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.
Hydrology is an earth science. It encompasses
the occurrence, distribution, movement and properties of the waters of the earth and their
- the evaporation of inland water and its subsequent precipitation over land before returning to the ocean is one example.
- 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.
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.
For a system, a water budget can be developed to account for the hydrologic components.
a). Perfect plane
b). Completely impervious plane
c). Confined on all 4 sides with an outlet
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.
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
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.
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.
P – R – G – E – T = S
since evap. and transp. can be combined,
ET = P – R – G - S
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
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?
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
1). Rain 2) Snow 3). Drizzle and 4). Hail
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
c). Orographic formation: This is due to lifting of air masses over barriers such as mountains.
Measurement – There are basically two ways of measurement:
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
- Error Analysis:
Main physical parameters causing error are:
1). Evaporation, 2). Inclination, 3). Splashing
4). Adhesion, 5). Wind
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.
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.
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
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
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
1). the distance from the gauge to the centre of the representative area.
2). the size of the area
4). the nature of the rainfall of concern (storm event vs. mean, monthly or annual)
5). local storm characteristics.
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.
Year Rank Annual Tr(1/P) Probability
. . . . .
. . . . .
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
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.
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
Hence, the stream flow is composed of:
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.
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
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
- c). Multiply each area by the average
velocity in each segment.
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.
coefficients can be determined by correlation techniques.
bt = kt; where k is a recession constant normally reported in the range 0.85 to 0.98.
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
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
A hydrograph has 4 components:
1). Direct surface runoff
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.
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.
1) divide the hydrograph into units
2) subtract base flow from each segment
3) divide the answer by the total hydrograph.
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.
(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
- 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
- uses the routing equation
- uses the continuity equation
2). Hydraulic routing
- continuity equation, Q = AV
- dynamic equation, V = 1/n R2/3 S1/2
-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
I1 + I2/2 = O1 + O2/2 + S2 – S1/t
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.
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.
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
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.
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
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.
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 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.
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.
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.
Values of specific yield depend upon the soil particle size, shape and distribution of pores and degree of compaction of the soil.
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
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
Here, flow is assumed to be radial, the original water table is considered to be horizontal
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)
Integrating over the limits specified;
Qdx/x = 2πKf ydy
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
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?
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