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P R E C I P I T A T I O N
G E O L O G YP recipitation refers to all form of liquid or solid water particles that form in the atmosphere and then fall to the Earth’s surface.
Types of Precipitation:
HailSleetSnowRainDrizzle
Formation: The formation of precipitation may occur at temperatures above and below the freezing. Warm Precipitation-is formed at temperatures entirely above freezing.
Cold Precipitation-Involves ice and at stage of the process
Problems
1. Assuming rain falling vertically, express the catch of the gage inclined 20° from the vertical as a percentage of the catch for the same gage installed vertically.
Solution:
% catch = cos 20° (1/1) x 100%
% catch = 94%
2. With a z value of 300,000 mm / m , what are the rainfall rates, in mm/hr, indicated by z-R relationships with a and b values of
a. 200 and 1.6 ?b. 300 and 1.4 ?
Solution:
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z = aR
a. a = 200 b. a = 300b = 1.6 b = 1.4
z = aR z = aR
R = ( z/a) = 300,000 mm / m R = (z/a) = 300,000 mm/m 200 300
R = 96.62 mm/hr R = 138.95 mm/hr
3. Precipitation station X was inoperative for part of a month during a storm occurred. The respective storm totals at three surrounding stations, A,B and C were 98, 80, and 110mm. The normal annual precipitation amounts at stations X, A, B, and C are, respectively, 880, 1008, 842 and 1080 mm. Estimate the storm precipitation for station X.
Solution:
Px = 1/3 P + P + P Nx N N N
Px = 880/3 ( 98/1008 + 80/842 + 110/1080)
Px = 86.27 mm
4. The average annual precipitation for the four sub-basins constituting a large river is 73, 85, 112, and 101 cm. The areas are 930, 710, 1090 and 1680 km² , respectively. What is the average annual precipitation for the basin as a whole ?
Solution:
P A = P A + P A + P A + P A
A = ( 930 + 710 + 1090 + 1680) km² = 4410 km²
P = [ 73 (930) + 85 ( 710) + 112 ( 1090) + 101 (1680) ] / 4410 km²
P = 95.24 cm
5. Compute the mean annual precipitation for the data shown below. Use the arithmetic average and Theissen network. Then compare your answers. Which of the two is the most accurate ?
Gage Station Precipitation (mm) Area (m²) A 81.5 120,000 B 73.00 200,000 C 75.25 200,000 D 76.25 280,000
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By Arithmetic average:
P = P + P + P + P = ( 81.5 + 73 + 75.25 + 76.25 ) mm 4 4
P = 76.5 mm
By Theissen Method:
P = P A + P A + P A + P A + P A A
P = 81.5 ( 120000 ) + 73 ( 200000 ) + 75.25 ( 200000 ) + 76.25 ( 280000 ) 800,000
P = 75.98 mm
Theissen Method is more accurate than that of Arithmetic Average.
STREAMFLOW
WATER STAGE
Manual Gages
River Stage – is the elevation above some arbitrary zero datum of the water surface at
the station. The datum is sometimes taken as mean sea level but more often it is slightly
below the Point of zero flow in the stream.
The simplest way to measure the river stage is by means of staff gage, a scale set
so that a portion of it is immersed in water at all times.
If no suitable exist in a location accessible at all stages, a sectional staff gage may
be used.
Wire-weight gage – has a drum with circumference such that each revolution unwinds
30 cm of wire. A counter records the number of revolutions of the drum while a fixed
reference point indicates tenths of a millimeter on a scale around the circumference.
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Recording Gages
Continuous – chart recorder – motion of the float moves a pen across a long strip chart.
When the pen reaches the edges of the chart, it reverses direction and
records in the other direction across the chart.
Crest-stage Gage
- Crest gage provide low cost, supplementary records of crest stage at locations where
records are not justified and where manually read staff gages are inadequate. The
gage used by the U.S. Geological survey consists of a length of pipe containing a
graduated stick and a small amount of a ground cork.
Miscellaneous stage gages
Water or mercury-filled manometers can be used to indicate reservoir water levels or to
actuate recording devices.
Selection of Station Site
The relation between stage and discharge is controlled by the physical features of the
channel downstream from the gage. When controlling features are situated in a short
length of channel, a section control exists. If the gage discharge relation is governed by
the slope, size, and roughness of the channel over a considerable distance, the station is
under the cannel control
DISCHARGE
Current Meters
Price meters – is the most common current meter in the United States.
- Consist of six conical cups rotating about a vertical axis.
Propeller –type – Current meters employ a propeller turning about a horizontal axis.
The location between revolution per second N of the meter cups and water velocity V is
given by an equation of the form.
Where b is the constant of proportionality and is the starting velocity or velocity required
to overcome mechanical friction.
Current-Meter Measurements
A discharge measurement requires determination of sufficient point velocities to permit
computation of an average velocity in the stream. Cross sectional area multiplied by
average velocity gives the total discharge.
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V = a + bN
Stage-distance relations
Rating curve or stage-discharge relations
- A calibration curve where periodic meter measurements of flow and simultaneous
stage observations provides data.
- The dispersion of the measured data about the mean rating curve should be small
(generally less than two percent)
- A larger dispersion indicates either
a. That the control shirts more or less continuously with scour and deposition in bed
and banks or the growth of vegetation
b. That the water surface slope at the control varies as the result of varying
backwater from tides, reservoirs fluctuations, or variable tributary inflow
downstream.
c. That the measurements are not carefully made
Shifting control
- Discharge is usually estimated by noting the difference between the stage
at a time of a discharge measurement and the stage on the mean rating
curve w/c shows the same discharge.
- If the correction changes between measurements, a linear variation with
time is usually assumed.
The basic approach is presented by:
Where:
Q = discharge
S = slope
M = expected to be ½
F = fall, difference in water surface elevation between two fixed sections
and is usually measured by two conventional river gages
f/l =s
k = need not to be ½ and must be determined empirically.
Slope-stage-discharge relation
- Requires a base gage and an auxiliary gage
- Gages should be far enough apart for f to be at least 30cm to minimize the
effect of observational errors.
Normal fall rating
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q/qo = (s/so)m = (f/fo)k
- Can be used when fall varies over a wide range and is correlated with stage.
- Fo is determined from a curve or equation expressing the relation between
fall and stage.
Change in stage rating
- Eliminate the need for auxiliary gage. Slope is equal to Sb + Sr, where Sb is
the slope of the channel bottom (or the slope of the water surface in uniform
flow) and sr + dg/11dt
Extension of rating curves
- There is no completely satisfactory method for extrapolating a rating curve
beyond the highest measured discharged. It is often assumed that the equation
of the rating curve is
Effect of Ice on Stream
- When ice covers a stream, a new friction surface is formed and the stream
becomes a closed conduit w/ lower discharge because of the decreased
hydraulic radius.
- The underline of the ice sheet may be extremely rough if ice cakes are tilted
helter-skelter and then frozen together.
- If the stage falls, leaving the ice is as a bridge across the stream, the stage-
discharge characteristics return to those of a free stream.
- If the stream is frozen over, the flow is usually small since there will be little
snowmelt or other source of run-off w/ in the tributary area.
-
Frazil ice – the first ice to form in the turbulent streams, small crystal are suspended in the
turbulent flow.
Anchor ice – frazil ice collecting on rocks on the streambed.
Other Methods Obtaining the Stream-flow Data
- The discharge at dams can be determined from calibration of the spill way,
sluiceway, and turbine gates.
- If a record of gate and turbine operation is maintained, the discharged can be
computed.
- On small streams, flow measurements may be made w/ weirs or flumes.
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Q = k (g –a )b
Slope-area computation
- Procedure used in estimating flow by application of hydraulic principles.
Chezy-manning formula – ordinarily used to compute discharge
Where:
n = roughness coefficient
Average value for natural stream is about 0.035, an error of 0.001
represents about 3% in discharge
Moving-Boat-methods – a boat traverses the streams at a constant speed on a course normal to
the flow.
Ultrasonic and electro magnetic methods – can provide continuous discharge measurement.
Ultrasonic method - sonic pulses are emitted from transducers on opposite banks and located on
a line about 450 from the direction of flow.
- The procedure is said to be capable of accuracies with in plus or minus 2%.
Planning and stream flow network
- The design of a stream flow network is a problem both of statistical sampling over
area and of the locations where data and most likely to be needed.
THREE TYPES OF STATIONS
Operational Station – required for stream flow forecasting, project operation, water allocation,
etc.
Special Stations – installed to secure data for a project investigation, special studies or research
location is determined by the special need, and they are operated until the
study is completed.
Basic Data Stations – operated to obtained data for future used. The time and nature of this
future use are usually unknown when the station is established.
Bench Mark Stations – should be maintained permanently on all streams that are substantially
unaffected by people.
INTERPRETATION OF STREAMFLOW
Water Years
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Q = 1/n ( AR2/8S1/2)
It is desirable to treat annual stream flow data in such way that the flood season is not
divided between successive years. Various water years have been used for special purposes: the
U.S. Geological Survey uses the water year, October 1 to September 30, for data publication.
Water years are customarily designated by the ending year, i.e., water year 1975 ends September
30, 1975.
Hydrographs
- is a graph of stage versus time
- a graph showing the change over time in the amount of water flowing down a river
Many different methods of plotting are used, depending on the purpose of the chart.
Monthly and annual mean or total flow is used to display the record of past run-off at a
station.
For detailed analysis, discharge hydrographs are plotted by computing instantaneous flow
values from the water stage recorder chart. The visual shape of the hydrograph is
determined by the scales used, and in any particular study it is good practice to use the
same scales for all floods on a given basin.
Mean Daily Flow
- Stream flow data are usually published in the form of mean daily flows from
midnight to midnight.
Adjustment of Stream flow Data
Before publication, stream flow data should be carefully reviewed and adjusted for errors
resulting from instrumental and observational deficiencies until they are as accurate a
presentation of the flow as it is possible to make.
Storage reservoirs, diversions, levees, etc., cause changes in either total flow volume or
rate of flow or both. An analysis of the effects on the record at a given stations requires a
careful search to determine the number and size of reservoirs, the number and quantity of
diversions, and the date of their construction.
Land-use changes, urbanization, deforestation, or reforestation affect streamflow and
cause apparent shifts in the flow record.
Mean Annual Runoff
The figure is a map of mean annual runoff in the United States.
Stream flow Variations
1. Variations in total runoff from year to year
2. Seasonal variations in runoff
3. Variations of daily rates of runoff throughout the year
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SAMPLE PROBLEM
Problem No. 1
Determine the flow through a trapezoidal concrete lined canal having a side slope of 3H
to 4H and bottom width of 2m if the depth of floe is 2 m. The channel is laid on a slope of 3m
per 2 km. Use n = 0.013
x x
2m 4 5 y
3
2m
Sol:
Q = A(1/n)R2/3S1/2
y = (5/4)2
y = 2.5 m
By ratio and proportion:
(2.5/5) = (x/3)
x = 1.5
A = 7 m2
R = A/P
P = 2 + 2(2.5) = 7 m
R = 7m2/7m = 1 m
S = 3/2000 = 0.0015
Q = (7)(1/0.013)(1)2/3(0.0015)1/2
Q = 20.85 m3/s ans
Problem No. 2
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A =
[2 + 2(1.5)] + 2
(2)2
Given the stream section shown in table and the following measurements, calculate the
total discharge and the average velocity throughout the station.
Measurement
Station
Distance
across stream
in (ft)
Width
Δw(ft)
Depth
D(ft)
Mean
velocity
V(ft/s)
Area
ΔwxD
Discharge
cfs
A 0 7 0 0 0 0.0
B 14 13 1.1 0.43 14.3 6.15
C 26 12 2.6 0.61 31.2 19.03
D 38 11.5 3.5 1.54 40.25 61.99
E 49 11.5 3.2 1.21 36.8 44.53
F 61 14.5 3.1 1.13 44.95 50.79
Total = 182.49 cfs
SOL:
Δw = 0.5 x (49 – 38) + 0.5 x (61 – 49) = 11.5 ft
Q = Δw + Di x vi =
Q = 11.5 x 3.2 x 1.21
Q = 44.53 cfs
Average velocity = 0.98 ft/s
Problem No. 3
For particular stream, estimate the floe rate (runoff for this case) using the following data
for velocities measured at two depths (0.2 & 0.8 of the total) and the cross-sectional area
corresponds to the velocity measures.
Section Sample
Depths
1 2 3 4 5
Velocity
(m/s)
0.5D 0.4 0.8 1.2 1.0 0.6
0.8D 0.3 0.6 1.3 1.2 0.6
Area(m2) 3 6 10 8 4
Q = ΣQ = ΣAV = ΣA(V0.2 + V0.8)/2
Q = 3(0.35) + 6(0.70) + 10(1.25) + 8(1.1) + 4(0.6)
Q = 1.05 + 4.20 + 12.50 + 8.8 + 2.4
Q = 29.00 m3/s
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EVAPORATION AND TRANSPIRATION
1. EVAPORATION
is restricted to the net rate of vapor transport to the atmosphere.
FACTORS CONTROLLING THE EVAPORATION PROCESS
Meteorological Factors. If natural evaporation is viewed as an energy-exchange process,
it can be demonstrated that radiation is by far the most important single factor and that
the term solar evaporation is basically applicable.
The rate of evaporation is influence by solar radiation, air temperature,
vapor pressure, wind, and minimally by atmospheric pressure.
Nature of Evaporating Surface. All surface exposed to precipitation, such as immersion,
bulb lines, and pave streets are potentially evaporation surfaces.
The Rate of Evaporation from soil surfaces is limited by the availability of
water or evaporation opportunity.
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Effects of Water Quality. The effect of salinity, or dissolved solids, is brought about by
the reduced vapor pressure of the solution. The vapor pressure of sea water (35,000 ppm
dissolved salts) is about 2% less than that of pure water at the same temperature.
WATER-BUDGET DETERMINATION OF RESERVOIR EVAPORATION
The direct measurement of evaporation under field conditions is not feasible, at
least in the sense that one is able to measure river stage, rain fall, etc.
Where:
E – evaporation
S – storage
I – surface inflow
O - surface outflow
Og – subsurface seepage
At lake Hefner at Oklahoma-correction as large as 10 mm per month were
required for changes in density. It was found that daily evaporation from lake Hefner,
Oklahoma, could be reliably computed from a water budget. Lake Hefner was selected,
after a survey of more than 100 lakes and reservoir as one of the 3 or 4 best meeting
water-budget requirements.
ENERGY-BUDGET DETERMINATIONS OF RESERVOIR EVAPORATIONS
The energy-budget approach, like the water budget, employs a continuity equation
and solves for evaporation as the residual required to maintain a balance.
The energy-budget equation for a lake or reservoir may be expressed as:
Where:
Qn – net (all-wave)
Qh – sensible-heat transfer (conduction) to the atmosphere
Qe – energy used for evaporation
QӨ – increase in energy stored in the water body
Qv - advected energy (net energy content of inflow and outflow elements)
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E = (S1-S2) + I + P - O – Og
Qn – Qh – Qe = QӨ – Qv
Another approach to the determination of net radiation involves the application of the
energy-budget to an insulated evaporation pan (radiation integrator).
Where:
Qir – the incident minus reflected all-wave radiation for the tank
Qir – adjacent lake
Qn – net radiation
Ťo – absolute temperature.
σ – stefan-boltzman constant (5.67 x 10-8 W m –2 K-4)
ε = 0.97
The energy content per gram of water (with respect to 00C) is the product of its specific
heat and Celsius temperature. Assuming the values of density and specific heat of water are
1000 kg/cu.m and 4.19 x 10-3MJ/ kgoC.
COMBINATION METHODS OF ESTIMATING RESERVOIR EVAPORATION
When it becomes necessary to take advection and energy storage into account, the
adjusted lake evaporation EL, can be computed from
Where:
E - evaporation
(Qv - Qθ) - net advection
ESTIMATION OF RESERVOIR EVAPORATION FROM PAN EVAPORATION AND
RELATED METEOROLOGICAL DATA
Pan Observations. There are 3 types of exposures employed for pan installations -
sunken, floating, and surface – and divergent views on the best exposure persist.
Sunken pans collect more trash; they are difficult to install, clean and repair; leaks are not
easily detected; and height of vegetation adjacent to the pan is quite ethical.
Evaporation from Pan Floating in the lake more nearly approximates evaporation from the
lake than from an – on – shore installation, but even so, the boundary effects are appreciable.
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Qir = Qir = Qn + εσ(Ťo)4
Qv – Qo = 4.19/A (ITI + PTp – OTo – OgTg – ETE + S1T1 – S2T2)
EL = E + α (Qv - Qθ)
Pans exposed above ground experience greater evaporation than sunken pans, primarily
because of the radiant energy intercepted by the wide walls, and heat exchange.
Pan Evaporation And Meteorological Factors.
1) To increase our knowledge of evaporation.
2) To estimate missing pan records.
3) To estimate data for station at which pan observations are not made.
4) To test the reliability and representativeness of observe data
5) To aid the study of lake pan relations.
Pan Coefficients. The ratio of the annual lake-to-pan evaporation. Water- budge, energy-
budget, and aerodynamic techniques can be use to estimate evaporation from existing reservoirs
and lakes.
Effects of advected energy on pan evaporation. Observations demonstrate that the
sensible-heat transfer through the pan can be appreciable and may flow in either direction,
depending upon water and air temperatures.
Making the further assumption that the pan coefficient is 0.7 when air and pan water
temperatures are equal:
Where:
p - pressure (kpa)
vp - velocity(km/day)
Ta - air temperature
To - water-surface temperature
Ep - pan evaporation
2. TRANSPIRATION
Only minute portion of the water absorb by the roots systems of plants remain in the
plants tissues; virtually all is discharge to the atmosphere as vapor through transpiration.
This is the process where an air enters the leaf and the waters escapes through the open
stomata.
FACTORS AFFECTING TRANSPIRATION
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E = 0.7[Ep ± 0.0064pαp (0.37 + 0.00255 vp){To – Ta}0.88]
The rate of transpiration is largely independent of plant type, provided there is adequate
soil water and the surface is entirely covered by the vegetation about 95 % of daily transpiration
occurs during daylight hours, compared with 75 to 90% for soil evaporation.
Wilting Point – moisture content at which permanent wilting of plants occurs.
The relative transpiration is not proportional to cover density, however, for 2 reasons;
1) An isolated plant receives radiation on the side facing the sun which would fall on an
adjacent plant were there solid cover.
2) A portion of radiation reaching the ground is subsequently transmitted to the plants
(oasis effect).
PLANT TYPES:
Xerophytes – dessert species which have fewer stomata per unit area and less surface
area expose to radiation.
Phreatophytes – have root systems reaching to the water table and transpire at rates
largely independent of moisture contents in the zone of aeration.
Mesophytes – have some ability to reduce transpiration during periods of growth.
Hydophytes – cannot pumps water into the atmosphere at rates in excess of those
controlled by available radiant and sensible energy.
MEASUREMENT OF TRANSPIRATION
Measured by Phytometer, a large vessel filled with soil in which one or more
plants are rooted.
3. EVAPOTRANSPIRATION
Total evaporation (or evapotranspiration) - the evaporation from all water, soil,
snow, ice, vegetation, and other surface plus transpiration.
Consumptive use is the total evaporation from an area plus the water use directly in
building plant tissue.
The concept of potential evapotranspiration introduced by Thornthwaite is widely
used. He defined the term as “water loss which will occur if at no time there is a
deficiency of water in the soil for the use of vegetation”.
LYSIMETER DETERMINATION OF EVAPOTRANSPIRATION
Many observations of evapotranspiration are made in soil containers variously known
as tanks, evapotranspirometer, and lysimeters. The first 2 terms customarily refer to
containers with sealed bottoms, while there has been an attempt to restrict the word
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lysimeter to containers with pervious bottoms or with a mechanism for maintaining
negative pressure at the bottom.
EQUATIONS FOR EVAPORATION COMPUTATIONS
The following equation for α, the portion of net advected energy contributing to
evaporation:
Where:
p - pressure, Kpa
To - water temperature
v4 - 4-m wind movement, km/day
The derivation of an equation for αp applicable to the Class A pan involves
additional assumptions, and the result is not particularly suited to computer use. The
following equation is an adequate approximation:
Where:
To - water temperature, ºC
vp - wind movement, km/day
ANOTHER EQUATION FOR SOLVING EVAPORATION
A commonly used empirical equation has been developed by Meyer. This equation takes
the form:
Where:
E = daily evaporation in inches depth
eo and ea = as previously defined but in units of inches of Hg
W = the wind velocity in mph measured about 25 ft above the water surface
C = an empirical coefficient
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αp = 0.34 + 0.0117To - 3.5 x 10-7(To + 17.8)3 + 0.0135(vp)0.36
For daily data on an ordinary lake, C will be about 0.36. For wet soil surfaces, small
puddles, and shallow pans, the value of C is approximately 0.50
Example Problem: Find the daily evaporating E from a lake for a day during which the
following mean values were obtained: air temperature 87˚F, water temperature 63˚F, wind speed
10 mph, and relative humidity 20%.
Table. Water Vapor Pressure at Various Temperatures
Temp(˚F)
Vapor Pressure
(in. Hg)
32 0.18
40 0.25
50 0.36
60 0.52
70 0.74
80 1.03
90 1.42
100 1.94
Interpolating from Table 3-1, we find that
60 63 70 (70 - 63) / (70 - 60) = (0.74 - x) / (0.74 - 0.52)
0.52 x 0.74 eo = x = 0.58 in Hg
80 87 90 (90 - 87) / (90 - 80) = (1.42 - x) / (1.42 - 1.03)
1.03 x 1.42 ea = x = 1.3 in. Hg (0.20) = 0.26 in. Hg
Where: C = 0.36; W = 10 mph
E = 0.36 (0.58 - 0.26)
E = 0.23 in/day x 2.54 cm/in x 10 mm/12 m = 5.85 mm/day
Example Problem: For a given month, a 300-acre lake has 15 ft3/s of inflow, 13 ft3/s outflow,
and a total storage increase of 16 ac-ft. A USGS gage next to the lake recorded a total of 1.3 in.
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precipitation for the lake for the month. Assuming that infiltration is insignificant for the lake,
determine the evaporation loss, in inches, over the lake.
Given: I = 15 ft3/s
O = 13 ft3/s
P = 1.3 in.
∆S = 16 ac-ft
Req: E = ?
Solution:
I = 2.975 ft. x 12 in./ft = 35.70 in.
O = 2.578 ft. x 12 in./ft = 30.94 in.
E = I - O + P - ∆S
E = (35.70) - (30.94) + (1.3) - (0.64) in.
E = 5.42 in.
Example Problem: A clear lake has a surface area of 708,000 m2. In May, the building brook
flows into the lakes at an average rate of 1.5 m3/s. The Meandering River flows out of clear lake
at an average rate of 1.25 m3/s. The evaporation rate was measured as 14.0 cm/mo. A total of
22.5 cm of precipitation fell in May. Seepage losses are negligible. The average depth in the lake
on May 1 was 19 m. What was the average depth on May 30th?
Given: I = 1.5 m3/s
O = 1.25 m3/s
P = 22.5 cm/mo.
E = 14.0 cm/mo.
Req: ∆S = ?
Solution:
P = (22.5 cm/mo)(1m/100 cm)(1 month) = 0.225 m
E = (14.0 cm/mo)(1m/100 cm) (1 month) = 0.14 m
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∆S = I - O + P - E
∆S = 5.49 - 4.58 + 0.225 - 0.14 = 0.995 m
The new average depth on May 30th would be: 19 + 0.995 = 19.995 m
Example Problem : Determine the evaporation rate per month of a lake 500,000m2, with a total
average inflow of 3.5 m3/s, outflow downstream is 3.2 m3/s. Before the large content is 180,700
m3/month, due to higher inflow it has increased to 220,000 m3/month. Precipitation rate is 15 mm
/month.
Given:
I = 3.5 m3/s
O = 3.2 m3/s
S2 = 220,000 m3/month
S1 = 180,700 m3/month
P = 15 mm/month
Solution:
E = P + I - O - ∆S
E = 15 mm (1/1000)(500,000 m2) + 3.5 m3/s(60)(60)(24)(30) - (3.2 m3/s)(60)(60)(24)(30)
- (220,000(m3/month - 180,700 m3/month)
E = 7500 m3/month + 9,072,000 m3/month - 8,294,400 m3/month - 39,300 m3/month
E =
E = 1.492 m/month
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SUBSURFACE WATER
6-1 Occurrence of subsurface
Water table – is the lows of points( in unconfined materials) where hydrostatic pressure equals atmospheric pressure.
Vadose zone – Above the water table, soils pores may contain either air or water.
Phreatic zone – Below the water level interstices are filled with water.-Local saturated zones sometimes exists as perched ground water above an impervious layer of limited extent.- Sometimes ground water is overlain by an impervious stratum to form confined or artesian water.
Soil water- ranging to 10 m below the soil surface.
6-2 Soil-Water Relationship
Soil moisture may be present as gravity water in transit in the large pore spaces as capillary water in the smaller pores, as hygroscopic moisture adhering in a thin film to soil grains and as water vapor.
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MOVEMENT OF GROUNDWATER
In 1856 Darcy confirmed the applicability of principles of fluid flow in capillary tubes, developed several years earlier by Hagen and Poiseveille, to the flow of water in permeable media.
Darcy’s Law:
where
v = velocity of flowk = coeffocient having the units of vs = slope of the hydraulic gradient
where
q = product of area and velocity p = porosity
K = coefficient of permeability or the hydraulic conductivity
where
K = intrinsic permeability of the medium w = specific weight of the fluid μ = absolute viscosity C = factor involving the shape, packing, porosity, and other characteristics of
the medium d = average pore size of the medium
It is convenient to use the transmissibility (T) to represent the flow rate per day through unit area under unit hydraulic gradient:
where
T = transmissibilityK = hydraulic conductivityY = saturated thickness of the aquifer
whereB = the width of the aquifer
6-7 DETERMINATION OF PERMEABILITY
Laboratory measurements of permeability are made with permeameters. A sample of materials is subjected to water under a known head, and the flow through the sample in a known time is measured.
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V = ks
q = kpAs = KAs
K = k w/ μ = Cd2 w/ μ
T = KY
q = TBs
6-8 SOURCES OF GROUNDWATER
Connate Water – present in the rock at its formation and is frequently highly saline.
Juvenile Water – formed chemically within the earth and brought to the surface in intrusive rocks, occurs in small quantities.
Connate and juvenile waters are sometimes important sources of undesirable minerals in groundwater.
Influent Streams – streams contributing to groundwater. Such streams are frequently ephemerical.
6-9 DISCHARGE OF GROUNDWATER
Effluent Streams – streams intersecting the water tale and receiving groundwater flow.
Spring or Seep – form where an aquifer intersects the earth’s surface.
Types of Springs:
1. Water hole or dimple spring2. Anticipating spring3. Perched spring4. Spring from solution channel
Phreatophytes – plants deriving their water from groundwater, often have root systems extending to depths of 12m or more.
6-10 EQUILIBRIUM HYDRAULICS OF WELLS
Flow toward the well through a cylindrical surface at radius x must equal the discharge of the well.
From Darcy’s Law:
Where
2 ΠxyK = area of cylinderdy = slope of the water tabledx
Integrating with respect to x from r, to r2 and y from h1 to h2:
where
22 | P a g e
q = 2 ΠxyK dy/dx
q = ΠK (h1 2 – h22) / In (r1/ r2)
h = height of water table above the base of the aquifer at distance r from the pumped well
In = logarithm to the base e
Assuming that the drawdown (Z) to be small compared with the saturated thickness
(h1 ≈ h2 ≈ Y):
Equations (6-9) and (6-9a) can be used to estimate T or K given q and z, provided that the assumption of equilibrium is satisfied.
6-11 NONEQUILIBRIUM HYDRAULICS OF WELLS
In 1935 theis is presented a formula based on the heat flow and which accounts for the effect of time and the storage characteristics of an aquifer.
Where:
Zr = drawdown of an observation well at distance r from the pumped well q = flow in cubic meters per dayT = transmissibility in cubic meters per day per meter
Where:
t = times in days since pumping began
Sc = storage constant of the aquifer, or the volume of water removed from a column of aquifer 1 meter square when the water table or pielometer surface is lowered 1 meter.
The integral in equation, written as W(u) and called the well function of u, can be evaluated from the series
Equation can be solved graphically by plotting a type curve of u versus W (u) on
logarithmic paper.
23 | P a g e
t = q In(r1/ r2) / 2Π (Z2 – Z1)
Zr = q / 4ΠT ∫ e -u du/u
U = r2Sc / 4Tt
W(u) = -0.5772 – In u + u – u2 /2.21 + u3 / 3.31
When u is small, the terms of equation (6-12) following in u are small and may be neglected. Equation (6-11) indicates that u will be small when t is large, and in this case a modified solution of the theis method is possible by writing:
Where:ΔZ = change in drawdown between t, and t2.
The drawdown z is plotted on an arithmetic scale against time to on a logarithmic scale. If ΔZ is taken as the change in drawdown during one log cycle, Log10 (t2/t1) = 1, and T is determined from equation 96-14), when z = 0,
Where:
to = intercept (in days) if the straight – line portion of the curve is extended z – 0
6-12 BOUNDARY EFFECT
When several are close together, their cones of depression may overlap or interfer.
When wells are located too close together, the flow from the wells is impaired and the drawdowns increased.
Method of Images – devised by Lord Kelvin for electrostatic theory.- is a convenient way to treat boundary problems.
Resultant cone of depression – found by subtracting the drawdown caused by image well from that caused by real well (assuming mo boundaries)
Boundaries across which no flow is transmitted, such as faults can be represented by pumping wells.
6-13 AQUIFER ANALYSIS
Hele-shaw Apparatus
- consisting of closely spaced glass plates with a viscous fluid between them- often convenient for solving two-dimentional ground flow problems.
Three – dimensional problems are commonly treated with a digital; or analog computer
Analog computer – consist of a network of resistors and capacitors
24 | P a g e
r2 / t = 4Th / Sc
T = 2.3q / 4TΔZ log t2 / t1
Sc = 2.25 Tto /r2
Current is analogous to flow and voltage to potential Permeability is simulated by the reciprocal or resistance Elaborate analogs representing large aquifer have been constructed
In finite-difference form using the grid notation of fig. 6-14 this becomes
Where: a = the grid size
Potential of a Groundwater Reservoir
6-14 SAFE YIELD
Safe yield – is defined by Menzer [16] as “The rate at which water can be withdrawn for human use without depleting the supply t such an extent that withdrawal at this rate is no longer economically feasible”- Kazmann has suggested that it be abandoned because of its frequent
interpretation as a permanent limitation on the permissible withdrawal.- Must be recognized as a quantity determined for a specific set of controlling
conditions and subject to change as a result of changing economic or physical condition.
The safe yield of a grounded basin is governed by many factors, one of the most important being the quantity of water available. This hydrologic limitation is often expressed by the equation
Where:
G = safe yieldP = precipitation on the area tributary tot he aquiferQs = surface stream flow from the same areaET = evapotranspirationQg = net groundwater inflow to the are Sg = change in groundwater Ss = change in surface storage
25 | P a g e
Y ← 0 →
3
2 1 4
5
↑
↓0
x
h2 + h4 - 2h1 + h3 + h5 - 2h1 = h2 + h4 + h3 - 4h1
G = P – Qs – ET + Qg – ΔSg – ΔSs
Hint: if the equation is evaluation on a mean annual basis, __ Sg will usually be zero
Mining – the permanent withdrawal of groundwater from storage- the term being used in the same sense as for mineral resources.
6-15 SEAWATER INTRUSION
Ghyben – Herzberg Lens – the lens of fresh water floating on salt water
About 1/40 unit of fresh water is requires above sea level for each unit of fresh water below sea level to maintain hydrostatic equilibrium.
6-16 ARTIFICIAL RECHARGE
Methods employed for artificial recharge1. Storing floodwaters in reservoirs constructed over permeable areas2. Storing floodwaters in reservoirs for later release into the stream channel at rates
approximating the percolation capacity of the channel.3. Diverting streamflow to spreading areas located in a highly permeable formation.4. Excavating recharge basins to reach permeable formations 5. Pumping water through recharge wells into the aquifer6. Over irrigating in areas of high permeability7. Construction of wells adjacent to a stream to induce percolation from steam flow.
6-17 ARTESIAN AQUIFERS
If the permeability of the aquiclude confining an Artesian Aquifer is 0.04 m/day and the hydraulic gradient is unity, the daily seepage would amount to 40,900 m3/km2.
Hantush - has demonstrated a procedure which accounts for such leakage in the analysis of pumping tests on artesian aquifers.
Artesian aquifers demonstrate considerable compressibility.
6-18 TIME EFFECTS IN GROUNDWATER
Flow rates in the groundwater are normally extremely slow, and considerable time may be involved in groundwater phenomena.
Werner – suggested that several hundreds years might be required for a sudden increase in water level in the recharge area of an extensive artesian aquifer to be transmitted through the aquifer.
26 | P a g e
Jacob – found that the water levels on long island were related to an effective precipitation which was the sum of the rainfalls for the previous 25 years.
McDonald and Langbein – found long-term fluctuations in streamflow in the Columbia basin which they believe are related to groundwater fluctuations.
STREAMFLOW HYDROGRAPHS
Engineering hydrology is concerned primarily with three characteristics of streamflow: 1. monthly and annual volumes available for storage and use 2. low-flow rates which restricts in-stream uses of water3. floods
Hydrograph – is a continuous graph showing the properties of streamflow with respect to time
CHARACTERISTICS OF THE HYDROGRAPH
7-1 Components of RunoffThree main routes of travel:1. overland flow ( or surface runoff)
- is that water which travels over the ground surface to channel 2. interflow (or subsurface storm runoff)
- the water which infiltrates the soil surface that move laterally through the upper soil layers until it enters a stream channel
- moves more slowly than the surface runoff and reaches the streams later 3. groundwater flow ( also called base flow and dry-weather flow)
- precipitation that percolate downward until it reaches the water table - is the discharge groundwater accretion into the streams, if the water table intersects the
stream channels of the basin
The hydrograph differs for each type of basin. The total flow is divided only into parts:
a. storm or direct runoff – is presumed to consist of surface runoff and a substantial portion of the interflow
b. base flow – is considered to be largely groundwater
7-2 Streamflow Recessions
27 | P a g e
A typical hydrograph resulting from an isolated period of rainfall consist of a rising limb, crest segment, and falling segment, or recession.
where: q0 - is the flow at any time q1 - is the flow one time unit laterKr - is a recession constant which is les than unity
where:qt - is the flow t after q0
e - is the napierian base α - -ln Kr
where:St - is the storage remaining in the basin at time t
7-3 Hydrograph Separation
Hydrograph separation or hydrograph analysis – the division of hydrograph into direct and groundwater runoff as a basis for subsequent analysis
where:N = time in daysA = is the drainage area (in sq. km)b = is the coefficient maybe taken as 0.87-4 Analysis of Complex Hydrographs
- This type of event easier to analyze than the complex hydrographs resulting from two or more closely spaced bursts of rainfall.
- Some other methods have been developed may also have some advantages where groundwater is a relatively important component of runoff and reaches the stream fairly quickly.
7-5 Determination of Total Runoff there is a need to determine the total streamflow resulting from a particular storm or
group of storms this can be done by computing the total volume of flow occurring during a period
beginning and ending with the same discharge and encompassing the rise under consideration, making certain that groundwater recession conditions prevail at both times.
HYDROGRAPH SYNTHESIS
7-6 The Elemental HydrographSurface detention – is the temporary storage where the rainfall goes since sheet flow over the surface cannot occur without a finite depth of water on the surface.
28 | P a g e
q1 = q0Kr
qt = q0Ktr = q0e
-αt
St = - qt / ln Kr = qt /α
N = bA 0.2
7-7 The Unit-Hydrograph Concept
- The unit hydrograph is a typical hydrograph for the basin- unit hydrograph can be defined as the, hydrograph of one centimeter or millimeter of
direct runoff from a storm of specified duration
1. Duration of rain- theoretically, the ideal unit hydrograph has a duration approaching zero, the
instantaneous unit hydrograph.2. Time-intensity pattern
- if one attempted to derive a separate unit hydrograph for each possible time-intensity pattern, an infinite number of unit hydrographs would be required
- unit hydrograph can be based only on an assumption of uniform intensity of runoff3. Arial distribution runoff
- the areal pattern of runoff can cause variations in hydrograph shape- if the area of high runoff is near the basin outlet, a rapid rise and sharp peak usually result- higher runoff in the upstream portion of the basin produces a slow rise and a lower
broader peak4. amount of runoff
- Inherent in the unit hydrograph concept is the assumption that ordinates of flow are proportional to volume of runoff for all storms of a given duration that the time bases of all such hydrographs are equals
7-8 Derivation of Unit Hydrographs
The unit hydrograph is best derived from the hydrograph of the storm of reasonably uniform intensity, duration of desired length, and a relatively large runoff volume.The steps are:1. separate the base flow from direct runoff2. determine the volume of direct runoff, and the ordinates of the direct runoff hydrograph3. then, divide by the runoff depth4. the adjusted unit ordinates form a unit hydrograph The proper procedure is to compute the average peak flow and time to peak. The average unit hydrograph is the sketched to confirm to the shape of the other graphs,
passing through the computed average peak, and having the required unit volume.
7-9 Derivation of Unit Hydrograph from Complex Storms
- If individual bursts of rain in the storm result in well-defined peaks, it is possible to separate the hydrographs of several bursts and to treat these hydrographs as independent storms. See Fig. 7-4.
- If the reconstructed hydrographs does not agree with the observed hydrograph, the assumed unit hydrograph is modified and the process repeated until a unit hydrograph which seems to give the best fit is determined.
7-10 The Conversion of Unit Hydrograph Duration
- There is frequently a need to convert an existing unit hydrograph for one storm duration to another-shorter to better scope with spatial and intensity variations, or longer to reduce required computations and possibly in recognition of the coarseness of available data.
S-curve Method (or summation-curve method)
- allows the construction of a unit hydrograph of any duration - assumed that a unit hydrograph of duration D is known and that we wish to generate a
unit hydrograph for the same watershed with duration D’
29 | P a g e
- is the hydrograph that would result from an infinite series of unit runoff increments
where: qe = equilibrium at flow (m3/s)
Instantaneous unit hydrograph (IUH)
- the flow is determined by weighing the antecedent rainfall excess, where the weight applied to rainfall occurring to τ hour ago is the IUH ordinate τ hour after the beginning of rainfall.
where: f(τ) = IUH ordinate at time τi e = intensity of rainfall at time (t – τ) τ = time in the past.
- By analogy with the S-curve technique for deriving a short-duration unit hydrograph, it will be seen that the IUH ordinates are the function of the slope of the S-curve and its duration:
f(τ) = tR (dq/dt)
- the required ordinate at any time is simply the average flow during the previous tR hr.
7-11 Synthetic Unit Hydrographs
- This requires a relation the physical geometry of the area and the resulting hydrographs three approaches the have venues formulas relating hydrograph features to basin characteristics, transposition of unit hydrograph and storage routine.
- in a study of basins in the Appalachian Mountain region, Snyder [14] found the basin log (in hours) to be a function of basins size and shape:
where : L - main stream distance from outlet to divide and Lc is the stream distance from outlet to a point opposite the basin centroid.
- Snyder found that tha unit-hydrograph peak qp could be estimated from:
where: A - drainage area. The coefficient Cp range from 0.15 to 0.19 with A in square kilometers qp in m3 / sec
- Snyder adopted as the time base of unit hydrograph (days )
- For any duration tR he used an adjusted lag
30 | P a g e
qe = 2.78A / tR
qt = ∫ f(τ) i e(t – τ) d τ
tp = C (LLc)0.3
qp = CpA / tp
T = 3 + 3 ( tp / 24)
- from Eq. 7-8 a general expression for basin lag might be expected to take the form
where : s - weighted channel slope
7-12 Application Of Units Hydrographs - The unit hydrograph has been a main stay of the hydrologist even though some of the
techniques may offer more flexibility and accuracy in many applications. - Care should be taken not to apply the unit hydrograph without considering the advantages
and the disadvantages of the other techniques.
7-13 Hydrograph of Overland Flow
- overland flow is supplied by rainfall and depleted by infiltration.- Overland flow is spatially varied and unsteady. It may be turbulent laminar or a
combination and depths maybe subcritical or supercritical.
Problem 1. (SNYDERS METHOD)
Use Snyder’s method to develop a unit hydrograph for the area of 100 mi square described below. Sketch the approximate shape. What duration rainfall does this correspond to?
Ct = 1.8 L = 18 miCp = 0.6 Lc = 10 mi
SOLUTION
tP = Ct(LLc)0.3 = 1.8(18-10)0.3 hr
tP = 8.6 hr.
Qp = 640CpA/tp
Qp = 640(0.6)(100)/8.6Qp = 4465 cfs.
For a small shed, Tb = 4tp (app.)
Tb = 4(8.6)Tb = 34.4 hr.
D = tp / 5.5D = 8.6/5.5D = 1.6 hr
Problem 2. (UNIT HYDROGRAPH)
31 | P a g e
tpR = tp + [ ( tR – tr)/4 ]
tp = Ct (LLc / √ s) n
Convert the direct runoff hydrograph shown into (a) a 2-hr Unit Hydrograph the Rainfall hyetograph is given i8n the figure and the diameter index for the storm was 0.5in/hr. The base flow in the channel was 100 cfs constant. What are Tp and tb for the storm?
TIME (hr) Q(cfs) Q-BF(cfs) 2-hr UH, Q
0 100 0 01 100 0 02 300 200 1003 700 600 3004 1000 900 4505 800 700 3506 600 500 2507 400 300 1508 300 200 1009 200 100 5010 100 0 011 100 0 0
The 2 hr UH graph as shown in figure . Tb, the time of this storm is, is 9 hr and the time to peak tp, measured from the center of mass of rainfall is 2 hr.
Problem 3. (S-curve method)
Convert the following tabulated 2-hr unit hydrograph to a 3-hr unit hydrograph using the S-curve method.
TIME(hr)
2-HRUHORDINATE(cfs)
012345678910
0752503002752001007550250
Solution:
TIME(hr)
2-HR UH 2-HR LAGGED UH’S SUM
0123
075250300
0 75
075250375
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4567891011
2752001007550250
250 0300 75275 250 0200 300 75100 275 250 0 75 200 300 7550 100 275 250 025 75 200 300 75
525575625650675675675675
TIME(hr)
S-CURVE
ORDINATE
S-CURVE LAGGED
3 HR
DIFFERENCE D/D’ 3-HR UH
ORDINATE
01234567891011
075250375525575625650675675675675
075250375525575625650675
07525037545032525012510050250
2/32/32/32/32/32/32/32/32/32/32/32/3
050.0166.7250.0300.0216.7166.783.366.733.316.7
0
33 | P a g e
RELATION BETWEEN PRECIPITION AND RUNOFF
Primarily variations in precipitation control the flow of steam.
THE PHENOMENA OF RUNOFF
Discussion w/c follow will be limited to process w/c retain water in the catchments until it
is removed by evapotranspiration.
SURFACE RETETION
A part of storm precipitation retained on or above the ground surface.
Types of Surface Retention:
1. Interception - first part of storm stored.
2. Depression storage - continues falling rain.
3. Evaporation - during the storm.
4. Interceptometers - are placed at random in an attempt to measure average interception
for an area.
Application of the data requires detained knowledge of the cover density over the area
of interest.
The Interceptometers are place on its project.
34 | P a g e
Trimble and Weitzman found that mixed hardwood about 50 years old and typical of
considerations are in the Southern Appalachian Mountain interest about 20% of the rainfall both
summer and winter.
It is likely that net interception is nearer 18% for storm with rainfall in the order of 13-
mm.
Qualitative, it can be said that annual interception by a well-developed forest canopy is
10 to 20 percent of the rainfall
The storage capacity of the canopy is range from 0.8 to 1.5-mm.
Horton derived a series of empirical formulas for estimating interception (per storm)
by various types of cover. Applying these formulas to 25-mm storm and assuming normal cover
density give values of interception.
Assuming sufficient rainfall to satisfy the interception-storage capacity, an equation for Vi
total-storm interception.
Where;
Si –storage capacity per unit of projected area,
E –evaporation rate,
tR –duration of rainfall.
Assuming that the interception given is approached exponentially as the rainfall increase
from zero to some high value, then
Where;
e –napierian base,
P –amount of rain,
k –equal to 1/ (Si + EtR)
Blind drainage is individual depressions of appreciable area relative to the drainage basin➨
under consideration.
The flood hydrograph moderate to;
Stock ponds
Terraces
35 | P a g e
Vi =Si + EtR
Vi =(Si + EtR )(1– e –kP)
Vs =Sd (1– e -kpe)
Contour.
The volume of water in depression storage Vs can be expressed as
Where;
Sd –depression-storage of the basin, Sd –the value most basin lie between 10 and 50mm.
k – has the value 1/ Sd,
Pe – value of precipitation
The eq. neglects evaporation from depression storage during, a factor w/c is usually unimportant.
RUNOFF MECHANISMS
Most hydrologists considered that all storm runoff was generated by this mechanism.
“Hortonian overland flow” is one of several mechanisms, and that is not necessarily the
dominantone.
Infiltration
Two Phenomena
1. Infiltration is the passage of water through the soil surface into the soil.
2. Percolation is the gravity flow of water within the soil.
Infiltration Capacity is the max. Rate w/c water can enter the soil at a particular pt. under a
given set of conditions.
Infiltration capacity depends on many factors such as soil type, moisture content, organic
matter, vegetative cover, and season.
Of the soil characteristics affecting infiltration, noncapillary porosity is perhaps the most
important.
Infiltration-capacity formulas curve approximate form;
Where;
ƒp –is at a max. ƒo at the beginning of a storm and approaches a low.
36 | P a g e
ƒp= ƒc + (ƒo – ƒc)e-kt
ƒc –constant rate of the soil profile becomes saturated.
k –is empirical constant.
t –time from beginning of rainfall.
The equation is applicable only when Is ≥ƒp throughout the storm. Phillip suggested the eq.;
Infiltration capacity depends on many factors such as soil
type, moisture content, organic matter, vegetative cover, and season.
Of the soil characteristics affecting infiltration, noncapillary porosity is perhaps the most important.
Infiltrometer is a tube or other boundary designed to isolate a section of soil.
Saturation overland flow
Primary at the base of slope marginal to stream channels.
Saturation storm flow
Discussed in chapter 7.
THE RUNOFF CYCLE is the descriptive term applied to that portion of the hydrologic cycle
between incident precipitation over land areas and subsequent discharge of this water through
stream channels or evapotranspiration.
ESTIMATING THE VOLUME OF STORM RUNOFF
As a fixed percentage of rainfall is the most commonly used method in design of urban
storm-drainage facilities, highway culverts, and small water-control.
INITIAL MOISTURE CONDITIONS
1. The moisture conditions of the catchment at the onset of the storm.
2. The storm characteristics rainfall amount, intensity, and duration.
In other words, the soil moisture should decrease logarithmically w/ time during periods of
no precipitation;
37 | P a g e
ƒp = ( bt -1/2 )/2 + a
It= Iokt
I1=kIo Io – I1= Io (1 – k)
Where;
It – reduce value t days later,
Io – is the initial value of the antecedent-precipitation index.
k –is a recession factor ranging normally between 0.85 and 0.98.
STORM ANALYSIS
The storm rainfall w/c produced runoff being considered only included.
Shower occurring before the rain storm should be excluded from the storm rainfall and
included in the antecedent-precipitation index.
Runoff also depends upon rainfall intensity, an average intensity as reflected by amount and
duration is usually adequate.(≥250km2 Rainfall). In this case duration can be estimated with
suffusion accuracy from 6-hr rainfall data.
Multivariate Relations for Total Storm Runoff It is convenient to calculate computed by subtracting storm rainfall by the storm runoff.
Development Three Variable Relation
1. Plotting antecedent precipitation versus recharge.
2. Labeling the points with week number.
3. Fitting a smooth family curves representing the week.
Analytical Correlation Technique/method it have result generally slightly better tan the
graphical solution.
Where;
Ips – runoff index (approx. first quadrant of coaxial plot).
Is – fixed function of week range between +1 and –1.
I – antecedent-precipitation index.
E – base of napierian logarithmic.
P – storm rainfall.
Q – direct runoff.
a, b c, d and n –statistically derived coefficients.
38 | P a g e
Ips=c + (a+dIs) e-bl Q= (Pn +I
nps)
1/n - Ips
RELATION FOR INCREMENTAL STORM RUNOFF
In other to determine increments of runoff throughout a storm for application of a unity
Hydrograph may be used the accumulated in equation
INFILTRATION APPROACH TO RUNOFF ESTIMATES
It assumes that the surface runoff from is equal to the portion of rainfall that is not
disposed of through;
1. Interception and depression storage
2. Evaporating during the storm
3. Infiltration
When the supply rate, Is, is at or in infiltrating capacity, surface runoff is equivalent to the
storm rainfall less surface retention and the area under capacity curve.
If is <ƒ p, the increment of soil mixture is lea than assume and the drop in the infiltration
curve correspondingly less.
Applicable infiltration capacity curve-varies from point depending on soils, vegetation, and
antecedent moisture.
INFILTRATION INDEXES
Infiltration index sample, and the approach is cloaked in an aura of logic.
The W index is the average infiltration rate during the time rainfall intensity exceeds the
capacity rate, i.e.,
Where:
t- time
P- total precipitation
Qs- surface runoff
S –effective surface retention.
ESTAMITING SNOWMELT RUNOFF
Plays important role in the hydrology in some areas.
39 | P a g e
W = F/t = (1/t)(P – Qs – S)
Reliable predictions of the rate of melt and release of liquid water from snow pack are
requisite to the efficient design and operation of water resources projects and the issuance
of river forecast and warnings.
Modeling is simplified if interest is restricted to periods of melt when the pack is ripe
(saturated and isothermal at 0℃).
PHYSICAL OF SNOWMELT
Snow and evaporation (including sublimation) are both thermodynamic processes, and
both are amenable to the energy-balance approach.
Percentage of the solar and diffuse radiation reflected by the surface. .
1. The energy for snowmelt is derived
2. Net radiation
3. Conducting and water vapor from the transfer of sensible heat from the overlying
air.
4. Condensation of water vapor from the overlying air,
5. Conduction from the underlying soil,
6. Heat supplied by incident rainfall.
Two processes can be described by similar equations for melt;
Where;
kh = ke –exchange coefficient.
To =℃
e =0.611kPa.
The melt (mm) from rain given by;
Where;
P- rainfall in mm.
Tw –wet-bulb temp. in ℃.
344 –is the latent heat of fusion in joules/gram.
4.19 – is the specific heat of water in joules/gram/ ℃.
40 | P a g e
Mh= kh (Tɑ –To) ѵ Me = ke (eɑ – eo) ѵ
Mr = 4.19 P Tw / 334
ESTIMATING SNOWMELT RATES AND COSEQUENT RUNOFF
Air temperature is the single most reliable index to snowmelt.
PRECIPITATION-RUNOFF RELATIONS
A simple plotting of annual precipitation versus annual runoff will often display good
correlation, particularly in areas where the major portion of the precipitation falls in the
winter months.
Example:
The Infiltration rate for excess rain on a small area was observed to be 4.5 in/hr at the
beginning of rain, and it decreased exponentially to an equilibrium of 0.5 in/hr after 10 hr, A
total of 30in of water infiltrates during the 10-hr internal, Determine the value of k in,
HOSTORS EQUATION: F = Fc + (Fo-Fc)e -kt
Given:
t = 10 in
Fc = 0.5 in/hr
Fo = 4.5 in/hr
F = depth = 30in
Time 10hr
Solution:
F = Fc + (Fo-Fc)e -kt
3in/hr =0.5in/hr + (4.5in/hr + 0.5in/hr)e –k(10hr)
3in/hr - 0.5in/hr = (4.5in/hr + 0.5in/hr)e –k(10hr)
2.5in/hr =(4in/hr) e –k(10hr)
2.5in/hr = e –k(10hr
(4in/hr)
ln 0.625 = -k(10hr)
-k =0.47 /10
41 | P a g e
=3in/hr
- 0.047
k=
Example:
The initial infiltration capacity of a watershed is estimate as 1.5 in/hr, and the time
constant is taken to the 0.35 hr -1. The equilibrium capacity fc is 0.2in/hr. Use the Horton’s
Equation to find a.) the values of fo at t =10min, 30min, 1hr, 2hr, and 6hr, b.) The total values of
infiltration over the 6hr period.
Given:
fc = 0.2in/hr
fo = 1.5in/hr
k = 0.35 hr -1
t = 10min (1hr/60min) =0.167hr
Solution:
a.)f = fc + (fo-fc)e –kt
Substituting Values:
f = 0.2in/hr + (1.5 -0.2) e –0.35 hr -1 (0.167 hr)
f =
b.) V= ∫f dt ; but t1 =0hr,t2 =6hrs
=∫ (0.2 +1.3 e -0.35t)dt
= [0.2 + (1.3/0.35)e -0.35t]6₀
= [0.2 ₋ (1.3/0.35)e -2.1] - [0.2 - (1.3/0.35)e0]
=
42 | P a g e
1.43in/hr
4.46 in over the
watershed
HYDROLOGIC ROUTING
Wave movement in naturals channels traditionally has been treated in design and prediction by applying HYDROLOGIC ROUTING procedures. Such procedures solve the continuity equation (or storage equation) for an extended reach of the river, usually bounded by selected gauged points.
STREAMFLOW ROUTING- is a general term applied to methods used to predict unsteadily flow in streams.
Given the flow at an upstream point, routing can be used to compute the flow at a downstream point. The principles at routing apply also to computation at the effect of a reservoir on the slope of a flood wave. Hydrologic storage occurs not only in channels and reservoirs but also as water flowing over the ground surface.
9-1 WAVE MOVEMENT
One of the simplest wave form is the monoclinal rising wave in a uniform channel. Such a wave consists of an initial steady flow, a period pf uniformity increasing flow, and a continuing steady flow at the higher rate. From the flow of continuity and assuming negligible effects from any changes in wave shape, the difference between inflow and outflow must equal the change in storage within the reach:
Where U and V are velocities of the wave and water, respectively and A is the cross-sectional area of the channel. Solving EQ. (9-1)for the wave velocity and substituting the discharge q for AV gives
The velocity of a monoclinal wave is thus a function of the area-discharge relation for the stream. Since velocity usually increases with stage, area-discharge curves are usually concave upward.
Where B is the channel top with. Equation (9-3) is known as seddons law after the man who first demonstrated its validity on the Mississippi River.
From the Chezy formula for flow in a wide, open channel (assuming depth equal to hydrologic radius)
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U( A2 – A1) = A2V2 – A1V1
U = dq/ dA = 1/B dq/dy
V = cy1/2 s1/2
And
Where S is water-surface slope. Differentiating give
Substituting Equation (9-6) into Equation (9-3),
The derived ratio between water and wave velocities depends on channel shape and the flow formula used, and involves the implicit assumption that discharge is a single-valued function of area.
9-2 WAVES IN NATURAL CHANNELS
Simple mathematical treatment of flood waves is necessarily limited to uniform channels with fairly regular cross-section. The hydrologist must deal with non-uniform channels of complex section with non-uniform slope and varying roughness. Most flood waves are generated by non-uniform lateral inflow along all the channels at the stream system. Thus natural flood waves are considerably more complex than the simplified cases which yield to mathematical analysis, but theoretical treatment is particularly useful in studies of surges in canals, impulse waves in still water (including seiches and tides), and waves released from dams.
Natural flood waves are generally intermediate between pure translation and pondage, which occurs in a broad reservoir or lake. Most natural flood waves move under friction control and have time bases considerably exceeding the dimensions of the stream system.
9-3 THE STORAGE EQUATIONThe continuing equation maybe expressed as
or
Where I is inflow rate, O is outflow rate, S is storage ( all for a specific reach of a stream), and t is time. To provide a form move convenient for hydrologic routing, it is commonly assumed that the average of the flows at times t1 and t2, the beginning and end of the routing period, t equals the average flow during the period:
Most storage routing methods are based on Eq. (9-10). It is assumed that I1, I2, O1 and S1 are two unknown, a second relation between storage and flow is needed to complete a solution. The assumption that ( I1 + I2/2 = I implies that hydrograph is a straight line during the routing period t.
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Q = ∆V = UBY = CBY1/2S1/2
dq/ dy = 3/2 CBY1/2S1/2 = 3/2 BV
U = 3/2V
I – O = ds /dt
∆s =S2 – S1 = ∫ Idt - ∫ O dt
(I2 + I2)/2 t – ( O1 + O2) /2 t = S2 - S1
9-4 DETERMINATION OF STORAGE
Before a relation between storage and flow can be established, it is necessary to determine the volume of water in the stream at various times. The obvious method for finding storage is to compute volumes in two channel from cross-sections by using the prismoidal formula. The water surface is usually assumed to be leveled between cross-sections. Storage elevation curves for reservoirs are usually computed by plain metering the area enclosed within successive by the contour interval gives the increment of volume from the midpoint of the next higher interval.
9-5 TREATMENT OF LOCAL INFLOW
One of the most annoying problems in flood routing is the treatment of local inflow which enters the reach between the inflow and the outflow stations.
If the local inflow occurs primarily near the downstream end of the reach, it may be subtracted from the outflow before storage is computed.
9-6 RESERVOIR ROUTING
A reservoir in which the discharge is a function of water-surface elevation offers the simplest of all routing situations.
Routing in a reservoir with gated outlets depends on the method of operation. A general equation is obtained by modifying Eq. (9-10) to
STORAGE COMPUTATIONS
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I1 + I2 + ( 2S1/t – O1 ) = 2S2/t + O2
(I1 + I2) / 2 t – ( O1 + O2) / 2 t – Ort = S2 – S1
1. Sample problem
Inflow and outflow for a reservoir are depicted in fig. E4.1 (a).a.) Determine the average storage for each one day period (t =1 day). Graph storage vs. time for the reservoir for the event. Assume that So = 0 (the reservoir is
initially empty.b.)What is the (approximate) maximum storage reached during this storm event?
SOLUTION
a.) the rate of change in storm in storage is equal to inflow minus outflow. First we tabulate values of I and Q and take their difference. Storage is equal to the area between the inflow and outflow curves. Or S = ƒ (I – Q) dt.This integral can be simply approximates by S = (I – Q) t.
Q (cfs)
10,000
5,000
0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Time (Days)
where I and Q are average for each day. This method id used in Example 2.1 to determine
volumes under hydrographs. To minimize error, I and Q values are averaged at noon each day.
Time (day) I(cfs) Q(cfs) S/t(cfs)0.5 500 250 2502.5 3500 1000 25003.5 9000 3000 60004.5 9750 4500 52505.5 8000 5750 22505.5 4500 6000 -15006.5 2250 5250 -30007.5 1250 4250 -30008.5 250 3250 -3000
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Inflow
Storage change for day 3
Outflow
9.5 0 2500 -250010.5 0 1500 -150011.5 0 1000 -100012.5 0 750 -75013.5 0 0 0 Using t = 1 day, storage at the end od the first day, S1, S2, is S1 = So + (I1 – Q1) t
= 0+ (250 cfs) (1 day) (24 hr ) (3600s) ( ac ) (Day ) (Hr ) (43,560 ft²)
= 496 ac-ft
for day 2, cumulative storage becomes
S2 = So + S1 + (I2 – Q2) t,
50,000
25,000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14Time day
S2 = 0 + 496 + (2500) (24) (3600) (1/ 43560) ac-ft
= 5455 ac-ft
The procedure is shown completed in the following table and the storage curve in Fig. E4.1(b).
Time Storage (Day) (Ac-ft)
1 4962 54553 17,3564 27,7695 32,2326 29,2567 23,3068 17,3569 11,40510 6,44611 3,47112 1,48813 014 0
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b) The maximum storage, as seen from the table in figure, is 32,232 ac-ft. This occurs at day 5 for this event, as seen from the equation.
dS =1 – Qdt
Smax will occur when dS/dt equal zero. At this point, 1 = Q which occurs at day 5 on their inflow-outflow hydrographs.
Muskingum Routing
Route the inflow hydrograph tabulated in the following table through a river reach for whigh x = 0.2 and k = 2 days. Use a routing period t= 1 day and assume that inflow equal outflow for the first day.
TIME (day) INFLOW (cfs)
1 4,0002 7,0003 11,0004 17,0005 22,0006 27,0007 30,0008 28,0009 25,00010 23,00011 20,00012 17,00013 14,00014 11,00015 8,00016 5,00017 4,00018 4,00019 4,00020 4,000
SOLUTION
First we determine the coefficients C0 , C1 and C2 for the reach (Eq. 4.9):
Co = - Kx + 0.5 ∆ t
D
C1 = Kx + 0.5 ∆t,
D
C2 = K – Kx – 0.5 ∆t
D
D = K – Kx + 0.5 ∆t
For K = 2 days ,∆t = 1 day , and x = 0.2
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D = 2 – 2(0.2) + 0.5 (1)
= 2.1
Co = - (2)(0.2) + (0.5)(1)
(2.1)
= 0.0476,
C1 = (2)(0.2) + (0.5)(1)
(2.1)
= 0.4286,
C2 = 2 –(2)(0.2) – (0.5)(1)
(2.1)
= 0.5238,
We may check our computations by seeing if the coefficients sum to 1;
(0.0476) + (0.5238) = 1.0000
We substitute these values into Eq. (4.6) to obtain
02 = (0.0476)12 + (0.4286)11 + (0.5238)01
For 1 = 1 day.
01 = 11 = 4000 cfs.
For 1 = 2 days.
02 = (0.0476)(7000) + (0.4286)(4000) + (0.5238)(4000)
= 4143 cfs.
For t = 3 days.
03 = (0.0476)(11,000) + (0.4286)(7000) + (0.5238)(4143)
= 5694 cfs.
This Procedure is shown completed for t = 1 to t = 20 days in the following table.\
TIME INFLOW (cfs) OUTFLOW (cfs)
1 4,000 4,0002 7,000 4,1433 11,000 5,6944 17,000 8,5065 22,000 12,7896 27,000 17,4137 30,000 22,1218 28,000 25,7789 25,000 26,69310 23,000 25,79211 20,000 24,31912 17,000 22,12013 14,000 19,53914 11,000 16,75815 8,000 13,97316 5,000 10,93417 4,000 8,061
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18 4,000 6,12719 4,000 5,11420 4,000 4,583
DETERMINATION OF THE MUSKINGUM ROUTING COEFFICIENTS
The values listed in the table for inflow, outflow and storage were measured for a particular
reach of a river. Determine the coefficients K and x for use in the Muskingum routing equations
for this reach.
SOLUTION
To determine Muskingum coefficients, we guess a value of x and then plot xI + (1 – x) Q vs. S.
The plot that comes closest to being a straight line is chosen to determine the coefficient values.
The average value is x = 0.2.
Avg. Avg.TIME INFLOW OUTFLOW STORAGE(Days) (cfs) (cfs) (cfs – days)
1 59 42 172 93 70 403 129 76 944 205 142 1575 210 183 1846 234 185 2337 325 213 3458 554 293 6069 627 397 83610 526 487 87511 432 533 77412 400 487 68713 388 446 62914 270 400 49915 162 360 30116 124 230 19517 102 140 15718 81 115 12319 60 93 9020 51 71 70
600 x = 0.1 600 x = 0.2 600 x = 0.3
500 500 500
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400 400 400
300 300 300
200 200 200
100 100 100
0 0 0
0 500 1000 0 500 1000 0 500 1000
for a natural stream. Therefore, we assume that x must lie between 0.1 and 0.3 using the values listed in the following table.
(xI + (I – x) Q) (cfs)STORAGE __________________________________(cfs – days) x = 0.1 x = 0.2 x = 0.3
17 43 45 4740 72 74 7794 81 86 92157 148 155 161184 186 188 191233 190 195 200345 224 235 247606 319 345 371836 420 443 466875 491 495 499774 523 513 503687 478 470 461629 440 434 429499 387 374 361301 340 320 301195 219 209 198157 136 132 1298123 112 108 10590 89 86 8370 69 67 65
STORAGE INDICATION ROUTING
The design inflow hydrograph shown in Fig. E4.5(a), developed for a commercial area, is to be
routed through a reservoir. Assume that initially the reservoir is empty (S0= 0) and there is no
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initial outflow (00 = 0). Using the depth , storage, and outflow relationships given in the table,
route the hydrograph through the reservoir. What is the maximum height reached in the reservoir
for this inflow? Use ∆t = 10 min.
DEPTH (ft) STORAGE (ac-ft) OUTFLOW (cfs)
0 0 01.0 1.0 152.0 2.0 323.0 3.0 554.0 4.0 905.0 5.0 1256.0 6.0 1587.0 7.5 1858.0 10.5 2109.0 12.0 23010.0 13.5 25011.0 20.0 27012.0 22.0 290
SOLUTIONFirst, we develop a storage indication curve for the reservoir. This is a plot of (2S/ ∆t) + Q vs. Q. For instance, at Q = 90 cfs, S = 4.0 as-ft and
2 (4.0 ac-ft) 43, 560 dt 2 2S + Q = ac + 90 cfs = 671 cfs.∆t (10 min) (60 s / min)
Graphical and tabulated results for the storage indication curve are given and the following table.
Q 2S/∆t + Q (cfs) (cfs)
0 015 16032 32255 49190 671125 851158 1,029185 1,274210 1,735230 1,972250 2,210270 3,174290 3,484
Equation (4.13) states:
(In + In + 1) = 2Sn – Qn = 2Sn+1 + Qn+1
∆t ∆t
HYDRAULIC ROUTING
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Hydraulic Routing - Based on a solution of the energy or momentum equations is an alternate
to the hydrologic methods.
Solution of these equations is a relatively complex task and generally
requires the use of a compute.
Governing Equations
Hydraulic routing rests on three assumptions
1. Water density is constant;
2. Stream length affected by the flood wave is many times greater than the depth
of flow;
3. Flow is essentially one-dimensional.
(Waves satisfying these assumptions are called shallow-water or translatory waves)
Principle of the Conservation of volume (mass) (eq.10-1);
∫ X2X1 [A(x,t2) - A(x,t1)]dx = ∫ t2
t1 [Q(x1,t) + I (t) - Q(x2,t)]dt
Principle of Conservation of momentum (eq.10-2);
∫ X2X1[Q(x,t2) - Q(x,t1)]dx = g ∫ t2
t1 [Qv (x1,t) - Qv(x2,t)]dt + g ∫ t2t1 [J(x1,t) - J(x2,t) +
∫X2X1 Jv
X dx ] dt + g ∫ t2t1 [ ∫X2
X1 Asodx - ∫X2X1 Asfdx]dt
Where:
A(x,t)- cross-sectional area of flow K-conveyance
x-distance v-average velocity
y-depth B- top width of flow
J- first moment of A about the water surface R-hydraulic radius
JvX - rate of the change of J n- Manning roughness coefficient
So- channel of the bottom slope g-acceleration of gravity
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J= ∫y0 (y-z) B (x,z) dz Jv
X = ∫y0 (y-z) ∂B/∂x (x,z) dz
sf = QIQI/K2 K = 1.49/n (AR⅔)
Sf - frictional slope z-dummy integration variable
(all other forms of the shallow-water equation are derived from eq.10-1 and eq.10-2)
Major assumptions made in deriving the shallow-water equations:
1. hydrostatic pressure distribution;
2. one-dimensional flow;
3. fixed channel geometry,
4. small channel slope,
5. uniform velocity distribution; and
6. Friction losses in unsteady flow which can be approximated by the losses in steady
uniform flow.
Numerical Techniques
Numerical scheme – the assumptions and the numerical methods used to define
the discrete equation.
Type of scheme:
1. Consistent – in the limit, the discrete equations become the continuous
equations as time and distance steps approach zero.
2. Convergent – the solution of the discrete equations approaches the solution of
the continuous equations as the time distance steps approach zero.
3. Conservative – it mimics the conservation properties of the governing
equations.
4. Dissipative – the attenuation of a wave in the discrete solution is greater then
the attenuation of a wave of similar length in the continuous solution.
5. Non dissipative – it shows no attenuation.
Stability of a scheme relates to the growth of round-off errors in the computations.
Unstable – if the round-off errors become so large that the solution is destroyed.
Conditionally stable – stable for a range of time and distance steps.
Unconditionally stable – stable for all finite time and distance steps.
Routing with Complete Equations
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Boundary conditions must be specified to permit a solution at the boundary points. With
supercritical flow, velocity exceeds celerity and the trajectories have slopes of the same sign. The
interval of dependence of the downstream point is wholly within the solution domain, and no
boundary condition is needed. If one is specified, the solution will be unstable. Conversely, two
boundary condition must be supplied at the upstream boundary where the interval of dependence
falls outside the solution domain.
Explicit and Implicit schemes.
Explicit scheme – permits solution of the equations node by node using information within one
or two distance steps from the unknown node.
- one for which the order of computation of the unknown at the nodes may be
arbitrary.
Implicit scheme – requiring a simultaneous solution for all unknown value.
- one for which the order of the computations is fixed, including
schemes requiring simultaneous solution of the equations.
- the computation must proceed from the upstream boundary to the downstream
boundary.
Where:
u - the wave velocity
u∆t/∆x - Courant number
Courant condition- it requires that the time step be less than the time required for a small
amount disturbance to transverse ∆x.
- does not apply to implicit schemes but the time step may be limited by
stability.
Courant number – must be less than unity for explicit schemes.
Diffusing or Lax scheme – not consistent with the governing equation since the limit of the
discrete equations contains terms dependent on ∆t and ∆x.
- useful in representing flows containing an abrupt water.
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∆t ≤ ∆x/u
Preissman weighted four-point scheme – an implicit scheme widely used for various forms.
Kinematic Routing
- Frequently used as the overland flow and stream flow routing component of
catchments models.
- Major forces affecting unsteady flow are pressure, gravity and friction.
- the kinematic flow approximation results if all factors other than friction and
gravity are ignored in the motion Equation
- Care must be exercised in kinematic routing to avoid unrealistic results.
- Rapid decreases in channel slope or flow capacity, or an increase in roughness
may increase the neglected terms sufficiently to invalidate the kinematic
assumption.
- Recent urban runoff model uses the Schaake modification for kinematic routing.
Zero-Inertia Routing
- A less radical simplification of the complete equations results if only the inertial
terms are neglected in the momentum equation.
- Approximation has been applied to border irrigation
applications and routing in streams
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So =Sf = Q2/K2
∂y/∂x = So - Sf
