Civil Engineering Dept.

CE 331 Engineering Statics

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”

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.

(From Purdue University)

From EPA

Elements of Hydrologic Cycle:

- Evaporation, E

- Transpiration, T

- Precipitation, P

- Surface runoff, R

- Groundwater flow, G, and,

- Infiltration, I

System Concept in Hydrologic Cycle:

Oceans

Aquifers

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

Hydrologic Budget

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

Precipitation Watershed outflow Inflow Change in Storage

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

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

I – O = ∂S / ∂t

Breaking system into individual component as shown in schematic figure:

P – E + ([(R_in + G_in)] – (R_out + G_out)] – T = dS/dt

Where:

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

E: Evaporation (L/T)

R_in, G_in: Inflow from surface and groundwater (L/T)

R_out, G_out: Outflow from surface and groundwater (L/T)

S: Storage (L) and,

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

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

Breaking the above into surface and subsurface balance equations:

Surface: P + R_in – R_out + G_up – I - E - T = ∆ S_s

Subsurface: I + G_in - G_out – G_up = ∆ S_g

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

P - R - G - E - T = D S / ∆t

Balance Equation for Open Bodies with Short Duration:

And, for an open water bodies, short duration,

I - O = D S/D t

Where,

I = inflow volume per unit time

O = outflow per unit time

Balance Equation for Urban Drainage:

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

P - I - R - D = 0

P = precipitation

I = infiltration

R = direct runoff

D = Combination of interception and depression storage

Illustrative Example:

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

Solution:

Combining E and T, then ET = P – R

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

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

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

Equivalent depth = volume of water / area of watershed

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

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

Illustrative Example: (From Viessman 2003)

The drainage area of a river in a city is 11,839 km². If the mean annual runoff is determined to be 144.4 m³/s and the average annual rainfall is 1.08 m, estimate the ET losses for the area. Assume negligible changes in groundwater flow and storage (i.e. G and ΔS = 0) .

Solution:

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

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

ET = P – R = 1.08 – 0.38 = 0.7 m

Precipitation:

Illustrative Example:

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

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

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

D S/D t = 230 – 285 = - 55 cfs

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

= - 18.182 acre-ft

b) S₂ = 40 – 18.182 = 21.82 AF

Illustrative Example:

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

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

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

D S/D t = 230 – 285 = - 55

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

= - 18.182 acre-ft

b) S₂ = 40 – 18.182 = 21.82 AF

Precipitation:

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

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

a) Cooling which condensate moist air to near saturation,

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

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

Distribution of Precipitation:

Measurement of Precipitation:

Errors:

· Splashing

· Moisture deficiency of gage

· Wind blowing

2. Tipping Bucket Gage:

Counter

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

-

To measuring tube

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

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

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

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

Methods of Estimating Areal Precipitation in the Ground:

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

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

Average PPT = ∑ a_i p_i / A_T

Illustrative Method:

Observed Precipitation (L)	Area of Polygon (L^2)	Precipitation x Area (L^3)
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 A_n ) / (A₁ + A₂ + A₃ + …. + A_n )

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

Average PPT = ∑ a_i p_i / A_T

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

Illustrative Example:

Ave PPT = 65610/ 666

= 98.51 inch

Adjusted values

Alternative Method:

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

_{m n n}

P = 1/A ∑ A_j ( ∑ d_ij ^–b )^-1 ∑ d _ij ^–b P _i

_{j = 1 i = 1 i = 1}

Where: A_j is the area of the j^th sub area, A is the total area, d_ij is the distance from the center of the j^th area to the i^th 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 ∑ A_j P_i

Which is simply the arithmetic mean.

1	2 O d2,18	3	4	5 O d5,18	6	7
8	9	10	11	12	13	14
15	16	17	18	19	20	21
22 O d22,18	23	24	25	26 O d26,18	27	28

O: Precipitation Gage

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

4 ₄

P₁₈ = [ ∑ d^-2 _i,18 P_i / ( ∑ d^-2 _i,18 )

i _{=1 i =1}

Illustrative Example:

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

Y = 1.5 km

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

1 d= 3 km P = 0.2 “

2 X Unknown PPT

d = 4.5 km 3 P = 0.15^”

P₂ = ∑ _{1 → 3} d ^-2 _{i , 2} P_i / (∑d ^-2 _{i , 2})

P₂ = [d^-2_1,2] P₁ / [d ^-2 _{1 , 2} + d ^-2 _3,2] + [d^-2_3,2] P₃ / [d ^-2 _{1 , 2} + d ^-2 _3,2]

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

Methods of Estimating Missing Precipitation:

a) Normal Ratio Method:

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

n

P_x = ∑ w_i P_i

i=1

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

w_i = A_x / n A_i

Where A_i is the average annual rainfall at gage i, A_x is the average annual rainfall at station x in question.

Combining the above two equations,

n

P_x = (A_X / n) ∑ P_i / A_i

_i =1

Illustrative Example:

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

Gage	Average Annual Rainfall (cm)	Total Annual Rainfall (cm)
A	32	2.2
B	28	2.0
C	25	2.0
D	35	2.4
X	26	?

Solution:

P_x = w_A P_A + w_B P_{B +} W_C P_C + W_D P_D

= (A_x / n A_A) P_A + (A_x / n A_B) P_B + (A_x / n A_C) P_C + (A_x / n A_D) P_D

= (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: P_x = 1.877 cm

b) Quadrant Method:

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

4

w_i = ∑ ( 1 / d²_i )

i=1

and the missing data is calculated as,

n n

P_x = ∑w_i . P_i / ∑ w_i

i=1 i=1

Gauge Consistency:

Double Mass Curve

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

Station A (mm)

P_A

P_B

12 Station Mean P_B (mm)

Method of Correction:

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

P_a = (m_a/m_o) p_o where: m_a (adjusted) & m_o (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 P_A2 and P_B2 are correct. So, if m₂ is the correct slope, the slope m₁ should be removed from P_A1 and replaced by m₂ by using the formula: p _A1 = (m₂ / m₁) P _A1 where p_A1 is the adjusted data.

p₁ = (m₂ / m₁) P₁ p₁ = (m₁ / m₂) P₂

Illustrative Example (McCuen 1998)

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

year	E	F	G	H	Total ∑E+F+G	Cumulative E+F+G	Cumulative H	h
1979	22	26	23	28	71	71	28	24.7
80	21	26	25	33	72	143	61	29.1
81	27	31	28	38	86	229	99	33.5
82	25	29	29	31	83	312	130
83	19	22	23	24	64	376	154
84	24	25	26	28	75	451	182
85	17	19	20	22	56	507	204
86	21	22	23	26	66	573	230

∞

Evaporation:

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

Factors Affecting Evaporation:

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

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

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

R = e / e_s

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 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/V_O = (Z/Z_O)^0.15

Where V is the wind speed in mi/hr at Z height, V_O is the wind speed at height Z_O 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.

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 –e_a)

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 e_s is the saturated vapor pressure at water surface,

The rate of evaporation from a lake can be calculated using empirical laws,

……………………… Meyer (1944)

Where, E: Lake evaporation (inches / day)

e_s – 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

………………. Dunne ((1978)

Where, R_h: The relative humidity in %

e: Vapor pressure of air (mill bars)

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

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 (@63^oF= air temp) = 0.58 in. Hg

e_s (@87^oF = water temp) = 1.29 in. Hg

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: Converting wind speed to km/day = (10 mph) (24hr/d) (1.6 km/mi)

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.

Some Empirical Equations:

Author	Equation	Explanation
Dalton	E (i/mo) = C(eO-ea)	C=15 for small, shallow water and 11 for large deep water
Meyer	E (i/mo) = 11 (1+0.1 ug) (eO-ea)	ea measured 30 above ground surface
Horton	E (i/mo) = 0.4 [2 – exp(-2u)]) (eO-ea)	u = speed of wind
Penman	E (i/day) = 0.35 (1+0.24 u2) (eO-ea)	u2 = wind speed 2 meters above surface
Harbeck	E(i/day)=0.001813 u (eO-ea)[1-0.03(Ta-Tw)]	Ta = Average Temp ^OC + 1.9^OC 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,

E_L = 0.7 [E_P + 0.00051 P α_P (0.37 + 0.0041 u_P) (T₀ –T_a)^0.88

T_o Outer face Temp of pan ^O F

Lake Evap. in/day T_a Air Temp. in ^o F

Windspeed in mi/day

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

Atmospheric pressure in in-Hg

α_P = 0.13 + 0.0065T₀ – (6.0 x 10^-8 T₀³) + 0.016 u_P^0.36 (Use Chart below to obtain α_P)

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 α_P .

- From wind speed u_P 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 α_P in the lower left chart.

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

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

- Turn right to read lake evaporation.

NO SCALE

Infiltration:

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

Elements of Infiltration

Infiltration Capacity:

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

Empirical Models of Infiltration:

Horton Model:

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

Horton Infiltration Formula:

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

f = f_∞ + (f₀ – f_∞) e^-Kt

Initial infiltration capacity (L/T)

Final infiltration capacity (equilibrium capacity)

Infiltration capacity at a given time t

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

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

Observations:

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

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

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

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

- Infiltration capacity depends on soil type (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,

f = 10 + (50) e^-0.4t

Integrating,

V = ∫[10 + 50 e^-0.4t] dt

V = [10t -(50)e^-0.4t│_{0 to 10} = [(10)(10) –(50 / 0.4)e^-4] + 50 = 147.7cm

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

Correction of Horton Curve:

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

Problem!!

τ

0 t₁

Equivalent time τ for the actual

Accumulated infiltration

τ

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

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

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

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

∆F = ∫_{t to ∆t} f (t) dt

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

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

Then f₀ (new) = f_∞ + (f₀ – f_∞) e ^{- K t} which is simply the Horton formula

But, if 2) i . dt < ∆F, then τ < ∆t

Then f₀ (new) = f_∞ + (f₀ – f_∞) e ^{- K τ}

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

Steps:

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

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

Horton Curve

( f_i + f_i+1 ) / 2

Δt

f_i f_i+1

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

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

Horton Curve Mass Curve = ∫ f(t)

t t

Actual infiltrated water (total rain @ t

Illustrative Example: (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	Intensity (in/hr)
0-10	3.5
10-20	3.0
20-30	8.0
30-40	5.0
40-50	1.5
50-60	2.4
60-70	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)	f (in/hr) (2)	∆t (min) (3)	Average f (4)	∆F (in)^(1) (5)	Σ∆F (in) (6)
0	11.66
		10	8.94	1.49	1.49
10	6.21
		10	4.86	0.81	2.3
20	2.5
		10	2.83	0.47	2.77
30	2.16
		10	1.83	0.31	3.08
40	1.49
		10	1.33	0.22	3.30
50	1.16
		10	1.08	0.18	3.48
60	0.99
		10	0.95	0.16	3.64
70	0.91

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

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:

12
10 7.5
8
6
4
2

0 1 2 3 4

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)	Time from beginning of storm (2)	f using t^’ (3)
0	20	7.5
10	30	4.14
20	40	2.47
30	50	1.65
40	60	1.24
50	70	1.03

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) (in) (7)	Excess Rain (8)
0	-	-	-	-	-	-	-
10	-	-	-	-	-	-	-
20	7.5
		10	5.82	0.97	8.0	1.33	0.36
30	4.14
		10	3.31	0.55	5.0	0.83	0.28
40	2.47
		10	2.06	0.34	1.5	0.25	-0.09
50	1.65
		10	1.45	0.24	2.4	0.4	0.16 – 0.09 =0.07
60	1.24
		10	1.13	0.19	1.5	0.25	0.06
70	1.03

(1) ∆F = (10/60) [5.82] = 0.97

12

Approximate Scale!!

10		Horton Curve
9
8
7				Modified Curve
6
5
4
3
2
1

0 10 20 30 40 50 60 70

Runoff:

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.

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:

Rectangular Weir:

V-Notch Weir:

Broad - Crested Weir:

Overflow Spillways;

Stage-Discharge Relations (Rating Curves)

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 ( S^1/2 )

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.

F

V²₂

2 g

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.

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)

Discharge →

log Q/Q₀

log F/F₀

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

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.

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.

Time of concentration

t

Physical Analogy:

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, and, as the stream stage starts to fall, the water from the banks start to drain back into the stream.

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 where the base flow rate is the same as the beginning of the direct runoff.

- 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 = A^0.2

Where 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 (flow reversal) discussed above.

Hydrograph Time Relationships:

Travel time: The time it takes for the direct runoff originating at a point in the channel 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.

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, t_b = t_s + t_c

Where t_b is the time base, t_s is the duration of rain and t_c 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 mi^2.

Solution:

Discharge m³/s

N = (1000)^0.2 = 3.98 days

3.98 d

Time : 4 6 8 10 12 14 16

Q: 30 45 70 84 74 50 30

Base Flow (Straight.Line): 30 30 30 30 30 30 30

D.R.O (Str.Line) : 0 15 40 54 44 20 0

Base (Arc L): 30 25 23 20 35 50 -

D.R.O. (Arc L ) 0 20 47 64 39 30 -

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

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 rainfall-producing 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.

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

Q

Δt t

- Measure the total volume direct runoff

Volume of direct runoff :

= ∑ [ (Q^T_i+1 + Q^T_i) / 2 - (Q^B_i+1 + Q^B_i) / 2 ] x ∆t

Q^B_i

= V_RO = Area Under Curve (R.O.)

- Divide the ordinates of the total runoff (Q^T - Q^B) by the total

volume of direct runoff measured above to obtain the

ordinates of the unit hydrograph (Q_UH = Q_S / V_RO). Where V_RO

is Vol. of runoff measured in (L) and

Q_S = Q^T - Q^B

-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 mi². Use straight-line separation and construct a hydrograph representing the following complex rain pattern.

t (hrs) Q (cfs) Base Flow (cfs) D.R.O. (cfs)

1 75 75 0

2 110 75 35

3 205 75 130

4 305 75 230

5 280 75 205

6 205 75 130

7 130 75 55

8 75 75 0

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) Q_U.H = Q/0.61

x 1000

1 0 0

2 35 57.4

3 130 213.1

4 230 377.0

5 205 336.1 Ordinates of the 2-hr unit hydrograph

6 130 213.1

7 55 90.2

8 0 0

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) Q_U.H Q_UH x 1.1 Q_UH x 2.1 Q_UH x 1.8 Hydrograph Ordinates

0 0 0 - - 0 Hydrograph resulting from the

1 57.4 63.1 - - 63.1 complex storm

2 213.1 234.4 0 - 234.4

3 377.0 414.7 120.5 - 535.2

4 336.1 369.7 447.5 0 817.2

5 213.1 234.4 791.7 103.3 1129.4

6 90.2 99.2 705.8 383.6 1089.4 No Scale

7 0 0 447.5 678.6 1126.1

8 0 0 189.4 605.0 794.4

9 0 0 0 383.6 383.6

10 0 0 0 162.4 162.4

11 0

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

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.

2-hr U.H.

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.

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): 0 1 2 3 4 5 6

Q (m³/s): 0 0 1.42 8.5 11.3 5.66 1.45 0

To construct S-Curve follow the following schematic calculations:

Given: 2-hr unit hydrograph Required: 2-hr S-curve

Time (hr) 2-hr U.H. S-Curve additions 2-hr S-Curve

0 0 - -

1 a - a

2 b - b

3 c A = (a) c + a

4 d B= (b) → d + (B=b) = d + b

5 e C = (c + A) → e + (C=c+a) = e + c + a

6 f D = (d + B) → f + (D=d+b) = f + d + b

7 g E = (e + C) → g + (E=e+c+a) = g + e + c + a

8 h F = (f + D) → h + (F=f+d+b) = h + f + d + b

9 i G

10 j H

11 k I = (i + G) → k + i + G

Inserting numbers:

Time (hr) 2-hr U.H. S-Curve additions 2-hr S-Curve

0 0 - -

1 1.42 - = 1.42

2 8.5 - = 8.5

3 11.3 = (1.42) 11.3+1.42 = 12.72

4 5.66 = (8.5) 5.66+8.5 = 14.16

5 1.45 = (11.3+1.42) 1.45+11.3+1.42 = 14.17

6 0 = (5.66+8.5) 0+5.66+8.5 = 14.16

7 = (1.45+11.3+1.42) 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 Lagged S-Curve ∆S 1-hr UH = (Column 4) (2/1)

(1) (2) (3) (4)

0 0 0 0

1 1.42 0 1.42 2.84

2 8.5 1.42 7.08 14.16

3 12.72 8.5 4.22 8.44

4 14.16 12.72 1.44 2.88

5 14.17 14.16 0 0

Multiply ∆S by the factor (2/1) to obtain 1-hr UH

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.

Storage Routing:

The rate of change of storage is defined by the continuity equation:

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 t₁. At t > t₁ , outflow exceeds inflow and the reservoir empties by an amount equal to B.

The rate of change in storage is defined by the equation:

I – O = ∂S / ∂t

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 →

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 = CH^3/2.
• 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.

Illustrative Example:

The inflow hydrograph of a given reservoir is as follows:

I (cfs)

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 H^3/2

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)

50

0 50 100 150 200

Step 1: Calculate outflow O and 2s/∆t + O for ∆t = 0.5 day

Table 1:

Water Surface El. (ft) (1)	Head H ^(1) (ft) (2)	Storage(From Data) (ft^3 x 10^6) (3)	Outflow^(2) (cfs) (4)	2s/∆t + O (cfs) (5)
50.0	0.0	40.00	0.00	160.00
50.5	0.5	73.70	37.1	331.9
51.0	1.0	107.50	105.0	535.00
51.5	1.5	141.25	192.0	757.00
52.0	2.0	175.00	297.0	997.00
52.5	2.5	208.75	415.00	1250

O = C L H^3/2

(1) Elevation above the crest

(2) 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) (1)	Inflow (cfs) (2)	I1 + I2 (cfs) (3)	Outflow (cfs) (4)	Storage, s (ft^3 x 10^6) (5)	2s1/∆t – O1 (cfs) (6)	2s2/∆t + O2 (cfs) (7)	Water Elevation (ft) (8)
---	0	0	0	40	160	160	50.0
0	0	50	0	40	160	210	50.15
0.5	50	150	6.10	52.5	203.9	353.90	50.55
1.0	100	250	42.83	86.95	304.97	554.97	51.05
1.5	150	350	112.97	128.04	399.17	749.17	51.52
2.0	200	300	196.77	159.05	439.43	739.43	51.46
2.5	100	100	185.23	135.67	357.43	457.43	50.81
3.0	0	0

From: H = 0.55 → O = CLH^3/2 From: (I₁+I₂) + (2s/∆t-O₁)→(150) + 203.9) = 353.9 From Table 1 @ 353.9 cfs

= 42.83

From: 2s/∆t + 6.1 = 353.9 → s = 86.95

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 T_r and find the intensity of rain that will be equaled or exceeded once every T_r. 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.

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

Runoff coefficients for Various Areas:

Business: Downtown Area 0.70 – 0.95

Business: Neighborhood Areas 0.50 – 0.70

Residential: Single-family areas 0.30 - 0.50

Multi units, detached 0.40 - 0.60

Multi units, attached 0.60 - 0.70

Residential (suburban) 0.25 – 0.40

Apartment dwelling areas 0.50 – 0.70

Industrial: Light areas 0.50 – 0.80

Industrial: Heavy areas 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

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.

500 ft 2000 ft

From IDF chart, for 50 year design and t_c = 18 minutes, i = 7 inches / hr.

Assume Business area controls, then t_c = 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: t_c = 40 + 8 = 48 minutes

From chart, for 50 year design and t_c = 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.

Illustrative example:

A composite residential and park areas is proposed for construction. The inlet times to the manholes for areas A_A to A_C are 20, 35, 41 and 53 minutes respectively. The coefficients C for areas A_A to A_C 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.

Inlet 1:

Contribution from area A = 20 minutes, then from IDF chart @ 10-yr storm and @ t_C = 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 (t_c)_B > (t_c)_A , contribution from area B to inlet 2 controls, then from the IDF @ 10-yr storm and @ t_C = 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 (t_c)_B > (t_c)_A and (t_c)_B > (t_c)_C , contribution from area B to inlet 3 controls, then from the IDF @ 10-yr storm, and @ t_c = 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 (t_C)_D > (t_c)_A , (t_c)_D > (t_c)_B and (t_c)_D > (t_c)_C , contribution from area B to inlet 3 controls, then from the IDF @ 10-yr storm, and @ t_D = 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

Civil Engineering Dept.

CE 331 Engineering Hydrology

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 generals 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.

Porosity: It is the ratio of the volume of interconnected voids within the soil to the total volume of the soil. If V_V designates the volume of “interconnected” voids, V_S is the bulk volume and V_T is the total volume of the soil media, the porosity θ is defined as:

θ = V_V / V_T = (V_T-V_S) / V_T = 1 – V_S/V_T

Volumetric Water content: It is the ratio of the volume of water to total volume of soil sample.

θ_W = V_W / V_T

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 = V_W / V_V

Moisture Content: It is the ratio of the weight of water to the weight of solid.

W = W_W / W_S

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.

V_V V_a

V_W V_T

V_S V_S

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 = V_w / V = (200-180) / 130 = 0.15

2) Total porosity Volume of voids / Total volume = V_V / V = (130-69) / 130 = 0.47

3) Degree of saturation = θ / θ_s = (V_w / V) / (V_V / V ) = 0.15 / 0.47 = 0.32

4) Moisture content = W_W / W_S = (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, S_S 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:

S = S_S b

Where b is the thickness of the aquifer. For unconfined aquifers, S is given the term “specific yield", S_y. It is defined as the amount of water released from a column of aquifer per unit area per unit decline of head. S_y 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 water-holding 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.

Unit Cross Section of Aquifer

Piezometric Surface 1

Unit decline of head

S

Confined Aquifer Water-Table Aquifer

Typical values of S: S ranges from 10^-4 to 10^-6 and S_y 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 = V_v / (A_T)(Δh) where area A_T (Area of aquifer) = 2.5 x 10⁷ / 30 = 8.3 x 10⁵ m²

Then Volume of water recovered V_v = S x A_T x Δh =(0.005) (8.3 x 10⁵) (20) = 8.3 x 10⁴ m³.

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

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 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. Or,

Introducing the term intrinsic permeability, k,

K = k γ / μ

K = k ν / 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. Various formulas relate k to properties of porous medium. One of them is,

k = C d² where k is measured in cm² and d in cm

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

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:

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:

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 of the medium. The average “real” velocity is, therefore is,

v = q / θ

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) Takes place from higher head to lower head and not necessarily from higher pressure to lower one.

2) 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.

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

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 x-direction.

L

H₁

∂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 linearize the equation that describes the flow in unconfined aquifers (Dupuit assumption).

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/∂y^{2 +} ∂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.

H₂

For one-dimensional flow: ∂²h²/∂x² = 0

Integrating twice and applying the boundary condition h = H₁ @ x = 0 & h = H₂ @ x = L

h² = [(H₂² – H₁^{2 )}/ 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.

Example:

The upstream elevation of the upstream 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 and the length is 5 km, and the hydraulic conductivity of the aquifer is 8.357 m/day. Find the discharge across the aquifer.

Solution:

Q = - K (h x b) dh / dx

Q ∫ dx = - K b ∫ h dh

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

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.

Assumed equipotential

Surfaces (Dupuit assumption)

Cylindrical Cut

Considering cylinder around the well extending from r_w to R where r_w 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. The boundary condition can now be stated as,

h = h_w at: r = r_w 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:

Q_w = 2πr h K dh/dr Cross Sectional Area

= 2πr K d/dr (h²/2)

Employing the boundary condition h = h_w @ r = r_w and h = H at r = R

H² – h²_w = Q / π K ln (R/r_w)

For small drawdown s_w (drawdown at the well) , H² – h²_w = (H – h_w) (H + h_w) = s_w (2H) The Solution become,

s_w = Q/2πT [ln (R/r_w)]

where T is the coefficient of transmissibility KH discussed earlier. The equation only applies 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 = s_w -Q/2πT [ln (r / r_w)]

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 R_w to R and bounded by the upper and lower impermeable boundaries,

Q

Q_w = 2πr b K dh/dr im

After integration, the equation for piezometric head become, 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:

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

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 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.

Cone of

Depression

Due to pumping with

Infinite aquifer (no

Source)

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.

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 r_i 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 r_i to the point in question and the drawdown to the pumping well.

s(x,y) = Σ_{i = 1,n} s_i + s_PW

Pumping Near Hydro-Geologic Boundaries (Impermeable Boundary)

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.

Image well

The solution of problems involving pumping near impermeable boundary 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 impermeable boundary infinite in extent at y axis.

x

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 r_i 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/r_i) = (Q/2 π T) ln (R²/rr_i)

= (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. Writing the above equation in terms of h_r and h_w,

h_r – h_w = (Q/2 π T) ln (r/r_w) + (Q/2 π T) ln (r_i/r_w)

= (Q/2 π T) ln (rr_i/r_w²)

And for r ≈ r_i

h_r – h_w = (Q/ π T) ln (r²/r_w²)^1/2 then

h_r at any point = h_w + (Q/ π T) ln (r/r_w) which is twice the drawdown produced by single well from infinite aquifer.

d) Image Well System:

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,

Drawdown at any point in the real domain is,

s_P = s_r + s₁ – s₂ – s₃

= (Q/2πT) [ln ( R/r) + ln (R/r_i1) – ln (R/r_i2) –ln (R/r_i3)

= (Q/2πT) [ln ( r_i2r_i3 / rr_i1)

= (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,

s_P (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]}

s_P = s_r – s₁ + s₂ – s₃

s_P = = (Q/2πT) [ln ( R/r) - ln (R/r_i1) + ln (R/r_i2) – ln (R/r_i3)

= (Q/2πT) [ln ( r_i1r_i3 / rr_i2)

s_P = s_r – s₁ - s₂ + s

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 (page 28) as,

∂s/∂t = α [∂²s/∂r² + (1/r)∂s / ∂r]

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 = bK for confined aquifers and = ĤK for unconfined aquifers where Ĥ represents an “average” saturated thickness.

Observations:

• 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 t^1/2, 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 S_y>>S), 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.

The Transient Groundwater Equation:

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 W(u) u W(u) u W(u) u W(u)

1.00E-15	33.96	9.00E-12	24.86	1.00E-07	15.24	9.00E-04	6.44
2.00E-15	33.27	1.00E-11	24.75	2.00E-07	14.85	1.00E-03	6.33
3.00E-15	32.86	2.00E-11	24.06	3.00E-07	14.44	2.00E-03	5.64
4.00E-15	32.58	4.00E-11	23.36	4.00E-07	14.15	3.00E-03	5.23
5.00E-15	32.35	5.00E-11	23.14	5.00E-07	13.93	4.00E-03	4.95
6.00E-15	32.17	6.00E-11	22.96	6.00E-07	13.75	5.00E-03	4.73
7.00E-15	32.02	7.00E-11	22.81	7.00E-07	13.6	6.00E-03	4.54
8.00E-15	31.88	8.00E-11	22.67	8.00E-07	13.46	7.00E-03	4.39
9.00E-15	31.76	9.00E-11	22.55	9.00E-07	13.34	8.00E-03	4.26
1.00E-14	31.66	1.00E-10	22.45	1.00E-06	13.24	9.00E-03	4.14
2.00E-14	30.97	3.00E-10	21.35	2.00E-06	12.55	1.00E-02	4.04
3.00E-14	30.56	4.00E-10	21.06	3.00E-06	12.14	2.00E-02	3.35
4.00E-14	30.27	5.00E-10	20.84	4.00E-06	11.85	3.00E-02	2.96
5.00E-14	30.05	6.00E-10	20.66	5.00E-06	11.63	4.00E-02	2.68
6.00E-14	29.87	7.00E-10	20.5	6.00E-06	11.45	5.00E-02	2.47
7.00E-14	29.71	8.00E-10	20.37	7.00E-06	11.29	6.00E-02	2.3
8.00E-14	29.58	9.00E-10	20.25	8.00E-06	11.16	7.00E-02	2.15
9.00E-14	29.46	1.00E-09	20.15	9.00E-06	11.04	8.00E-02	2.03
1.00E-13	29.36	2.00E-09	19.45	1.00E-05	10.94	9.00E-02	1.92
2.00E-13	28.66	3.00E-09	19.05	2.00E-05	10.24	1.00E-01	1.82
3.00E-13	28.26	4.00E-09	18.76	3.00E-05	9.84	2.00E-01	1.22
4.00E-13	27.97	5.00E-09	18.54	4.00E-05	9.55	3.00E-01	0.91
5.00E-13	27.75	6.00E-09	18.35	5.00E-05	9.33	4.00E-01	0.7
6.00E-13	27.56	7.00E-09	18.2	6.00E-05	9.14	5.00E-01	0.56
7.00E-13	27.41	8.00E-09	18.07	7.00E-05	8.99	6.00E-01	0.45
8.00E-13	27.28	9.00E-09	17.95	8.00E-05	8.86	7.00E-01	0.37
9.00E-13	27.16	1.00E-08	17.84	9.00E-05	8.74	8.00E-01	0.31
1.00E-12	27.05	2.00E-08	17.15	1.00E-04	8.63	9.00E-01	0.26
2.00E-12	26.36	3.00E-08	16.74	2.00E-04	7.94	1.00E+00	0.219
3.00E-12	25.96	4.00E-08	16.46	3.00E-04	7.53	2.00E+00	0.049
4.00E-12	25.67	5.00E-08	16.23	4.00E-04	7.25	3.00E+00	0.013
5.00E-12	25.44	6.00E-08	16.05	5.00E-04	7.02	4.00E+00	0.0038
6.00E-12	25.26	7.00E-08	15.9	6.00E-04	6.84	5.00E+00	0.0011
7.00E-12	25.11	8.00E-08	15.76	7.00E-04	6.69	6.00E+00	0.00036
8.00E-12	24.97	9.00E-08	15.65	8.00E-04	6.55	7.00E+00	0.00012

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 two straight parallel streams 20 meters apart. The well is located at the middle. Pumping starts at Q = 1000 m³/d. If T = 500 m³/d , find the drawdown at the well after 10 days of pumping. Assume R = 500 m and r_PW = 0.5 m and S_ya = 0.1.

Impermeable

The transient drawdown equation is,

s = Q/4πT W(u), where,

u = r²S / 4Tt,

Taking four image wells from each side,

s_T = s_PW + (- s_iRW20 - s_iRW40 + s_iPW60 + s_iPW80 )_LEFT + ( s_iPW20 - _iRW40 – _iRW60 + s_iPW80 )_RIGHT

= s_PW - 2 s_iRW40 + 2 s_iPW80, the rest cancel each other.

The drawdown at the well is,

s_T = Q/4πT [ W(u_PW) - 2 W(u_iRW40) + 2 W(u_iRW80)]

u_PW = (0.5)² (0.1) / (4)(500)(10) =0.00000125,

u_iPW40 =(40)² (0.1 / (4) (500)(10) = 0.0008

u_iPW80 =(80)² (0.1) / (4) (500)(10) = 0.032

The total drawdown become,

s_T = (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: s_T = 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 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 distance-drawdown 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:

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)^*

u

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 that has the intercept

2.303 Q/4πT

and the slope 2.246Tt / r²S. 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), 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^*.

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 88.9 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 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 S_ya = (4) (0.814) (0.1)/ 6.2 = 0.053

0.1

10.0

1.0

Example:

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

t (min) s (meter)	1.0 0.200	2.0 0.300	3.0 0.370	4.0 0.415	5.0 0.450	6.0 0.485	8.0 0.530	10.0 0.570	12.0 0.600	14.0 0.635	18.0 0.670	24.0 0.720
t (min) s (meter)	30.0 0.760	40.0 0.810	50.0 0.850	60.0 0.875	80.0 0.925	100.0 0.965	120.0 1.000	150.0 1.045	180.0 1.070	210.0 1.100	240.0 1.200

r = 61 m and Q = 1.894 m³ /min

Plotting the given data on a semi log paper,

∆s = 0.4 m

Per Log cycle

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:

Calculate the limit of the pseudo-steady state region around the well if Q = 100 m³ / hr, S_ya = 0.09, T = 36 m² / hr and t = 10 hrs.

r² / 4αt = r² / 4 (T/S_ya)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-t_i)] for t>t_i

Or, in general,

s = 1/4πT Σ ∆Q_ii+1 W [r²S/4T(t-t_i )]

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.

t₄

DD due Q₁ @ t’ DD due Q₂ – Q₁@ t” DD at t^’’’ due to – Q₂ Recovery

If Q₂ is zero then, s = (Q/4πT) W [ r²S/4T (t-t_i)]

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

Buildup

Recovery