consisting of ring tube inserted into the soil. Water is poured and maintained at constant level and replenishment rate is measured. A better method for measuring infiltration rate is through the use of hydrograph analysis. Hydrographs will be discussed in detail later.

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King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

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. At the beginning of rainfall, the soil surface possesses a high degree of soil absorption capacity. At this point, the infiltration capacity is maximum. As more water enters the soil, the infiltration capacity is reduced with time and eventually reaching a constant value.

Initial Initial infiltration infiltration capacity capacity (Very hungry for dinner!)

f f

Rain (FOOD!) Rain

Final infiltration capacity Final infiltration capacity (Cant eat anymore but can Horton Curve munch forever!)

(L/T) (L/T)

Horton Curve) Amount of infiltrated water

time

time

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 always “ponded” Horton equation requires evaluation of f0, f∞ and K .which can be derived from infiltration tests

27

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

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 the capacity f (see figure below). Depression Storage Hyetographs of rain intensity vs. time

Horton Curve

f = f∞ + (f0 – f∞) e-Kt

-

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 (such as porosity and pore-size distribution), moisture content, vegetative cover, 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 yields: f = 10 + (50) e-0.4t Integrating yields: V = ∫[10 + 50 e-0.4t] dt V = [10t - (50/0.4)e-0.4t│0 to 10 = [(10)(10) - (50/0.4)e-4] – [ 0 - (50/.4) (1) = 222.7cm Volume in = (222.7 / 100 cm/m)100 km) (1000 m/km)2 = 2.227 x 108 m3 28

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

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) Infiltration capacity decreases even though rain intensity is low.

Problem!!

‫مشكله‬

τ 0

t1

Equivalent time τ for the actual Accumulated infiltration

The accumulated water, which is found here by integrating Horton formula between 0 and t1 , does not represent the actual infiltrated water because the actual rain is less than what Horton formula predicted. ∆t

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:

29

King Fahd University of Petroleum & Minerals

If:

Dr. Rashid Allayla

1) i . dt ≥ ∆F, then τ = ∆t

Then: f0 (new) = f∞ + (f0 – f∞) e - K t

where i is the rain intensity between t to ∆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)

f f Vs F (mass) curve

Horton Curve F@ τ

f1

Volume under curve = F

Fig A

F1

τ

Fig C t

Fig B

revised Horton curve

f1

τ

F

Fig D

t Actual infiltrated water (total rain @ t)

30

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Illustrative Example: (Example 2.14 Kupta) Given: A storm pattern shown below on a watershed that has the following elements of Horton Curve: f0 = 11.66 in/hr, h∞ = 0.83 and k = 0.07 Required: Find the revised Capacity curve and the excess rain. T, minutes 0-10 10-20 20-30 30-40 40-50 50-60 60-70

Intensity (in/hr) 3.5 3.0 8.0 5.0 1.5 2.4 1.5

Solution: The infiltration capacity curve (Horton Curve); f = 0.83 + 10.83 e-0.07t Steps: 1) Find the cumulative infiltration under Horton Curve by finding the area under Horton curve. This is accomplished by multiplying ∆t (column 3) by average f (column 4), finding ∆F (column 5) and adding the cumulative values of ∆F ’s (column 6):

Time (minutes) (1) 0

f (in/hr) (2) 11.66

10

6.21

20

2.5

30

2.16

40

1.49

50

1.16

60

0.99

70

0.91

∆t (min) (3)

Average f (4)

∆F (in)(1) (5)

Σ∆F (in) (6)

10

8.94

1.49

1.49

10

4.86

0.81

2.3

10

2.83

0.47

2.77

10

1.83

0.31

3.08

10

1.33

0.22

3.30

10

1.08

0.18

3.48

10

0.95

0.16

3.64

(1) ∆F = (10)/(60 min/hr) [8.94) = 1.49

31

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

2) During the first 20 minutes, the amount of rain fell is < the infiltration capacity as prescribed by the Horton Curve. The amount of rain is: = 3.5 (10 min /60 m/h) + 3 (10 min / 60 min/hr) = 1.08 in Plot accumulated infiltration F against f to establish mass curve. For example plot f = 11.66 correspond to F = 0, f = 6.21 correspond to F = 1.49 and so on. The resulting cumulative infiltration curve is shown below: f 12 10 7.5 8 6 4 2 0

1

2

3

4

F 3) Horton Curve can now be shifted by setting f0 = 7.5 instead of 11.66 occurring at t’ = 0 where t’ is the time counted 20 minutes after the start of the rain. The modified Horton equation become: f = 0.83 + (7.5 - 0.83) e-0.07 t’

f @ t’ = 20

4) Using the revised equation, the infiltration capacity for different times is calculated as shown in the following table: t’ (min) (1) 0 10 20 30 40 50

Time from beginning of storm

(2) 20 30 40 50 60 70

f using t’ (3) 7.5 4.14 2.47 1.65 1.24 1.03

32

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

5) Compute the excess rain as follows: Time (min) (1)

Revised Curve (in/hr) (2)

∆t (min) (3)

Ave f (in/hr) (4)

Σ ∆F(1) (in) (5)

Rainfall intensity i (6)

(i) (∆t) (2) (in) (7)

Excess Rain (3) (8)

0 10 20

7.5

-

-

-

-

-

-

10

5.82

0.97

8.0

1.33

0.36

10

3.31

0.55

5.0

0.83

0.28

10

2.06

0.34

1.5

0.25

-0.09

10

1.45

0.24

2.4

0.4

0.16 – 0.09 =0.07

10

1.13

0.19

1.5

0.25

0.06

30

4.14

40

2.47

50

1.65

60

1.24

70

1.03

(1): ∆F = (10/60) [5.82] = 0.97 (2): (8)(10/60) = 1.33 (3): (8-5.82) (10/60) =0.36 12 10

Horton Curve

9

Excess Rain

8 7

Modified Horton Curve

6 5 4

3.5

3

3.0

2 1

0

10

20

30

40

50

60

70

33

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

PHI INDEX, Φ: A simple way of approximating the amount of direct surface runoff is through the use of ΦIndex. It is defined as the average rainfall intensity above which the volume of excess rain equals direct runoff. The value of Φ is adjusted (up or down) such that the computed direct runoff equal that of excess rain.

Method of Estimating Φ: a) b) c) d) e) f)

Estimate the volume of direct runoff (Vo) Plot rainfall intensity vs. time. Make initial value of Φ. Compute the excess rain volume above Φ (V). Check if V=Vo. Adjust Φ until the volume until V=Vo.

Unfortunately the Φ-Index determined from a single storm is not generally applicable to other storms.

Runoff: 34

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Surface runoff is a term used to describe the flow of water, from rain, snowmelt, or other sources, over the land surface. Runoff is a major component of the water cycle Measurement of Streamflow: Measuring Stream Gage: Stream gage is an automated equipment housed in the an enclosed gauging station where stream stage can be continuously monitored and reported to high accuracy. The equipment is powered by linking battery-powered stage recorders with satellite radios that enable transmission of stage data to computers. Approximate this segment of Channel as rectangle, measure the velocities @ y = 0.2 and 0.8 of the depth and multiply the average by the cross sectional area of the segment Bottom of channel

From USGS

Measurement of Discharge: a) Floating Devices, Current Meter (From USGS) The USGS Type AA current meter is commonly known as the Price-type current meter. This current meter is suspended in the water using a cable with sounding weight or wading rod (taking the tail section off) and will accurately measure streamflow velocities from 0.1 to 25 feet per second (0.025 to 7.6 meters per second). The main features of this meter are the uniquely designed bucket wheel shaft bearings and the two post contact chamber. The bucket wheel has six conical shaped cups, is five inches in diameter.

From USGS

b) Weirs:

35

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Rectangular Weir: 1 Equation:

Kinetic Energy is Neglected

2 H

L h h

H + V21/2g = h + V22/2g V2 = SQRT[2g(H-h)]

H

Flow through dh of area Ldh is: dA = L dh dQ=Ldh SQRT[2g(H-h)] Integrating from 0→H

Q = 2/3 Cd L (2g)1/2 H3/2 Where: Cd: Discharge coefficient

V-Notch Weir: θ H

dA = b dh dA = 2h tan θ/2 dh Since: V2 = SQRT[2g(H-h)] dQ = Cd (2h tan θ/2 dh) SQRT[2g(H-h)]

h

H H

θ

dA = b dh

Integrating from 0→H

Q = 8/15 Cd (2g)1/2 tan θ / 2 H5/2 Where: Cd: Discharge coefficient and θ is the V-notch angle

Broad - Crested Weir: 36

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

H = h + v2/2g Q = Cd L h SQRT[2g(H-h)] Q = Cd √2g L (2/3 H) SQRT[H-2/3 h) Then:

h

h

H

h H

Q = 0.385 Cd (2g)1/2 b H3/2 Where: Cd: Discharge coefficient

L h

Overflow Spillways:

+x Ho +y

Q = 2/3 Cd L (2g)1/2 Ho3/2

.

Where: Cd: Discharge coefficient

37

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Stage-Discharge Relations (Rating Curves) (READ ONLY) Stage-discharge relation is a plot of on arithmetic graph with discharge on the horizontal axis and the corresponding gauge height on the vertical axis. The relation of stage to discharge is not always unique relationship. The rise and the fall of flood waves and the backwater development caused by intersecting streams and other forms disturbances may effect the slope of the energy line which in turn effects the discharge. Therefore, in order to obtain a correct correlation between stage and discharge, an auxiliary stage near the main station is required. Constant Fall Rating Curve: If the slope of the energy line is approximately the same as the slope of the water surface, the discharge is proportional to the square root of the water surface (Manning & Chezy Equations): Q = f ( S1/2 ) The ratio of any two discharges Q and Q0 at a given station corresponding to the same stage but different slopes S and S0, then, Q/Q0 = (S/S0)1/2 From above, it is evident that if we establish an auxiliary stage downstream from the main gage, then: Q/Q0 = (F/F0)m Where F0 is constant fall that corresponds to Q0, m is constant between 0.4 and 0.6.

V21 /2g

F

Energy Line Water Surface

V22 2g

Datum Bottom of Channel

y

Methodology: a) Make series of current meter measurements to calculate y Vs.Q corresponding to fall F. b) By interpolation, draw a curve that corresponds to F=1. This is the boundary that separates F/F0 > 0 from F/F0 < 0 values. 38

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

c) It is now possible to Find Q0 from reading base stage. d) Plot correction curve Q/Q0 vs. F/F0. e) By measuring Q and F for a given base stage, find the corresponding Q0 from the correction curve. f) Plot the data Q/Q0 vs. F/F0 on log – log scale to obtain m. g) The above equation can now be used to obtain any value of Q with the help of simultaneous measurement at the base gage and the auxiliary gage. Gage Height (Stage)

F/F0 > 0

1.5 1.2

0.5

F/F0 < 0

0.7 1.3 1.4

0.4

Curve corresponds to F=1

0.3

1.0 1.1

0.8 Q/Q0

0.6

Q

F/ F0

Q0

Discharge → log Q/Q0

m 1

log F/F0

39

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Illustrative Example: For a given river, the following data were obtained by stream gauging. Estimate the flow rate when the main gage reads 30.0 ft and auxiliary gage reads 28.0 ft. Main Stage, ft

Auxiliary Stage

Q (cfs)

30

29.0

250

30

27

470

Solution: Q1 / Q2 = [(F1/L) / (F2/L)]m 250/470 = [(30-29)/(30-27)] m → 0.5319 = 0.3333 m → m = log (0.5319) / log (0.3333) = 0.5746 250/Q3 = (1/2)0.5746 → Q3 = 250 / 0.5 0.5746 = 372.32 cfs

40

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Rainfall Runoff Relationships: Hydrograph Analysis: A hydrograph is a graphical representation of discharge verses time. It describes the changes of flow rate in a given catchments area following a rain event. The hydrograph is constructed from monitoring stream stages and the stage-discharge relationship of the stream located at the outlet of the watershed in question. Rain

Basin

Outlet Q

T Rising limb: f (rain, basin)

Crest: All water reached outlet

Drainage from underground

Components of Hydrograph: -

Surface Runoff: It is the component of rain, which flows overland eventually reaching stream or channels. Interflow: It is the component of rain, which moves at shallow depths underground eventually reaching stream channels. Channel Precipitation: It is the component of rain, which falls directly on the stream. Base flow: It is part of stream water that is contributed by groundwater and delayed subsurface runoff.

The part of precipitation that contributes directly to direct runoff is often referred to as effective runoff. The time to peak of hydrograph depends on rain characteristics, which include intensity and duration, and characteristics of watershed, which include size, slope, shape and storage capacity. The last segment of the hydrograph is the recession curve which depends on the characteristics of the watershed and independent of the rain characteristics.

41

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Storm Duration

Point of inflection

Direct R.O. i

Recession Curve (independent of rain characteristics

Q The Crest

Hydrograph here represents period of no direct runoff and water coming into stream reflects discharge from groundwater. The shape depends characteristics of the watershed. Groundwater depletion

Rising limb Crest (Concentration Curve)

Beginning of new storm Channel PPT

Interflow Groundwater flow

t

Time of concentration

Time Base

Physical Analogy:

Direct R.O. Channel PPT Interflow

GW Flow

Watershed

Q

42

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Base Flow Separation: When a hydrograph of a particular storm is constructed, it becomes necessary to separate the hydrograph into at least two components, namely base flow and direct runoff. Separation of base flow from direct runoff is not always an easy task because, following a large storm, the river stage rises rapidly at a rate faster than the rise of water table, as a consequence, the flow reverses direction. As the stream stage starts to fall, the water from the banks start to drain back into the stream.

After Storm

Stream Stage Rising Stream Stage Falling

Base Flow = + Before Storm

Negative base flow

Separation Techniques: -

Straight-line separation: It is the simplest base flow separation. The separator is a horizontal line drawn from the point of the start of the direct runoff to a point in the recession curve. At this point, it is assumed that the base flow starts and the direct runoff ends.

-

Fixed arc-length Separation: This procedure extends the recession curve of the

previous rain to a point directly below the peak then a straight line is drawn to a point N days after the peak where N is calculated as, N = A0.2 N is the time in days and A is the drainage area in square miles. The reason for extending the recession curve and, in effect, reducing the base flow is that contribution from groundwater is reduced considerably due to the bank storage effect. Hydrograph Time Relationships: Travel time: The time it takes for the direct runoff originating at a point in the channel inside the watershed to reach the outlet of the watershed. Access –rainfall release time (time of concentration): The time it takes for the last drop of rain to

reach the outlet of the watershed. It is the same as the time difference between the end of rain and the cessation of direct runoff. Time of concentration is the same for a given watershed regardless of the characteristic of the storm. This is the bases for constructing unit hydrographs.

43

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Time base: It is the time between the beginning and end of direct runoff. Ideally, time base is

the same if rain characteristics are the same for a given catchment area. The equation for time base is, tb = ts + tc Where tb is the time base, ts is the duration of rain and tc is the time of concentration. Example: Shown is a hydrograph of a particular rain in a particular basin. Separate the direct runoff using straight-line separation and Arc length separation and find the direct runoff. Assume the drainage area = 1000 mi2. Solution: N = (1000)0.2 = 3.98 days Discharge m3/s 90 80 70 60 50 40 30 20 10

3.98 d 0

0

2

Time : Q: Base Flow (Straight.Line): D.R.O (Str.Line) : Base (Arc L): D.R.O. (Arc L )

4

4 30 30 0 30 0

6 45 30 15 25 20

6

8 70 30 40 23 47

8 t Day 10

10 84 30 54 20 64

12 74 30 44 35 39

14 50 30 20 50 30

12

14

16

18

16 30 30 0 -

Vol. of D.R.O (SL) = {(1/2) [(0 + 15) + (15 + 40) + (40 + 54) + (54 + 44) + (44 + 20) + (20 + 0)}(2 dayx86400 s/d) = 2.989 x 10 7 m3 Amount of direct runoff in meters = 2.989 x 107 / (1000 mi2 x 2590000 m2 /mi2 ) = 0.012 m Volume of D.R.O. (AL) = {(1/2) [(0 + 20) + (20 + 47) + (47 + 64) + (64 + 39) + (39 + 30)} (2 day x 86400) = 3.1968 x 107 m3 Amount of direct runoff in meters = 0.01234 m

44

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

The Unit Hydrograph: Observation has proven that, since the physical characteristics of a watershed (such as shape, size, slope and topography) remain unchanged over different cycles of rainstorms, one would expect that the resulting hydrographs produced by similar storms would be similar. It is therefore, safe to assume that, for a given watershed, “different storms of similar duration would produce the same hydrograph base and the resulting peak flows would vary directly with the rain intensity and the volume of direct runoff”. Sherman (1932) introduced the concept of unit hydrograph. He defined it as the hydrograph of a given watershed resulting from effective rainfall producing 1 inch of direct runoff distributed uniformly over the basin. The assumptions inherent in the unit hydrograph theory are: -

-

There is direct proportionality between effective rainfall and surface runoff. An effective rainfall of two units of T duration would produce twice the runoff of one unit rainfall of the same duration. The effective rainfall is uniformly distributed within its duration. The effective rainfall is uniformly distributed throughout the basin. Once a unit hydrograph for a catchment area is derived, it can be used as a base to represent the response of the catchment area to different storms. If two successive rains of T duration produce two different direct runoffs, then the hydrograph produced is the sum of the two runoff producing hydrographs of 2T duration. The ordinates of direct runoff of common-base hydrographs are directly proportional to the amount of direct runoff represented by each hydrograph.

There are inherent weaknesses in the unit hydrograph theory as well. It is obvious that if the catchment area is saturated prior to the rain, much of the rain will contribute to direct runoff and any extra rainfall in the same time period will produce proportionally extra runoff. This is obviously not the case for catchment areas of high initial moisture deficiency. Another weakness in the theory pertains to the assumption of uniform time and areal distribution of rain in the basin. It is obvious that this assumption fails for both complex rains and for catchment areas having nonhomogeneous composition. Derivation of Unit Hydrograph: Before proceeding to develop representative unit hydrograph for a certain catchment area, it is essential that the unit hydrograph be a product of many rainfallproducing hydrographs resulting from various storms of the same duration and different magnitudes, rather than the product of a single storm. This is necessary because, for reasons previously mentioned, similar rains may not necessarily produce similar hydrographs. In addition, the intensity and distribution of rainfall over the catchment area must be uniform and simple.

45

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Steps: - Separate direct runoff from groundwater using one of the methods described earlier. - Measure the total volume direct runoff Q

QTi

Volume of direct runoff :

QBi

= ∑ [ (QTi+1 + QTi) / 2 - (QBi+1 + QBi) / 2 ] x ∆t = VRO = Area Under Curve (R.O.) Δt

- Divide the ordinates of the total runoff (QT - QB) by the total volume of direct runoff measured above to obtain the ordinates of the unit hydrograph (QUH = QS / VRO). Where VRO is Volume of runoff measured in (L) and

t

QS = QT - QB -The effective duration of the unit hydrograph is the same as the original hydrograph.

It is important to note that the unit hydrograph represents a given basin and rain of specific duration. Since most storms do not have the same characteristics of the unit hydrograph, it is necessary to construct another unit hydrograph having the same rain characteristics before any attempt to construct the hydrograph. Construction of Unit Hydrographs: Example: Construct a unit hydrograph from the following hydrograph. The area is 2 mi 2. Use straight-line separation and construct a hydrograph representing the following complex rain pattern. Q

i

2 hr effective rain

NO SCALE!

400

Losses =0.1 1.2” 2.2” 1.8” NO SCALE!

300 Q cfs 200 100

0

1

2

3

4

5

6

t (hrs)

Q (cfs)

Base Flow (cfs)

1 2 3 4 5 6 7 8

75 110 205 305 280 205 130 75

75 75 75 75 75 75 75 75

7

t ( hr)

8

t ( hr) 2

4

6

D.R.O. (cfs) 0 35 130 230 205 130 55 0

46

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Volume of direct runoff = (1/2) [(0+35) + (35+130) + (130+230) + (230+205) + (205+130) + (130+55) + (55+0)] (3600 s/hr) = 2.826 x 106 ft3 Effective runoff in inches = (2.826 x 106 ft3) (12 inches / ft) / [(2 mi2) (5280 ft/mi)2] = 0.61 inches Now calculate the ordinates of U.H. by dividing the ordinates of the hydrograph by 0.61” t (hrs)

D.R.O. (cfs) x 1000 0 35 130 230 205 130 55 0

1 2 3 4 5 6 7 8

QU.H = Q/0.61 0 57.4 213.1 377.0 336.1 213.1 90.2 0

Ordinates of the 2-hr unit hydrograph

The hydrograph of the complex rain is found by multiplying the ordinates of U.H. by the effective rain (rainfall minus the losses) and shifting the resulting hydrographs by two hours and adding ordinates. t (hrs)

QU.H

QUH x 1.1

0 1 2 3 4 5 6 7 8 9 10 11

0 57.4 213.1 377.0 336.1 213.1 90.2 0 0 0 0

QUH x 2.1 0 63.1 234.4 414.7 369.7 234.4 99.2 0 0 0 0

QUH x 1.8

Hydrograph Ordinates

0 120.5 447.5 791.7 705.8 447.5 189.4 0 0

0 103.3 383.6 678.6 605.0 383.6 162.4

0 63.1 234.4 535.2 817.2 1129.4 1089.4 1126.1 794.4 383.6 162.4 0

Hydrograph resulting from the complex storm

No Scale

Important Note: Add base flow (75 cfs) to ordinates to obtain full hydrograph.

NO SCALE! Final hydrograph without base flow

NO SCALE! UH

U.H. x 1.1 U.H. x 2.1 U.H. x 1.8

0

2

4 ….

47

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Conversion of Small T Hydrograph to larger T Unit Hydrograph: Given: t-hr U.H. Required: T-hr unit hydrograph (T > t ) Method: Lag unit t-hr unit hydrograph by t to obtain T-hr hydrograph.

Add t-hr unit hydrograph to another t-hr unit hydrograph but shifted by t-hours. The two unit hydrographs will produce a hydrograph of 2t duration totaling two inches of rain. The new hydrograph represents rain intensity of 1/t (because intensity is 2”/2t). The new 2t unit hydrograph is obtained by dividing ordinates by 2. Resulting 4-hr hydrograph 2-hr U.H.

Two 2-hr U.H. 2 hours shift t

Given: T-hr U.H. Required: t-hr unit hydrograph (T > t ) Method: Use S-Curve described below.

S curve: S-Curve is a hydrograph resulting from addition of infinite series of unit hydrographs of specific duration, each lagged by its duration t. It is only necessary to add a limited number of unit hydrographs to reach equilibrium of constant direct runoff discharge. If T is the time base of the unit hydrograph and t is the duration then only T/t unit hydrographs need to combined to produce equilibrium. I x t = 1 inch

Q

T-hr S Curve

ΔQ

Q T-hr U.H.’s

Ordinates of U.H. of thr duration = (T/t) ΔQ

t

T

t

48

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Example on the Use of S-Curve to Obtain Unit Hydrograph of Different Duration: The following unit hydrograph results from 2–hr storm. Determine the 1-hr unit hydrograph. Time (hr): Q (m3/s): 0

0 0

1 1.42

2 8.5

3 11.3

4 5.66

5 1.45

6 0

To construct S-Curve follow the following schematic calculations: Given: 2-hr unit hydrograph Required: 2-hr S-curve Time (hr) 0 1 2 3 4 5 6 7 8 9 10 11

2-hr U.H. 0 a b c d e f g h i j k

S-Curve additions A = (a) B= (b) C = (c + A) D = (d + B) E = (e + C) F = (f + D) G H I = (i + G)

Inserting numbers: Time (hr) 2-hr U.H. 0 0 1 1.42 2 8.5 3 11.3 4 5.66 5 1.45 6 0 7

2-hr S-Curve a b c+a → d + (B=b) = d + b → e + (C=c+a) = e + c + a → f + (D=d+b) = f + d + b → g + (E=e+c+a) = g + e + c + a → h + (F=f+d+b) = h + f + d + b →

S-Curve additions = (1.42) = (8.5) = (11.3+1.42) = (5.66+8.5) = (1.45+11.3+1.42)

k+i+G

2-hr S-Curve = 1.42 = 8.5 11.3+1.42 = 12.72 5.66+8.5 = 14.16 1.45+11.3+1.42 = 14.17 0+5.66+8.5 = 14.16 1.45+11.3+1.42 = 14.17

The 1 hour unit hydrograph is calculated by lagging the 2 – hr S-Curve by 1 hour and multiplying ordinates by 2/1 (old duration divided by new duration) Time

2-hr S-Curve

(1) 0 1 2 3 4 5

(2) 0 1.42 8.5 12.72 14.16

Lagged S-Curve (3) 0 1.42 8.5 12.72

14.17

∆S (4) 0 1.42 7.08 4.22 1.44 14.16

1-hr UH = (Column 4) (2/1) 0 2.84 14.16 8.44 2.88

Multiply ΔS by the factor (2/1) to obtain 1 hr UH 0

0

49

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Flood Routing: Flood routing is a “bookkeeping” technique that uses continuity to predict temporal and spatial variation through river reach or a reservoir. It is a mathematical description of the behavior of the flood waves that move through a point along the stream. The outcome of the routing is a discharge hydrograph Q = Q(t) measured at a prescribed point along a stream from a known hydrograph upstream or downstream and a known characteristics of the stream. Waves are generated as a result of inflow or outflow into the channel as a result of rain, channel failure, tides or releases from reservoirs.

Event Q = Q(t) Inflow

Q = Q(t) Outflow

Sudden opening of sluice gate Upstream flow wave

Downstream flow wave

Storage Routing: The rate of change of storage is defined by the continuity equation: Sudden closure of sluice gate

Reservoir Routing: Referring to the figure below, the inflow and outflow are plotted in the same graph. Area A represents the volume of water that is available in the reservoir at time t 1. At t > t1 , outflow exceeds inflow and the reservoir empties by an amount equal to B. Inflow

outflow

(Inflow Volume) A Storage in channel

Reservoir

(outflow volume) B Water released I = O here Peak of outflow

The rate of change in storage is defined by the equation: I – O = ∂S / ∂t 50

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

If the average rate of flow during a given time period is equal to the average flow at the beginning and the end of the period (i.e. short time period), the routing equation takes the following form: ( I1 + I2 ) ∆t - ( O1 + O2 ) ∆t = s2 – s1 2 2 Where subscripts 1 and 2 refer to the beginning and end of the time period. In the above equation all elements are known except O2 and s2 and a second equation is needed in order to solve for O and s. Routing procedure: For a large reservoir where water velocity is low, the pool level of the reservoir is near horizontal which allows a relationship between upstream head H and weir geometry to be established. The form of equation which was discussed earlier is: O = CLH 3/2 Where C is the weir coefficient, H is the upstream depth and L is the width of the weir. The active storage can be found by measuring the height of the reservoir and the planimetering the reservoir surface area from topographic maps. The storage and the outflow can now be established as shown in the following graph: Spillway Discharge O = CLH 3/2 →

s Vs. H (Measured) O Vs. H Calculated From Spillway formula

H Reservoir volume s →

The routing equation can be written as: (I 1 + I 2) + (2s 1 / ∆t – O 1) = (2s 2 / ∆t + O 2) Solution Method:          

Establish 2s / ∆t + O as a function of H from H vs. s data and from the spillway discharge formula O = CH3/2. At the beginning of routing period, all values in the left-hand side are known and the values at the right-hand side are computed. From the computed value of 2s/∆t + O, a new value of H is interpreted from 2s/∆t + O vs. H data. From the computed value of 2s/∆t + O, and O, find s. Using Spillway formula, find O from H. From the new values of s and O find 2s/∆t – O. Add (I1 + I2) and 2s/∆t – O to obtain a new value of 2s/∆t + O. Repeat step 3-7 to find new values of s, O, H, 2s/∆t – O and 2s/∆t + O. Obtain a new value of s from the calculated value of 2s/∆t + O curve established earlier. From the values of 2s/∆t + O and O, a new value of s is calculated.

51

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Illustrative Example: The inflow hydrograph of a given reservoir is as follows: I (cfs) 200 ( cfs) 150

100

50

00

1.0

2.0

3.0 Time (day)

Time (days):

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

Flow (cfs):

0

50

100

150

200

100

66.67

33.33

0.00

The crest discharge formula is described by the following formula: O = C L H3/2 Where: O is the outflow, L is the length and H is the head above the crest. The characteristics of the spillway are as follows: C = 3, L (Length of the spillway) = 35 ft and H0 (Initial crest height of the spillway) = 50 ft. Stage Vs. Storage is as follows: Stage (ft) 208.75 52.0

175.00 141.25

51.5

(ft)

107.50

51.0 73.70

50.5

Storage ( x 106 ft3)

.

50

0

50.0 50

50

100

150

200

52

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Step 1: Calculate outflow O and 2s/∆t + O for ∆t = 0.5 day Table 1: Water Surface El.

Head H (1)

(ft) (1)

(ft) (2)

50.0 50.5 51.0 51.5 52.0 52.5

0.0 0.5 1.0 1.5 2.0 2.5

Storage(From Data)

x (3)

Outflow (2) (cfs) (4)

2s/∆t + O (cfs) (5)

40.00 73.70 107.50 141.25 175.00 208.75

0.00 37.1 105.0 192.0 297.0 415.00

160.00 331.9 535.00 757.00 997.00 1250

(ft3

106)

O = C L H3/2

(1) (2)

Elevation above the crest From spillway formula

Step 2: Plot O vs. Stage from values obtained in column 4:

0

50

100

150

200

250

300

350

400

450

500

Step 2: Conduct Flood Routing Computation to Time Vs. Elevation / Storage: Table 2: Time (Day)

Inflow (cfs)

--0 0.5 1.0 1.5 2.0 2.5 3.0

0 0 50 100 150 200 100 0

(1)

(2)

I1 + I2 (cfs) (3) 0 50 150 250 350 300 100 0

From: H = 0.55 → O = CLH3/2 = 42.83

Outflow (cfs) (4) 0 0 6.10 42.83 112.97 196.77 185.23

Storage, s (ft3 x 106) (5) 40 40 52.5 86.95 128.04 159.05 135.67

2s1/∆t – O1 (cfs) (6) 160 160 203.9 304.97 399.17 439.43 357.43

2s2/∆t + O2 (cfs) (7) 160 210 353.90 554.97 749.17 739.43 457.43

Water Elevation

(ft) (8) 50.0 50.15 50.55 51.05 51.52 51.46 50.81

From: (I1+I2) + (2s/∆t-O1)→(150) + 203.9) = 353.9 From Table 1 @ 353.9 cfs From: 2s/∆t + 6.1 = 353.9 → s = 86.95

53

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Hydrographs of Basin Outflow: The Rational Method: In some engineering designs such as design storm water-sewer system, estimates of peak flow rate from a small basin is required. Hydrologic methods including the rational method is employed to estimate peak flow used for such designs. The underlying assumption of the rational method is that a catchment area has a specific time of concentration which is defined as the time needed for water to travel from the most remote point of the area to travel through the outlet. The basic equation of the rational method is, Q = CiA Where: Q is the peak flow, C is runoff (rational) coefficient and A is the area of the watershed. Note that if Q is in m3 / s, i in mm / hr, A in km2, the equation should be written as, Q = 0.278 CiA Although the rational equation is dimensionally consistent, it yields correct values for Q in cfs, i in inches per hour and A in acres. The figure below is provided by the US Department of Transportation for frequency correction factor. The table shows some typical values of the rational coefficient C for various surface areas. Note that the runoff coefficient varies from 0 to 1 which embodies a number of variables that include rainfall duration and intensity, soil type, shape and slope of the watershed, design frequency, amount of depression storage and interception Application of Rational Method: 1) Estimate the time of concentration which is the time required for water to travel from the most remote area to reach the outlet. For combination of various routes, t c is taken as the longest time of travel to the outlet. 2) Estimate the runoff coefficient C. 3) Select a return period Tr and find the intensity of rain that will be equaled or exceeded once every Tr. This is obtained from IDF curves (Published by US National Weather Service See Figure) 4) Determine peak flow from the rational formula. This value is then used for design of storm systems.

54

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Dr. Rashid Allayla

Intensity-Duration-Frequency Curves (US National Weather Service):

Runoff coefficients for Various Areas: Business: Downtown Area Business: Neighborhood Areas

0.70 – 0.95 0.50 – 0.70

Residential: Single-family areas Multi units, detached Multi units, attached Residential (suburban) Apartment dwelling areas

0.30 - 0.50 0.40 - 0.60 0.60 - 0.70 0.25 – 0.40 0.50 – 0.70

Industrial: Light areas Industrial: Heavy areas

0.50 – 0.80 0.60 - 0.90

Park, cemeteries

0.10 – 0.25

Playgrounds

0.20 – 0.35

Railroad yard areas

0.20 – 0.40

Unimproved areas

0.10 – 0.30

55

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Illustrative Example: Find the 50 year design storm for a the following composite area: A business downtown area with C = 0.8 in the left and a park with C = 0.2 in the right. The lateral time of flow in the left portion is 10 minutes and in the right is 40 minutes. The travel time in the gutter is 8 minutes. 10 minutes

10 minutes

Business

Park

40 minutes

8 minutes

2000 ft

18 minutes

500 ft

2000 ft

From IDF chart, for 50 year design and tc = 18 minutes, i = 7 inches / hr. Assume Business area controls, then tc = 18 minutes Area of the watershed = Area of the business section + effected portion of the park Area of the business section = (500)(2000) / (43560 ft2/acre) = 22.96 acres Area of the park ={ {[(10/40) (2000) + (18/40) (2000)]/2} (2000) } / (43560) =32.14 acres Q = C1 i1 A1 + C2 i2 A2 = (0.8) (7inches/hr)(22.96 acre) + (0.3) (7) (32.14) = 196.07 cfs Assume the park controls, then: tc = 40 + 8 = 48 minutes From chart, for 50 year design and tc = 48 minutes, i = 4.8 inches / hr Area of the park = (2000) (2000)/43560 = 91.83 Q = (0.8) (4.8) (22.96) + (0.2) (4.8) (91.83) = 176.32 Since Q business > Q Park , then Q = 196.07 cfs is used for drainage design.

56

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Illustrative example: A composite residential and park areas is proposed for construction. The inlet times to the manholes for areas AA to AC are 20, 35, 41 and 53 minutes respectively. The coefficients C for areas AA to AC are 0.8, 0.3, 0.6 and 0.25 respectively. The time of travel between manhole 1 to 2 is 5 minutes, 2 to 3 is 7 minutes and between 3 and 4 is 10 minutes. Find the 10-year peak flow at each inlet that will be used for storm design.

AA = 3.3 Acres AB = 4 Acres AC = 4.5 Acres A4

AD = 5.4 Acres

4

1 2

3

Inlet 1: Contribution from area A = 20 minutes, then from IDF chart @ 10-yr storm and @ tC = 20 minutes, i A = 5.6 in/hr Q1 = CiA =(0.8) (5.6 in/hr) (3.3 acres) = 14.78 cfs Inlet 2: Contribution from area A = 20 minutes + 5 minutes = 25 minutes Contribution from area B = 35 minutes Therefore, since (tc)B > (tc)A , contribution from area B to inlet 2 controls, then from the IDF @ 10yr storm and @ tC = 35, i B = 4.6 in/hr Q2 = CiA =(0.8) (4.6 in/hr) (3.3 acres) + (0.3) (4.6 in/hr) (4 acres) = 17.66 cfs Inlet 3: Contribution from area A = 20 minutes + 5 minutes + 7 minutes = 32 minutes Contribution from area B = 35 minutes + 7 minutes = 42 minutes Contribution from area C = 41 minutes Therefore, since (tc)B > (tc)A and (tc)B > (tc)C , contribution from area B to inlet 3 controls, then from the IDF @ 10-yr storm, and @ tc = 42 minutes, i B = 4.0 in/hr Q3 = CiA =(0.8) (4 in/hr) (3.3 acre) + (0.3) (4 in/hr) (4 acres) + (0.6) (4 in/hr) (4.5 acre) = 26.16 cfs

Inlet 4: Contribution from area A = 20 minutes + 5 minutes + 7 minutes + 10 minutes = 42 minutes Contribution from area B = 35 minutes + 7 minutes + 10 minutes = 52 minutes Contribution from area C = 41 minutes + 10 minutes = 51 minutes Contribution from area D = 53 minutes Therefore, since (tC)D > (tc)A , (tc)D > (tc)B and (tc)D > (tc)C , contribution from area B to inlet 3 controls, then from the IDF @ 10-yr storm, and @ tD = 53 minutes, i B = 3.5 in/hr Q4 = CiA = (0.8) (3.5 in/hr) (3.3 acre) + (0.3) (3.5 in/hr) (4 acres) + (0.6) (3.5 in/hr) (4.5 acre) + (0.25) (3.5 in/hr) (5.4 acre) = 27.61 cfs

57

King Fahd University of Petroleum & Minerals

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Civil Engineering Dept. CE 331 Engineering Hydrology Dr. Rashid Allayla

58

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Groundwater Hydrology Introductions & Definitions: Porous Media A medium of interconnected pores that is capable of transmitting fluid. These pores often referred to as “interstices”. Interstices can be fully saturated or partially saturated. Aquifer is any porous formation that can transmit water at economic (usable) quantity. It comes from the Latin words aqua (means water) and Ferro (to bear). Classification of Aquifers: A confined aquifer also known as artesian aquifer is a formation containing transmittable quantities water that is under pressure greater than atmospheric pressure. Confined aquifer is bounded from above and from below by impermeable formations. The pressure at this type of aquifers is greater than atmospheric. Unconfined aquifer, also known as water-table aquifer is an aquifer with free water surface that is open to atmosphere. The upper surface of the zone of saturation is the water table. It is the upper portion of the saturated zone and is open to atmosphere. Unconfined aquifers that contain Aquitard and and/or Aquiclude may contain additional water tables. A piezometric surface is the contour of water elevation of wells tapping confined aquifers. The contours provide indications of the direction of groundwater flow in the aquifers. If the piezometric surface falls below the top formation of confined aquifers (as shown in the left figure next page), the aquifer at that point is a water-table aquifer and the surface becomes water-table surface. It is easy to confuse water-table surface with piezometric surface when both confined and unconfined aquifers exist on to of each other. But , in general, the two surfaces do not coincide. A piezometer is a device, which indicates the water pressure head at a point in the confined aquifer. It consists of casing that is open in both sides and fits tightly against the geological formation making up the aquifer. The height to which water rises in the piezometer is the water pressure head.

Piezometric Surface Upper Rock Formation Piezometer Direction of Flow Lower Rock Formation

Direction Of Flow

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King Fahd University of Petroleum & Minerals

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Piezometric Surface H = P/γ + Z

Piezometer

nn

Water Table P/γ = 0

P/ Confined Aquifer

Unconfined Aquifer Aquifer Z

P/γ

Porosity: It is the ratio of the volume of interconnected voids within the soil to the total volume of the soil. If VV designates the volume of “interconnected” voids, VS is the bulk volume and VT is the total volume of the soil media, the porosity θ is defined as: θ = VV / VT = (VT-VS) / VT = 1 – VS/VT Volumetric Water content: It is the ratio of the volume of water to total volume of soil sample. θW = VW / VT Degree of Saturation: It is the ratio of the volume of water to the volume of voids. The soil is said to be saturated when the degree of saturation is 100%. S = VW / VV Moisture Content: It is the ratio of the weight of water to the weight of solid. W = WW / WS Effective Porosity: It is the ratio of the volume of interconnected voids that allow free water flow to the total volume of soil sample. The value of effective porosity is always less than soil porosity because of the existence of micro voids that do not allow free flow of water. The difference between effective porosity and porosity is more pronounced in silt and clay than in medium to coarse sand. VV

Va VW VS

VT

VS

60

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

EXAMPLE: A 130 cm3 soil sample weighs 200 grams when it was at field moisture. After drying the sample, it weighed 180 grams. The volume of solid is 69 cm3 Find: 1) Volumetric water content, 2) Total porosity and 3) Degree of Saturation and 4) moisture content. 1) Volumetric water content = Volume of Water/Total Volume = Vw / V = (200-180) / 130 = 0.15 2) Total porosity Volume of voids / Total volume = VV / V = (130-69) / 130 = 0.47 3) Degree of saturation = θ / θs = (Vw / V) / (VV / V ) = 0.15 / 0.47 = 0.32 4) Moisture content = WW / WS = (200-180) / 180 = 0.111 Distribution of Pressure: The force that balances the gravity force and holds water in equilibrium is, ∂P / ∂ z = - γ Where P is the pressure and γ is the unit weight of water. Integrating, P = - γz + C Selecting water table as the datum, pressure above water table is negative and pressure below is positive. One can ask how does water stay in voids and not drain. The answer is because of Surface Tension. Specific Storage, SS and Storage coefficient, S: Specific storage is the amount of water in storage (aquifer) that is released from a unit volume of aquifer per unit decline of head. It has the dimension of (1/L). For two dimensional aquifer analysis, a more useful storage parameter is the one that integrates the storage over the depth of the aquifer. This is called storage coefficient. For confined aquifers it is defined as the amount of water released from aquifer per unit area of aquifer per unit decline of head: S = SS b

Vw/(VTx1/L)

Where b is the thickness of the aquifer. For unconfined aquifers, S is given the term “specific yield", Sy. It is defined as the amount of water released from a column of aquifer per unit area per unit decline of head. Sy is, therefore, a measure of how much water can drain away from the rock when subjected to gravity force versus how much water the rock actually holds. Since surface retention is proportional to the waterholding capacity of soil particles, the grain size of the soil plays important role in determining the magnitude of specific yield. The smaller the particle, the larger the surface area, the larger the surface tension, the less the specific yield. This is the reason why pumping from sandy aquifer yields more water than aquifers made of clay.

61

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Dr. Rashid Allayla

Unit Cross Section of Aquifer Piezometric Surface 1 Amount of head decline S

Confined Aquifer

Piezometric Surface

WT 1 WT 2 1.unit area

∆h

Sy

S

Confined Aquifer

Water-Table Aquifer

Typical values of S: S ranges from 10-4 to 10-6 and Sy ranges from 0.2 to 0.3 (Bear ‘72)

EXAMPLE: The average volume of a confined aquifer is 2.5 x 107 m3. The storage coefficient of the aquifer at a location where the thickness b = 30 m is 0.005. Estimate the volume of water recovered by reducing the pressure head by 20 meters. Using the definition of storage coefficient S as the volume of water released from storage per unit area of aquifer per unit decline of head : S = Vv / (AT)(Δh) where area AT (Area of aquifer) = 2.5 x 107 / 30 = 8.3 x 105 m2 Then Volume of water recovered Vv = S x AT x Δh =(0.005) (8.3 x 105) (20) = 8.3 x 104 m3.

62

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Groundwater Motion: Darcy’s Empirical Equation: The French engineer Henry Darcy introduced Darcy’s Law in 1856 when he was investigating flow under foundations in the French city of Dijon. The law is a generalized relationship between the flow of fluid and the change in head in porous media under saturated conditions. Darcy concluded that Q, measured in volume of water per unit time, is directly proportional to the cross-sectional area of the porous medium transmitting water, the difference in head between the entrance and the exit and inversely proportional to the length separating the points. This law is valid for any Newtonian fluid and, with some modification; it could describe flow in unsaturated zone. Darcy’s Experiment: Henry Darcy conducted an experiment to determine the relationship between discharge and head variation across a porous medium. In general, flow of pore water in soils is driven from positions of higher total head towards positions of lower total head. The level of the datum is arbitrary. It is the differences in total head that determines discharge. Introducing the hydraulic gradient as the rate of change of total head along the direction of flow. ▼h= ∆h / L

∆h hout L

hin

Q (Discharge)

Area Porous Media

Porous Ceramic

For a one dimensional flow perpendicular to the cross sectional area, Darcy found that, Q = - A K Δh / L Where the minus sign is placed in the right side because the head loss (∆h = h2 – h1) is negative and Q must be positive. In the equation, Q is the volumetric flow rate (measured in L3 / T), A is the flow area perpendicular to the flow direction (m2 or ft2), K is the hydraulic conductivity of the porous medium (measured in L / T), L is the flow path length (measured in L), h is the hydraulic

63

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head (measured in L) and ∆h is the change in h over the path L. The hydraulic head at a specific point, h is the sum of the pressure head and the elevation, or h = (p/γ + z) Darcy’s Formula in Three-dimensional Form: q = -K ▼h where ▼h (Del h) is (∂h/∂x) i + (∂h/∂y) j + (∂h/∂z) k and q is the vector velocity discharge equal to Q/A. Introducing the term intrinsic permeability, k, K=kγ/μ K=kν/g Where K is the hydraulic conductivity (L/T), k is the intrinsic permeability, γ is the unit weight of water and μ and ν are the dynamic kinematic viscosities of water. In the above formula, it is apparent that K depends on both fluid properties (γ / μ ) and soil property (k). Various formulas relate k to properties of porous medium. One of them is, k = C d2 where k is measured in cm2 and d depends on dimensional property of soil. In the above equation, C is a dimensionless constant that varies from 45 for clay sand to 140 for pure sand. From the above definition, k has the dimension L2. Intrinsic permeability pertains to the relative ease with which a porous medium can transmit a liquid under a hydraulic gradient. It is apparent from the definition of k that it is a property of the porous medium and is independent of the nature of the liquid or the potential field, whereas the hydraulic conductivity K is dependent on both soil medium (represented by k) and fluid properties (represented by viscosity μ & unit weight γ. Units: The standard unit used by hydrologists for hydraulic conductivity, K is meter per day. In laboratory, the standard unit is in gallons per day through area of a porous medium measured in ft2. In the field, hydraulic conductivity is measured as the discharge of water through a cross-area of an aquifer one foot thick and one mile wide under a hydraulic gradient of 1 ft/mile (Bear 1979). The value of k in length units is very small. One practical unit employed to represent k is Darcy, which is defined as, 1 Darcy = 9.87 x 10-9 cm2 ≡ 1.062 x 10-11 ft2

64

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Laboratory Measurement of Hydraulic Conductivity: Measurement of hydraulic conductivity through field aquifer tests provides more reliable results than in laboratory tests because, field tests involve large magnitudes of porous medium compared to small laboratory samples. Field tests will be discussed when aquifer testing method is covered. In laboratories, hydraulic conductivity is measured using Permeameters. Two commonly used permeameters are shown below; one is constant head where the difference between water levels reflects the head loss between inlet and outlet of the soil sample. In the falling-head permeameters, the rate of discharge through the sample decreases with time as the driving head decreases. The equations used for the two measurements are, a) Constant Head Permeameters:

K = QL / ∆h

Constant Head Inflow

Variable Head inflow dh

∆h h0 h(t)

L

Soil Sample

65

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

b) Falling Head Permeameters: Darcy states that: Q = - K Δh(t) / L But: Q = a dh(t) / dt a dh(t) / dt = - KA (Δh(t) /L then: dt = - a L dh(t) / [Δh(t) KA] Integrating both sides from t=0 to t=t Then: t = - (a L / KA) ln h(t)||From h0 to ht t = - (aL / KA) [ln h(t) – ln h0] t = (aL / KA) [ln h0 – ln h(t) = (aL/KA) ln[h0/ h(t)] Which gives: K = (a L / t A) ln[h0/ h(t)] Darcy’s law shown above is limited to flow that is one–dimensional and homogeneous incompressible medium. Introducing the concept of specific discharge q, Darcy’s law becomes, q = K Δh / L In the above, q is also referred to as the Darcy flux. It is fictitious form of velocity because the equation assumes that the discharge occurs throughout the cross sectional area of soil in spite of the fact that solid particles constitute a major portion of the cross sectional area. The portion of the area available to flow is equal to θA, where θ is the porosity (as defined as the volume of pore relative to the total volume of the medium). The average “real” velocity is, therefore is, v=q/θ

V

q

The energy loss Δh is due to the friction between the moving water and the walls of the solid. The hydraulic gradient is simply the difference between the head at inlet and head at the outlet divided by the distance between the inlet and the outlet. Note that the flow prescribed by Darcy’s law, 1) 2)

Takes place from higher head to lower head and not necessarily from higher pressure to lower one. Darcy’s law specifies linear relationship between velocity and hydraulic head. This relationship is valid only for small Reynolds number. At high Reynolds number, viscous forces do not govern the flow and the hydraulic gradient will have higher order terms.

66

King Fahd University of Petroleum & Minerals

Dr. Rashid Allayla

Non-Homogeneity (Heterogeneity) and Anisotropy: A medium is non-homogeneous when elements such as hydraulic conductivity vary with space. Rarely is an aquifer actually homogeneous, and due to the difficulties in solving non-homogeneous aquifer system, flow nets are often employed to transform such system before employing such solutions. A medium is an anisotropic if the hydraulic conductivity is directional. In non-homogeneous aquifers, Darcy’s equation is still valid because the value of K(x,y,z) is still a scalar and lies outside the head gradient. Non-homogeneity in aquifers is often the result of stratification of the aquifer. The individual stratum may be homogeneous but the values of K may vary between one layer to the other making the whole system non-homogeneous

Homogeneous / Isotropic

Nonhomogeneous / Isotropic

Homogeneous / Anisotropic

Nonhomogeneous / Anisotropic

Nonhomogeneouity: Area A

Area B

(KA )x ≠ (KA )x (KA )y ≠ (KA )y

67

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Groundwater Flow Equation: The derivation of groundwater equation is based on the conservation of mass between one point in flow to another coupled with the application of Darcy’s law. For incompressible flow, the continuity equation states that, ∂2h / ∂x2 + ∂2h / ∂y2 + ∂2h / ∂z2 = 0, and for in one-dimensional flow, the equation is, ∂2h / ∂x2 = 0 a) Steady One-Dimensional Flow in Confined Aquifer: Assume a flow situation shown below. The head upstream is H1 and the head downstream is H2. The hydraulic conductivity is K and the length of flow is L. The distribution of head as function of x is found using the above Laplace equation in xdirection.

Piezometric head Area = A

H1

H2 K

L ∂2h / ∂x2 = 0 Integrating twice, h =C1 x + C2 Applying the boundary condition h = H1 @ x=0 and h = H2 @ x=L h = [(H2 – H1) / L] x + H1 The discharge is found by employing Darcy’s law Q = - KA dh/dx Q = KA (H1 – H2) / L Note that for flow in unconfined aquifer, there is no direct solution available even in one dimension. The reason is that the water table is part of the unknown upper boundary and the location of the water table is required (a priori) before one can proceed to solve the equation. This would lead us to make further simplification to linearized the equation that describes the flow in unconfined aquifers (Dupuit assumption).

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b) Steady One-Dimensional Flow in Unconfined Aquifer: Dupuit Assumption: The equation describing steady flow of water in unconfined aquifer assuming K is independent of position is, ∂2h2/∂x2 + ∂2h2/∂y2 + ∂2h2/∂z2= 0 The solution of the above equation yields the location of the water table h(x,y,z) which is function of position and, for unsteady flow, a function of time. However, the location of the water table is required (a priori) before one can proceed to solve the equation. The difficulties associated with solution of this equation lead us to seek some approximations. This is known as the Dupuit approximation.

qs H1 H2

Actual water table

h = f (x, z)

Parabolic distribution

h = f (x)

For one-dimensional flow: ∂2h2/∂x2 = 0 Integrating twice and applying the boundary condition: h = H1 @ x = 0 & h = H2 @ x = L h2 = [(H22 – H12 )/ L] x + H12 The discharge is found by employing Darcy’s law Q = -K (hx1) dh/dx. Differentiating: 2h dh/dx = [(H22 – H12) / L] -2 Q / K = [(H22 – H12) / L] Q = K/2L (H12 – H22) Measured in L3 / time per unit width of aquifer.

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Example: The upstream elevation of the water-table of the unconfined aquifer shown below is 85 meters above horizontal impermeable boundary and the elevation of the downstream water-table is 74 meters. The width of the aquifer is 200 meters, the length is 5 km, and the hydraulic conductivity of the aquifer is 8.357 m/day. Find the discharge across the aquifer.

Water-table 85 m

Cross sectional Area = A = h x 200 74 m

Head h 5000 m 200 m

Solution: Q = - K (h x b) dh / dx Q ∫x = 0 → 5000 (dx) = - K b ∫h = 85 → 74 (h) dh X= 5000

Qx│

=+Kb

X=0

h2 /

h=85

2│

h=74

Q = ( 8.357 m/d) (200 m) [ 0.5 (852 – 742) / (5000 m) = (835.7) (1749) = 292.33 m3/day Verify the answer using the equation derived in previous page: Q = K /2L [H21 – H22] = (8.357 m/day) / (10000) [ 852 – 742 ] = (0.0008357)(7225 – 5476) (200) = 292.33 m3/day OK

70

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Steady State Solution of Groundwater Equation (Radial Symmetry): 1) Steady Flow to A Well in Unconfined Aquifer (Radial Symmetry): Figure below shows a well fully penetrating the saturated unconfined aquifer. The assumption inherent here is that the aquifer is radially symmetric, homogeneous and isotropic. Q

Observation well Q

s s

r

Assumed equipotential Surfaces (Dupuit assumption)

∆r

h

H

Cylindrical Cut

Considering cylinder around the well extending from rw to R where rw is the radius of the well and R is the radius of influence (which is defined as the distance from the well to the point where drawdown is negligible), the linearized form of differential equation for steady, radial flow in unconfined aquifers is defined from the following differential equation: ∂2h/∂r2 + (1/r)∂h / ∂r = 0 Which is the radial form of Laplace equation ▼2 h = 0. The boundary condition can now be stated as, h = hw at: r = rw and: h = H at r = R In the above: H is the static head before the start of pumping and R is the radius of influence. From Darcy’s Law, the discharge across a cylinder (shown in the above figure) becomes: Qw = 2πr h K dh/dr

Cross Sectional Area (cylinder)

= 2πr K d/dr (h2/2) Employing the boundary condition h = hw @ r = rw and h = H at r = R H2 – h2w = Q / π K ln (R/rw)

= sw

hw ≈ H

For small drawdown sw (drawdown at the well) , H2 – h2w = (H – hw) (H + hw) = sw (2H) The Solution become, sw = Q/2πT [ln (R/rw)] 71

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where T is the known as the coefficient of transmissibility KH. Note that the equation can only be applied to aquifers that are infinite in their lateral extension. To find the drawdown between two points in the aquifer, the equation become, s1 – s2 = Q/2πT [ln (r1/r2)] The drawdown at any point in the unconfined aquifer is, s = sw -Q/2πT [ln (r / rw)] It is apparent from the above equation that, to obtain maximum discharge, the well would have to penetrate to the impermeable boundary. However, this is not economically feasible. Studies found that wells, which penetrate a depth greater than two-thirds of the saturated zone, would not yield significantly more discharge. b) Steady Flow to a Well in Confined Aquifers Considering a cylinder around the well extending from Rw to R and bounded by the upper and lower impermeable boundaries, Qw = 2πr b K dh/dr

Note: we substituted thickness b for h

After integration, the equation for piezometric head become,

Q

b

H – h = Q/2πbK ln (R/r) or, H – h = Q/2πT ln (R/r) The above equation can be applied to any two points r1 and r2 s1 – s2 = Q/2πT [ln (r1/r2)]

Example:

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A 0.5 meter well fully penetrates an unconfined aquifer of 100 meters in depth. Two observation wells located 50 and 100 meters apart have drawdown of 5 meters and 4.5 meters. If K is 5 meters per day what is the discharge? Solution: h12 – h22 = Q / πK ln (r1 / r2) h1 = 100 - 5 = 95,

h2 = 9025

h2 = 100 – 4.5 = 95.5, h2 = 9120.25 9025 – 9120.25 = Q / (π) (5) ln (50/100) Then Q = 9525 x π x 5 x 0.6931 = 2158.54 m3/day PW

OW 1

OW 2

100 m

h1

h2

Example: A 0.5 meter well is pumping from confined aquifer of 50 meter in depth at a rate of 1000 m3 / day. After a long time of pumping, the drawdown at the well is 4 meters and the drawdown at r = 10 m is 3 meters. Find the hydraulic conductivity of the aquifer. Solution: s – sw = - (Q / 2πT ) [ln (r/rw)] = (1000 m3/d/ 2πT) ln (10/0.5) 3 – 4 = (1000 / 2πT) ln (20) Then: T = (1000) (2.996) / 4.28 = 699.93 Then K = 699.93 / 50 = 14 m/d

Pumping Near Hydro-Geologic Boundaries (Recharge Source) Consider pumping around fully penetrating stream (recharge boundary). When pumping occurs, the drawdown will increase and the cone of depression will expand until it hits the recharge source. At this point on, the cone of depression cannot spread

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beyond the recharge source and no drawdown will take place beyond this point. The drawdown at any point in the system is calculated by placing an imaginary recharge (production) well pumping from an infinite aquifer and placed at the exact opposite distance to the recharge source. The recharge well operates simultaneously and at the same pumping rate as the real well. Remember that after this point in time, the flow into the well is no longer radially symmetric because the source of water is the stream. Buildup due to recharge Stream level

Qin

Stream level

Qout

Resultant cone Of depression

a

Drawdown Due to pumping with Infinite aquifer (no Source)

-a

Cone of Depression Due to pumping with Infinite aquifer (no Source)

Recharge source

Recharge source Static Line (nonPumping)

Water level

The solution of problems involving pumping near sources in an aquifer can be illustrated as follows: It is required to find the drawdown at any point in an aquifer for steady flow to a well located at point P(x0,0). The line x = 0 is a contentious stream boundary infinite in extent at y axis.

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Recharge boundary

Image well -Q

ri

Dr. Rashid Allayla

P (x,y)

r

(-x0 , 0)

Pumping well Q

(x0 , 0)

Image region

Real region

The drawdown at any point in the real region in the aquifer P(x,y) is the sum of drawdown of two wells, each operating in a fictitious infinite field. The equation of the drawdown is the sum of the drawdown due to pumping well (+Q) and recharge well (-Q). If r is the distance to the real well and ri is the distance to the imaginary well, then the drawdown at any point is, s(x,y) = (Q/2πT) ln (R/r) + (-Q/2 π T) ln (R/r i) = (Q/2πT) ln (r i /r) = (Q/4πT) ln {[ (x + x0)2 + y2] / [ (x - x0)2 + y2]} where R is the radius of influence of the well. Note that the assumption here is that R > x0 otherwise the recharge boundary will have no effect on drawdown. The superposition in side view is illustrated by the first graph in the previous page. The procedure illustrated in the above example is known as the method of images. Note that the gradient of the head (or drawdown) is zero at any point in the recharge boundary (line x = 0). If more than one recharge boundary exists in the system, the method of superposition still applies and the total drawdown will be the sum of all drawdown due to imaginary wells with distances ri to the point in question and the drawdown to the pumping well. s(x,y) = Σi = 1,n si + sPW

Pumping Near Hydro-Geologic Boundaries (Impermeable Boundary)

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Consider pumping around fully penetrating impermeable boundary. When pumping occurs, the drawdown will increase and the cone of depression will expand until it hits the impermeable boundary. At this point on, the cone of depression cannot spread beyond the impervious boundary and the rate of drawdown accelerates. The drawdown is calculated by placing a fictitious pumping (discharge) well placed at the exact opposite distance to the impervious boundary. The imaginary discharge well operates simultaneously and at the same pumping rate as the real well. Remember that, after this point in time, and just like the case in recharge boundary, the flow into the well is not radially symmetric flow. Buildup of drawdown due to impermeable QOUT

Resultant cone Of depression with impermeable boundary

Real region

a

Qout

Static line (no pumping)

Drawdown

Due to pumping with Infinite aquifer with no boundary)

Image well Image region

-a

Impermeable Boundary

Static Line (noPumping)

Water table

Impermeable Boundary

The solution of problems involving pumping near impermeable boundary in an aquifer can be illustrated as follows:

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It is required to find the drawdown at any point in an aquifer for steady flow to a well located at point P(x0,0). The line x = 0 is a contentious impermeable boundary infinite in extent at y axis.

y

Impermeable boundary

P (x,y)

ri

r

Pumping Well (+Q)

Pumping well (+Q)

x

(x0 , 0)

(-x0 , 0)

Image region

Real region

The drawdown at any point in the real region in the aquifer P(x,y) is the sum of drawdown of two wells, each operating in a fictitious infinite field. The equation of the drawdown is the sum of the drawdown due to pumping well (+Q) and recharge well (-Q). If r is the distance to the real well and ri is the distance to the imaginary well, then the drawdown at any point is, s(x,y) = (Q/2 π T) ln (R/r) + (Q/2 π T) ln (R/ri) = (Q/2 π T) ln (R2/rri) = (Q/2 π T) ln { R2/ {[ (x - x0)2 + y2]1/2 [ (x + x0)2 + y2]1/2} Where R is the radius of influence of the well. Note that the assumption here is that R > x0 otherwise the impermeable boundary will have no effect on drawdown. The superposition in side view is illustrated by the first graph in the previous page.

d) Image Well System:

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Method of images also applicable in a system of wells that is bounded by two types of boundaries at angles less than 180 degrees. Some are shown in the following illustrations, O Real pumping well o Image pumping well X Image recharge well

(+b, -a)3 IWX -Q

I W X (-b, -a)2

-Q

O. P (x,y)

b

PW O +Q

a

I W 0 (b, -a)1

+Q

Drawdown at any point in the real domain is,

sP = sr + s1 – s2 – s3 = (Q/2πT) [ln ( R/r) + ln (R/ri1) – ln (R/ri2) –ln (R/ri3) = (Q/2πT) [ln ( ri2ri3 / rri1) = (Q/2πT) {ln [(b+x)2 + (y+a)2]1/2 [(b+x)2 + (y-a)2]1/2 / [(x-b)2 + (a-y)2]1/2 + [(x-b)2 + (y+a)2]1/2 The equation becomes, sP (x,y) = (Q/4πT) ln { [(b+x)2 + (y+a)2] [(b+x)2 + (y-a)2] / [(x-b)2 + (a-y)2] + [(x-b)2 + (y+a)2]}

78

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0 Real pumping well o Image pumping well x Image recharge well

-Q

. P (x,y)

x (-b , a)3

PW

b

0 +Q a

+Q

-Q

o (-b , -a)2

x (b , -a)1

sP = sr – s1 + s2 – s3 sP = = (Q/2πT) [ln ( R/r) - ln (R/ri1) + ln (R/ri2) – ln (R/ri3) = (Q/2πT) [ln ( ri1ri3 / rri2)

0 Real pumping well o Image pumping well x Image recharge well

. P (x,y)

+Q o (-b , a)3

b

PW 0 +Q a

-Q

x (-b , -a)2

-Q x (b , -a)1

sP = sr – s1 – s2 + s3

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Transient Well Hydraulics Groundwater flow in confined and unconfined aquifers is transient (variable with time) when the piezometric surface or water table position changes with time. The solution is obtained by solving the linearized form of Boussinesq equation, which employs Dupuit assumption. In the following analysis, the aquifer is assumed homogeneous, isotropic, for the case of confined aquifers, the thickness is constant, and in both cases, storativity of the aquifer is constant. Furthermore, the release of water from the aquifer is immediate upon decline of head. It is important to note that, upon employment of Boussinesq equation, it is apparent that the restrictive nature of Boussinesq equation is more pronounced in unconfined aquifers than in confined aquifers. The linearized form of differential equation of flow of groundwater in radial symmetry was presented earlier as, ∂s/∂t = α [∂2s/∂r2 + (1/r)∂s / ∂r]

s is the drawdown

The equation of drawdown in transient flow conditions becomes, s = Q/4πT ʃ U→∞ e-U / u du and u = r2 S/4Tt Or, in short s = Q/4πT W (u) which is the non-equilibrium equation describing transient flow of groundwater to a well developed by Theis (1935). The well function W (u) is can be expanded in an infinite series as follows, W (u) = -0.5772 – ln u + u – u2/2.2! +u3/3.3! – u4/4.4! + …………………… For small values of u (say u<0.01, i.e. for small r and/or large time), the W function can be approximated by the first two terms and the equation can be approximated as, s (r,t) = Q/4πT [ -0.5772 + ln(1/u)] Substituting for u = r2 S/4Tt, the approximate equation become, s (r,t) = Q/4πT ln (2.246Tt / r2S) The above equation approximates the drawdown in both confined and unconfined aquifers with S represents storage coefficient for confined aquifers and apparent specific yield for unconfined aquifers. In addition, T would be = b K for confined aquifers and = ĤK for unconfined aquifers where Ĥ represents an “average” saturated thickness.

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Observations Concerning Well Function Equation: s = Q/4πT W (u) , u = r2S/4Tt   

 

 

  

Transmissibility T: Effects the shape of the drawdown. If transmissibility is high, the drawdown is narrow and deep. If the transmissibility is low, the drawdown is wide and shallow. Storativity S: Effects the amplitude of the drawdown. The lower the storativity, the deeper the drawdown. The well function equation developed above can be applied to unconfined aquifers in the most restrictive sense. The assumptions inherent in the application of this equation to unconfined aquifers are that gravity drainage is instantaneous and no delayed yield occurs. The other assumption is that flow is horizontal and drawdown is too small compared to the overall saturated thickness of the unconfined aquifer. In this approach, which is by far the simplest, is to use the same equation as the confined aquifer equation but with different arguments. For instance, the transmissibility Kb is replaced by the value KH where H either is the initial saturated thickness or “averaged” thickness of unconfined saturated zone. The exact solution of the integral equation predicts that the cone of depression around the well develops instantaneously and extends infinitely. Since u is a function of r, then as r → ∞, u → ∞ and W (u) → 0. This makes s → 0 and for all practical purposes, the drawdown becomes negligible for a finite radius. Mathematically, however, the equation suggests that the radius of influence R grows infinitely with time. Since R is function of t1/2, the cone of influence expands rapidly initially and slows down thereafter. The equation of R tells us that, given the same pumping conditions the cone of influence is larger in aquifers with small S compared to those of large S. This means that in an unconfined aquifer (which has much larger S than confined aquifers), the cone of influence does not have to expand as fast as the cone of influence in confined aquifers to account for the same value of Q. Physical limit of cone of influence is replenishment source (such as rivers, lakes or sea). Mathematically, steady state conditions will never prevail because u is not constant. A condition known as pseudo-steady state occur by setting u equal or less than an arbitrary value of 0.01 if attention is restricted to regions around the well and pumping takes place at sufficiently long time. Remember that pseudo-steady state does not imply steady state conditions where s does not vary with time, it rather imply that, at regions not far from the well, the rate of fall of the piezometric surface or the water table is the same everywhere at this region and the rate of change in drawdown is independent of distance r.Pseudo-Steady state is defined by the formula: s (r,t) = Q/4πT ln (2.246Tt / r2S)

The Transient Groundwater Equation: 81

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s = Q/4 π T [W (u)] W (u) = -0.5772 – ln u + u – u2/2.2! +u3/3.3! – u4/4.4! + ……………………

u = r2 S/4Tt

For Psydo-Steady State: u ‹ 0.01 Then s = Q/4πT [ - 0.5772 + ln(1/u)]

Values of u vs. W (u) (After Wenzel 1942): u 1.00E-15 2.00E-15 3.00E-15 4.00E-15 5.00E-15 6.00E-15 7.00E-15 8.00E-15 9.00E-15 1.00E-14 2.00E-14 3.00E-14 4.00E-14 5.00E-14 6.00E-14 7.00E-14 8.00E-14 9.00E-14 1.00E-13 2.00E-13 3.00E-13 4.00E-13 5.00E-13 6.00E-13 7.00E-13 8.00E-13 9.00E-13 1.00E-12 2.00E-12 3.00E-12 4.00E-12 5.00E-12 6.00E-12 7.00E-12 8.00E-12

W(u) 33.96 33.27 32.86 32.58 32.35 32.17 32.02 31.88 31.76 31.66 30.97 30.56 30.27 30.05 29.87 29.71 29.58 29.46 29.36 28.66 28.26 27.97 27.75 27.56 27.41 27.28 27.16 27.05 26.36 25.96 25.67 25.44 25.26 25.11 24.97

u 9.00E-12 1.00E-11 2.00E-11 4.00E-11 5.00E-11 6.00E-11 7.00E-11 8.00E-11 9.00E-11 1.00E-10 3.00E-10 4.00E-10 5.00E-10 6.00E-10 7.00E-10 8.00E-10 9.00E-10 1.00E-09 2.00E-09 3.00E-09 4.00E-09 5.00E-09 6.00E-09 7.00E-09 8.00E-09 9.00E-09 1.00E-08 2.00E-08 3.00E-08 4.00E-08 5.00E-08 6.00E-08 7.00E-08 8.00E-08 9.00E-08

W(u) 24.86 24.75 24.06 23.36 23.14 22.96 22.81 22.67 22.55 22.45 21.35 21.06 20.84 20.66 20.5 20.37 20.25 20.15 19.45 19.05 18.76 18.54 18.35 18.2 18.07 17.95 17.84 17.15 16.74 16.46 16.23 16.05 15.9 15.76 15.65

u 1.00E-07 2.00E-07 3.00E-07 4.00E-07 5.00E-07 6.00E-07 7.00E-07 8.00E-07 9.00E-07 1.00E-06 2.00E-06 3.00E-06 4.00E-06 5.00E-06 6.00E-06 7.00E-06 8.00E-06 9.00E-06 1.00E-05 2.00E-05 3.00E-05 4.00E-05 5.00E-05 6.00E-05 7.00E-05 8.00E-05 9.00E-05 1.00E-04 2.00E-04 3.00E-04 4.00E-04 5.00E-04 6.00E-04 7.00E-04 8.00E-04

W(u) 15.24 14.85 14.44 14.15 13.93 13.75 13.6 13.46 13.34 13.24 12.55 12.14 11.85 11.63 11.45 11.29 11.16 11.04 10.94 10.24 9.84 9.55 9.33 9.14 8.99 8.86 8.74 8.63 7.94 7.53 7.25 7.02 6.84 6.69 6.55

u W(u) 9.00E-04 6.44 1.00E-03 6.33 2.00E-03 5.64 3.00E-03 5.23 4.00E-03 4.95 5.00E-03 4.73 6.00E-03 4.54 7.00E-03 4.39 8.00E-03 4.26 9.00E-03 4.14 1.00E-02 4.04 2.00E-02 3.35 3.00E-02 2.96 4.00E-02 2.68 5.00E-02 2.47 6.00E-02 2.3 7.00E-02 2.15 8.00E-02 2.03 9.00E-02 1.92 1.00E-01 1.82 2.00E-01 1.22 3.00E-01 0.91 4.00E-01 0.7 5.00E-01 0.56 6.00E-01 0.45 7.00E-01 0.37 8.00E-01 0.31 9.00E-01 0.26 1.00E+00 0.219 2.00E+00 0.049 3.00E+00 0.013 4.00E+00 0.0038 5.00E+00 0.0011 6.00E+00 0.00036 7.00E+00 0.00012

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EXAMPLE Well is being pumped at constant rate of 0.004 m3/s. The transmissibility of the aquifer is 0.0025 m2/s the distance to observation well is r = 100 meters and the storage coefficient is 0.00087. Find the drawdown in the observation well after 20 hours of pumping using the Theis well function and pseudo-steady state approximation and compare the results. Solution: u = r2S /4tT = (100)2 (0.00087) / [(4)(20x3600)(0.0025) = 0.0121 From u vs. W(u) table @ u=0.0121 , W(0.0121 = 3.895 s = Q/4πT (3.895) = [0.004 / (4π)(0.0025)] [3.895] = 0.4959 m Using the approximate solution s = Q/4πT [ -0.5772 + ln(1/u)] s = (0.004) / [(4 π)(0.0025)] [-0.5772 + ln (1/0.0121] = 0.4886 m The two answers are approximately the same because u is very small. EXAMPLE: After 2 hours of pumping in an aquifer, it was observed that the radius at which drawdown is negligible is 400 meters. At what radius would the drawdown be negligible after 5 hours of pumping? Assume pseudo-steady state prevail. At pseudo-steady state,

1/u

s = Q/4πT [ -0.5772 + ln (4tT/R2S) ≡ Q/4πT [ln [(0.561)( 4tT/R2S)] s = Q/4πT [ ln (2.246 (Tt / R2S)) 0 = ln [(2.246 t / R2) (T/S)] 1 = (2.246) (2)/160000) T/S

T/S = 2.81 x 10-5

For 5 hrs of pumping, 0 = ln [2.246 (5) / R2] (2.81 x 10-5) Then, R2 = (5) (2.246) / (2.81 x 10-5)

Then: R = 632.46 m

EXAMPLE: A pumping well is bounded by a straight stream and a parallel impermeable boundary as shown that are 20 meters apart. The well is located at the middle. Pumping starts at Q = 1000 m3/d. If T = 500 m3/d , find the 83

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drawdown at the well after 10 days of pumping. Assume R = 500 m and rPW = 0.5 m and Sy = 0.1. Image impermeable

Image recharge 20m

PWi

Impermeabl e

Recharge

The transient drawdown equation is, s = Q/4πT W(u), where, u = r2S / 4Tt, Taking four image wells from each side, sT = sPW + (- siRW20 - siRW40 + siPW60 + siPW80 )LEFT + ( siPW20 -

iRW40

– iRW60 + siPW80 )RIGHT

= sPW - 2 siRW40 + 2 siPW80, the rest cancel each other. The drawdown at the well is, sT = Q/4πT [ W(uPW) - 2 W(uiRW40) + 2 W(uiRW80)] uPW = (0.5)2 (0.1) / (4)(500)(10) =0.00000125, uiPW40 =(40)2 (0.1 / (4) (500)(10) = 0.0008 uiPW80 =(80)2 (0.1) / (4) (500)(10) = 0.032 The total drawdown become, sT = (1000) / (4 π) (500) [13.03555 – (2) (6.5545) + (2) (2.8845)] = 0.1592 [13.03555-13.109+5.769] Drawdown at the well after 100 days of pumping: sT = 0.91 m

Aquifer Tests Aquifer tests are field tests that are performed to determine field data under controlled conditions. The outcome of the field tests are the hydrologic parameters of the aquifer such as storage coefficient, apparent specific yield, hydraulic conductivity 84

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and transmissibility of the aquifer. The obtained values of the tested aquifer can be used for designing well field and predicting future drawdown. They can also be used for assessing groundwater supply, and estimating inflow and outflow to and from groundwater basins. The outcome of aquifer properties obtained from field tests will eventually be compared to the values obtained from theoretical considerations. For this reason, it is important that elements such as boundary conditions, initial conditions and the physical characteristics of the chosen sites match closely the ones assumed in theory. The other important consideration is to minimize uncertainties associated with the selection of pumping wells and the placement of observation wells. For example, data obtained from partially penetrating wells are difficult and uncertain to analyze than the fully penetrating wells. In addition, the proper choice of the number and the location of observation wells would minimize uncertainties associated with measuring hydraulic parameters of the aquifer. In order to ensure that the water level properly represents the average piezometric head below water table, the casing in the observation wells should be perforated and should penetrate the entire saturated zone. In confined aquifers, piezometer should be sealed properly in order to ensure against water transport from one stratum to another. During aquifer test analysis, the number and the appropriate spacing of observation wells play an important part to insure the integrity of data collected. According to Walton (1987), no less than three observation wells should be selected and spaced at one logarithmic cycle from one another. Because the radius of influence expands more rapidly in confined aquifers than in unconfined aquifers, observation wells should be placed at greater distances from each other than in unconfined aquifers. When hydro geologic boundary is present, and in order to minimize their effect, pumping well should be placed at least one saturated thickness away from the boundary. In addition, the observation wells should be spaced along a line through the production well and parallel to the boundary. This is made to minimize the effect of the boundary on distancedrawdown data. Finally, the hydrologic data collected for unconfined aquifers using aquifer test analysis represent mean values of that portion of aquifer bounded by the initial saturated zone and the of cone of depression and does not represent the entire saturated thickness. Furthermore, the best estimates of the aquifer properties can be obtained from tests that are conducted over a long time when delayed drainage is not a factor influencing the apparent specific yield.

a) Analysis of Aquifer Test Data Using Theis Solution: This method utilizes type curve matching technique. This method involves matching logarithmic time-drawdown or distance-drawdown graphs obtained from the field measured values of time, drawdown or distance, drawdown data with the theoretical logarithmic well function curves (type curves) of W(u) vs. u. The procedure is illustrated as follows:

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1. At a given time or at a given location obtain measured field data of drawdown and plot drawdown s vs. r2 / t on a log-log paper. 2. Plot “type” curve of W(u) versus u choosing appropriate cycle(s) on a log-log paper of the same size. 3. Super-impose the two plots data by keeping the coordinate axis of the two curves parallel until best fit is obtained. 4. Select any arbitrary point in the graph. Read the values of W(u) *, u* and the corresponding values of s* and (r2/t)* 5. The values of T and S are calculated using the formulas: T = Q W(u)*/4πs* and S = 4Tu*/(r2/t)* s

W(u)

w(u*)*

s*

r2/t* r2/t u*

u

Log-Log Graphs

b) Analysis of Aquifer Test Data Using Cooper-Jacobs Method It was noted by Cooper and Jacob (1946) that for small u (large values of t or small values of r), the non-steady equation for s become, s (r,t) = Q/4πT (0.5772 – ln r2S/Tt) or, in terms of decimal log scale, the equation become, s (r,t) = 2.303 Q/4πT log (2.246Tt / r2S) Plotting the drawdown s versus the logarithm of time t would form a straight line. The x axis intercept is at t = t0 @ s = 0, therefore: 0 = 2.303 Q/4πT log (2.246Tt0 / r2S) Therefore:

For this to happen, 2.246Tt0 / r2S must = 1

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S = 2.246Tt0 / r2

Dr. Rashid Allayla

=1

and the slope over one cycle is: s (r,t) 10t - s (r,t) t = ∆s = 2.303 Q/4πT log [(2.246T10t / r2S)/ 2.246Tt / r2S)] Therefore: T = 2.303 Q/4π ∆s The procedure is summarized as follows: 1. For stationary reading, measure s at different times. 2. Plot s vs. log t on a semi-log paper. 3. Find the slope by reading the difference between two points across one log cycle ∆s*. 4. Find the horizontal intercept t0* on the log t axis. 5. The values of S and T are calculated from the formulas: S = 2.246Tt0* / r2, and T = 2.303 Q/4π ∆s*. An alternative method is to measure s at different observation wells (different r) at given time, plot s versus r on logarithmic paper, find the slope across one cycle (∆s*) and find the intercept r0* on the log r axis. The values of S and T are calculated from the formulas: S = 2.246Tt / r02 and T = 2.303 Q/4π∆s*.

s

∆s

t

t0

Semi-log graph

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Note that s versus t is a straight line as long as the assumption of small value of u is still valid. However, it is apparent that when time is small, this assumption can no longer be valid as shown in the above chart. It is also important to note that using non-equilibrium equation is valid for unconfined aquifers as well as long as the measured drawdown is small in relation to the overall saturated thickness of the unconfined aquifer. Example: In order to Illustrate the use of the methods discussed above, the following is a data collected from an observation well during a test in unconfined aquifer in Fort Collins, Colorado (McWhorter & Sunada 77): s (m) 0.025

0.050

0.055

0.110

0.170

0.180

0.220

0.300

0.370

0.450

0.53

r2/t

53.3

47.1

25.0

16.7

15.1

11.1

6.25

4.12

2.47

1.55

s (m) 0.620

0.640

0.650

r2/t

0.98

0.82

88.9

0.78

Plotting W(u) versus u, the coordinates of match point W(u) = 1.0 u = 0.1 s = 0.183 and r2 / t = 6.2 then T = (1.872)(1.0) / 4 π (0.183) = 0.814 m2/min and Sya = (4) (0.814) (0.1)/ 6.2 = 0.053

0.01

0.1

1.0

1.0

10.0 W(u)

s 1.0 0.1 Match Point 0.1 0.01

r2/t

1.0

10.0

100.0

Example: Data collected during a test of a confined aquifer by U.S. Geological Survey is as follows: 88

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Time: (min):

1

2

3

DD s (m):

0.2

0.3

0.370 0.415 0.450 0.485 0.530 0.570 0.850

Time: (min):

60

DD s (m):

0.875 0.925 0.965 1.000 1.045 1.070

80

4

100

120

5

150

6

8

10

50

180

r = 61 m and Q = 1.894 m3 /min Plotting the given data on a semi log paper,

1.2 S (m)

0.8

∆s = 0.4 m Per Log cycle

0.4

1

t

t0 = 0.4 min

10

100

0.0

The slope of the line is ∆s = 0.4, then T = (2.303) (1.894) / 4π (0.4) = 0.868 m2 / min The extrapolated t at s = 0 is 0.4 minute, then s = (2..246) (0.868) (0.4) / (61)2 = 2 x 10-4

EXAMPLE:

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Calculate the limit of the pseudo-steady state region around the well if Q = 100 m3 / hr, S = 0.09, T = 36 m2 / hr and t = 10 hrs. r2 / 4αt = r2 / 4 (T/S)t = 0.01 then r = [(0.01)(4)(36/0.09)(10)]1/2 = 12.65 m. The Pseudo-steady state does not apply beyond this limit

Drawdown with Variable Pumping Rates: The linearity principle of groundwater equation allows for the application of the principle of superposition of drawdown resulting from variation of pumping rates. The principle of superposition also applies to drawdown recovery following well shutdowns. If amount of water pumped changes from Q1 to Q2, the resulting drawdown is described by the following equation, s = (Q1/4πT) W(r2S/4Tt) + (Q2 – Q1) W [ r2S/4T (t-ti)]

for t>ti

Or, in general, s = 1/4πT Σ ∆Qii+1 W [r2S/4T(t-ti )] The equation states that the resulting drawdown is equal to the drawdown from t = 0 to t = t2 with Q = Q1 plus drawdown resulting from an “imaginary” well pumping at a rate of Q = ∆Q placed in the same position and pumping from t = t1 to t = t2.

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Q2

Q3

Q1 t2

t’

DD due Q1 @ t’

t4 t1

t’’

t’’’

DD at t’’’ due to – Q2

DD due Q2 – Q1@ t”

t3

Recovery

If Q2 is zero then, s = (Q/4πT) W [ r2S/4T (t-ti)] If Q2 is zero and r2S / 4t T < 0.01 then s = (Q/4πT ln [t /(t-t1)]

ti

Buildup

t

Recovery

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Example: A well discharging a confined aquifer with Q = 40.2 ft3/minute. The aquifer transmissibility is 2270 ft2/day and its storage coefficient is 4x10-4. The pumping lasted for 10 days and then shut down. Find the drawdown after: a) 15 days and 1 month. Solution: Drawdown after 15 days: s = Q/4πT W(r2S/4T(10)-Q/4πT W(r2S/4T(15-10) =Q/4πTW[(0.25)(0.0004)/[(4)(2270)(10)] -

Q/4πTW[(0.25)(0.0004) [4π(2270)] / [(4)(2270)(5)] = (40.2)(60x24) / [4π(2270)] W[(1.1x10-9) – W(2.2x10-9)]

From table: W(1.1 x10-9) = 20.08 W(2.2 x 10-9) = 19.37 Then: s = 6.3753 (20.08-19.37) = 4.53 ft

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