Engineering Hydrology

Engineering Hydrology
Faruque Abdullah
Lecturer
Dept. of Civil Engineering
Dhaka International University

Introduction
Hydrology: Hydrology is the science, which deals with the occurrence,
distribution and disposal of water on the planet earth; it is the science which
deals with various phases of the hydrologic cycle.
Hydrologic cycle: Hydrologic cycle is the water transfer cycle, which occurs
continuously in nature; three important phases of hydrologic cycle are: (a)
Evaporation and evapotranspiration (b) precipitation (c) runoff.

Scope of Hydrology:
 The maximum probable flood that may occur at a given site and frequency,
this is required for the safe design of drains and culverts, dams and
reservoirs, etc.
 The water yield from a basin-this is necessary for the design of dams,
municipal water supply, water power, river navigation, etc.
 The ground water development for which a knowledge of the hydrology of
the area, rainfall pattern, climate, etc. are required.
 The maximum intensity of storm and its frequency for the design of a
drainage project in the area.

Hydrologic Equation:
The hydrologic equation is simply the statement of the law of conservation of
matter and is given by,
I = O + ∆S
where,
I = Inflow
O = Outflow
∆S = Change in storage

Precipitation:
Precipitation is any form of moisture which falls to the earth. This includes rain,
snow, hail, sleet, etc.
Forms of precipitation:
Rain: The condensed water vapor of the atmosphere falling in drops from clouds.
Snow: Ice crystals resulting from sublimation (i.e. water vapor condensed to ice).
Hail: Small lumps of ice (>5 mm dia.) formed by alternate freezing and melting.
Dew: Moisture condensed from the atmosphere in small drops upon cool surfaces.
Sleet: Rain or melted snow that freezes into ice pellets before hitting the ground.

Rain Gauge:
Rainfall may be measured by a network of rain gauges which may either be of non-
recording or recording type.
Non-recording type:
 Usually used Symon’s rain gauge.
 Consists of funnel with a circular rim of 12.5 cm diameter and a glass bottle as
receiver.
 The cylindrical metal casing is fixed vertically to the masonry foundation with
the level rim 30 cm above the ground surface.
 The rain falling into the funnel is collected in the receiver and it can measure
1.25 cm of rain when full. The rainfall is measured in a special measuring glass
graduated in mm.

Recording Rain Gauge:
 It is also called self-recording, automatic or integrated rain gauge.
 This type of rain gauge has an automatic mechanical arrangement
consisting of a clockwork, a drum with a graph paper fixed around it and a
pencil point which draws the mass curve of rainfall.
 From this mass curve, depth of rainfall in a given time, the rate of intensity
of rainfall, can be determined.
 The gauge is installed on an area of 45 cm square concrete or masonry
platform.
 Three types: a) Tipping Bucket gauge, b) Weighing type rain gauge, c)
Float type rain gauge.

Fig. Tipping Bucket Gauge Fig. Weighing Type Bucket Gauge

a.a.r.:
The mean of yearly rainfall observed for a period of 35 consecutive years is
called the average annual rainfall (a.a.r.).
a.a.r. < 40 cm --- Arid climate
40 cm < a.a.r. < 75 cm --- Semi-arid climate
a.a.r. > 75 cm --- Humid climate
Isohyet:
A line joining the place having the same a.a.r. is called an isohyet.
Index of wetness:
The ratio of rainfall in a particular year to the a.a.r. is called the index of
wetness.

Estimates of missing data and adjustment of records:
 Station-year Method:
=> For two stations:
PA
a.a.r.of A
=
PB
a.a.r.of B
Here, PA = Certain year a.a.r. rainfall in station, A
PB = Certain year a.a.r. rainfall in station, B
=> For more than two stations:
PN =
1
N−1
[
PA
a.a.r.of A
x a.a.r. of N +
PB
a.a.r.of B
x a.a.r. of N + …….
+
P(N−1)
a.a.r.of (N−1)
x a.a.r. of (N-1)]

Problem-1: The rainfall of station A, B and C is 8.5, 6.7 and 9.0 cm
respectively. If the a.a.r. for the stations are 75, 84, 70 and 90 cm respectively.
Estimate the storm rainfall at station D.
Solution:
The average value of PD =
1
3
[
8.5
75
x 90 +
6.7
84
x 90 +
9.0
70
x 90] = 9.65 cm.

Problem-2: The rainfall of station A, B, C and D is 8.5, 6.7, 9.0 and 9.65 cm
respectively. If the a.a.r. for the stations A, B and C are 75, 84 and 70 cm
respectively. Estimate the storm rainfall at station D.
Solution:
The average value of PD =
1
3
[
8.5
75
x a.a.r of D +
6.7
84
x a.a.r of D +
9.0
70
x a.a.r of D ] = 9.65 cm.
=> a.a.r. of D = 90 cm

Mean Areal Depth of Precipitation:
 Arithmetic Average Method:
Pavg =
∑P
n
 Thiessen Polygon Method:
Pavg =
∑i=1
n
AiPi
∑A

 Isohyetal Method:
Pavg =
∑i=1
n
Ai−(i+1)Pi−(i+1)
∑Ai−(i+1)

Problem-3: Calculate the mean areal depth of precipitation from the
following figure by arithmetic mean and Thiessen polygon method.
A (46 cm) B (65 cm)
C (76 cm)D (80 cm)
E (70 cm)F (60 cm)
10 Km
10Km

A (46 cm) B (65 cm)
C (76 cm)D (80 cm)
E (70 cm)F (60 cm)
52 + 52 = 5 2
Area of inner square = (5 2 )2 = 50 km2
Area of square = 10 x 10 = 100 km2
Area of each corner triangle =
1
4
(100-50)
= 12.5 km2
Area of equilateral triangle = 0.5 x 10 x 10 sin 600= 25 3 km2
1
3
Area of equilateral triangle =
1
3
x 25 3 = 14.4 km2
Solution:

Station Area
A
km2
Precipitation
P
cm
A x P
km2
. cm
Pavg
cm
A 12.5+14.4 46 1238
Pavg =
∑i=1
n
AiPi
∑A
=
9517
143.2
= 66.3
B 12.5 65 813
C 12.5 76 950
D 12.5+14.4 80 2152
E 50 70 3500
F 14.4 60 864
n=6 143.2 ∑P = 397 ∑AP =
9517
By Arithmetic mean,
Pavg =
∑P
n
=
397
6
= 66.17 cm
By Thiessen Polygon Method:
A (46 cm) B (65 cm)
C (76 cm)D (80 cm)
E (70 cm)F (60 cm)

Problem-4: Determine the optimum number of rain-gauge considering the
error in the mean value of rainfall to 10% from the following basin.

Station Normal annual
rainfall, x
cm
Difference
(x-𝑥)
Difference2
(x − 𝑥)2
Satisfaction
parameters
𝑥, σ, Cv
A 88 -4.8 23 𝑥 =
∑ x
n
= 92.8
σ =
∑(x−𝑥)2
n−1
=
3767
5−1
= 30.7
Cv =
σ
𝑥
=
30.7
92.8
x 100
= 33.1 %
B 104 11.2 125.4
C 138 45.2 2040
D 78 -14.8 219
E 56 -36.8 1360
n = 5 ∑ x = 464 ∑(x − 𝑥)2
=
3767.4
N = (
𝐶 𝑣
𝑃
)2
= (
33.1
10
)2
= 11.09 ≈ 11 [P = error in the mean value of rainfall = 10%]
Additional rain gauge required = 11- 5 = 6.
Solution:

Hyetograph:
A hyetograph is a bar graph showing the
intensity of rainfall with respect to time.
Use:
 Determining the maximum intensities
of rainfall during a particular storm.
 Land drainage and design culverts.
Fig. Hyetograph

Problem-5: If there is 1000 items of storm in a record book and an engineer find
the rank number of the most dangerous storm as 50. Then find out,
1) What is the recurrence interval of the storm order 50 by California method?
2) Find the Frequency of occurrence/.
3) Will occur in the next 10 years?
4) May not occur in the next 10 years?
Solution:
1)Recurrence interval =
n
m
=
1000
50
= 20
2)Frequency, F =
1
T
x 100% = 5 %
3) P(N,0) = (1 − F)N = (1−0.05) 10
= 0.6 or 60 %
4) PEx = 1 - (1 − F)N = 1− (1−0.05) 10
= 0.4 or 40 %

Moving Averages Curve:
If the rainfall at a place over a number of years is plotted as a bar graph it
will not show any trends or cyclic patterns in the rainfall due to wide
variations in the consecutive years. In order to depict a general trend in the
rainfall pattern, the averages of three or five consecutive years are found out
progressively by moving the group averaged, one year at a time.

Water Losses:
 Interception loss- due to surface vegetation.
 Evaporation- (a) From water surface, (b) From soil surface
 Transpiration- From plant leaves.
 Evapotranspiration or consumptive uses- From irrigated or cropped land.
 Infiltration- into the soil at the ground surface.
 Watershed leakage- ground water movement from one basin to another or
into the sea.
Evaporation:
Losses of water from the free water surface and soil is known as
evaporation.
The factors affecting the rate of evaporation:
 Temperature
 Relative humidity
 Wind Velocity
 Surface area
 Barometer pressure and salinity of the water.

The Storage Equation:
P + I ± 𝑂𝑔 = E + O + ± S
Where,
P = Precipitation
I = Surface Inflow
𝑂𝑔 = Subsurface inflow or outflow
E = Evaporation
O = Surface outflow
S = Change in surface water storage.
Pan Coefficient:
Pan coefficient is the ratio between lake evaporation and pan evaporation
Pan coefficient =
Lake evaporation
Pan evaporation
Experimental values for pan coefficients ranges from 0.67 to 0.82 with
an average of 0.7

Why we need to use pan-coefficient:
 Climatological and physical factors affecting the evaporation amount is
different.
 The surface area and depth of pan is too small with respect to lake or
reservoir which evaporation amount we need to calculate.
 Pan contains little volume of water compared to the big water body.
 The rate of evaporation of pan is greater than big water body.
 Effect of solar radiation, wind velocity or temperature is different for pan
and big water body.
In this way, the data obtained by pan evaporation is not correct, which can
be corrected by multiplying a co-efficient known as pan coefficient to
measure the evaporation from the pan.

Problem-6: The following are the monthly pan evaporation (Jan-Dec) at
Natore in 2018 are 16.7, 14.3, 17.8, 25.0, 28.6, 21.4, 16.7, 16.7, 16.7, 21.4,
16.7, 16.7 respectively. In January 2.80 km2 and at the end of December 2.55
km2 area is used for irrigation purposes. Calculate the volume of water loss.
Solution:
Mean water spread area, Aave =
1
3
[ A1 + A2 + A1A2 [Cone formula]
=
1
3
[ 2.8 + 2.55 + 2.8 x 2.55 ] = 2.673 km2
Annual loss of water = 228.7 cm
Annual volume of water loss = (2.673 x 106) x
228.7
100
x 0.7 = 4.29 x 106 m3
= 4.29 Mm3.

Problem-7: The total observed runoff during a storm of 6 h duration with a
uniform intensity of 15 mm/h is 21.6 Mm3. If the total area of the basin is
300 km2, find the average infiltration and runoff coefficient.
Solution:
Infiltration loss, FP = Rainfall (P) – Runoff (R)
= 15 x 6 -
21.6 x 106
300 x 106 x 1000 = 90 – 72 mm = 18 mm
Average infiltration, fave =
FP
t
=
18
6
= 3 mm/h
Runoff coefficient =
Runoff
Rainfall
=
72
90
= 0.8

Problem-8: What is the evaporation if 4.75 liters of water is removed from
an evaporation pan of diameter 122 cm and the simultaneous rainfall
measurement is 8.8 mm.
Solution:
Evaporation loss =
4.75
1000
π
4
x (
122
100
)2
x 1000 mm
= 3.83 mm

Measures to Reduce Lake Evaporation:
 Storage reservoirs of more depth and less surface area.
 By growing tall trees like Causerina on the windward side of the
reservoirs to act as a wind breaker.
 By spraying certain chemical that will form a film on the surface of water.
 By allowing flow of water, temperature is reduced and evaporation is
reduced.

 By removing water loving weeds and plants like Phreatophytes.
 By providing mechanical covering like thin polythene sheets.
 By developing underground reservoir.
 If the reservoir is surrounded by huge trees and forest, the evaporation
loss will be less due to cooler environment.
Infiltration:
Water entering the soil at ground surface is called infiltration. It replenish
the soil moisture deficiency and the excess moves downward by the force of
gravity.
Infiltration Capacity:
The maximum rate at which the soil in any given condition is capable if
absorbing water is called its infiltration capacity.

Fig. Infiltration Curve (Horton)

Problem-9: The Horton’s infiltration equation for a basin is given by f = 6 + 16 e−2t
where, f is in mm/h and t is in hours. Determine the depth of infiltration for the first 45
minutes and the average infiltration rate for the first 75 minutes.
Solution:
Comparing Horton’s infiltration equation f = fc + (fo - fc) e−kt
with the given equation,
fc = 6, fo - fc = 16 or fo = 16 + 6 = 22 and k = 2.
For the 1st 45 minutes,
F = 0
45
60
f. dt = 0
0.75
(6 + 16 e−2t
)dt = [6t − 8e−2t
]0
0.75
= 10.715 mm
For the 1st 75 minutes,
F = 0
75
60
f. dt = 0
1.25
(6 + 16 e−2t
)dt = [6t − 8e−2t
]0
1.25
= 14.843 mm
favg =
F
t
=
14.843
1.25
= 11.874 mm/h.

Infiltration Indices:
Three types of index are:
1) Φ – index: It is defined as that rate of rainfall above which the
rainfall volume equals the runoff volume.
2) w – index: It is the average infiltration rate during the time rainfall
intensity exceeds the infiltration capacity rate.
3) favg - index: In this method an average infiltration loss is assumed
throughout the storm, for the period i > f.

Problem-10: The rates of rainfall for the successive 30 min period of a 3-hour storm are:
1.6, 3.6, 5.0, 2.8, 2.2, 1.0 cm/h. The corresponding surface runoff is 3.6 cm. Establish the
φ – index and w – index.
Solution:
We know,
∑(i – φ) t = Pnet
=> [(1.6 – φ) +(3.6 – φ) + (5.0 – φ) + (2.8 – φ) + ( 2.2 – φ) + (1 – φ)]
30
60
= 3.6
=> φ = 1.5 cm/h.
As 1 cm/h is less then φ – index then,
=> [(1.6 – φ) + (3.6 – φ) + (5.0 – φ) + (2.8 – φ) + ( 2.2 – φ)]
30
60
= 3.6
=> φ = 1.6 cm/h.
P = (1.6 + 3.6 + 5 + 2.8 +2.2) x
30
60
= 8.1 cm w – index =
P−Q
tR
=
8.1−3.6
3
= 1.5 cm/h.

Problem-11: A small watershed consists of 1.5 km2
of cultivated area (c = 0.2), 2.5 km2
of forest (c = 0.1) and 1.0 km2
of grass cover (c = 0.35). There is a fall of 20 m in a
watercourse of length 2 km. The IDF relation for the area is given by I =
80 T0.2
(t+12)0.5, I in
cm/h, T –yr., t- min. Estimate the peak rate of runoff for a 25 year frequency.
Solution:
tc ≈ 0.02 L0.8S−0.4 = 0.02 (2000)0.8 (
20
2000
)−0.4 = 55 min = t
I =
80 x 250.2
(55+12)0.5 = 18.6 cm/h
Q = 2.78 I (∑CA) = 2.78 x 18.6 x (1.5 x 0.2 + 2.5 x 0.1 + 1 x 0.35) Cumec.

Problem-12: A culvert is proposed across a stream draining an area of 185 hectares. The
catchment has a slope of 0.004 and the length of travel for water is 1150 m. Estimate the
25 year flood if the rainfall is given by,
I =
1000 T0.2
(t+20)0.7
Where, I in mm/h, T –yr., t- min. Assume a runoff coefficient of 0.35.
Solution:
tc ≈ 0.02 L0.8
S−0.4
= 0.02 (1150)0.8
(0.004)−0.4
= 51 min = t
I =
1000 x 250.2
(51+20)0.7 = 96.32 mm/h = 9.6 cm/h
Q = 2.78 I (∑CA) = 2.78 x 9.6 x (0.35 x 1.85) = 17.28 Cumec.

Hydrograph:
A hydrograph is a graph showing discharge (i.e. stream flow at the
concentration point) versus time past a specific point in a river, channel or
conduit carrying flow. The rate of flow is typically expressed as cumec or
cusec.
Use:
 Design of sewerage
 Estimation of Peak flood.
 Design of Dam
 Development of flood forecasting and warning system.
 To calculate the maximum flood for designing of spillway.

Unit Hydrograph:
The unit hydrograph is defined as the hydrograph of storm runoff resulting
from an isolated rainfall of some unit duration occurring uniformly over the
entire area of the catchment produces a unit volume of runoff.
Use:
 The UH theory can’t be applied catchment area greater than 5000 km2.
 The UH can’t be applied to very small catchments with area less than
2km2.
 This theory can’t be applied when the major portion of the storm is in the
form of snow.
 This theory is not very accurate. The accuracy obtained is ± 10%.

Problem-13: Convert the following 2-hr UH to 3-hr UH using S-curve method and
mark the peak flood.
Time 0 1 2 3 4 5 6 7 8 9 10
2-hr UH ordinate (Cusec) 0 75 250 300 275 200 100 75 50 25 0

Time
(h)
2-hr
UGO
(cusec)
S-Curve Ordinates S-Curve
Ordinates
(cusec)
(2) + (3)
Lagged
S-curve
(cusec)
S-curve
Difference
(cusec)
(4) - (5)
3-hr UGO
(6) x
2
3
1 2 3 4 5 6 7
0 0 0 0 0
1 75 75 75 50
2 250 0 250 250 166.7
3 300 75 375 0 375 250
4 275 250 0 525 75 450 300
5 200 300 75 575 250 325 216
6 100 275 250 0 625 375 250 166.7
7 75 200 300 75 650 525 125 83.3
8 50 100 275 250 0 675 575 100 66.7
9 25 75 200 300 75 675 625 50 33.3
10 0 50 100 275 250 0 675 650 25 16.7
11 25 75 200 300 75 675 675 0 0

Problem-14: Convert the following 4-hr UH to 2-hr UH using S-curve method mark the
peak flood.
Time 0 2 4 6 8 10 12 14 16 18 20 22 24
4-hr UH ordinate (Cusec) 0 25 100 160 190 170 110 70 30 20 6 1.5 0

Time
(h)
4-hr UGO
(cusec)
S-Curve Ordinates S-Curve
Ordinates
(cusec)
(2) + (3)
Lagged
S-curve
(cusec)
S-curve
Difference
(cusec)
(4) - (5)
2-hr UGO
(6) x
4
2
1 2 3 4 5 6 7
0 0 0 0 0
2 25 25 0 25 50
4 100 0 100 25 75 150
6 160 25 185 100 85 170
8 190 100 0 290 185 105 210
10 170 160 25 335 290 65 130
12 110 190 100 0 400 335 45 90
14 70 170 160 25 425 400 25 50
16 30 110 190 100 0 430 425 5 10
18 20 70 170 160 25 445 430 15 30
20 6 30 110 190 100 0 436 445 -9 -18
22 1.5 20 70 170 160 25 446.5 436 10.5 21
24 0 6 30 110 190 100 436 446.5 -10.5 -21

Problem-15: The design storm of a water shed has the depth of rainfall of 4.9 and 3.9 cm
for the consecutive 1 hr. periods. The 1-hr UG can be approximated by a triangle of base
6 h with a peak of 50 cumec occurring after 2-hr from the beginning. Compute the flood
hydrograph assuming an average loss rate of 9 mm/h and constant base flow of 10 cumec.
What is the area of water shed and its coefficient of runoff.
Solution:

Time
(h)
UGO DRO due to rainfall
excess, cumec
________________
4.9-0.9 3.9-0.9
=4 cm =3 cm
Total B.F. TRO
0 0 0 0 10 10
1 25 100 0 100 10 110
2 50 200 75 275 10 285
3 37.5 150 150 300 10 310
4 25 100 112.5 212.5 10 222.5
5 12.5 50 75 125 10 135
6 0 0 37.5 37.5 10 47.5
7 0 0 10 10
Peak Flood
Volume of water basin = Area of UG
=>A x
1
100
= 0.5 x (6 x 60 x 60) x
50
106
=> A = 54 km2
Runoff coefficient, C =
R
P
=
4.9−0.9 +(3.9−0.9)
4.9+3.9
= 0.795

Stream Gauging:
The most satisfactory determination of the runoff from a catchment is by measuring
the discharge of the stream draining it, which is termed as stream gauging. A gauging
station is the place or section on a stream where discharge measurements are made.
Some of the usual methods of stream gauging are given below:
 Venturiflumes or standing wave flumes (critical depth meter) for small channel.
 Weirs or anticuts. [Q = CLH
3
2]
 Slope-area method: Q = AV
V = C RS
V =
1
n
R
2
3s
1
2

 Contracted area method:
Q = CdA1 2g(∆H + ha)
 Sluiceway, spillway and power conduits.
 Salt-Concentration method.
Q =
C−C2
C2−C1
q
 Area-Velocity Method:
Q = AV
The area of cross-section of flow may be determined by sounding and plotting the
profile. The mean velocity may be determined by making velocity measurements.

 Floats
 Velocity rods
 Current meter
[d is the mean depth in the strip]
Surface
Sub-Surface
Measures surface velocity, Vs;
V ≈ 0.85 Vs
Measures main velocity, V
Pigmy
Propeller
Price
One point method:
V = V0.6d
Two- point method:
V =
V0.2d+V0.8d
2

Price Current Meter:
The relationship between the revolution per
second (N, rps) of the meter and the velocity
of flow past the meter (V, m/sec) has to be
first established or if the rating equation is
given by maker it has to be verified. The
process of calibration of the meter is called
‘Rating of the current meter’. The rating
equation is of the form,
V = aN + b
Fig. Price Current Meter

Fig. Current Meter Rating Curve

Problem-16: Calculate the velocity by one point method and two point method using
current meter parameter, a = 0.3 and b = 0.05 from the following data:
Solution:
Using two-point method:
V =
V0.2d+V0.8d
2
=
0.26+0.18
2
= 0.22 m/sec
Depth Revolutions Time (sec)
0.2d 35 50
0.6d 30 55
0.8d 25 60
Depth Revolutions Time (sec) N
rps
V = aN + b
0.2d 35 50 0.7 0.26
0.6d 30 55 0.55 0.22
0.8d 25 60 0.42 0.18
Using One-point method:
V = V0.6d = 0.22 m/sec

 Over 70% of the earth's surface
is covered in water.
 But of that water, just 1% is
readily available for human use,
and of that 1%, 99% of it is
stored beneath our feet as
groundwater.
 We all rely on groundwater in
some way, so it's important that
we understand this vital
resource.

What is Groundwater?
Groundwater is the water found underground in the cracks and
spaces in soil, sand and rock. It is stored in and moves slowly
through geologic formations of soil, sand and rocks called aquifers.

How much do we depend on groundwater?
Groundwater supplies drinking water for 90% of the total BD.
Population
Groundwater helps grow our food. 70%-90% of groundwater is
used for irrigation to grow crops.
Groundwater is an important component in many industrial
processes.
Groundwater is a source of recharge for lakes, rivers, and
wetlands.

Advantages of Ground Water(GW) over surface water
01. GW is clean, colorless, and almost free from turbidities.
02. GW is free from Pathogenic organism which causes disease.
03. GW is rich with several minerals which are beneficial to health.
04. For most cases GW needs no treatment.
05. GW have uniform temperature in all most all the year.
06. About 99% drinkable water is stored as ground water.
07. GW is available for irrigation in dry season when surface water
in not available.
08. Ground water storage is free from atomic attack.

Aquifer:
An aquifer is an underground layer of water bearing permeable rock, rock features
or unconsolidated materials ( gravel, sand or slit) from which groundwater can be
extracted using a water well.

Aquiclude:
The opposite of an aquifer. An aquiclude is a subsurface rock, soil or sediment unit
that does not yield useful quantities of water. It may be porous and capable of
containing water but the transmission rate is so poor that cannot be considered to be
a water source. Clay and shale are typical aquiclude.

Aquifuge:
An aquifuge is an impermeable formation neither containing nor transmitting water.
Example: Solid granite, basalt, etc.

Aquitard:
An aquitard is a geological unit that is permeable enough to transmit water in
significant quantities when viewed over large areas and long periods but its
permeability is not sufficient to justify production well being placed in it.

Artesian Aquifer:
An aquifer that is bounded above and below by impermeable rock or sediment
layers. The water in the aquifer is also under pressure that, when the aquifer is
tapped by a well, the water rises up the well bore to a level that is above the top of
the aquifer. The water may or may not flow onto the land surface.

Perched Aquifer:
A perched water table (or perched aquifer) is an aquifer that occurs above the
regional water table, in the vadose zone. This occurs when there is an impermeable
layer of rock or sediment (aquiclude) or relatively impermeable layer (aquitard)
above the main water table/aquifer but below the surface of the land.

Darcy’s Law:
Darcy’s Law defines groundwater flow:
Q = kA
dh
d𝑥
Where,
Q = Discharge
k = Hydraulic conductivity
A = The cross-sectional area of flow
dh
dx
= The gradient of pressure

Problem-22: Find the amount of ground water flow in m3
/day, through fine sand when
cross-section of the soil sample 85 m x 25 m and head loss 2 m for a travelling path
length of 800 m. Use the hydraulic conductivity k = 5 m/day.
Solution:
Q = kA
dh
dt
= 5 x (85 x 25) x
2
800
= 26.6 m3
/day

Steady state flow to a unconfined aquifer:
Q =
πk(H2−hw
2
)
ln(
R
rw
)

Problem-23: A well with a radius of 0.5 m, completely penetrates an unconfined aquifer
of thickness 50 m and k = 30 m/day. The well is pumped so that the water level in the
well remains at 40 m above the bottom. Assuming that pumping has essentially no effect
on water table at R = 500 m. What is the steady-state discharge?
Solution:
Q =
πk(H2−hw
2
)
ln(
R
rw
)
=
π x 30 x (502−402)
ln(
500
0.5
)
= 12279.387 m3/day
= 0.1421225 m3/s = 142.12 L/s.

Steady state flow to a confined aquifer:
Q =
2πT(h1−h2)
ln(
r1
r2
)
Q =
2πT(s2−s1)
ln(
r1
r2
)
Q =
2πTSw
ln(
R
rw
)

Problem-24: A well with a diameter of 0.5 m penetrates fully into a confined aquifer of
thickness 20 m and hydraulic conductivity 8.2 x 10−4
m/s. What is the maximum yield
expected from this well if the drawdown in the well is not to exceed 3 m. The radius of
influence may be taken as 260 m.
Solution:
Q =
2πTSw
ln(
R
rw
)
=
2π x 0.0164 x 3
ln(
260
0.25
)
= 0.04943m3/s
= 49.43 L/s.
Transmissivity, T = kb
= 8.2 x 10−4 x 20
= 0.0164 m2/s

Design Flood:
A design flood is the flood discharge adopted for the design of a structure after
careful consideration of economic and hydrologic factors.
Standard Project Flood (SPF):
This is the estimate of the flood likely to occur from the most severe
combination of the meteorological and hydrological conditions, which are
reasonably characteristic of the drainage basin being considered, but excluding
extremely rare combination.

Problem-25: A coffer dam is designed for a 25 year flood and constructed. If it takes 5
years to complete the construction of main dam, what is the risk that the coffer dam may
before the end of the construction period? What return period in the design of coffer dam
would have reduced the risk to 10%?
Solution:
The risk of failure is given by,
R = 1 – (1−
1
T
)
N
Here, T = 25 years and N = 5 years. Therefore,
R = 1 – (1−
1
25
)
5
= 0.1846 = 18.46 %
If the risk is to be reduced to 10%, we have
0.1 = 1 – (1−
1
T
)
5
, which gives
T = 47.96 years say 50 years.

Problem-26: The analysis of a 30 year flood data at a point on a river yielded x =
1200 m3/s and Sx = 650 m3/s. For what discharge would you design the structure
at this point to provide 95% assurance that the structure would not fail in the next
50 years? For N= 30 years, yn = 0.53622 and σn = 1.11238.
Solution:
Assurance = 95%
Risk = 100-95 = 5% = 0.05
R = 1 – (1−
1
T
)
N
=> 0.05 = 1 – (1−
1
T
)
50
=> T = 975.3 years.

We know,
yT = -ln ln(
975.3
974.3
) = 6.88223
KT =
yT−yn
σn
=
6.88223−0.53622
1.11238
= 5.705
Design flood,
xT = x + KTSx
= 1200 + 5.705 x 650
= 4908.25 m3/s
⸫ The structure has to be designed for a discharge of 4910 m3
/s.

Problem-27: From the analysis of available data on annual flood peaks of a small
stream for a period of 35 years, the 50 year and 100 year flood have been
estimated to be 660 m3/s and 740 m3/s using Gumbel’s method. Estimate the 200
year flood for the stream. For n = 35 years, yn = 0.54034 and σn = 1.12847.
Solution:
Using Gumbel’s method,
Y50 = -ln ln (
50
49
) = 3.90194
K50 =
3.90194−0.54034
1.12847
= 2.9789
Y100 = -ln ln (
100
99
) = 4.60015
K100 =
4.60015−0.54034
1.12847
= 3.59762

x50 = x + K50Sx
660 = x + 2.9789 Sx
Solving equation (1) & (2), we get
x = 274.83 m3/s and Sx = 129.3 m3/s.
Now,
Y200 = -ln ln (
200
199
) = 5.29581
K50 =
5.29581−0.54034
1.12847
= 4.2141
The 200 year flood for the stream would be 820 m3
/s.
x100 = x + K100Sx
740 = x + 3.59762 Sx1 2
x200 = x + K200Sx
= 274.83 + 4.2141 x 129.3 = 819.71 m3/s.

Well Diameter:
The size of the well diameter should be properly chosen since it significantly
affects the cost of well construction.
The diameter must be chosen to give the desired percentage of open are in the
screen (15 to 18%), so that entrance velocities near the screen do not exceed 3 to 6
cm/sec. So as to reduce the well losses and hence the drawdown.
Depuit’s Equation:
Q ∝
1
log10
𝑅
𝑟 𝑤
For, R = 300 m, a 60 cm well yields only 25% more than a 15 cm well and 12%
more than a 30 cm well. Which shows that Drilling a large diameter well will not
necessarily mean proportionally large yields.

Anticipated
well yield
(lpm)
Nominal size
of pump bowl
(cm)
Size of well casing
Minimum
(cm)
Optimum
(cm)
400 10 12.5 15
400-600 12.5 15 20
600-1400 15 20 25
1400-2200 20 25 30
2200-3000 25 30 35
3000-4500 30 35 40
4500-6000 35 40 50
6000-10000 40 50 60

Well Depth:
 The depth of a well and the number of aquifers have to penetrate is usually
determined from the lithological log of the area and confirmed from electrical
resistivity and drilling logs.
 An experienced driller can decide the depth at which drilling can be stopped
after being advised by the hydrologist who analyses the samples collected during
the drilling.
 The well is usually drilled up to bottom of the aquifer so that the full aquifer
thickness is available, permitting greater well yield.

Design of screen well:
 Screen length: In homogenous artesian aquifer about 70-80% or ¾ of the aquifer
thickness is screened. The screen should best positioned at equal distance
between the top and bottom of the aquifer.
 In case the non-homogenous artesian aquifer, it is best to screen the most
permeable strata.
 Theory and experience have shown that screening the bottom one-third of the
aquifer provides the optimum design.
 The principle of design in a non-homogenous water table aquifer are the same as
is the case of non-homogenous artesian aquifer.

 Design of slot size:
 The size of slots depends upon the gradation and size of the formation
material.
 In case of naturally developed wells the slot size is taken as 40 to 70% of the
size of formation materials.
 If the slot size selected on this basis becomes smaller than 0.75mm, then it calls
for an artificial gravel pack.
 Artificial gravel pack is required when the aquifer material is homogeneous
with Uniformity co-efficient less than 3.00 and effective grain size less than
0.25 mm.

 Screen Diameter:
After the length of the screen (depending upon the aquifer thickness) and the
slot size ( based on the size and gradation of the aquifer materials) have been
selected , the screen diameter is determined so that the entrance velocities near
the Screen will not exceed 3 to 6 cm/sec to prevent incrustation and corrosion
and to minimize friction losses.

 Open wells-Advantages:
 Storage capacity of water is available in the well itself.
 Do nor require sophisticated equipment and skilled personnel for
construction.
 Can be easily operated by installing a centrifugal pump at different settings
for low and high water levels.
 Can be revitalized by deepening by blasting or by putting a few vertical bores
at the bottom or horizontal or inclined bores on the sides to intercept the
water bearing strata.

 Open wells-Disadvantages:
 Large space is required for the well and for excavated material lying on the
surface like a big mound.
 Construction is slow and laborious.
 Subjected to high fluctuations of water table during different seasons.
 Susceptibility to dry up in years of drought.
 High cost of construction as the depth increases in hard rock areas.
 Deep seated aquifer cannot be economically trapped.
 Uncertainty of tapping water of good quality.
 Susceptibility for contamination or pollution unless sealed from surface water
ingress.

 Tube wells (bore well)-Advantages:
 Do not require much space.
 Can be constructed quickly.
 Fairly sustained yield of water can be obtained even in years of scantly
rainfall.
 Economical when deep-seated aquifer are encountered.
 Flowing artesian wells can sometimes be struck.
 Generally good quality of water is trapped.

 Tube wells (bore well)-Disadvantages:
 Required costly and complicated drilling equipment and machinery.
 Required skilled workers and great care to drill and complete the tube wells.
 Installation of costly turbine or submersible pumps is required.
 Possibility of missing the fracture, fissures and joints in hard rock areas
resulting in many dry holes.

Thank you
For
Taking the Stress

Engineering Hydrology

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