7. Hydrologic Budget
It consists of inflows, outflows, and storage, presented by the following equation:
Inflow = Outflow +/- Changes in Storage
Inflows contribute or add water to the different parts of the hydrologic system, outflows
remove water from them, and storage is the retention of water by parts of the system.
Since, water movement is cyclical; therefore, an inflow for one part of the system is an
outflow to another.
As example, for an aquifer the percolation of water into the ground is the inflow to the
aquifer while discharge of groundwater from the aquifer to a stream is an outflow. Over
time, if inflows to the aquifer are greater than its outflows, the amount of water stored in
the aquifer will increase. Conversely, if the inflow to the aquifer is less than the
outflow, the amount of water stored decreases.
7
8. 8
Example: The following table gives values of measured discharges at a streamβ gauging site in a year. Upstream of
the gauging site a weir built across the stream diverts 3.0 Mm3 and 0.50 Mm3 of water per month for irrigation and
for use in an industry, respectively. The return flows from the irrigation is estimated as 0.8 Mm3 and from the
industry at 0.30 Mm3 reaching the stream upstream of the gauging site. Estimate the natural flow. If the catchment
area is 180 km2 and the average annual rainfall is 185 cm, determine the runoff-rainfall ratio.
Month
Gauged flow (Mm3)
1
2.0
2
1.5
3
0.8
4
0.6
5
2.1
6
8.0
7
18.0
8
22.0
9
14.0
10
9.0
11
7.0
12
3.0
SOLUTION: In a month the natural flow volume RN is obtained from Eq. (5.1) as
RN = (Ro β Vr) + Vd + E + EX + βS
Here E, EX and βS are assumed to be insignificant and of zero value.
Vr = Volume of return flow from irrigation, domestic water supply and industrial use = 0.80 + 0.30 = 1.10 Mm3
Vd = Volume diverted out of the stream for irrigation, domestic water supply and industrial use = 3.0 + 0.5 = 3.5 Mm3
The calculations are shown in the following Table:
Month 1 2 3 4 5 6 7 8 9 10 11 12
Ro
(Mm3)
2.0 1.5 0.8 0.6 2.1 8.0 18.0 22.0 14.0 9.0 7.0 3.0
Vd
(Mm3)
3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5
Vr
(Mm3)
1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1
RN(Mm3
)
4.4 3.9 3.2 3.0 4.5 10.4 20.4 24.4 16.4 11.4 9.4 5.4
Total RN = 116.8 ππ3
Annual natural flow volume = Annual runoff volume = 116.8 ππ3 Area of the catchment = 180 πΎπ2 = 1.80 Γ 108
Annual runoff depth =
1.168 Γ108
1.80 Γ 108 = 0.649 π = 64.9 ππ
π΄πππ’ππ ππππππππ = 185 ππ (π π’ππππ/π πππππππ) = 64.9/185 = 0.35
9. Watershed- Concept and Laws
Definition- Watershed is an isolated area with a
well-defined boundary line, draining the rainwater
to a single outlet. Within the boundary a watershed
contains various natural resources such as the soil,
water and natural vegetation. Also, there is a
network of stream system to drain the rainwater.
The stream network is also called drainage system
of watershed.
9
10. Rainfall- It is in liquid form (drops) falling from the clouds to the earthβs surface. The size
of water droplets is about 0.5 mm or a little bit bigger. The rate of rainfall varies from time to
time. A light rain ranges from 2.5mm/h, moderate rain from 2.5-7.5mm/h, and heavy rain
above 7.5mm/h. Rainfall is the most important component of the hydrologic cycle which
replenishes a large percentage of fresh water on earth. Rain and drizzle are beneficial for
plants.
Drizzle- It is also in liquid form, but the size of its droplets is less than 0.5mm diameter. Its
intensity is lesser than 2.5mm/h. It contributes moisture to the lower atmosphere effective for
cooling and generating warm air mass to create a cloud in the sky. Drizzle usually falls from
low stratus clouds and is frequently accompanied by fog.
10
11. Rainfall Measurement
Rain gauge- The rain gauge is the instrument used for rainfall
measurement. The measured rainfall is termed the point rainfall. The
point rainfall is used for determining the mean areal rainfall by using
various computing methods. The mean aerial rainfall can be used for
determining the volume of rainwater received over the surface area of the
watershed by multiplying the mean depth of rainfall and area of the
watershed/region.
Types of Rain gauge
Broadly, it is classified as
1. Non- recording type; and
2. Recording type rain gauge.
11
12. Non- Recording Type Rain gauge (Simon type) - It is the most common type of rain
gauge, consists of a 127mm diameter cylindrical vessel with a base width of 210mm diameter for making stability. At the
top, a funnel is provided with a circular brass rim which is exactly 127mm to fit into the vessel correctly. This funnel shank
is inserted in the receiving bottle placed below. The height of receiving bottle is 75 to 100mm. The bottle receives the
rainfall. The capacity of receiving bottle is to measure the rainfall depth is 100mm. During heavy rainfall, the rainfall
amount is likely to exceed the bottle capacity. In this condition, it is suggested to take the observations frequently, normally
3 to 4 times a day. The water collected in the receiving bottle is measured by a graduated measuring cylinder. The
measuring accuracy of the graduated cylinder is being up to 0. 1mm. The timing of rainfall measurement is uniformity did,
every day at 8:30 Am IST. For accurate measurement, the proper care, maintenance, and inspection of rain gauge should be
carried out during dry weather.
12
14. 14
Recording Type Rain gauge: It records the information about the
start and end of rainfall events taking place. With the help of this information, one
can determine the rainfall intensity and depth of the place under measurement. The
following rain gauges are commonly used as recording type rain gauges,
1. Float type rain gauge,
2. Weight type rain gauge, and
3. Tipping bucket-type rain gauge.
Float Type Rain Gauge- It is also known as a natural siphon type rain gauge. In
India, this rain gauge is adopted as the standard recording rain gauge. The working
of this rain gauge is similar to the weighing type rain gauge. In this, a funnel
receives the rainwater which is collected into a container equipped with a float at
the bottom. The position of float rises as the water level rises in the container
depending on rainwater coming into it. The movement of the float is transmitted to
a pen which traces a curve on the rain chart mounted on a clockwise rotating Drum.
When the float rises to the top of the container the siphon comes into action and
drains the total water from the container. At this stage, the pen traces a straight line.
If rainfall is continued and water is coming into the container, then further float
rises up and the pen traces the curve. This process is continued. If rainfall is stooped
the pen traces a horizontal line on the chart. The obtained curve is the mass curve.
The view of this rain gauge is shown in Figure.
15. 15
Weighing Type Rain Gauge:- It is the most common self-recording rain gauge, consists of a receiver
(bucket) supported by a spring/ lever balance or some other weighing mechanism. The movement of the bucket due to its
increased weight because of the accumulation of rainwater is transmitted to a pen, which traces a curve on the rain chart
wrapped on a clock. The obtained rainfall record in terms of the curve is a mass curve, i.e. the plot of cumulative rainfall vs
elapsed time. View of rain gage is shown in Fig-
Tipping Bucket Type Rain Gauge:- It is a
30cm size rain gauge, used as a recording type rain gauge. US weather
bureau uses this rain gauge for measuring the rainfall. Its construction
includes a 30cm diameter sharp-edged receiver. At its end, a funnel is
provided for directing the rainwater into the receiver. One pair of
buckets is pivoted on a fulcrum below the funnel in such a way that
when one bucket receives 0.25mm depth of rainfall, it tips and
empties its rainfall into the container, and immediately the second
bucket comes below the funnel (Fig..). The rainfall measurement is
recorded in terms of a number of tips made for a given rainfall event,
which is indicated on a dial actuated by an electrical circuit.
16. Common Errors in Rainfall Measurement
Few important errors in rainfall measurement by the rain gauge are
mentioned as under,
1. In non- recording type rain gauge (Symonsβs type) about 2% error is
introduced due to displacement of water level by measuring scale.
2. Possibility of errors due to initial wetting of dried surface of the catch
can or receiver.
3. The dents in catch can or receiver also introduces errors in
measurement.
4. A high temperature cause evaporation loss also signifies a kind of
errors in rainfall measurement. The errors may be up to 2%.
5. A high wind velocity deflects the rainfall to fall at the mouth of the
rain gauge, introduces an errors in rainfall catching. The research
revealed that at the wind velocity of 10mile per hour the catching of
rainfall is declined to the tune of about 17% while at 30mile per hour
it may be up to 60%.
16
18. Missing Rainfall Data
In normal course, sometime what happens, because of several reasons such as absence of
observer, instrumental fault etc there is short breaks in the rainfall records. In this condition
to fill the break the estimation of missing rainfall data is essentially required. The following
methods are commonly used for computing the value of missing rainfall data
1. Arithmetic Mean Method
2. Normal Ratio Method
Arithmetic Mean Method- This method follows following formula for determining the mean
aerial rainfall,
Normal Ratio Method- This method is used when normal annual rainfall at any of the index
station differs from the interpolation station by more than 10%. Missing rainfall data is
predicted by weighing the rainfall of index stations by the ratios of their normal annual
rainfall. Formula is given as under, 18
19. For 3 number of defined index rain gauge stations the above formula
is expanded as:
in which ππ₯ is the missing rainfall at rain gauge station β²π₯β² of a
given rainfall event, ππ is the precipitation for the same period and
same rainfall event of "ith" rain gauge station among group of index
stations, ππ₯ the normal annual rainfall (NAR) of station π₯ and
ππ the normal annual rainfall of 'ith' station. The solve example β
illustrates the procedure.
19
20. Mean Areal Rainfall
Average rainfall is the representative of large area, which is computed with
the help of rainfall data generated from well distributed rain gauge network
system of the watershed. The computing methods are elaborated as under,
1. Arithmetic or station average method
2. Thiessen Polygon Method
3. Isohyetal Method.
Arithmetic Average Method
This method computes arithmetic average of the rainfall by considering point
rainfall observations of all the rain gauge stations installed in the area. This
method computes accurate value when rainfall is uniformly distributed in the
entire area, as in this situation equal weightage of area is assigned to the
point rainfall data. Formula is given as under,
20
21. Problem (1)- In a watershed four rain gage stations namely a, B, C and D are installed for
recording rainfall data. The normal annual rainfall of these four stations is 75, 60, 70.5 and
87 cm, respectively. The rain gauge station A does not have the annual rainfall observation
for one year during total length of record, because of disorder of the rain gauge. Calculate
the missing value of rainfall data of rain gauge station A, if the annual rainfall recorded at
other three stations for that particular year was 85, 67.5 and 75 cm, respectively at B, C and
D, respectively.
Solution- The variation in normal rainfall data is more than 20% at all the four rain gauge
stations. In this condition, the normal ratio method for computing the missing value of
annual rainfall of station A is suitable. Accordingly, the formula for computing the missing
annual rainfall is given as under.
π1 =
π1
π β 1
π2
π2
+
π3
π3
+
π4
π4
in which, π2 = 85ππ; π3 = 67.5ππ; π4 = 75ππ, πππ π1 = 75ππ; π2 = 60ππ; π3 =
70.5ππ; π4 = 87ππ πππ π = 4. Substituting these values in above formula and solving , we
have, π1=
75
4β1
85
60
+
67.5
70.5
+
75
87
= 81ππ π¨ππ. 21
22. Mean Areal Rainfall
Average rainfall is the representative of large area, which is computed with the
help of rainfall data generated from well distributed rain gauge network system
of the watershed. The computing methods are elaborated as under,
1. Arithmetic or station average method
2. Thiessen Polygon Method
3. Isohyetal Method.
Arithmetic Average Method
This method computes arithmetic average of the rainfall by considering point
rainfall observations of all the rain gauge stations installed in the area. This
method computes accurate value when rainfall is uniformly distributed in the
entire area, as in this situation equal weightage of area is assigned to the point
rainfall data. Formula is given as under,
ΰ΄€
π =
1
π
ΰ·
π=1
π
ππ
22
23. where ΰ΄€
π is the mean rainfall is over an area, P is the point rainfall at
individual station i, and n is the total number of stations.
Solve problem (1) illustrates the computation procedure.
Problem (1)- In a topographically homogeneous watershed total four number
of non- recording and one
recording type rain gauges have been installed for recording the rainfall
measurements. The point rainfall of four non- recording type rain gauge
stations have been observed to the tune of 250,175,225 and 270mm,
respectively during a given rainfall event. Determine the mean areal rainfall
of the watershed for the said rainfall event.
Solution- The mean areal rainfall of the watershed can be computed by using
the simple arithmetic mean method, given as under:
ππ =
π1+π2 +π3+π4
π
=
250+175+225+270
4
= 230ππ Ans.
23
24. Thiessen Polygon Method
This is a graphical method for computing MAP. It computes by weighing the relative area of
each rain gauge station equipped in the watershed. It follows the concept that rainfall varies by
its intensity and duration, spatially. Therefore, the rainfall recorded by each station should be
weighed as per the influencing area (polygons). This method computes better for the areas
having flat topography and size ranging from 500 to 5000 km2. Computing steps are described
as under,
1. Plot the locations of rain gauge stations on the map of the area drawn to a scale.
2. Join each station by a straight line.
3. Draw perpendicular bisectors of each line. These bisectors form polygons around each
station. The area enclosed within the polygon is the effective area for the station. For a rain
gauge station close to the boundary, the boundary lines form its effective area.
4. Determine the effective area of each rain gauge station. For this, the planimeter can be used.
5. Calculate MAP by using the following formula,
ΰ΄€
π =
Οπ=1
π
ππ. π΄π
π΄
in which, Pi is the rainfall depth of rain gauge station i and A is the total area of watershed.
24
25. Problem (2)- Compute the mean areal rainfall of the watershed by using Theissen Polygone Method.
The details are cited below.
Measured rainfall (cm) 10.5 11.56 9.57 10.50 11.63
Area of enclosed polygon
(sqkm)
15.0 23.5 35.9 8.50 12.35
Rain gauge station A B C D E
Solution- In Thiessen Polygon Method the following formula
is used for computing the value of mean areal rainfall depth.
Rain
gauge
station
Measured
rainfall
(cm)
Enclosed
area of
polygon
(sqkm)
Rainfall x enclosed
area of
polygon (cm.sqkm)
Col.II x Col.IV
I II III IV
A 10.5 15.0 157.50
B 11.56 23.5 271.66
C 9.57 35.9 343.56
D 10.50 8.5 89.25
E 11.63 12.35 143.63
Total 95.25 1005.60
in which, ππ is the rainfall depth for rain gauge station i
and π΄π is the area of polygon enclosed by the rain gauge
station i and A is the total area of watershed. Computation
is shown below.
Therefore, mean areal rainfall= 1005.60/95.25=10.56 cm Ans.
25
ΰ΄€
π =
Οπ=1
π
ππ. π΄π
π΄
32. Example: For the watershed region shown in the figure,
determine the average depth of rainfall by using Thiessen
Method ?
A=11 cm , B= 14 cm ,
C =15 Cm , D=12.5 cm
32
39. 39
5. POPULATION FORECASTING
Design of water supply and sanitation scheme is based on the projected population of a particular city,
estimated for the design period. Any underestimated value will make system inadequate for the purpose
intended; similarly overestimated value will make it costly. Change in the population of the city over the
years occurs, and the system should be designed considering of the population at the end of the design
period.
Factors affecting changes in population are:
ο· increase due to births.
ο· decrease due to deaths.
ο· increase/ decrease due to migration.
ο· increase due to annexation.
The present and past population record for the city can be obtained from the census population records.
After collecting these population figures, the population at the end of design period is predicted using
various methods as suitable for that city considering the growth pattern followed by the city.
40. 40
1. ARITHMETICAL INCREASE METHOD
This method is suitable for large and old city with considerable development. If it is used for small, average or
comparatively new cities, it will give low result than actual value. In this method the average increase in
population per decade is calculated from the past census reports. This increase is added to the present
population to find out the population of the next decade. Thus, it is assumed that the population is increasing at
constant rate.
Hence,
ππ
ππ‘
= πΆi.e. rate of change of population with respect to time is constant.
Therefore, Population after nth decade will be ππ = π + π. πΆ
Where, ππ is the population after n decade and P is present population.
41. 41
Example:1
Predict the population for the year 2021, 2031, and 2041 from the following population data.
Year Population Increment
1961 858545 -
1971 1015672 157127
1981 1201553 185881
1991 1691538 489985
2001 2077820 386282
2011 2585862 508042
π΄π£πππππ πππππππππ‘ = 345463
ππππ’πππ‘πππ ππ π¦πππ 2021 ππ , π2021 = 2585862 + 345463 x 1 = 2931325
πππππππππ¦, π2031 = 2585862 + 345463 x 2 = 3276788
π2041 = 2585862 + 345463 x 3 = 3622251
42. 42
1. GEOMETRICAL INCREASE METHOD
(OR GEOMETRICAL PROGRESSION METHOD)
In this method the percentage increase in population from decade to decade is assumed to remain constant.
Geometric mean increase is used to find out the future increment in population. Since this method gives higher
values and hence should be applied for a new industrial town at the beginning of development for only few
decades. The population at theend of nth decade βππβ can be estimated as:
ππ = π (1 + πΌπΊ/100) π
Where, πΌπΊ = ππππππ‘πππ ππππ % ,
π = ππππ πππ‘ ππππ’πππ‘πππ ,
π = ππ. ππ πππππππ .
43. 43
Example : 2
Considering data given in example 1 predict the population for the year 2021, 2031, and 2041 using
geometrical progression method.
Solution
Year Population Increment Geometrical increase
Rate of growth
1961 858545 -
1971 1015672 157127 (157127/858545)= 0.18
1981 1201553 185881 (185881/1015672)= 0.18
1991 1691538 489985 (489985/1201553)= 0.40
2001 2077820 386282 (386282/1691538)= 0.23
2011 2585862 508042 (508042/2077820)= 0.24
πΊπππππ‘πππ ππππ πΌπΊ = (0.18 π₯ 0.18 π₯ 0.40 π₯ 0.23 π₯ 0.24)π/π = 0.235 π. π. , 23.5%
ππππ’πππ‘πππ ππ π¦πππ 2021 ππ , π2021 = 2585862 x (1 + 0.235)1
= 3193540πππππππππ¦ πππ π¦πππ 2031 πππ 2041 πππ ππ πππππ’πππ‘ππ ππ¦,
π2031 = 2585862 π₯ (1 + 0.235)2 = 3944021
π2041 = 2585862 π₯ (1 + 0.235)3 = 4870866
44. 44
1. INCREMENTAL INCREASE METHOD
This method is modification of arithmetical increase method, and it is suitable for an average size town
under normal condition where the growth rate is found to be in increasing order. While adopting this
method the increase in increment is considered for calculating future population. The incremental
increase is determined for each decade from the past population and the average value is added to the
present population along with the average rate of increase.
Hence, population after ππ‘β decade is ππ = π + π. π + {π (π + 1)/2}. π Where, ππ
= ππππ’πππ‘πππ πππ‘ππ ππ‘β ππππππ
π = π΄π£πππππ πππππππ π
π = πΌππππππππ‘ππ πππππππ π
45. 45
Example : 3
Considering data given in example 1 predict the population for the year 2021, 2031, and 2041using
incremental increase method.
Year Populatio
n
Increase
(X)
Incremental
increase (Y)
1961 858545 - -
1971 1015672 157127 -
1981 1201553 185881 +28754
1991 1691538 489985 +304104
2001 2077820 386282 -103703
2011 2585862 508042 +121760
Total 1727317 350915
Average 345463 87729
46. 46
CALCULATION OF STORAGE VOLUME-FLOW MASS CURVE
Consider a reservoir on the stream whose mass curve is plotted in Fig. If it is assumed that the reservoir is full at the
beginning of a dry period, i.e. when the inflow rate is less than the withdrawal (demand) rate, the maxi- mum amount of
water drawn from the storage is the cumulative difference between supply and demand volumes from the beginning of the dry
sea- son. Thus, the storage re- quired S is S = maximum of (Ξ£VD β Ξ£Vs)
47. 47
where VD = demand volume, VS = supply volume. The storage, S which is the maximum
cumulative deficiency in any dry season is obtained as the maximum difference in the
ordinate between mass curves of supply and demand. The minimum storage volume required
by a reservoir is the largest of such S values over different dry periods. Consider the line CD
of slope Qd drawn tangential to the mass curve at a high point on a ridge. This represents a
constant rate of withdrawal Qd from a reservoir and is called demand line. If the reservoir is
full at C (at time tc) then from point C to E the demand is larger than the supply rate as the
slope of the flowβmass curve is smaller than the demand line CD. Thus the reservoir will be
depleting and the lowest capacity is reached at E. The difference in the ordinates between the
demand line CD and a line EF drawn parallel to it and tangential to the mass curve at E (S1 in
Fig.) repre- sents the volume of water needed as storage to meet the demand from the time the
reservoir was full. If the flow data for a large time period is available, the demand lines are
drawn tangentially at various other ridges (e.g. Cβ Dβ in Fig.) and the largest of the storages
obtained is selected as the minimum storage required by a reservoir. Example 5.9 explains
this use of the mass curve. Example 5.10 indicates, storage calculations by arithmetic
calculations by adopting the mass-curve principle.
48. 48
Example: The following table gives the mean monthly flows in a river during1981. Calculate the minimum
storage required to maintain a demand rate of 40 π3/π ππ.
Month Jan Feb Mar Apr May June July Aug Sept Oct Nov Dec
Mean
Flow
(m3/s)
60 45 35 25 15 22 50 80 105 90 80 70
SOLUTION: From the given data the monthly flow volume and accumulated volumes and calculated as in Table. The
actual number of days in the month are used in calculating of the monthly flow volume. Volumes are calculated in
units of cumec. day (= 8.64 Γ 104
).
Month Mean flow (m3/s) Monthly flow
volume
(cumec-day)
Accumulate
dvolume
(cumec-day)
Jan 60 1860 1860
Feb 45 1260 3120
Mar 35 1085 4205
April 25 750 4955
May 15 465 5420
June 22 660 6080
July 50 1550 7630
Aug 80 2480 10,110
Sep 105 3150 13,260
Oct 90 2790 16,050
Nov 80 2400 18,450
Dec 70 2170 20,620
Table : Calculation of Mass CurveβExample 5.9
A mass curve of accumulated flow volume against time is
plotted (Fig.). In this figure all the months are assumed to be of
average duration of 30.4 days. A demand line.
49. 49
with slope of 40 π3
/sec is drawn tangential to the hump at the beginning of the curve; line AB in Fig.. A line parallel to
this line is drawn tangential to the mass curve at the valley portion; line AβBβ. The vertical distance S1 be- tween these
parallel lines is the minimum storage required to maintain the demand. The value of S1 is found to be 2100 cumec. Days=
181.4 million π3
.
50. 50
Work out the previous Example through arithmetic calculation without the use of mass curve. What is the maximum
constant demand that can be sustained by this river?
Month Mean
inflow
rate
(m3/s)
Inflow
volume
(cumec
.day)
Dema
nrate
(m3/s
)
Deman
volume
(cumec
.day)
Departure
[co1. 3-
col. 5)
Cum.
excess
deman
d
volume
(cumec
.day)
Cum.
excess
inflow
volume
(cumec
.day)
Jan 60 1860 40 1240 620 620
Feb 45 1260 40 1120 140 760
Mar 35 1085 40 1240 β155 β155
Apr 25 750 40 1200 β450 β605
May 15 465 40 1240 β775 β1380
Jun 22 660 40 1200 β540 β1920
July 50 1550 40 1240 310 310
Aug 80 2480 40 1240 1240 1550
Sept 105 3150 40 1200 1950 3500
Oct 90 2790 40 1240 1550 5050
Nov 80 2400 40 1200 1200 6250
Dec 70 2170 40 1240 930 7180
Monthly
mean = 1718.3
SOLUTIOH: The inflow and demand volumes of each month are calculated as in Table5.9. Column 6
indicating the departure of the inflow volume from the demand. The negative values indicate the excess of
demand over the inflow and these have to be met by the storage. Column 7 indicates the cumulative excess
demand (i.e., the cumulative excess negative departures). This column indicates the depletion of storage, the
first entry of negative value indicates the beginning of dry period and the last value the end of the dry period.
Col. 8 indicates the filling up of storage and spill over (if any). Each dry period and each filling up stage is to
be calculated separately as indicated in Table 5.9.The maximum value in Col. 7 represents the minimum
storage necessary to meet the demand pattern. In the present case, there is only one dry period, and the
storage requirement is 1920 cumec.day. Note that the difference between this value and the value of 2100
cumec.day obtained by using the mass curve is due to the curvilinear variation of inflow volumes obtained
by drawing a smooth mass curve. The arithmetic calculation assumes a linear variation of the mass curve
ordinates between two adjacent time units. [Note: It is usual to take data pertaining to a number of N full
years. When the analysis of the given data series of length N causes the first entry in Col. 7 to be a negative
value and the last entry is also a negative value, then the calculation of the maximum deficit may pose some
confusion. In such cases, repeating the data sequence by one more data cycle of N years in continuation with
the last entry would overcome this confusion. (See Sec. 5.7, item 2.) There are many other combinations of
factors that may cause confusion in interpretation of the results and as such the use of Sequent Peak
Algorithm described in Sec. 5.7 is recommended as the foolproof method that can be used with confidence in
all situations.]Column 8 indicates the cumulative excess inflow volume from each demand withdrawal from
the storage. This indicates the filling up of the reservoir and volume in excess of the provided storage (in the
present case 1920 cumec.day) represent spill over. The calculation of this column is necessary to know
whether the reservoir fills up after a depletion by meeting a critical demand and if so, when? In the present
case the cumulative excess inflow volume will reach +1920 cumec.day in the beginning of September. The
reservoir will be full after that time and will be spilling till end of February.
Average of the inflow volume per month = Annual inflow volume/12 = average monthly demand that can be sustained by this river
= 1718.33 cumec.day.
51. 51
Example:For a proposed reservoir, the following data were calculated. The prior water rights required the release of
natural flow or 5 π3/sec, whichever is less. Assuming an average reservoir area of 20 πΎπ2, estimate the storage required
to meet these demands. (Assume the runoff coefficient of the area submerged by the reservoir = 0.5.)
Month Mean flow
(m3/s)
Deman
(million m3)
Monthly evaporation
(cm)
Monthly rainfall
(cm)
Jan 25 22.0 12 2
Feb 20 23.0 13 2
Mar . 15 24.0 17 1
April 10 26.0 18 1
May 4 26.0 20 1
June 9 26.0 16 13
July 100 16.0 12 24
Aug 108 16.0 12 19
Sept 80 16.0 12 19
Oct 40 16.0 12 1
Nov 30 16.0 11 6
Dec 30 22.0 17 2
SOLUTION: Use actual number of days in a month for
calculating the monthly flow and an average month of 30.4 days
for prior right release.
Prior right release = 5 Γ 30.4 Γ 8.64 Γ 104 = 13.1 π π3 when Q
> 5.0 m3/s.
Evaporation volume = E Γ 20 Γ 106 = 0.2 E π π3
100
Rainfall volume = P Γ (1 β 0.5) Γ 20 = 0.1 P π π3
100
Inflow volume: I Γ (No. of days in the month) Γ 8.64 Γ 104 π3
The calculations are shown in Table 5.6 and the required storage
capacity is 64.5 π π3 . The mass-curve method assumes a
definite sequence of events and this is its major drawback. In
practice, the runoff is subject to considerable time variations
and definite sequential occurrences represent only an idealized
situation. The mass-curve analysis is thus adequate for small
projects or preliminary studies of large storage projects. The
latter ones require sophisticated methods such as time-series
analysis of data for the final design.
72. RUNOFF CALCULA
TIONS
To estimate the magnitude of a flood peak the following alternative methods are available:
1.Unit-hydrograph technique
2.Empirical method
3.Semi-Empirical method (such rational method).
There many empirical or Semi-Empirical formulae used to estimate the runoff discharge from catchment area. These formulae can be classified into three
categories;
1. Formulae consider the area only into calculation, like Dickens, Ryves, Ingles and others. The formulae take forms as Q=CAn
; n exponent is
almost Λ1.
2. Formulae consider Area and some other factors such as Craig , Lillie and Rhinds (Taking velocity , and may be intensity, depth or max,
depth of rainfall).
3. Formulae consider the recurrence interval ,like Fullers , Hortons , Pettis and other.
The use of a particular method depends upon (i) the desired objective, (ii) the available data, and (iii) the importance of the project.
Above all , two methods depend on semi-empirical bases are preferable for storm design and have a wide use by the designer. The Rational method and the
SCS-CN method . Further the Rational formula is only applicable to small-size (< 50 km2
) catchments and the unit-hydrograph method is normally restricted to
moderate-size catchments with areas less than 5000 km2
RATIONAL METHOD
Consider a rainfall of uniform intensity and duration occurring over a basin in a time taken for a drop of water from the farthest part of the catchment to reach
the outlet that called tc = time of concentration, it is obvious that if the rainfall continues beyond tc, the runoff will be constant and at the peak value. The peak
value o f the runoff is given by;
7
2
73. ππ = πΆ I A ; for π‘ > π‘π β β β β β β β β β (1)
Where,
C = coefficient of runoff = (runoff/rainfall), A = area of the catchment and i = intensity of rainfall.
The simulation of above equation is quoted from the following; To quantify peak runoff discharge ππ,
74. 1
Water Supply Part 3
GROUND WATER FLOW
Hydraulic head, Darcyβs law, determining hydraulic conductivity,
heterogeneity and anisotropy, and applications
Reference :Mohammad N. Almasri
An-Najah National University
76. Hydraulic Head
βͺ Total head at a point is the summation of the
pressure head and elevation head;
βͺ Total head also equals the distance between ground
surface and the datum minus the depthto water in the
well;
7
6
77. Hydraulic Head
Ground water moves in
the direction of
decreasing total head,
which may or may notbe
in the direction of
decreasing pressure head
The direction of the
slope of the water table
or the potentiometric
surfaceis also important
because it indicates the
direction of groundwater
movement
7
7
78. Causes of Ground Water Movement
Hydraulic head (h) = elevation + pressure
Water flows from zones of high
hydraulic head to low
hydraulic head
7
8
79. Hydraulic Head
Example
In an aquifer, the ground surface is at 1,000 m abovesea
level, the depth to the water table is 25 m, and the water
table height above the measurement point is 50 m.
Calculate:
1. The total hydraulic head at the point of
measurement
2. The pressure head, and
3. The elevation head
7
9
80. Hydraulic Head
Example Solution
Hydraulic head = distance from the water tableto the
mean sea level = 1,000 β 25 = 975 m
Pressure head = distance from the water tableto the
point of measurement = 50 m
Elevation head = ground surface elevation β depth to
water table β pressure head = 1,000 β25 β 50 = 925
m
8
0
82. Hydraulic Gradient
Simple Example
Water table elevation was measured at two locations
with a distance of 1,000 ft. If the measured elevations
were 100 and 99 ft, then what is the direction of the
groundwater flow andwhat is the hydraulic gradient?
86. Example
13
Three piezometers monitor
water levels in a
confined aquifer.
Piezometer A is located
3,000ft due south of
piezometer B.
Piezometer C is located
2,000 ft
due west of piezometer B.
The surface elevations of A, B,
and C are 480, 610 and 545ft,
respectively.
The depth to water in A is 40ft,
in B is 140 ft, and in C is85 ft.
Determine the direction of
groundwater flow through
the triangle ABC and
hydraulic
calculate the
gradient
91. Darcyβs Law
βͺ Circular cylinder (A);
βͺ Filled with sand;
βͺ Each end outfitted with inflow and outflow tubes;
βͺ Water to flow through the sand until saturation and the inflow rate Q
equals the outflow rate;
βͺ The fluid heads are h1 and h2 [using a pair of manometers];
βͺ The experiments carried out by Darcy showed that Q/A is directly
proportional to h1 β h2 and inversely proportional to βl
92. Darcyβs Law
Observed that flow rate in sand columns
increased or decreased uniformly as the
hydraulic head difference between two
manometers increased or decreased
94. Darcyβs Law
21
Q: Total flow (L3/T)
q: (Q/A) Darcy flux (L/T)
A: Cross-sectional area of
the flow (L2)
K: Hydraulic conductivity
(L/T)
dh: Head difference (L) dl:
Increment distance (L)dh/dl:
Hydraulic gradient
(L/L)
95. Velocity
Since the flow occurs across the pores that can transmit
water and part of the pore space is occupied by stagnant
water, then velocity equals thespecific discharge divided
by the effective porosity
v: velocity (L/T) ne:
is the effective
porosity
99. Darcyβs Law
Solved Example β Unconfined
We have
K = 10 m/day, h1 = 20 m, h2 = 19 m, L = 1,000 mThen
q = - (10) [(20-19)/(1,000)] = - 0.01 m/day
100. Darcyβs Law
Solved Example β Unconfined
q is negative and in the opposite direction of x!!
This is correct since the flow is in the direction of
decreasing head or the direction of hydraulic gradient
or from high to low head
101. Darcyβs Law
Solved Example β Unconfined
Q = q A
What is the area of the flow?
The area is the saturated thickness. This thicknessvaries
from point to point
Take the average height [A = 0.5 (h1 + h2) W]where
W is the width of the aquifer (assume unit
width if not given; W = 1 m)
Then A = 0.5(20 + 19) (1) = 19.5 m2
102. Darcyβs Law
Solved Example β Unconfined
Then
Q = 0.01 x 19.5 = 0.195 m3/day
For the velocity
We know that
v = q/ne = 0.01 /0.2 = 0.05 m/day
103. Darcyβs Law
Solved Example β Unconfined
We know that transmissivity (T) is: Hydraulic
conductivity x aquifer thickness
= 10 X 19.5 = 195 m2/day
105. Darcyβs Law
Solved Example β
Confined
q = - (10) [(20-19)/(1,000)] = - 0.01 m/dayQ = q
A
What is the area perpendicular to the flow?
The area is the thickness of the aquifer which is b.This
thickness is constant as far as the aquifer istotally
saturated.
A = b W where W is the width of the aquifer
(assume unit width if not given; W = 1 m)
Then A = 10 x 1 = 10 m2
106. Darcyβs Law
Solved Example β
Confined
Then
Q = 0.01 x 10 = 0.1 m3/dayfor
the velocity
We know that
v = q/ne = 0.01 /0.2 = 0.05 m/day
107. Darcyβs Law
Solved Example β
Confined
We know that transmissivity (T) is: Hydraulic conductivity
x aquifer thickness
= 10 X 10 = 100 m2/day
109. Applications
Unconfined Aquifers
βͺ Flow area (A) is NOT
constant (h W)
βͺ Total flow (Q) is
constant in the
aquifer
βͺ If Q is constant but Ais
not, then gradient is
NOT constant
10
9
110. Applications
Unconfined Aquifers
We notice that in unconfined aquifers
βͺ The flow is not horizontal near the water table
βͺ There is a curvature in the water table
11
0
111. Applications
Unconfined Aquifers
Dupuit assumptions
βͺ For small water-table gradients, the streamlinesare
horizontal and the equipotential lines are vertical
βͺ The hydraulic gradient equals the slope of thewater
table
11
1
116. Applications
Unconfined Aquifers with Recharge
Applying Darcyβs law,
we get the flow per
unit width
There is a water divideat
which the flow is zero
and the water table is
at the maximum value
117. Two rivers are located 1,000 m apart and fully penetrate an
unconfined aquifer of K=0.5 m/day. The mean annualrainfall
and evaporation are 15 cm/yr and 10 cm/yr, respectively. The
water elevations in rivers 1 and 2 are 20 and 18 m, respectively
a) Determine the location and height of the water divide
b) Determine the flow across the aquifer per m width ofeach
river
123. Introduction
βͺ What is well hydraulics?
To understand the processes in effect when one ormore wells
are pumping from an aquifer. This for instance considers the
analysis of drawdown due topumping with time and distance
βͺ Importance of well hydraulics Groundwater
withdrawal from aquifers are
important to meet the water demand. Therefore,
we need to understand well hydraulics to design apumping
strategy that is sufficient to furnish the adequate amounts of
water
4
124. Basic Assumptions
βͺ The potentiometric surface of the aquifer is horizontalprior
to the start of the pumping
βͺ The aquifer is homogeneous and isotropic
βͺ All flow is radial toward the well
βͺ Groundwater flow is horizontal
βͺ Darcyβs law is valid
βͺ The pumping well fully penetrates the aquifer
5
125. Steady versus Transient
Steady state implies that the drawdown is afunction
of location only
Transient state implies that the drawdown is afunction
of location and time
Thus
h = f(r) in case of steady state
h = f(r,t) in case of transient state
6
127. Steady Radial Flow to a Well in Confined
Aquifers
When water is pumped from a confined aquifer, the
pumpage creates a drawdown in the piezometric surface
that induces hydraulic gradient toward the well
Drawdown at a given point is the distance by which the
water level is lowered. A drawdown curve shows the
variation of drawdown with distance from the well
The induced flow moves horizontally toward the well
The flow is radial to the well
8
128. Steady Radial Flow to a Well in Confined
Aquifers
Apply Darcyβs law to derive the flow equation thatrelates
drawdown with pumping:
Water pumped out of the aquifer
Flow throughthe aquifer
9
129. Steady Radial Flow to a Well in Confined
Aquifers
Rearranging and integrating for the boundaryconditions at
the well h = hw and r = rw
and
at the edge of the aquifer h = h0 and r = r0yields (with
the negative sign neglected):
10
130. Steady Radial Flow to a Well in Confined
Aquifers
or in a general form
11
131. Steady Radial Flow to a Well in Confined
Aquifers
Thiem equation
where r1 and r2 are the distances and h1 and h2 are the
heads of the respective observation wells
12
132. Steady Radial Flow to a Well in Confined
Aquifers
From a practical standpoint, the drawdown srather
than the head h is measured so:
13
133. Example [1] β Steady State ConfinedAquifer
A well in a confined aquifer is pumped at a rate of 220 gal/min
Measurement of drawdown in two observation wells shows that
after 1,270 min of pumping, no further drawdown is
occurring
Well 1 is 26 ft from the pumping well and has a head of 29.34ft
above the top of the aquifer
Well 2 is 73 ft from the pumping well and has a head of 32.56ft
above the top of the aquifer.
Use the Thiem equation to find the aquifer transmissivity
14
135. Example [2] β Steady State ConfinedAquifer
A well is producing water from a confined aquiferat a
rate of 42943.6 ft3/day
An observation well 100 feet from the pumping well
shows a drawdown of 5 feet, while an observation well
20 feet from the pumping wellshows a drawdown of 16
feet
What is the drawdown at the pumping well whenthe
radius of the well is 8 inches?
Consider that the aquifer thickness is 100 ft
16
139. Steady Radial Flow to a Well in
Unconfined Aquifers
The flow equation is similar for that of confinedaquifers
except we use h instead of b
20
140. Steady Radial Flow to a Well in
Unconfined Aquifers
Rearranging and integrating for the boundary conditions
at the well, h = hw and r = rw, and atthe edge of the
aquifer, h = h0 and r = r0, yields:
21
141. Steady Radial Flow to a Well in
Unconfined Aquifers
Converting to heads (h1 and h2) and radii at twoobservation
wells at locations r1 and r2:
22
142. Steady Radial Flow to a Well in
Unconfined Aquifers
Rearranging to solve for the hydraulic conductivity:
23
143. Steady Radial Flow to a Well in
Unconfined Aquifers
The following form is a useful one for computingh as a
function of r:
where r0 is the radius of influence at which thedrawdown
is negligible
24
144. Example [1] β Steady State UnconfinedAquifer
In a fully-penetrating well, the equilibrium drawdown is
30 feet with a constant pumpingrate of 20 gpm
The aquifer is unconfined and the saturated
thickness is 100 feet
What is the steady-state drawdown measured in the well
when the constant discharge is 10 gpm?
25
149. Steady Radial Flow in Leaky Aquifers
s(r): drawdown T:
transmissivity
bΒ΄: thickness of the confining bedK0( ):
modified Bessel function
B: leakage factor
KΒ΄: hydraulic conductivity of the confining bed
30
150. Steady Radial Flow in Leaky Aquifers
For small values of r/B then
For r/B > 1 then
R = 1.123 B is known as the apparent radius of influence
31
151. Example
Steady State Leaky Aquifer
A leaky aquifer has a transmissivity of 400 ft2/d
The aquitard has a conductivity of 0.02 ft/d and a
thickness of 20 ft
A well of radius 10 inches pumps at a rate of 9,600ft3/d
What is the drawdown at the well and its apparentradius of
influence?
32
152. Solution
We have
where T=400 ft2/day, Kβ=0.02
ft/day, bβ=20 ft, r=10 inches, and
Q=9,600 ft3/day. Substitute in
the above equation gives:
33
156. Profile of pressure zones
(Walski, et al. 2001 figure 7.20)
Customers must be served from separate pressure zones
(Walski, et al. 2001 figure 7.17)
156
158. Hardy-Cross Method of Water Distribution
Design
Definitions
β Pipe sections or links are the most
abundant elements in the network.
β’ These sections are constant in diameter and may contain fittings and other
appurtenances.
β’ Pipes are the largest capital investment in the distribution system.
β Node refers to either end of a pipe.
β’ Two categories of nodes are junction nodes and fixed-grade nodes.
β’ Nodes where the inflow or outflow is known are referred to as junction
nodes. These nodes have lumped demand, which may vary with time.
β’ Nodes to which a reservoir is attached are referred to as fixed-grade nodes.
These nodes can take the form of tanks or large constant- pressure mains.
Schematic network illustrating the use of a pressure
reducing valve
(Walski, et al. 2001 figure 3.29)
158
159. Steps for the Hardy-Cross Method
Street
Number
Building
Description
Without Fire Demand With Fire Demand (worst
building)
Steps for Setting Up and Solving a Water Distribution
System using the Hardy-Cross Method
1. Set up grid network to resemble planned flow
distribution pattern.
159
1** 5 A, 1 S MGD 0.059 ft3/sec 0.092 MGD 2.00 ft3/sec
3.10
2 3 A 0.019 0.030 0.019 0.030
3 18 A 0.12 0.18 0.12 0.18
4 3 A, 1 O, 1 C 0.056 0.086 0.056 0.086
5 2 A 0.013 0.020 0.013 0.020
6 7 A 0.045 0.070 0.045 0.070
7 10 A 0.065 0.100 0.065 0.100
8 2 A 0.013 0.020 0.013 0.020
9 7 A 0.045 0.070 0.045 0.070
10 8 A 0.052 0.080 0.052 0.080
11 No buildings 0.0 0.0 0.0 0.0
160. Steps for the Hardy-Cross Method
4. Assume internally consistent distribution of flow,
i.e., at any given node and for the overall water
distribution system:
ο flow entering node = ο flow leaving node
Steps for the Hardy-Cross Method
3. Add up the flow used in the neighborhood without
fire demand and distribute it out the nodes where
known outflow is required. Repeat for fire demand.
Total without Fire Demand = 0.75 cfs Influent = 2.5
MGD = 3.87 cfs
Left Over to Other Neighborhoods = 3.12 cfs
Distribute 50/50 to two outflow nodes = 1.56 cfs
(arbitrary for this problem β would be based on known
βdownstreamβ requirements).
160
161. Steps for the Hardy-Cross Method
6. For each of the
pipes leaving the inflow node, put
the water demand for that street at
the node at the end of the pipe.
Steps for the Hardy-Cross Method
5. For the inflow node,
split the flow among the
pipes leaving that node
(there will be no
additional outflow since
no water has been used
by the neighborhood as
yet).
Inflow = ο Outflows
10
162. Steps for the Hardy-Cross Method
For the above node and the next pipes in the distribution system,
subtract the water used on the street (and aggregated at the node)
from the water flowing down the pipe. Pass the remaining water
along to one or more of the pipes connected to that node.
Repeat Steps 6 and 7 for each pipe and node in the distribution
system. The calculation can be checked by seeing if the total
water outflow from the system equals the total inflow to the
system, as well as checking each node to see if inflow equals
outflow.
162
163. Steps for the Hardy-Cross Method
Select initial pipe sizes (assume a velocity of 3 ft/sec for normal flow
with no fire demand). With a known/assumed flow and an assumed
velocity, use the continuity equation (Q = VA) to calculate the cross-
sectional area of flow. (when conducting computer design, set
diameters to minimum allowable diameters for each type of
neighborhood according to local regulations)
Pipe
Number
1
2
3
4
5
6
7
8
9
10
11
Flow (ft3/sec) Velocity
(ft/sec)
3.0
3.0
3.0
3.0
3.0
3.0
3.0
3.0
3.0
3.0
3.0
Area (ft2) Diameter
(ft)
0.91
0.35
0.63
0.91
0.65
0.63
0.31
0.68
0.60
0.49
0.49
Diameter Actual D
(in) (in)
10.9 12
4.2 6
7.5 8
10.9 12
7.8 8
7.5 8
3.7 4
8.2 10
7.2 8
5.9 6
5.9 6
1.94
0.288
0.922
1.930
1.000
0.922
0.220
1.100
0.852
0.562
0.562
0.65
0.10
0.31
0.64
0.33
0.31
0.07
0.37
0.28
0.19
0.19
Steps for the Hardy-Cross Method
9b. Check each node to see if inflow equals outflow.
163
164. Steps for the Hardy-Cross Method
Select initial pipe sizes (assume a velocity of 3 ft/sec for normal flow
with no fire demand). With a known/assumed flow and an assumed
velocity, use the continuity equation (Q = VA) to calculate the cross-
sectional area of flow. (when conducting computer design, set
diameters to minimum allowable diameters for each type of
neighborhood according to local regulations)
Pipe
Number
1
2
3
4
5
6
7
8
9
10
11
Flow (ft3/sec) Velocity
(ft/sec)
3.0
3.0
3.0
3.0
3.0
3.0
3.0
3.0
3.0
3.0
3.0
Area (ft2) Diameter
(ft)
0.91
0.35
0.63
0.91
0.65
0.63
0.31
0.68
0.60
0.49
0.49
Diameter Actual D
(in) (in)
10.9 12
4.2 6
7.5 8
10.9 12
7.8 8
7.5 8
3.7 4
8.2 10
7.2 8
5.9 6
5.9 6
1.94
0.288
0.922
1.930
1.000
0.922
0.220
1.100
0.852
0.562
0.562
0.65
0.10
0.31
0.64
0.33
0.31
0.07
0.37
0.28
0.19
0.19
Steps for the Hardy-Cross Method
9b. Check each node to see if inflow equals outflow.
164
165. Steps for the Hardy-Cross Method
12. Paying attention to sign (+/-),
compute the head loss in each element/pipe of the
system (such as by using Darcy- Weisbach or
Hazen-Williams).
Hazen β Williams
οΆ1.85
ο·
ο¨ 0.432CD 2.63
Darcy β Weisbach
L ο¦V 2 οΆ
D ο§ 2g ο·
165
166. Pressure comparison for 6-, 8-, 12-, and 16- inch pipes
(Walski, et al. 2001 figure 7.4)
Extended period simulation (EPS) runs showing low
pressure due to elevation or system capacity problem
(Walski, et al. 2001 figure 7.3)
166
167. Chemical Reactions in Water
Distribution Systems
Many of the water distribution models include
water quality subcomponents.
These are used to determine the age of the water in
the distribution system and the mixing of
water from different sources at the different
locations.
This information is used to calculate the resulting
concentrations of conservative and
nonconservative compounds in the water.
Knowledge of the fate of disinfectants in the
distribution systems is very important to
ensure safe drinking water: we need to ensure
that sufficient concentrations of the
disinfectants exist at all locations in the
system.
Head loss comparison for 6-, 8-, 12-, and 16- inch pipes
(Walski, et al. 2001 figure 7.5)
167
168. Disinfectant reactions occurring within a typical
distribution system pipe
N.O.M. = natural organic matter
(Walski, et al. 2001 figure 2.24)
Conceptual illustration of concentration vs. time for zero,
first, and second-order decay reactions
(Walski, et al. 2001 figure 2.23)
168
