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### AGE 411: Hydrology

Introduction

Hydrology is an earth science. It encompasses

the occurrence, distribution, movement and properties of the waters of the earth and their

environmental relationships.

Introduction Contnd:Hydrologic Cycle.

- Is a continuous process by which water is transported from the oceans to the atmosphere, to the land and back to the sea.
- Many such cycles exist
- the evaporation of inland water and its subsequent precipitation over land before returning to the ocean is one example.

Hydrologic Cycle Contnd.

- the driving force for the global water transport system is provided by the sun, which furnishes the energy required for evaporation.

The complete water cycle is global in nature.

World water problems require studies on regional, national, international, continental and global scales.

PROBLEM? – Practical significance of the fact that the total supply of fresh water available to the earth is limited and very small compared with the small water content of the ocean has received little attention. Thus water flowing in one country can not be available at the same time for use in other regions of the world.

Hydrologic Cycle Contnd.

- Water resources are a global problem with local roots. Therefore, developing techniques to control weather must receive careful attention, since climatologically changes in one area can profoundly affect the hydrology and therefore the water resources of other regions.

Introduction Contd.

- The Hydrologic Budget:
Because the total quantity of water available

to the earth is finite and indestructible,

the global hydrologic system may be looked

upon as closed.

However, open hydrologic subsystems are abundant and these are usually the types analyzed by the engineering hydrologist.

Introduction Contd.

- Hydrologic Budget Contd.
For a system, a water budget can be developed to account for the hydrologic components.

Assumptions:

a). Perfect plane

b). Completely impervious plane

c). Confined on all 4 sides with an outlet

Introduction Contd.

- Hydrologic Budget Contd.
As a perfect plane, it means there is no depressions in which water can be stored. If rain input is applied, and output designed as outflow, surface runoff will be developed at A.

Then the hydrologic budget for an open system can be represented by:

I – Q = ds/dt

Where; I = inflow/unit time, Q = outflow/unit time

ds/dt = change in storage in the system/unit time.

Introduction Contd.

- Hydrologic Budget Contd.
Until a minimal depth is accumulated on the

Surface, outflow can not occur, but as storm intensifies, the depth retained on the surface increases. At the cessation of input, water held on the surface becomes the outflow from the system.

For this example, all input becomes output,

neglecting the small quantity bonded in the surface

Introduction Contd.

- Hydrologic Budget Contd.
This is always misleading as will be seen later that the terms in the equation can not always be adequately or easily quantified.

A more generalized version explains all the

components of the hydrologic cycle.

Hydrologic budget for a region can be written as:

P – R – G – E – T = S ( Basic equation)

where; P = precipitation, R = Runoff,

G = Groundwater flow, E = Evaporation,

T = Transpiration, S = storage.

Introduction Contd.

- Application:
Question: In a given year, 12,560km2

Watershed received 50.8cms of precipitation

The annual rate of flow measured in the river draining the area was found to be 20.57cms. Make a rough estimate of the combined amounts of water evaporated

and transpired from the region during the year.

Introduction Contd.

- Solution:
P – R – G – E – T = S

since evap. and transp. can be combined,

ET = P – R – G - S

Assumptions:

1). Drainage area is quite large, G component may be assumed Zero.

2). That S = 0

Then ET = P – R = 50.8 – 20.57 = 30.23cm/year

Introduction Contd.

- Question (2):
The storage in a river reach at a given time is 19,734.m3. At the same time, the inflow is 14.15m3/s and the outflow is

19.81m3/s. One hr. later, the inflow rate is

19.81m3/s and outflow is 20.942m3/s.

Determine the change in storage in the reach that occurred during the hour. Is the storage at the end of the hour > or < the original value?

Introduction Contd.

- Solution to Question(2):
I – O = S/T ( continuity for the time interval)

(I1 +I2)/2 – (O1 + O2)/2 = (S2 – S1)/t

14.15+19.81–19.81 + 20.94 = s2 -19,734m3

3600sec

Precipitation

- Occurs primarily in the form of:
1). Rain 2) Snow 3). Drizzle and 4). Hail

1). Rainfall

- An extremely variable quantity.
- Problem? – How to deal with variability
- Rainfall is random and stochastic
- Rainfall is variable both spatially and temporarily i.e. varies in space and time.

Precipitation Contd.

- Rainfall Contd.
- Formation: There are three types of formation:
a). Convective occurrence – this is primarily due to a lifting mechanism by which warm light air rises and mixes with colder denser air.

b). Cyclonic formation: This occurs when lifting mechanism occurs because of air rising and meeting with low pressure area

eg: cyclone.

c). Orographic formation: This is due to lifting of air masses over barriers such as mountains.

Precipitation Contd.

- Rainfall Contd.
Measurement – There are basically two ways of measurement:

- Non – recording gauges.
What is obtained is the depth of rain that occurs on periodic basis eg:- standard rain gauge with no time history of the event.

2). Recording gauges

This gives rainfall rate because it gives a continuous record of the event with time.

i). Tipping bucket; ii). Weighing bucket, iii).Radar type

Data useful from recording gauges.

-good for rainfall runoff modeling

- gives better information due to rainfall variability

Precipitation Contd.

- Rainfall Contd.
- Error Analysis:

Main physical parameters causing error are:

1). Evaporation, 2). Inclination, 3). Splashing

4). Adhesion, 5). Wind

- Given any basin, there is problem of estimating point rainfall and basin precipitation. There is also problem of estimating Basin precipitation based on the data available in the Basin.
- Point rainfall is needed for:
a). Estimating missing data – It may mean record not collected due to measurement problems.

b). Constructing depth-duration frequency curves called DDF and Intensity-duration frequency curves called IDF.

Precipitation Contd.

- Rainfall contd.
- Estimating Point Rainfall:
In estimating point rainfall, the method of weighted averages is often used. This gives depth of rainfall that is smaller than the greatest amount and larger than the smallest amount of the area.

i.e. lies between Pmax and Pmin of the area.

Precipitation Contd.

- Estimating Point Rainfall Contd.
Method: Consider that rainfall is to be calculated for point A. Establish a set of axes running through A and determine the absolute coordinates of the nearest surrounding points BCDEF. The estimated precipitation at A is determined as a weighted average of the other five points. The weights are reciprocals of the sums of the squares of X and Y, that is

D2 = X2 + y2 and W = 1/D2

The estimated rainfall at the point of interest is given by: ( P x W )/ W

Precipitation Contd.

- Example of Point Rainfall Determination:
Point Rainfall X Y X2 + Y2 D2 W P x W

( mm ) (m) (m)

A - - - - - - -

B 160 40 20

C 180 10 16

D 150 30 20

E 200 30 30

F 170 20 20

Sums W P x W

Estimated Precipitation of A = PxW / W

Precipitation Contd.

- Method of Estimating Point Rainfall:
1). Draw X and Y axis through the interested point

2). Mark up the distances along the X and Y axis.

3). Put them down (distances along the ordinates)

4). Calculate D2 = X2 + Y2

5). Calculate W = 1/D2

6). Have the sum of (PxW) i.e. (PxW)

7). Rainfall at the point = (PxW)/W

Aerial Precipitation

- For most hydrologic analyses, it is important to know the aerial distribution of precipitation. Usually, the average depth for representative portions of the watershed are determined and used for this purpose.
- The most direct approach is to use the arithmetic average of gauged quantities. This procedure is satisfactory if gauges are uniformly distributed and the topography is flat.
- Other commonly used methods are: isohyetal and Thiessen methods.

Aerial Precipitation Contd.

- The reliability of rainfall measured at one gauge in representing the average depth over a surrounding area is a function of:
1). the distance from the gauge to the centre of the representative area.

2). the size of the area

3). topography

4). the nature of the rainfall of concern (storm event vs. mean, monthly or annual)

5). local storm characteristics.

Probabilities of Handling Rainfall Data.

- Rain is random (Stochastic) – element of uncertainty.
- To get the probability, we use:
a). Plotting position- to give an idea of the possibility.

b). Carry out probability analyses.

c). Frequency analyses.

These give probability of P, associated with the magnitude of an event equal or exceeded in a given period of time.

The probability concept gives a useful way of presenting hydrologic data. To calculate the probability of an event, one often uses the ‘Weibull’s formula in the form:

P = m/n+1, where: m = rank or order number, n = number of years of record. From the probability, the return period is calculated.

Tr = 1/P; or n+1/m; where: Tr is return period.

Example on Probability of Event

- Given a Rainfall data, say for 30 years:
Year Rank Annual Tr(1/P) Probability

totals

. . . . .

. . . . .

- On a log probability paper, plot annual totals against probability; one can then estimate the once in 50 years or 100 years or 75 years etc.
- Note: Rank the values from max. to min.
- This is useful when you want values for draught periods i.e. for irrigation requirements etc.

Question on Probability of Event

- The following are annual rainfall totals for 33 years in mm. 108, 113, 133, 180, 115, 163, 320, 110, 230, 118, 123, 115, 118, 100, 145, 200, 238, 230, 153, 245, 160, 463, 115, 118, 100,145, 200, 238, 230, 153, 245, 160, 468.
- Calculate return periods for the 33 years record. Plotting probabilities and return periods on the graph provided, estimate return periods for: 120, 180, 215 and 370 annual totals.

Consistency of data

- This is used to determine if there is any trend in the rainfall data.
- To do this, take as many number of gauging stations as possible from a Basin.
- Taken seven of the gauging stations as base group.
- Test each one of the stations against the base group.

Consistency of data Contd.

- How to test?
1). With annual rainfall data, take the 7

stations and find the annual means for

the number of years record available.

2). Plot cumulative annual data of station 1

against the cumulative annual data of the

base group.

3). Find the slope. If it is not a straight line, then the data is inconsistent, due to certain conditions that are not known.

4). Note where the curve changes, take the slope of the curve before the change and the slope after the change.

AF = Slope of change after/ slope of change before

AF = average factor

5). Multiply each of the annual values for the station by AF.

6). Repeat the process for each station.

Stream Flow

- Rainfall can be considered as input function and stream flow as output function.
- Surface water hydrology deals with the transfer of water along the earth surface. Satisfactory surface water flow rate and quantity are highly important to such fields as municipal and industrial water supply, flood control, stream flow forecasting, reservoir design, navigation, irrigation, drainage, water quality control, water based recreation and fish and wildlife management.

Stream Flow Contd.

- The relationship between rainfall and runoff is usually complex and is influenced by various storm pattern, antecedent and basin characteristics. Because of these complexities and the frequent paucity of adequate data, many approximate formulae have been developed to relate rainfall and runoff.
- After depression storage, then comes detention storage. Detention storage is the formation of temporary thin films of water over the land surface. When detention storage occurs, then overland flow occurs and goes into the stream.

Stream Flow Contd.

- Overland flow is described in the form:
Q = C yn

where; Q is the flow (m3/s),

C is Basin Characteristics

y is depth of overland flow

n is Manning’s coefficient of

roughness.

Hence, the stream flow is composed of:

- rainfall; 2). Interflow and 3). Subsurface flow.
They all produce what is known as stream flow hydrograph.

Basically, the hydrograph is affected by: 1). Rainfall

2). Basin characteristics and 3). Antecedent soil moisture.

The higher the Antecedent moisture, the more the runoff.

Stream Flow Measurement

- 1). The most common method is the slope
area method based on Manning’s

equation V = 1/n R2/3 S1/2, from where

we get; Q = AV

2). The second method is by chemical gauging

Stream Flow Measurement Contd.

3). The third method is by stream gauging

- a). Break up the channel up to about 20 to 30 segments

- b). Find the average velocity between

each segment

- c). Multiply each area by the average

velocity in each segment.

Basin Characteristics Affecting Runoff

- The nature of stream flow in a region is a function of the hydrological input to that region and the physical, vegetative and climatic characteristics of that region.
- Sr = f (hydrological, physical, vegetative, climatic)
- The character of the soil determines to a large extent the storage system into which precipitated water will enter.

Basin Characteristics Affecting Runoff Contd.

- Streams are classified as being young, mature or old on the basis of their ability to erode channel materials.
- Young streams are highly active and usually flow rapidly or that they are continually cutting their channels.
- Mature streams are those in which the channel slope has been reduced to the point where flow velocities are just able to transport incoming sediments and where the channel depth is no longer being modified by erosion.

Basin Characteristics Affecting Runoff Contd.

- A stream is classified as old when the channels in its system have become graded. The flow velocities of old streams are low due to gentle slopes that prevail.

Antecedent Precipitation

- Soil moisture relationships normally have a soil moisture index as the variable, since direct measurements of actual antecedent soil moisture are not generally practical.
- Indexes that have been inserted are groundwater flow at the beginning of the storm, antecedent precipitation and basin evaporation.
- Groundwater values should be weighted to reflect the effects of precipitation occurring within a few days of the storm, because soil moisture changes from previous rains may be important.

Antecedent Precipitation Contd.

- Pan evaporation measurements can be employed to form soil moisture figures, since evaporation is related to soil moisture depletion.
- Antecedent precipitation indexes (API) have probably received the widest use because precipitation is readily measured and related directly to moisture deficiency of the basin.

Antecedent Precipitation Contd.

- A typical Antecedent precipitation index is:
- Pa = aP0 + bP1 + cP2 ……………
where: Pa is antecedent precipitation index

P0, P1, P2…. = the amounts of rainfall during the present period and for two preceding periods in question. This index links annual rainfall and runoff values. Coefficients a, b, and c are found by trial and error or other fitting techniques to produce the best correlation between runoff and the antecedent precipitation index. The sum of the coefficients must be 1.

Antecedent Precipitation Contd.

- For an individual storm, the following API is proposed for use:
- Pa = b1P1 +b2P2 + ……. btPt where subscripts of P refer to precipitation which occurs many days prior to the given storm and the constants b 1 are assumed to be a function of t, values for the
coefficients can be determined by correlation techniques.

- In daily evaluation of the index, bt is considered to be related to t by:
bt = kt; where k is a recession constant normally reported in the range 0.85 to 0.98.

Antecedent Precipitation Contd.

- The initial value of the API (Pa0) is coupled to the API t days later (Pat) by:
- Pat = Pa0kt
- To evaluate the index of a particular day based on that of the preceding one, the above equation becomes;
- Pa = k Pao because t=1

Question on Antecedent Precipitation

- Precipitation depth Pi for a 14 day period are listed below. The API on April 1 is 0.00
use k = 0.9 to determine the API for each successful day.

Date (i) 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Prep. (Pi) 0.0 0.0 0.5 0.7 0.2 0.1 0.0 0.1 0.3 0.0 0.0 0.6 0.0 0.0

Antecedent Precipitation Contd.

- Solution:
- Equation of API reduces to:
APIi = k(APIi-1) + Pi

Date (i) 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Prep. (Pi) 0.0 0.0 0.5 0.7 0.2 0.1 0.0 0.1 0.3 0.0 0.0 0.6 0.0 0.0

APIi 0.0 0.0 0.5 1.15 1.24 1.22 1.10 1.09 1.28 1.15 1.04 1.54 1.39 1.25

Hydrograph Analysis

- A knowledge of the size and time distribution of stream flow is essential to many aspects of water management and environmental planning.
- A hydrograph is a continuous graph showing the properties of stream flow with respect to time, normally obtained by means of a continuous strip recorder which indicates stage versus time and is then transformed to a discharge hydrograph by application of a rating curve.

Hydrograph Analysis Contd.

- Components of Hydrograph
A hydrograph has 4 components:

1). Direct surface runoff

2). Interflow

3). Groundwater or base flow

4). Channel precipitation

The rising portion of the hydrograph is known as concentration curve.

The region in the vicinity of the peak is called the crest segment and the falling portion is the recession.

In most hydrograph analysis, interflow and channel precipitation are grouped with surface runoff rather than been treated independently.

Base Flow Separation Techniques

Several methods for base flow separation are used when the actual amount of base flow is unknown.

During large storms, the maximum rate of discharge is first slightly affected by base flow and inaccuracies in separation are fortunately not important. The separation techniques are given below:

1). The simplest base flow separation technique is to draw a straight line from the point at which surface runoff begins, point A in the figure below to an intersection with the hydrograph recession as shown at point B.

2). A second method projects the initial recession curve downwards from A to C which lies directly below the peak rate of flow. Then point D on the hydrograph representing N days after the peak, is connected to point C by a straight line defining the ground water component. The choice of N is based on the formula:

N = A0.2 where; N = the time in days and A is the drainage area in m2.

Base Flow Separation Techniques Contd.

- 3). A third procedure employs a base flow recession curve that is fitted to the hydrograph and then computed in a time decreasing direction. From the diagram shown, point F where the computed curve begins to deviate from the actual hydrograph marks the end of direct runoff. The curve is projected backwards arbitrarily to some point E below the inflection point and its shape from A to E is arbitrarily assigned.

Base Flow Separation Technique Contd.

- 4). A fourth widely used method is to draw a line between A and F.
- Note: All of these methods are approximate since the separation hydrographs are partly subjective procedures.

Time Base of the Hydrograph

- The time base of a hydrograph is considered to be the time from which the concentration curve begins until the direct runoff component reaches zero.
- An equation for the time base may be written as:
- Tb = ts + tc; where: Tb is time base of the hydrograph, ts is the duration of runoff producing rain and tc is time of concentration.
- The hydrograph time base related to both storm and basin characteristics is a basic element of unit hydrograph theory.

UNIT HYDROGRAPH

- Unit hydrograph represents a unit direct runoff for a specified duration over the entire watershed.
- Concept:
1) divide the hydrograph into units

2) subtract base flow from each segment

3) divide the answer by the total hydrograph.

Time Lag of Hydrograph

- Time lag – The difference in time between the centre of mass of effective rainfall and the centre of mass of runoff produced.
- Time lag is influenced by the shape of the drainage area, the slope of the main channel geometry and storm pattern.

Tutorial Question on Unit Hydrograph

- Actual discharge for a flood hydrograph passing the point of concentration for a 600 ha drainage basin are given below. The flood was produced by a uniform rainfall of 5cms/hr which started at 9.00am and abruptly ended at 11.00am and resulted in 5cms of direct surface runoff. The base flow prior to , during and after the storm was 100cm3/sec.
Time 8 9 10 11 12 1pm 2 3 4 5 6 7

Discharge 100 100 300 500 700 800 600 400 300 200 100 100

a). At what time did direct surface runoff begin and cease?

b). Derive the 2 – hr unit hydrograph ordinates for each time listed.

Example on Unit Hydrograph Contd.

- Solution: a) Direct surface runoff began at 9.00am and ceased at 6.00pm.
- b) Time flow base direct 2-hr unit
(cm3) flow runoff hydrograph(÷5)

8 100 100 0 0

9 100 100 0 0

10 300 100 200 40

11 500 100 400 80

12 700 100 600 120

1pm 800 100 700 140

2 600 100 500 100

3 400 100 300 60

4 300 100 200 40

5 200 100 100 20

6 100 100 0 0

7 100 100 0 0

Flood Routing

- gives some information on the movement of the flood wave along a river reach

- takes into effect, the storage within a river reach or channel system

- always done in river systems and reservoirs

- The two methods used for routing are:
- 1). Hydrologic methods
- uses the routing equation

- uses the continuity equation

2). Hydraulic routing

- continuity equation, Q = AV

- dynamic equation, V = 1/n R2/3 S1/2

Flood Routing Contd.

- Hydrologic River Routing:
-Hydrologic River Routing are all founded upon the equation of continuity in the form:

I – O = ds/dt; I is inflow; O is

outflow; ds/dt is rate of change

in storage within the

reach.

I1 + I2/2 = O1 + O2/2 + S2 – S1/t

Flood Routing Contd

- Muskingum Method:
Storage in a stable river reach can be expected to depend primarily on the discharge into and out of a reach and upon the hydraulic characteristics of the channel section.

The storage within the reach can be expressed in the form:

S = b/a (xIm/n + (1-x)Om/n) --------------- (1)

Constants a and n reflect the storage discharge characteristics and b and m mirror the storage volume characteristics. The factor x defines the relative weights given to inflow and outflow within the reach.

The method assumes that m/n =1 and let b/a =k then (1) becomes: S = k( xI + (1-x) O)------------ (2)

k = storage time constant for each reach, x is weighing factor that varies between 0 and 0.5 for a given reach.

Flood Routing Contd.

- Muskingum Method Contd.
Application of the equation has shown that k is reasonably close to the travel time within the reach and x will average about 0.2.

If k and x are known, routing procedure begins by dividing time into a number of equal increaments t and express equation (1) in finite difference form.

Flood Routing Contd.

- Muskingum Method Contd.
The routing time t is normally assigned any convenient value between the limits k/3 to k. Using subscipts 1 and 2 to denote the beginning and the ending times for the interval t, the storage change in the river reach during the routing interval is:

S2 – S1 = k(x(I2 – I1) + (1-x)(O2-O1))……(6)

Substituting into I1+I2/2 – O1+O2/2 = S2-S1/2 results in Muskingum routing equation:

O2 + C0I2 + C1I1 + C2O1

Flood Routing Contd.

- Muskingum Method Contd.
where:

C0 = (-k x +0.5t)/( k – k x + 0.5t)

C1 = (kx + 0.5t)/ (k – k x + 0.5t)

C2 =( k – kx – 0.5t)/ (k – k x + 0.5t)

Note: k and t must have the same time units and the coefficients must sum up to 1.0

Since I1and I2 are known for every time increament, routing is accomplished by solving equation (6) for successive time increament using O2 as O1 for the next time increament.

Flood Routing Contd.

- Tutorial Question:
Perform the flood routing for a reach of river given x = 0.2 and k = 2 days. The inflow hydrograph using t = 1 day is shown in the table below. Assume equal and outflow rate on the 16th. From the inflow hydrograph given, calculate the outflow rate on the 26th.

Date Inflow CoI2 C1I1 C2O1 Comp. outflow

- 4260 - - - 4260
- 7646 364 1823 2232 4419
- 11,167 532 3272 2315 6110
- 16,730
- 21590
- 20,950
- 26,570
- 46,000
- 59,960
- 57,740
- 47,890
- 34,460
- 21,660
- 34,680
- 45,180

Ground Water Hydrology

- The quantification of the volume and rate of flow of groundwater in various regions is an exceedingly difficult task because volume and flow rates are determined to a considerable extent by the geology of the region.
- The character and arrangements of rocks and soils are important factors, and these are often highly variable within a ground water reservoir.
- An additional difficulty is the inability to measure directly many critical geologic and hydraulic reservoir characteristics.

Ground Water Hydrology Contd.

- Ground Water Flow:
Understanding the movement of ground water requires a knowledge of the time and space dependency of the flow, the nature of the porous medium and fluid and the boundaries of the flow system.

Ground water flows are usually 3-dimensional, unfortunately, the solution of such problem by analytical methods is intensely complex unless the system is symmetric.

Fluid properties such as velocity, pressure, temperature, density, and viscosity often vary in time and space.

When time dependency occurs, the issue is termed an unsteady flow problem, and solutions are always difficult.

Ground Water Hydrology Contd.

- On the other hand, situations where space dependency alone exists are steady state flow problems.
- Porous media through which ground water flow may be classified as:
1). Isotropic medium – This has uniform properties in all directions from a given point.

2). Anisotropic medium – This has one or more properties that depend on a given direction, eg: permeability of the medium might be greater along a horizontal plane than along a vertical one.

3). Heterogeneous medium – This has non uniform properties of anisotropy or isotropy.

4). Homogeneous media – These are uniform in their characteristics.

Ground Water Hydrology Contd.

- Subsurface Distribution of Water:
Ground water distribution may be generally categorized into zones of aeration and saturation. The saturated zone is one in which all voids are filled with water under hydrostatic pressure. In the zone of aeration, the interstices are partly filled with air and partly filled with water.

Ground Water Hydrology Contd.

The saturated zone is commonly called the “ GROUND WATER ZONE’’. The zone of aeration may ideally be subdivided into several sub zones:

1). Soil water zone – begins at the ground surface and extends downwards through the major root band. Its total depth is variable and dependent upon soil type and vegetation. The zone remains unsaturated except during periods of heavy infiltration. Three categories of water may be encountered in the region:

Hygroscopic water – which is adsorbed from the air

Capillary water – held by surface tension

Gravitational water – excess soil water draining through the soil.

Ground Water Hydrology Contd.

2). Intermediate zone – This belt extends from the bottom of the soil water zone to the top of the capillary fringe and may change from non existence to serious hundred meters in thickness. The zone is essentially a connecting link between a near ground surface region and the near water table region through which infiltrating fluids must pass.

3). Capillary zone – A capillary zone extends from the water table to a height determined by the capillary rise that can be generated in the soil. The capillary band thickness is a function of soil texture and may fluctuate not only from region to region but also within a local area.

4). Saturated Zone – In the saturated zone, groundwater fills the pores spaces completely and porosity is therefore a direct measure of storage volume. Part of this water can not be removed by pumping or drainage because of molecular and surface tension forces. This is called specific retention. Specific retention is the ratio of water retained against gravity drainage to gross volume of the soil.

Ground Water Hydrology Contd.

- A knowledge of the distribution and nature of geohydrologic units such as aquifers, aquifuges and aquicludes is essential to proper planning for development or management of ground water supplies.
- An aquifer is a water bearing stratum or formation that is capable of transmitting water in quantities sufficient to permit development. Aquifers may be considered falling into two categories: confined and unconfined depending upon whether or not a water table or a free water surface exists under atmospheric pressure.
- An aquifuge is a formation that is impermeable and devoid of water.
- An aquiclude is an impervious stratum.

Ground Water Hydrology Contd.

- Water that can be drained from a soil by gravity is known as specific yield. It is expressed as the ratio of the volume of water that can be drained by gravity to the gross volume of soil.
Values of specific yield depend upon the soil particle size, shape and distribution of pores and degree of compaction of the soil.

Ground Water Hydrology Contd.

- Darcy’s Law:
Darcy’s law for fluid flow through a permeable bed is stated as:

Q = - KA dh/dx

A is total x-sectional area including the space occupied by the porous material

K is the hydraulic conductivity of the material

Q is the flow across the controlled area

h = z + p/ + C

h – the piezometric head

z – the elevation above datum

p – hydrostatic pressure

C – an arbitrary constant

Ground Water Hydrology Contd.

- If the specific q = (Q/A) is substituted for:
q = - k d/dx ( z + p/ ), Darcy’s law is used in groundwater flow problems. It is limited in applicability to cases where the Reynolds # is on the order of 1.

For Reynolds # 1, Darcy’s law is invalid.

Reynolds # NR = qd/

q = specific discharge

d = mean drain diameter

= fluid density

= dynamic viscosity

Ground Water Hydrology Contd.

- Steady Unconfined Radial Flow towards a Well:
Here, flow is assumed to be radial, the original water table is considered to be horizontal

- The well is presumed to fully penetrate the aquifer of infinite aerial extent
- Steady state condition must prevail.

Ground Water Hydrology Contd.

- Steady Unconfined Radial Flow Contd.
The flow towards the well at any distance x away must equal the product of the cylindrical element of area at that section and the flow velocity. With the Darcy’s law, this becomes:

Q = 2 πx y Kf dy/dx

where: 2πxy = the area through any cylindrical shell in m2 with the well as its axis.

Kf = hydraulic conductivity (m/sec)

dy/dx = water table gradient at any distance x (m)

Q = well discharge (m3/Sec)

Ground Water Hydrology Cont (d.

Integrating over the limits specified;

r2 h2

Qdx/x = 2πKf ydy

r1 h1

Q ln r2/r1 = 2πKf ( h22 – h12 )/2

Q = Kfπ ( h22 – h12 )/ ln ( r2/r1 )

Converting Kf to field unit of gpd/m2 and ln to log;

Kf = 96.327 Qlog (r2/r1)/ h22 – h12

Ground Water Hydrology Contd.

- Tutorial Question:
A 45.7cm well fully penetrate an unconfined aquifer of 30.47m depth. Two observation wells located 30.47m and 71.62m from the pumped well are known to have draw downs of 6.76m and 6.40m respectively. If the flow is steady and Kf = 4.99gpd/m2. What will be the discharge?

I

Ground Water Hydrology Contd.

- Solution:
Using the equation:

Qln r2/r1 = 2πKf(h22 – h12)/2

Q = πKf (h22 – h12)/ ln (r2/r1)

Converting Kf to the field unit, and loge to log10

Kf = 96.327Q log10 (r2/r1) / h22 - h12)

Q 2.40 gpm

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