NATURAL HYDROGRAPH
Base flow separation technique, Qb:
a. Empirical formula
b. Master depletion curve
c. Simple straight lines
Topic 4
EXPLAIN
HYDROGRAPH
ANALYSIS
CONSTRUCT
CALCULATE
CONCEPT OF UNIT HYDROGRAPH
UNIT HYDROGRAPH
Volume Of Direct Runoff From Direct
Runoff Hydrograph.
DETERMINE
The Change Of Unit Hydrograph
Duration Using:
a. S-curve
b. Superposition Concept.
HYDROGRAPH
• A hydrograph is a graph showing the rate
of flow (discharge) versus time past a
specific point in a river, or other channel or
conduit carrying flow. The rate of flow is
typically expressed in cubic meters or
cubic feet per second (cms or cfs).
DEFINITION OF NATURAL HYDROGRAPH
• A natural hydrograph is one recorded at a
stream gauging site and is a finger-print of
the upstream drainage area’s response to
rainfall
CONCEPT OF UNIT HYDROGRAPH
• A Hydrograph is a graph showing changes in the
discharge of river over a period of time.
• It can also refer to a graph showing the volume
of water reaching particular outfall or location in
a sewerage network, graph are commonly used
in the design of sewerage,more specillcally, the
design of surface water sewegare systems and
combination system.
A hydrograph consists of:
a) A rising limb or
concentration curve
b) A peak or crest
segment
c) Falling limb or
recession curve
• The properties or characteristics of a unit hydrograph are
a) the volume under a unit hydrograph is equal
to 1 unit (1cm or 1in) rainfall excess
b) If the duration of two rainfall excess events is equal,
without regard to their respective rainfall intensities,
they must result in the same hydrograph time base.
c) The unit hydrograph results in a linear system
whereby the direct runoff for a storm of a specified
duration is directly proportional to the rainfall excess
amount or volume
d) Rainfall distribution for all equal duration storms is
identical in space and time.
Calculate direct runoff from stream flow
& base flow data
• Direct runoff = stream flow – base flow
Schedule below shows observation data from stream flow with base flow
depth in catchment's area 250 km2. from that data , get direct runoff
magnitude.
example
Time (hour)
Stream flow, Q (m3/s)
Base Flow, (m3/s)
0
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
1.37
1.25
1.12
5.00
12.00
15.60
17.15
14.40
8.80
6.80
5.50
4.10
2.75
2.00
1.20
0.65
0.15
0.32
0.62
1.25
1.28
1.35
1.42
1.65
1.82
1.88
2.05
2.55
1.91
1.53
1.15
0.28
Calculate Direct Runoff Depth Base On Volume
DRO And Cathment Area.
• DRO Depth,rd =
DIMANA;
Vd
A
Qn
Δt
Vd
A
= RUNOFF VOLUME =  Qn Δt
= CATCHMENT WIDE
= HYDROGRAPH ORDINATE
= HYDROGRAPH TIME INTERVAL
Calculate runoff volume and excess rain depth
to a 7 acres catchment's area. Schedule below
show a direct runoff data resultant from one
precipitation
EXAMPLE
Time
(minute)
0
2
4
6
8
10
Flow, Q (cfs)
0
0
9
21
17
13
Convert the duration of unit hydrograph from
short to long (superposition)
Convert The Duration Of Unit Hydrograph From Short
To Long (Superposition Method)- graphic
Time 4hr-UH
(Hour) (m3/s)
0
0
4
3
8
33
12
96
16
63
20
35
24
25
28
12
32
6
36
3
4j-UH (m3/s)
The following are the ordinates of a 4-hour unit hydrograph. Derive the ordinates
of a 12 - hour unit hydrograph
250
Combined
4hr-UH
200
Lagged
150
4hr-UH
4hr-UH
100
12hr-UH
50
0
0
4
8
12
16
20
24
Time (Hour)
28
32
36
40
44
Convert The Duration Of Unit Hydrograph From Short
To Long (Superposition Method)- table
Time
4j-UH
(Hour)
(m3/s)
0
0
4
3
0
8
33
3
12
96
16
Lagged
Lagged
(+) 4j-UH
12j-UH
0
0
3
1
0
36
12
33
3
132
44
63
96
33
192
64
20
35
63
96
194
64.67
24
25
35
63
123
41
28
12
25
35
72
24
32
6
12
25
43
14.33
36
3
6
12
21
7
3
6
9
3
3
3
1
40
44
Example :
• The following are the ordinates of a 3-hour unit
hydrograph. Derive the ordinates of a 6-hour unit
hydrograph and plot the graph.
Solution
Time
(Hour)
0
4j-UH
(m3/s)
0
3
1.5
6
(+) 4j-UH
12j-UH
0
0
0
2
0.75
4.5
1.5
6
3.00
9
8.6
4.5
13
6.55
12
12.0
8.6
21
10.30
15
9.4
12.0
21
10.70
18
4.6
9.4
14
7.00
21
2.3
4.6
7
3.45
24
0.8
2.3
3
1.55
0.8
1
0.40
27
30
Lagged
25
20
15
3hr-UH
lagged 3hr-UH
Combined 3hr-UH
6hr-UH
10
5
0
0
3
6
9
12
15
18
21
24
27
Question 1:
• Determine the 4hr-UH from the 2hr-UH given in the table
using the superposition principle method
Changing a Short Duration Unit
Hydrograph to Long Duration using
S-Curved method
Example
• The ordinates of a 4-hour unit hydrograph for a particular
basin are given below. Determine the ordinates of the
S-curve hydrograph and there from the ordinates of the
6-hour unit hydrograph
Solution
Time
4hr-UH
(Hour)
0
0
2
25
4
100
6
160
8
190
10
170
12
110
14
70
16
30
18
20
20
6
22
1.5
24
0
S- Curve
Addition
S- Curve
Ordinates
Lagged
S- Curve
S- Curve
difference
6hr - UH
Time
(Hour)
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
4hr-UH
0
25
100
160
190
170
110
70
30
20
6
1.5
0
S- Curve S- Curve Lagged S- Curve
6hr - UH
Addition Ordinates S- Curve difference
0
25
100
185
290
355
400
425
430
445
436
446.5
436
0
25
100
185
290
355
400
425
430
445
436
446.5
436
446.5
436
0
25
100
185
290
355
400
425
430
445
436
446.5
0
25
100
185
265
255
215
135
75
45
11
16.5
-9
10.5
-10.5
0.00
16.67
66.67
123.33
176.67
170.00
143.33
90.00
50.00
30.00
7.33
11.00
-6.00
7.00
-7.00
Question 1
• Determine the 3hr-UH from the 1hr-UH given in the table
using the S-curve method.
Time
(Hour)
3hr-UH
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
0
66
134
266
534
466
400
334
266
200
134
66
0
0
0
S- Curve
Addition
S- Curve
Ordinates
0
66
134
266
600
600
666
934
866
866
1068
932
866
1068
932
Lagged
S- Curve
S- Curve
difference
1hr - UH
0
66
134
266
600
600
666
934
866
866
1068
932
866
1068
0
66
68
132
334
0
66
268
-68
0
202
-136
-66
202
-136
0
198
204
396
1002
0
198
804
-204
0
606
-408
-198
606
-408
Changing a Long Duration Unit
0
Hydrograph66 to Short Duration using
134
S-Curved
method
266
600
600
666
934
866
866
1068
932
Question
• The following are the ordinates of a 4hr –
UH. Derive the ordinates of a 3hr-UH
Derive the S-curve for 4hr-UH
Hydrograph Unit from Stream
Discharge Data
Example
• Table below show runoff from hydrograph data, for a
50km2 catchment . Calculate the unit hydrograph
(UH)10mm. Assume that the basic flow is 5.0m3/s
Solution
Mendapatkan Hidrograf Jumlah
• Penggunaan UH10 bagi kawasan tadahan
yang sama dengan contoh 1.
• Diberikan ribut yang lain dengan tempoh
yang sama bagi kawasan tadahan yang
sama iaitu 50km2. Anggapkan aliran dasar
7 m3/s
Latihan
• Cerapan kadaralir bagi satu kejadian hujan berkesan 12 jam bagi
kawasan seluas 1150km2 adalah seperti yang ditunjukkan.
Dapatkan hidrograf unit (UH)10mm untuk kawasan ini.
• Diberi UH 4jam dan dapatkan UH -12jam
dengan menggubakan kaeadah tindihan
REFERENCES
• http://www.ce.utexas.edu/prof/mckinney/c
e374k/Overheads/12-RunoffProcesses.pdf