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 φ - index and w- index. * * Explain the method for the determination of φ - 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
10
15
30
60
90
120
0.8
1.2
1.4
1.7
2.1
2.4
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.
1
2
3
Area of Thiessen polygon (ha)
450
750
300
Rain gauge reading 6 to 7 AM
0.8
1.2
2.0
7 to 8 AM
1.4
3.6
3.2
8 to 9 AM
6.2
5.8
5.6
9 to 10 AM
4.4
4.6
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
35-40 cm
30-35
25-30
20-25
15-20
10-15
below 10 cm
Area (ha)
40
80
170
310
480
670
250
Compute the φ –index from the following data: Total runoff
=
Estimated ground water contribution Area of basin
= =
6
3
6
3
77 x 10 m
2 x 10 m 250 km2
The rainfall distribution is as follows:
**
Hour
0-2
2-4
4-6
6-8
8-10
10-12
12-14
14-16
Rainfall(cm/hr)
2.5
5.0
5.0
3.5
2.0
2.0
1.5
1.5
The following rainfall distribution was measured during a 6-hour storm Time (hr) Rainfall
0
1 0.5
2 1.5
3 1.2
4 0.3
5 1.0
6 0.5
intensity (cm/hr) The runoff depth has been estimated at 2 cm. Calculate the φ- 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
Precipitation at z (mm)
10 station Average (mm)
1972
35
28
1973
37
29
1974 1975
39 35
31 27
1976
30
25
1977
25
21
1978
20
17
1979
24
21
1980
30
26
1981
31
31
1982
35
36
1983
38
39
1984
40
44
1985
28
32
1986
25
30
1987
21
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)
0-2
2-4
4-6
6-8
8-10
10-12
1.0
2.0
4.0
3.0
0.5
1.5
Rainfall intensity (cm/hr)
Runoff depth was 16 cm. Calculate the φ-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) 2
Area enclosed (km ) *
40
35
30
25
20
15
10
-
20
70
150
320
450
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
Sta. X
15 Sta. Avg
Year
Sta. X
15 Sta. Avg
1950
47
29
1965
36
34
1951
24
21
1966
35
28
1952
42
36
1967
28
23
1953
27
26
1968
29
33
1954
25
23
1969
32
33
1955
35
30
1970
39
35
1956
29
26
1971
25
26
1957
36
26
1972
30
29
1958
37
26
1973
23
28
1959
35
28
1974
37
34
1960
58
40
1975
34
33
1961 1962
41 34
26 24
1976 1977
30 28
35 26
1963
20
22
1978
27
25
1964
26
25
1979
34
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)
0
1
2
3
4
5
6
Accumulated Rainfall (cm)
0
0.05
1.65
3.55
5.65
6.80
7.75
Calculate the φ 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 φ
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.
*
2
A reservoir had an average surface area of 20 km during June 1992 . In that month , the 3
3
mean rate of inflow is 10m /s , out flow is 15 m /s, monthly rainfall is 10 cm and 3
change in storage is 16 Mm Assuming the seepage losses to be 1.8 cm, estimate the evaporation in that month . **
3
The average annual discharge at the outlet of a catchment is 0.5 m /s . The catchment is 2
situated in desert area (no vegetation) and the size is 800 Mm .The average annual precipitation is 200 mm / year (a) Compute the average annual evaporation from the catchment in mm / year. 2
In the catchment area an irrigation project covering 10 Mm is developed. After some 3
years the average discharge at the outlet of the catchment appears to be 0.175 m /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 Penman’s method for a given weather station locality
Yangon N16° 07'
Month
March
Temperature C°
36.2(max) and 19.8 (min)
Relative humidity (%)
64 (max)and 48 (min)
Mean wind speed
200 Km /day
Mean daily sunshine hours
9.6
Mean daily possible sunshine hours
12
6
*
Reflection coefficient (albedo)
0.05
Psychrometer constant
0.49
At a reservoir in a certain location, the following climatic month of June by Penman’s method, assuming that the lake evaporation is the same as P.E.T.
*
Latitude
28° N
Elevation
230m above MSL
Mean monthly temperature
33.5° C
Mean relative
52° %
humidity
Mean observed sunshine hour
9 hr
wind speed at 2m height
10 Km /hr
For an area (latitude 12° N) the mean monthly temperature are given Month
June
July
Aug
Sept
Oct
Temp(C°)
31.5
31.0
30.0
29.0
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
June
July
Aug
Sept
Oct
12°
8.68
8.94
8.76
8.26
8.31
Chapter 4
*
The following data were collected during a stream gaging operation in a river. Compute the discharge. Distance from bank (m)
Velocity (m/s)
Depth at 0.2d
at 0.8 d
0.0
0.0
0.0
0.0
1.5
1.3
0.6
0.4
3.0
2.5
0.9
0.6
4.5
1.7
0.7
0.5
6.0
1.0
0.6
0.4
7.5
0.4
0.4
0.3
9.0
0.0
0
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. Area (ft2)
Depth (ft)
1.72
263
1.5
1070
2.50
674
1.8
2700
3.47
1200
2.1
4900
4.02
1570
2.8
6600
4.26
1790
3.2
7700
5.08
2150
3.9
9450
5.61
2380
4.6
10700
5.98
2910
4.9
13100
6.70
3280
5.2
15100
6.83
3420
5.4
16100
7.80
3960
5.7
19000
8.75
4820
6.0
24100
9.21
5000
6.1
25000
9.90
5250
6.5
27300
14.50
8200
9.0
Stage (ft)
**
Discharge (ft2/s)
The following data were collected at a gauging station on a stream. Compute the
discharge by (a) the mid-section method Distance from one
(b) the mean-section method.
0
3
6
9
12
15
18
21
24
27
Water depth (m)
0
1.5
3.2
5.0
9.0
5.5
4.0
1.6
1.4
0
Mean velocity (m/s)
0
0.12
0.24
0.25
0.26
0.24
0.23
0.16
0.14
0
bank (m)
*
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
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
bank (m)
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
X, cm
Y, cm
Year
X, cm
Y,cm
1
150
124
9
135
116
2
141
123
10
184
151
3 4
184 205
134 178
11 12
119 150
104 113
5
131
127
13
192
164
6
222
158
14
179
133
7
181
147
15
156
140
8
133
106
16
182
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 3
the minimum storage required to maintain a demand rate of 90m /s. Month Jan
Feb
Mar
Apr
May June July
Aug
Sept Oct
Nov
Dec
Flow
60
40
30
25
300
200
150
100
90
80
60
200
(m3/s) **
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
2650
3010
2240
2630
3200
1000
950
1200
4150
3500
(cumecs) 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
Runoff(Mm ) 1.3
2.0
2.7
19.0 22.0 2.5
3
***
8.5
12.0
12.0
Aug
Sept
Oct
Nov Dec
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
Runoff
500 350
March Apri
May
June
July
650
300
650
7500 6000 3500 2500 600
600
Aug
Sept
Oct
Nov
Dec 700
(hect-m) (a) If the monthly demands are as under, determine the required storage capacity. Assume that the reservoir is full on Jan 1. Month
Jan
Feb
March Apri
May
June
July
Aug
Sept
Oct
Nov
Dec
Demand
5
6
7
8
10
5
6
4
8
10
12
5
(Cumecs) (b) If there is a uniform demand of 6 m 3/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 φ - 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
18
21
24
27
30
Q(cumec)
3.08
4.94
8.64
9.88
7.41
4.94
3.70
2.47
1.23
0
0
Develop a unit hydrograph of 6 hour duration. *
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 2
catchment area of 500 km . Time(hr)
0
3
Flow(m /s) 20
6
12
18
24
30
36
42
48
54
60
66
72
120
270
220
170
120
90
70
55
45
35
25
20
3
Assuming a constant base flow of 20 m /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
2
4
6
8
10
12
14
16
18
20
22
4 hr
150
500
610
450
320
220
140
80
40
10
0
0
UH *
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
½
1
1½
2
2½
3
3½
4
4½
5
Q(cumec)
404
1979
2343
2506
1460
453
381
274
173
0
0
2
A catchment of 5 km has rainfall of 5.0, 7.5 and 5.0 cm in three consecutive days. The average φ - 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
13.2
12.8
11.6
16.9
17.2
14.0
14.2
17.8
18.8
11.7
13.3
13.6
13.9
16.4
14.7
8.4
12.5
11.2
20.7
19.7
19.7
18.9
17.4
15.8
14.9
18.3
17.7
18.6
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
1950
1951
1952
1953
1954
1955
1956
1957
(Lakh cumecs)
1.38
1.25
1.92
1.45
1.65
1.43
1.84
1.74
Year
1958
1959
1960
1961
1962
1963
1964
1965
(Lakh cumecs)
1.32
1.86
1.20
1.82
1.70
1.95
1.60
1.32
Year
1966
1967
1968
1969
1.41
1.78
1.80
1.50
Discharge
Discharge
Discharge (Lakh cumecs) **
The maximum annual floods for 23 years are given below, arranged in the descending order Year
1960
1952
1970
1954
1972
1971
1968
1964
discharge(m3/s)
720
710
705
665
570
490
450
440
Year
1966
1973
1956
1957
1951
1965
1951
1961
discharge(m /s)
425
410
405
400
395
390
385
375
Year
1955
1963
1958
1962
1969
1959
1967
360
345
340
330
320
310
300
Flood
Flood 3
Flood discharge(m3/s)
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 3
below the undisturbed ground water table. When pumped at a steady rate of 1.5 m /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
1 0.4
2 0.32
3 0.38
4 0.43
5 0.49
6 0.52
8 0.57
10 0.61
14
18
22
26
30
40
50
60
80
067
0.71
0.75
0.78
0.81
0.85
0.89
0.94
0.97
100
140
180
240
1.02
1.07
1.12
1.15
12 0.64
Chapter 9
*** 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
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) 3
Inflow (m /s) 3
Outflow (m /s) *
0
6
12
18
24
30
36
42
48
54
60
66
5
20
50
50
32
22
15
10
7
5
5
5
5
6
12
29
38
35
29
23
17
13
9
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
6 AM
155
Noon
140
6 PM
127
MN
103
6 AM
95
Noon
76
6 PM MN
65 54
6 AM
40
Noon
30
6 PM
24
MN
20
2
3
4
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) 2 Area enclosed (km )
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 2 (km ) 20 70 150 320 450 600
Net Area 2 (km ) 20 50 80 170 130 150
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 φ index (constant loss rate) for the storm. Solution
Time from start (hr) Incremental rainfall (cm) Total rainfall direct runoff Total infiltration
1 0.05
2 1.6
3 1.95
6 0.95
= 7.75 cm = 3.5 cm = 7.75 – 3.5 = 4.25 cm
st
1 trial Assume te
φ1
= 6 hr =
infiltration loss = te
4.25 6
= 0.708 cm/hr
4.2 5
0.84 cm/hr
st
φ1 is not effective the 1 hr
2
nd
trial Total infiltration = 4.25 – 0.05 = 4.20 cm Assume te = 5 hr
φ2
=
infiltration loss = te
infiltration index φ = 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 – 15 – 190 = 35 mm pan coeff. for class A pan = 0.7 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 km 2 during June 1992. In that month, the 3 3 mean rate of inflow is 10m /s, outflow is 15m /s, monthly rainfall is 10cm and change in storage 3 is 16 Mm . Assuming the seepage losses to be 1.8 cm, estimate the evaporation in that month. Solution 2 3 6 2 Surface area = 20 km = 20 x (10 ) = 20 x 10 m
Inflow for June
3 = 10 m /s 6 2 x 30 x 24 x 3600 x 100 = 129.6 cm 20 x 10 m
outflow for June
15 m /s = x 30 x 24 x 3600 x 100 = 194.4 cm 20 x 106m2
storage change
16 x 10 m = 16 Mm = 6 20 x 10
seepage losses
= 1.8 cm ( assume )
rainfall (monthly)
= 10 cm
3
6
3
3
= 0.8 m x 100 = 80 cm
Evaporation for June = ∆S + I + P – O – O s ( Water balance method ) = 80 + 129.6 + 10 – 194.4 – 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 0.6 0.4 0.9 0.6 0.7 0.5 0.6 0.4 0.4 0.3 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 2
Discharge 3
mean velo. 0
(m )
(m /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
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 (ΣXY) – (ΣX) (ΣY) N (ΣX2) – (ΣX)2
=
16 (368896) – 2644 x 2180 2 16 (450264) – (2644)
= 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
ΣY - bΣX N
a =
2180 – 0.648 x 2644 16
=
Y = 29.10 + 0.648 X
Correlation coefficient
←
= 29.10
Regression line N (ΣXY) - (ΣX) (ΣY)
r =
22
22
22
22
[NΣX [NΣX ––(ΣX) (ΣX) ]][NΣY [Σ NY ––(ΣY) (Y) ]]
r
16 (368896) - 2644 x 2180
=
2
2
[16 x 450264 – (2644) ] [16 x 304354 – (2180) ] r
= 0.875 > 0.6
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 φ index of 1 cm/ hr. Take base flow = 4 cumecs. Solution Effective rainfall depth R = 15 – 0.5 – 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 0
3 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=
-3
30 x 10 =
2π bk (Z2 – Z1) Log (r 1 / r 2) 2π x 15 x k (1.5 – 1.0) Log (40/10) -4
k = 8.8 x 10 m/s T
= bk = 15 x 8.8 x 10
-4
= 1.32 x 10
Let Zw be the drawdown at the well face
Q=
2π bk (Zw – 1.0) Log (40/0.15)
2π x 15 x 8.8 x 10 -4 (Zw – 1.0) 30 x 10 = 5.586 -3
-2
2
m /s
7
Zw = 3.02 m
Let R be the radius of influence ( ie drawdown Z = 0 ) Q=
2π bk Zw Log (R/r w) -4
2π x 15 x 8.8 x 10 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 3 below the undisturbed ground water table. When pumped at a steady rate of 1.5 m /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 2
Q
=
1.5 60
=
πk (h 12 - h 2 ) Log (r 1 /r 2)
2
2
πk [ (80 – 2) – (80-4) ] Log 15/5
= 2.84 x 10-5 m/s
k
Let hw = the depth of water in the well
Q
=
πk (h 22 – h 2 ) w
Loge (r 2 /r w) 1.5 60
=
π x 2.84 x 10
-5
2
(78 – h 2 )
Loge (15/0.2)
hw
= 69.82 m
Zw
= 80 – 69.82 = 10.18 m
w
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 3 4 5
Discharge (cfs) 0 5 10 20 25 30 80 130
Storage (sfd) 0 30 50 80 110 138 160 190
2s t
+ Q , cfs 0 125 210 340 465 582 720 890
(2s/t) - Q
(2s/t) + Q
MN
Inflow (cfs) 20
300
340
outflow Q , cfs 20
NOON
50
328
370
21
MN
100
426
478
26
NOON
120
544
646
51
MN
80
572
744
86
NOON
40
560
692
66
MN
20
534
620
43
NOON
10
564
29
Maximum outflow discharge = 86 cfs at pool level = 30.5 ft
5 NOON 10
