Sample Questions Worked Out Examples for CE-04026 - most

MINISTRY OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF
TECHNICAL AND VOCATIONAL EDUCATION
Sample Questions & Worked Out Examples
for
CE-04026
ENGINEERING HYDROLOGY
B.Tech. (Second Year)
Civil Engineering
MINISTRY OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF
TECHNICAL AND VOCATIONAL EDUCATION
CE-04026
ENGINEERING HYDROLOGY
Sample Questions
B.Tech. (Second Year)
Civil Engineering
1
Problems for CE 04026 Engineering Hydrology
Chapter 1
*
*
*
*
*
*
*
*
What is the hydrologic cycle?
Sketch the hydrologic cycle.
Explain the hydrologic cycle.
Describe the liquid transport phases of the hydrologic cycle.
Name the vapour-transport phases of the hydrologic cycle.
What is a catchment?
Give a brief description of different components of the hydrologic cycle.
How can you get the catchment area?
Chapter 2
*
*
*
*
What are the different forms of precipitation and rainfall?
Distinguish between the precipitation and rainfall.
What are the different methods for the measurement of precipitation.
Describe various types of recording type rain gauges. What are the advantages and
disadvantages of these gauges?
* Explain the method for estimation of missing rainfall data.
* When is the normal ratio method used to fill in missing precipitation records? What is a
double mass analysis?
* What are different methods for the estimation of average rainfall depth over an area?
* Describe the methods for plotting the mass rainfall curve and the hyetograph.
* What is a double-mass curve? What is its use?
* Differentiate between the infiltration capacity and infiltration index.
* * Differentiate between П† - index and w- index.
* * Explain the method for the determination of П† - index.
* What are the various losses which occur in the precipitation to become runoff.
* Draw the intensity duration curve from the following data.
Duration (mts)
Precipitation (cm)
5
0.8
10
1.2
15
1.4
30
1.7
60
2.1
90
2.4
120
2.8
** Storm precipitation occurred from 6 AM to 10 AM on a particular day over a basin of
1500 ha area. The precipitation was measured by 3 rain gauges suitably located on the
basin. The rain gauge readings and the areas of the Thiessen polygons are as follows:
2
Rain gauge no.
Area of Thiessen polygon (ha)
Rain gauge reading 6 to 7 AM
7 to 8 AM
8 to 9 AM
9 to 10 AM
1
450
0.8
1.4
6.2
4.4
2
750
1.2
3.6
5.8
4.6
3
300
2.0
3.2
5.6
2.8
Compute and draw the storm hyetograph and mass rainfall curve of the basin.
* * The computation of an isohyet map of a 2000 ha basin following a 6 hr storm gave the
following data. Determine the average precipitation for the basin.
Isohyet
Area (ha)
35-40 cm
40
30-35
80
25-30
170
20-25
310
15-20
480
* * Compute the П†вЂ“index from the following data:
Total runoff
Estimated ground water contribution
Area of basin
10-15
670
below 10 cm
250
77 x 106 m3
2 x 106 m3
250 km2
=
=
=
The rainfall distribution is as follows:
Hour
Rainfall(cm/hr)
**
0-2
2.5
2-4
5.0
4-6
5.0
6-8
3.5
8-10
2.0
10-12
2.0
12-14
1.5
14-16
1.5
The following rainfall distribution was measured during a 6-hour storm
Time (hr)
Rainfall
intensity (cm/hr)
0
1
0.5
2
1.5
3
1.2
4
0.3
5
1.0
6
0.5
The runoff depth has been estimated at 2 cm. Calculate the П†- index.
*
The precipitation gage for station X was inoperative during part of the month of
January. During that same period, the precipitation depths measured at three index
stations A, B, and C were 25, 28, and 27 mm respectively. Estimate the missing
precipitation data at X, given the following average annual precipitation at X, A, B and
C: 285, 250, 225 and 275 mm, respectively.
*
The annual precipitation at station z and the average annual precipitation at 10
neighbouring stations are as follows:
3
Year
1972
1973
1974
1975
1976
1977
1978
1979
1980
1981
1982
1983
1984
1985
1986
1987
Precipitation at z (mm)
35
37
39
35
30
25
20
24
30
31
35
38
40
28
25
21
10 station Average (mm)
28
29
31
27
25
21
17
21
26
31
36
39
44
32
30
23
Use double-mass analysis to correct for any data inconsistencies at station z.
**
The following rainfall distribution was measured during a 12-h storm:
Time (hr)
Rainfall
intensity (cm/hr)
0-2
2-4
4-6
6-8
8-10
10-12
1.0
2.0
4.0
3.0
0.5
1.5
Runoff depth was 16 cm. Calculate the П†-index for this storm.
*** Using the data of above problem, calculate the w-index, assuming the sum of the
interception loss and depth of surface storage is 1 cm.
*
The isohyets for annual rainfall over a catchment were drawn and the area enclosed by
the isohyets are given below. Determine the average depth of annual rainfall over the
catchment.
Isohyet (cm)
Area enclosed (km2)
*
40
-
35
20
30
70
25
150
20
320
15
450
10
600
Precipitation station X was inoperative for part of a month during which a storm
occurred. The respective storm totals at three surrounding stations A, B and C were 98,
80 and 110 mm. The normal annual precipitation amounts at station X, A, B and C are,
respectively, 800, 1008, 842 and 1080 mm.
4
*** The annual precipitation at station X and the average annual precipitation at 15
surrounding stations are shown in the following table.
(a) Determine the consistency of the record at station X.
(b) In what year is a change in regime indicated?
(c) Compute the mean annual precipitation for station X for the entire 30 year period
without adjustment.
(d) Repeat (c) for station X at its 1979 site with the data adjusted for the change in
regime.
Yean
1950
1951
1952
1953
1954
1955
1956
1957
1958
1959
1960
1961
1962
1963
1964
Sta. X
47
24
42
27
25
35
29
36
37
35
58
41
34
20
26
15 Sta. Avg
29
21
36
26
23
30
26
26
26
28
40
26
24
22
25
Year
1965
1966
1967
1968
1969
1970
1971
1972
1973
1974
1975
1976
1977
1978
1979
Sta. X
36
35
28
29
32
39
25
30
23
37
34
30
28
27
34
15 Sta. Avg
34
28
23
33
33
35
26
29
28
34
33
35
26
25
35
* * A rain gage recorded the following accumulated rainfall during the storm. Draw the
mass rainfall curve and the hyetograph.
Time (AM)
Accumulated
rainfall (mm)
**
8:00
8:05
8:10
8:15
8:20
8:25
8:30
0
1
2
6
13
18
19
An isolated storm in a catchment produced a runoff of 3.5 cm. The mass curve of the
average rainfall depth over the catchment was as below:
Time (hr)
Accumulated
Rainfall (cm)
0
1
2
3
4
5
6
0
0.05
1.65
3.55
5.65
6.80
7.75
Calculate the П† index ( constant loss rate ) for the storm.
5
**
A 10 hour storm occurred over 18 sq.km basin . The hourly values of rainfall were as
follows 1.8,4.2,10.4,5.8,16.4,7.7,15.2,9.6,5.4, 1.2 cm. If the surface runoff was
observed to be 705 6 ha-m, determine the infiltration index П†
Time from
start (hour)
Incremental
rainfall (cm)
1
2
3
4
5
6
7
8
9
10
1.8
4.2
10.4
5.8
16.4
7.7
15.2
9.6
5.4
1.2
Chapter 3
*
A class A pan set up adjacent to a a late. the depth of water in the pan at beginning of
a certain week was 195 mm . In that week , there was a rainfall of 45 mm and 15 mm of
water removed from the pan to keep the water level within the specified depth range. If
the depth of the depth of the water in the pan at the end of the week was 190 mm,
estimate the lake evaporation in that week.
*
A reservoir had an average surface area of 20 km2 during June 1992 . In that month , the
mean rate of inflow is 10m3/s , out flow is 15 m3/s, monthly rainfall is 10 cm and
change in storage is 16 Mm3 Assuming the seepage losses to be 1.8 cm, estimate the
evaporation in that month .
* * The average annual discharge at the outlet of a catchment is 0.5 m3/s . The catchment is
situated in desert area (no vegetation) and the size is 800 Mm2 .The average annual
precipitation is 200 mm / year (a) Compute the average annual evaporation from the
catchment in mm / year.
In the catchment area an irrigation project covering 10 Mm2 is developed. After some
years the average discharge at the outlet of the catchment appears to be 0.175 m3/s.
(b) Compute the evapotranspiration from the irrigated area in mm/ year, assuming no
change in the evaporation from the rest of the catchment.
*
Determine the monthly evaporation (mm) from a free water surface using the PenmanвЂ™s
method for a given weather station
locality
Month
Temperature CВ°
Relative humidity (%)
Mean wind speed
Mean daily sunshine hours
Mean daily possible sunshine hours
Yangon N16В° 07'
March
36.2(max) and 19.8 (min)
64 (max)and 48 (min)
200 Km /day
9.6
12
6
Reflection coefficient (albedo)
Psychrometer constant
*
0.05
0.49
At a reservoir in a certain location, the following climatic month of June by PenmanвЂ™s
method, assuming that the lake evaporation is the same as P.E.T.
Latitude
Elevation
Mean monthly temperature
Mean relative humidity
Mean observed sunshine hour
wind speed at 2m height
*
28В°N
230m above MSL
33.5В° C
52В° %
9 hr
10 Km /hr
For an area (latitude 12В° N) the mean monthly temperature are given
Month
Temp(CВ°)
June
31.5
July
31.0
Aug
30.0
Sept
29.0
Oct
28.0
Calculate the seasonal consumptive use of water for the rice crop in the season (June to
October) by using the Blaney- Criddle method .
Monthly daytime hour percentage, P
N lat
12В°
June
8.68
July
8.94
Aug
8.76
Sept
8.26
Oct
8.31
Chapter 4
*
The following data were collected during a stream gaging operation in a river.
Compute the discharge.
Distance from bank
(m)
0.0
1.5
3.0
4.5
6.0
7.5
9.0
Velocity (m/s)
Depth
0.0
1.3
2.5
1.7
1.0
0.4
0.0
at 0.2d
0.0
0.6
0.9
0.7
0.6
0.4
0
at 0.8 d
0.0
0.4
0.6
0.5
0.4
0.3
0
7
***
Given below are data for a station rating curve Extend the relations and estimate the
flow at a stage of 14.5 ft by A D method and logarithmic method.
Stage (ft)
1.72
2.50
3.47
4.02
4.26
5.08
5.61
5.98
6.70
6.83
7.80
8.75
9.21
9.90
14.50
Area (ft2)
263
674
1200
1570
1790
2150
2380
2910
3280
3420
3960
4820
5000
5250
8200
Discharge (ft2/s)
1070
2700
4900
6600
7700
9450
10700
13100
15100
16100
19000
24100
25000
27300
Depth (ft)
1.5
1.8
2.1
2.8
3.2
3.9
4.6
4.9
5.2
5.4
5.7
6.0
6.1
6.5
9.0
**
The following data were collected at a gauging station on a stream. Compute the
discharge by (a) the mid-section method (b) the mean-section method.
Distance from one
0
bank (m)
Water depth (m)
0
Mean velocity (m/s) 0
3
6
9
12
15
18
21
24
27
1.5
0.12
3.2
0.24
5.0
0.25
9.0
0.26
5.5
0.24
4.0
0.23
1.6
0.16
1.4
0.14
0
0
*
Calculate the discharge of river from the following measurements made with a
flow meter.
Distance from one 0.0
15
30
45
60
75
90
105
bank (m)
depth of water (m)
0.0
0.8
1.2
1.5
1.8
1.5
0.9
0.0
average velocity (m/s) 0.0
0.15
0.24
0.30
0.36
0.33
0.24
0.0
Chapter 5
* * The data given below are the annual rainfall, X and annual runoff, Y for a certain river
catchment for 16 years. It has been decided to develop a linear relation between these
two variables so as to estimate runoff for those years where rainfall data only are
available.
8
Year
1
2
3
4
5
6
7
8
X, cm
150
141
184
205
131
222
181
133
Y, cm
124
123
134
178
127
158
147
106
Year
9
10
11
12
13
14
15
16
X, cm
135
184
119
150
192
179
156
182
Y,cm
116
151
104
113
164
133
140
162
Find the equation of regression line. Is the linear relationship appropriate for the
above data?
**
The following table gives the mean monthly flows in a river during a year. Calculate
the minimum storage required to maintain a demand rate of 90m3/s.
Month Jan
80
Flow
3
(m /s)
**
Feb
60
Mar
40
Apr
30
May June July
25
60
200
Aug Sept Oct
300 200 150
Nov Dec
100 90
The average annual discharge of a river for 11 years is as follows:
Year
1960
1961
1962
1963
1964
1965
1966
1967
1968
1969
1970
Discharge 1750
(cumecs)
2650
3010
2240
2630
3200
1000
950
1200
4150
3500
Determine the storage capacity required to meet a demand of 2000 cumecs throughout
the year.
* * * The runoff from a catchment area during successive months in a year is given below.
Determine the maximum capacity of the reservoir required if the entire volume of
water is to be drawn off at a uniform rate, without any loss of water over the spillway.
Month
Jan Feb March Apri May June July Aug Sept Oct Nov Dec
3
Runoff(Mm ) 1.3 2.0 2.7
8.5 12.0 12.0 19.0 22.0 2.5 2.2 1.9 1.7
* * * The average monthly runoff that flowed down a river during a critical year is given
below
9
Month
Jan Feb March Apri May June July Aug Sept Oct Nov
Runoff
500 350 650
600 300 650 7500 6000 3500 2500 600
(hect-m)
Dec
700
(a) If the monthly demands are as under, determine the required storage capacity. Assume
that the reservoir is full on Jan 1.
Month
Jan
Demand 5
(Cumecs)
Feb
6
March Apri May June July
7
5
8
10
5
Aug
6
Sept
4
Oct
8
Nov
10
Dec
12
(b) If there is a uniform demand of 6 m3/s, determine the required storage.
Chapter 6
* What is a hydrograph? What are its different segments?
* Explain various methods for the separation of base flow a hydrograph. Why the
separation of flow is required?
* Explain the procedure for the derivation of a unit hydrograph from an isolated storm
hydrograph.
* What is S-hydrograph? How would you derive a S-hydrograph? Discuss the procedure of
derivation of the unit hydrograph from a S-hydrograph.
* How would you obtain a storm hydrograph from a unit hydrograph?
* The ordinates of 3 hour unit hydrograph of a basin at 6 hour interval are below
0,3,5,9,11,7,5,4,2,1,0 Cumecs. Derive the storm hydrograph due to a 3 hour storm with a
total rainfall of 15 cm. Assume an initial loss of 0.5 cm and П† - index of 1 cm/hr. Take
base flow = 4 cumecs.
* A 3 hour duration unit hydrograph has the following ordinates:
Time(hour) 0
3
6
9
12
15
Q(cumec)
0
3.08 4.94 8.64 9.88 7.41
Develop a unit hydrograph of 6 hour duration.
18
4.94
21
3.70
24
2.47
27
1.23
30
0
* Find out the ordinates of storm hydrograph resulting from a 9 hour storm with rainfall of
2.0, 5.75 and 2.75 cm during subsequent 3 hour intervals. The ordinates of 3 hour U.H at
3 hour intervals are as follows:
0,100,355,510,380,300,200,225,165,120,85,55,30,22,10,0 (cumecs)
Assume an initial loss of 0.5 cm, an infiltration index of 0.25 cm/hr and a base flow of
10 cumecs.
10
*
Given below are observed flows from a storm of 6 hour duration on a stream with a
catchment area of 500 km2.
Time(hr)
0
3
Flow(m /s) 20
6
12 18 24 30
120 270 220 170 120
36
90
42
70
48
55
54
45
60
35
66
25
72
20
Assuming a constant base flow of 20 m3/s, derive the ordinates of a 6 hour unit
hydrograph.
*
Given below is the 4 hour UH for a basin. Compute the S-curve ordinate and find the
6 hour UH.
Hour 0
4 hr 0
UH
2
150
4
500
6
610
8
450
10
320
12
220
14
140
16
80
18
40
20
10
22
0
* Calculate the streamflow hydrograph for a storm of 6 inches excess rainfall, with 2
inches in the first half-hour, 3 inches in the second half-hour and 1 inch I the third
half-hour. Use the half-hour unit hydrograph and assume the base is constant at 500 cfs
throughout the flood.
Check that the depth of direct runoff is equal to the total excess precipitation (watershed
area = 7.03 sq.mile). The ordinates of half hour unit hydrograph are given below.
Time(hour) 0
Q(cumec)
0
ВЅ
404
1
1ВЅ
2
2ВЅ
3
1979 2343 2506 1460 453
3ВЅ
381
4
274
4ВЅ
173
5
0
* * * A catchment of 5 km2 has rainfall of 5.0, 7.5 and 5.0 cm in three consecutive days.
The average П† - index is 2.5 cm/day. The surface runoff extends over 7 days for each
rainfall of 1 day duration. Distribution graph percentage for each day are
5,15,35,25,10,6,4. Determine the ordinates of the storm hydrograph. Neglect base
flow.
Chapter 7
*
The annual rainfall for 10 years are as follows: 40, 35, 55, 65, 70, 25, 45, 30, 50 and 42
cm. Determine the rainfall which has a recurrence interval of 12 years.
*
The maximum values of 24 hr rainfall at a place from 1960 to 1980 are as follows:
11
12.7
11.7
19.7
13.2
13.3
19.7
12.8
13.6
18.9
11.6
13.9
17.4
16.9
16.4
15.8
17.2
14.7
14.9
14.0
8.4
18.3
14.2
12.5
17.7
17.8
11.2
18.6
18.8
20.7
19.2
Estimate the maximum rainfall having a recurrence interval of 10 years and 50 years.
**
The maximum annual observed floods for 20 years from 1950 to 1969 for a catchment
are given below. Determine the maximum flood with a recurrence interval of 30 years
by the following methods:
(a) Probability plotting, using a log-log paper
(b) Gumble's method
Year
Discharge
(Lakh cumecs)
Year
Discharge
(Lakh cumecs)
Year
Discharge
(Lakh cumecs)
**
1950
1951
1952
1953
1954
1955
1956
1957
1.38
1958
1.25
1959
1.92
1960
1.45
1961
1.65
1962
1.43
1963
1.84
1964
1.74
1965
1.32
1966
1.86
1967
1.20
1968
1.82
1969
1.70
1.95
1.60
1.32
1.41
1.78
1.80
1.50
The maximum annual floods for 23 years are given below, arranged in the descending
order
1960
Year
Flood
discharge(m3/s) 720
1952
1970
1954
1972
1971
1968
1964
710
705
665
570
490
450
440
1966
Year
Flood
discharge(m3/s) 425
1973
1956
1957
1951
1965
1951
1961
410
405
400
395
390
385
375
1955
Year
Flood
discharge(m3/s) 360
1963
1958
1962
1969
1959
1967
345
340
330
320
310
300
Find the magnitude of 100-Year flood, using Gumbel's method.
12
*
A well of diameter 30 cm fully penetrates a confined aquifer of thickness 15 m. When
pumped at a steady rate of 30 lps, the draw downs observed in wells at radial distances
of 10 m and 40 m, are 1.5 and 1.0 m respectively. Compute the radius of influence, the
permeability, the transmissibility and the draw down at the well.
*
A 0.4 m diameter well fully penetrates an unconfined aquifer whose bottom is 80 m
below the undisturbed ground water table. When pumped at a steady rate of 1.5 m3/min,
the draw downs observed in two observation wells at radial distances of 5 m and 15 m
are, respectively, 4 m and 2 m. Determine the drawn down in the well.
*
A well penetrates in the centre of an unconfined aquifer bounded externally by a circle
of radius 600m along which the height of water table is 8m. If at a distance of 10m from
the centre of well, the height of the water table is 7.5m when steady conditions are
established, determine the discharge of the well. Take k = 10-4 m/s.
* * A well fully penetrating a confined aquifer was pumped at a constant rate of 0.03
cusecs. During the pumping period, the draw down S in an observation well measured
at different instants of time are given below. If the distance of the observation well from
the pump well was 50m, determine the formation constants S and T by (a) Theis'
method (b) Cooper Jacob's method.
Time(t)minutes
Drawdown
0
0.0
14
067
100
1.02
1
0.4
18
0.71
140
1.07
2
0.32
22
0.75
180
1.12
3
0.38
26
0.78
240
1.15
4
0.43
30
0.81
5
0.49
40
0.85
6
0.52
50
0.89
8
0.57
60
0.94
10
0.61
80
0.97
12
0.64
Chapter 9
*** Tabulated below are the elevation storage and elevation discharge data for a small
reservoir.
Elevation(ft)
Storage(sfd)
Discharge(cfs)
0
0
0
5
30
5
10
50
10
15
80
20
20
110
25
25
138
30
30
160
80
35
190
130
From the inflow hydrograph shown below, Compute the maximum outflow discharge
and pool level to be expected. Assume initial outflow = 20 cfs
13
*
The following inflow and outflow hydrographs were observed in a river reach. Estimate
the values of k and x applicable to this reach for use in the Muskingum equation.
Time (hr)
Inflow (m3/s)
Outflow (m3/s)
*
0
5
5
6
20
6
12
50
12
18
50
29
24
32
38
30
22
35
36
15
29
42
10
23
48
7
17
54
5
13
60
5
9
66
5
7
The inflow hydrograph for a stream channel reach is tabulated below. Compute the
outflow hydrograph using Muskingum method of routing with k = 36 hr and x = 0.25.
Assume initial outflow as 30 cumecs.
Date
hr
Inflow (cumec)
1
6 AM
30
Noon
50
6 PM
86
MN
124
2
6 AM
155
Noon
140
6 PM
127
MN
103
3
6 AM
95
Noon
76
6 PM
65
MN
54
4
6 AM
40
Noon
30
6 PM
24
MN
20
MINISTRY OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF
TECHNICAL AND VOCATIONAL EDUCATION
CE-04026
ENGINEERING HYDROLOGY
Worked Out Examples
B.Tech. (Second Year)
Civil Engineering
1
CE04026 Engineering Hydrology
1.(2-7)* The isohyets for annual rainfall over a catchment were drawn and the area enclosed by
the isohyets are given below. Determine the average depth of annual rainfall over the catchment.
Isohyet (cm)
Area enclosed (km2)
40
-
35
20
30
70
25
150
20
320
15
450
10
600
Solution
Isohyet
(cm)
40
35
30
25
20
15
10
Area enclosed
(km2)
20
70
150
320
450
600
Net Area
(km2)
20
50
80
170
130
150
в€ґ Average depth of annual rainfall over the catchment
Average
depth (cm)
37.5
32.5
27.5
22.5
17.5
12.5
=
ppt. volume
750
1625
2200
3825
2275
1875
12550
12550
600
= 20.92 cm
2.(2-7)* Precipitation station X was inoperative for part of a month during which a storm
occured. The respective storm totals at three surrounding stations A, B and C were 98, 80 and 110
mm. The normal annual precipitation amounts at station X, A, B and C are, respectively, 800,
1008, 842 and 1080 mm. Estimate the storm precipitation for station X.
Solution
PA = 98 mm,
NA = 1008 mm,
PB = 80 mm,
NB = 842 mm,
PX =
NX
M
PA
PB
PC
+
+
NA
NB
NC
PX =
800
3
98
80
110
+
+
1008
842 1080
PX = 78.42 mm
PC = 110 mm,
NC = 1080 mm,
PX = ?
NX = 800 mm
2
3.(2-7)** An isolated storm in a catchment produced a runoff of 3.5 cm. The mass curve of the
average rainfall depth over the catchment was as below.
Time (hr)
Accumulated
Rainfall (cm)
0
1
2
3
4
5
6
0
0.05
1.65
3.55
5.65
6.80
7.75
4
2.1
5
1.15
Calculate the П† index (constant loss rate) for the storm.
Solution
Time from start (hr)
Incremental rainfall (cm)
1
0.05
Total rainfall
direct runoff
в€ґ Total infiltration
=
7.75 cm
=
3.5 cm
= 7.75 вЂ“ 3.5 = 4.25 cm
1st trial
Assume te
= 6 hr
П†1
=
infiltration loss =
te
2
1.6
3
1.95
4.25
6
= 0.708 cm/hr
4.2
5
0.84 cm/hr
6
0.95
П†1 is not effective the 1st hr
2nd trial
Total infiltration = 4.25 вЂ“ 0.05 = 4.20 cm
Assume te
= 5 hr
П†2
=
infiltration loss =
te
в€ґ infiltration index П† = 0.84 cm/hr
4.(3-7)* A class A pan set up adjacent to a lake. The depth of water in the pan at beginning of a
certain week was 195 mm. In that week, there was a rainfall of 45mm and 15mm of water
removed from the pan to keep the water level within the specified depth range. If the depth of the
water in the pan at the end of the week was 190 mm, estimate the lake evaporation in that week.
Solution
pan evaporation
= 195 + 45 вЂ“ 15 вЂ“ 190 = 35 mm
pan coeff. for class A pan = 0.7
в€ґ lake evaporation = pan evaporation x pan coefficient
= 35 x 0.7
= 25.5 mm
3
5.(3-7)* A reservoir had an average surface area of 20 km2 during June 1992. In that month, the
mean rate of inflow is 10m3/s, outflow is 15m3/s, monthly rainfall is 10cm and change in storage
is 16 Mm3. Assuming the seepage losses to be 1.8 cm, estimate the evaporation in that month.
Solution
Surface area = 20 km2 = 20 x (103) = 20 x 106 m2
Inflow for June
3
= 10 m /s
x 30 x 24 x 3600 x 100 = 129.6 cm
20 x 106m2
outflow for June
=
storage change
= 16 Mm3 =
seepage losses
= 1.8 cm ( assume )
rainfall (monthly)
= 10 cm
15 m3/s
x 30 x 24 x 3600 x 100 = 194.4 cm
20 x 106m2
16 x 106 m3
20 x 106
= 0.8 m x 100 = 80 cm
Evaporation for June = в€†S + I + P вЂ“ O вЂ“ Os ( Water balance method )
= 80 + 129.6 + 10 вЂ“ 194.4 вЂ“ 1.8
= 23.4 cm
6(4-7)* The following data were collected during a stream gaging operation in a river. Compute
the discharge.
Distance from bank
(m)
0.0
1.5
3.0
4.5
6.0
7.5
9.0
Depth
Velocity (m/s)
at 0.2 d
at 0.8 d
0.0
0.0
0.4
0.6
0.6
0.9
0.5
0.7
0.4
0.6
0.3
0.4
0.0
0.0
0.0
1.3
2.5
1.7
1.0
0.4
0.0
Solution
Distance
Width
Depth
Meter
depth
(m)
(m)
(m)
(m)
0
1.5
0
1.5
3.0
1.5
Velocity (m/s)
at point
0
1.3
0
0.3
1.0
0
0.6
0.4
2.5
0.5
2.0
0.9
0.6
Area
Discharge
mean
velo.
0
(m2)
(m3/s)
0
0
0.5
1.95
0.975
0.75
3.75
2.8125
4
4.5
1.5
1.7
0.34
1.36
0.7
0.5
0.60
2.55
1.53
6.0
1.5
1.0
0.2
0.8
0.6
0.4
0.50
1.5
0.75
7.5
1.5
0.4
0.1
0.3
0.4
0.3
0.35
0.6
0.21
9.0
0.75
0
0
0
0
0
6.2775
7.(5-7)** The data given below are the annual rainfall, X and annual runoff, Y for a certain river
catchment for 16 years. It has been decided to develop a linear relation between these two
variables so as to estimate runoff for those years where rainfall data only are available.
Year
1
2
3
4
X,cm
150
141
184
205
Y,cm
124
123
134
178
Year
5
6
7
8
X,cm
131
222
181
133
Y,cm
127
158
147
106
Year
9
10
11
12
X,cm
135
184
119
150
Y,cm
116
151
104
113
Year
13
14
15
16
X,cm
192
179
156
182
Y,cm
164
133
140
162
Find the equation of regression line. Is the linear relationship appropriate the above data?
Solution
Year
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
ОЈ
X, cm
150
141
184
205
131
222
181
133
135
184
119
150
192
179
156
182
2644
Y, cm
124
123
134
178
127
158
147
106
116
151
104
113
164
133
140
162
2180
X2
22500
19881
33856
42025
17161
49284
32761
17689
18225
33856
14161
22500
36864
32041
24336
33124
450264
Y = a + bX
b=
N (ОЈXY) вЂ“ (ОЈX) (ОЈY)
N (ОЈX2) вЂ“ (ОЈX)2
=
16 (368896) вЂ“ 2644 x 2180
16 (450264) вЂ“ (2644)2
= 0.648
Y2
15376
15129
17956
31864
16129
24964
21609
11236
13456
22801
10816
12769
26896
17689
19600
26244
304354
XY
18600
17343
24656
36490
16637
35076
26607
14098
15660
27784
12376
16950
31488
23807
21840
29484
368896
5
ОЈY - bОЈX
N
a =
=
в€ґ Y = 29.10 + 0.648 X
Correlation coefficient
r =
2180 вЂ“ 0.648 x 2644
16
= 29.10
в†ђ Regression line
N (ОЈXY) - (ОЈX) (ОЈY)
[NОЈX
[NОЈX22вЂ“вЂ“(ОЈX)
(ОЈX)22]][NОЈY
[ОЈNY22вЂ“вЂ“(ОЈY)
(Y)22]]
16 (368896) - 2644 x 2180
r
=
r
= 0.875 > 0.6
[16 x 450264 вЂ“ (2644)2] [16 x 304354 вЂ“ (2180)2 ]
в€ґ good correlation
Linear relationship is appropriate for above data.
8.(6-8)* The ordinates of a 3hr unit hydrograph of a basin at 6 hr interval are given below. 0, 3 , 5
, 9 , 11 , 7 , 5 , 4 , 2 , 1 , 0 cumecs. Derive the storm hydrograph due to a 3 hr storm with a total
rainfall of 15 cm. Assume an initial loss of 0.5 cm and a П† index of 1 cm/ hr. Take base flow = 4
cumecs.
Solution
Effective rainfall depth R = 15 вЂ“ 0.5 вЂ“ 1 x 3 = 11.5 cm
Time (hours)
0
6
12
18
24
30
36
42
48
54
60
Unit hydrograph
ordinates
(cumecs)
0
3
5
9
11
7
5
4
2
1
0
Direct runoff
ordinates
(cumecs)
0
34.5
57.5
103.5
115.0
80.5
57.5
46.0
23.0
11.5
0
Base
flow
4
4
4
4
4
4
4
4
4
4
4
Ordinate of storm
hydrograph
(cumecs)
4
38.5
61.5
107.5
119.0
84.5
61.5
50.0
27.0
15.5
4.0
6
9.(6-8)* A 3 hr duration unit hydrograph has the following ordinates:
Time (hr)
Q (cumec)
0
3
0 3.08
6
4.94
9
8.64
12
9.88
15
7.41
18
4.94
21
3.70
24
2.47
27
1.23
30
0
Develop a unit hydrograph of 6 hour duration
Solution
Time
(hr)
Ordinate of
3 hr unit
hydrograph
0
3.08
4.94
8.64
9.88
7.41
4.94
3.70
2.47
1.23
0
0
3
6
9
12
15
18
21
24
27
30
33
3 hr U.H
lagged 3hr
Combined
hydrograph
6 hr UH
0
3.08
4.94
8.64
9.88
7.41
4.94
3.70
2.47
1.23
0
0
3.08
8.02
13.58
18.52
17.29
12.35
8.64
6.17
3.70
1.23
0
0
1.54
4.01
6.79
9.26
8.65
6.18
4.32
3.09
1.85
0.62
0
10.(8-8)* A well of diameter 30 cm fully penetrates a confined aquifer of thickness 15m. When
pumped at a steady rate of 30 lps, the drawdowns observed in wells at radial distances of 10m and
40m, are 1.5 and 1.0 m respectively. Compute the radius of influence, the permeability, the
transmissibility and the drawdown at the well.
Solution
Q=
30 x 10-3 =
2ПЂ bk (Z2 вЂ“ Z1)
Log (r1 / r2)
2ПЂ x 15 x k (1.5 вЂ“ 1.0)
Log (40/10)
k = 8.8 x 10-4 m/s
T
= bk = 15 x 8.8 x 10-4 = 1.32 x 10-2 m2 /s
Let Zw be the drawdown at the well face
Q=
30 x 10-3 =
2ПЂ bk (Zw вЂ“ 1.0)
Log (40/0.15)
2ПЂ x 15 x 8.8 x 10-4 (Zw вЂ“ 1.0)
5.586
7
Zw = 3.02 m
Let R be the radius of influence ( ie drawdown Z = 0 )
Q=
2ПЂ bk Zw
Log (R/rw)
2ПЂ x 15 x 8.8 x 10-4 x 3.02
30 x 10 =
Log (R/0.15)
-3
R = 634.0 m
11.(8-8)* A 0.4 m diameter well fully penetrates an unconfined aquifer whose bottom is 80 m
below the undisturbed ground water table. When pumped at a steady rate of 1.5 m3/min, the
drawdowns observed in two observation wells at radial distance of 5m and 15m are,
respectively, 4m and 2m. Determine the drawdown in the well.
Solution
Q
=
1.5
60
=
2
ПЂk (h 12 - h 2 )
Log (r1 /r2)
ПЂk [ (80 вЂ“ 2)2 вЂ“ (80-4)2]
Log 15/5
= 2.84 x 10-5 m/s
k
Let hw = the depth of water in the well
Q
=
ПЂk (h 22 вЂ“ h 2w )
Loge (r2 /rw)
1.5
60
=
ПЂ x 2.84 x 10-5 (782 вЂ“ h 2w )
Loge (15/0.2)
hw
= 69.82 m
Zw
= 80 вЂ“ 69.82 = 10.18 m
8
12(9-8)*** Tabulated below are the elevation-storage and elevation-discharge data for a small
reservoir.
Elevation (ft)
0
5
10
15
20
25
30
35
Storage (sfd)
0
30
50
80
110
138
160
190
Discharge (cfs)
0
5
10
20
25
30
80
130
From the inflow hydrograph shown below, compute the maximum outflow discharge and pool
level to be expected. Assume initial outflow = 20 cfs.
Data
Hour
Inflow, cfs
1
MN
20
2
NOON
50
3
MN
100
NOON
120
4
MN
80
NOON
40
MN
20
Solution
t = 0.5 day
Elevation
0
5
10
15
20
25
30
35
Date
Hour
1
2
MN
NOON
MN
NOON
MN
NOON
MN
NOON
3
4
5
Discharge
(cfs)
0
5
10
20
25
30
80
130
Inflow
(cfs)
20
50
100
120
80
40
20
10
Storage
(sfd)
0
30
50
80
110
138
160
190
2s
t
(2s/t) - Q
(2s/t) + Q
300
328
426
544
572
560
534
340
370
478
646
744
692
620
564
в€ґ Maximum outflow discharge = 86 cfs
at pool level = 30.5 ft
+ Q , cfs
0
125
210
340
465
582
720
890
outflow
Q , cfs
20
21
26
51
86
66
43
29
5
NOON
10

Download Report