1. Chapter Five Hydrograph
A hydrograph is a graph showing stream flow at the concentration point with time.
Steps of development:
At the beginning, there is only base flow (i.e., the ground water contribution to the stream)
Gradually ,after the storm beginning, the initial losses like interception and infiltration are met and then the surface flow begins.
AB (Rising Limb) , The hydrograph gradually rises ,and reaches its peak value after a time tp
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3. P(Peak point ):point placed between inflection points (B, and C)
BC (Crest Segment) : Is the increase in the drain because of the gradual increase in the storage due to the channels above basin
P(Peak point ):point placed between inflection points (B, and C)
Inter Flow of the rain up to this point and after this there is gradual dragging of catchment storage.
By this time the ground water table has been built up by the more infiltrating water .
CD (falling Limb): it declines and there is a change of slope at the inflection point (c).
this represents for us the process of withdrawing water from the storage , which was stored in basin during the first phase of the hydrograph.
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4. tB :time of ground water contribution of stream flow (Base Flow ) .
If a second storm occurs now, again the hydrograph starts rising
till it reaches the new peak and then falls and the ground water withdrawal begins.
Hydrograph represent the three cases of runoff :
Surface Runoff.
Inter Flow .
Base Flow .
The hydrograph does not depend on rain- storm characteristics but is entirely dependent on the characteristics of basin .
Basin shape
Volume of basin
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5. Vegetation in the basin . Geological soil of basin .
Slop of basin .
Vegetation in the basin .
Geological soil of basin .
If there is another basins and catchments in the region.
Base Flow Separation :
Hydrograph surface Runoff get from total hydrograph by separated the fast flow from the slow runoff(base flow).
if we consider that Inter flow part of the surface runoff, it is located within the fast flow, and for this it is subtracted from the base runoff hydrograph we get the surface runoff hydrograph.
There are three methods used to separate base runoff :
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6. 1- The straight-line method :
It is separated runoff base by connect the beginning of surface runoff in a straight line on the falling side, which represents the end of direct runoff
There is experimental equation to determine the time period N (day) of the inflection Pi point to point B:
N = 0.83 A0.2
A = Area of drainage (km2)
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7. Second Method :
Start of surface runoff be extended until intersection with the coordinates of the peak point (point C) and This point to be linked with point B in a straight line. C
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8. Third method :
In this way, the yield curve of ground flow is extended back until it intersection with the line coming down from the inflection Pi point (line EF) and two points A and F are connect them and be drawn is roughly.
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9. Excess Rain hydrograph ( ERH) : Produces subtracted the losses from Total runoff hydrograph( TRH).
Both hydrographic ERH and DRH (Direct Rain hydrograph) represent the same total amount but the different in units, where the units of the (ERH) (cm / hr) .
when paint the (ERH) against time, the area under curve and you multiply by area basin the output represents the total volume of direct runoff and that It is at the same time represent the area of (DRH).
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10. From the data in the table draw the hydrograph .
Example 1/
Rain that valued 3.8,2.8 cm occurred during successive term sustainability of 4 hours and the amount of area on the 27 km2 and has produced the following hydrograph to flow in the discharge basin points, Determine the excess rain and the value of the index Ф?
Solution
From the data in the table draw the hydrograph .
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12. N = 0.83 (27) 0.2 = 1.6 day = 38 hr.
من خلال الرسم، طريقة الخط المستقيم تعطينا قيمة ثابتة للجريان القاعدي مقدارها 5 م3/ ثا
، إذن وقت الـ DRH الكلي من t = 0 و لغاية t = 48
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13. 5-5=0, 13-5=8
Volume of DRH = 6*60*60[0.5*8+0.5(8+21)+0.5(21+16)+0.5(16+11)+0.5(11+7)+
0.5(7+4)+0.5(4+2)+0.5(2)]= *106 m3
Depth of DRH = Runoff vol./ Area = *106 / 27*106 = 5.52 cm. المطرالصافي))Pnet
Total Rainfall = = 6.6 cm.
Time of Duration = 8 hr.
Ф index = (6.6 – 5.52) / 8 = cm/hr.
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14. H.W.
Rain that valued 7 cm occurred during successive term sustainability of 5 hours and the amount of area on the 29 km2 and has produced the following hydrograph to flow in the discharge basin points, Determine the excess rain and the value of the index Ф?
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15. Steps of unit hydrograph.
is defined as the hydrograph of runoff resulting from an isolated rainfall of some unit duration occurring uniformly over the entire area of the catchment, produces a (unit volume for 1 cm) of runoff.
Steps of unit hydrograph.
Separate the base flow from the total runoff (by the well-known base flow separation procedures).
From the ordinates of the total runoff hydrograph deduct the corresponding ordinates of base flow, to obtain the ordinates of direct runoff.
Divide the volume of direct runoff by the area of the drainage basin to obtain the net rain depth over the basin.
Divide each of the ordinates of direct runoff by the net rain depth to obtain the ordinates of the unit hydrograph.
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16. The steps for the derivation of unit hydrograph can be formulated as follows :
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17. Example: Determine peak hydrograph 3- UG standard hours, if you learned that the peak hydrograph flood caused due to 3 hour rain is 270 m3/s, the rain depth rate equal to 5.9 cm, suppose that losses rate equal to 0.3 cm/hour, the base flow value constant and equal to 20 m3/s?
Solution
Pnet = 5.9 – 3*0.3 = 5 cm
Peak for DRH = =250 m3/s
Peak for 3-UH = 250/5= 50 m3/s /cm
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18. APPLICATION OF UNIT HYDROGRAPH
The application of unit hydrograph consists of two aspects:
(i) From a unit hydrograph of a known duration to obtain a unit hydrograph of the desired duration,
(ii) From the unit hydrograph ,to obtain the flood hydrograph corresponding to a single storm or multiple storms. For design purposes, which with the help of unit hydrograph, gives a design flood hydrograph.
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19. Example / The successive three-hourly ordinates of a 6-hr UG for a particular basin are 0,15, 36, 30, 17.5, 8.5, 3, 0 m3, respectively. The flood peak observed due to a 6-hr storm was150 m3. Assuming a constant base flow of 6 m3 and an average storm loss of 6 mm/hr, determine the depth of storm rainfall and the stream flow at successive 3 hr interval.
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20. Example
The design storm of a water shed has the depths 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 hr 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/hr and constant base flow of 10 cumec. What its coefficient of runoff?
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21. Solution
(i) The flood hydrograph due to the two consecutive hourly storms is computed in Table:
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22. Example :The following information is coordinates 6 - hour record for the collector hydrograph, calculate the direct runoff hydrograph due to excess rain 3.5 cm, which occur during the 6-hour vertical coordinates?
Solution :
Time (hr.) 3 6 9 12 15 18 24 30 36 42 48 54 60 66
coordinate UH (m3/s) 25 50 85
125
160
185
110 16 8
Time (hr.) 3 6 9 12 15 18 24 30 36 42 48 54 60 66
coordinate UH (m3/s) 25 50 85
125
160
185
110 16 8
coordinate DRH (m3/s)
87.5
175
297.5
437.5
560
647.5
385
210
126 56 28
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23. Example /rain storms for each 6-hour for them to increase their value in the rain 3 and 2 cm alternately occurred one after the other, and that storm worth 2 cm rain occurred after storm 3 cm. Record hydrograph the collector and the 6-hour is given in the previous example. Compute the DRH output.
Solution:
time hr
coordinate UH (m3/s)
coordinate DRH – 3 cm (m3/s)
coordinate DRH – 2 cm (m3/s)
coordinate DRH – 5 cm (m3/s) 6 50
150 12
125
375
100
475 18
185
555
250
805 24
160
480
370
850 30
110
330
320
550 36 60
180
220
400 42
108
120
228 48 25 75 72
147 54 16 98 8 32 56
(66)
(2.7)
(8.1)
(16)
(24.1)
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25. Example: / Rain storm rate table below the collector in three successive periods and 6-hour values were, respectively, 3.5, 7.5 and 5.5 cm. Loss of the storm rain rate (index Ф) falling on the collector was guessed by 0.25 cm / hour. Use a 6-hour time for the vertical coordinates of the unit hydrograph in the table, direct runoff hydrograph. If the assumption that the runoff base flow value is 15 m 3 / s at the beginning and increasingly 2m 3 / sec every 12 hours until the end of the direct runoff hydrograph. Evaluated flood hydrograph output?
Time (hr.) 6 12 18 24 30 36 42 48 54 60 66
coordinate UH (m3/s) 75
130
200
180
110 25 16 8
2.7
Solution
: loess = 0.25* 6= 1.5 cm
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26. End coordinate DRH (3+4+5)
tim hr
coordinate UH
col . 2 * 2
col. 2 *6
first 6 hr
col. 2 *4
second 12 hr
End coordinate DRH (3+4+5)
base flow
flood hydrograph
(6+7) 1 2 3 4 5 6 7 8 15 50
100
115 12
125
250
300
550
565 18
185
370
750
200
1320 17
1337 24
160
320
1110
500
1930
1947 30
110
220
960
740
1920
1937 36 60
120
660
640
1420 19
1439 42 72
360
440
872
891 48 25
216
240
506
525 54 16 32
150
144
326 21
347 96
212
233 66
(2.7)
(5.4)
(48)
(64)
(117)
138
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27. Unit Hydrograph Derivation :
Is the process of finding a standard hydrograph coordinates By dividing the coordinates of the DRH on the excess rain or( Pnet )
The( Pnet) result from the summation of DRH multiply the time and divided by area of basin .
Example:
The following information is coordinates hydrograph rain storm basin of area of 432 km2 flow base value is 10 m3/s at the beginning and increasingly 0.5 m3/ s every 12 hours until the end of the direct runoff. Derive record hydrograph coordinates of 6 hours?
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28. Coordinate of hydrograph for 6 hr (m3/s)
Time of storm
(hr)
Coordinate of storm
(m3/s)
Base flow
Coordinate DRH (m3/s)
Coordinate of hydrograph for 6 hr (m3/s)
) 3÷ 4 col . ( 10 6 30 20
6.7 12
87.5
10.5 77
25.7 18
111.5
101
33.7 24
102.5 92
30.7 85 11 74
24.7 36 71 60 42 59 48 16
47.5
11.5 54 39
27.5
9.2
31.5
6.6 66 26 14
4.6 72
21.5
9.5
3.2 78
17.5
5.5
1.8 84 15
12.5
2.5
0.8 90
587 m3/s
Depth of DRH = (587*6*3600)/ (423*106)= 3 cm
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29. Example : The runoff data at a stream gauging station for a flood are given below. the drainage area is 40 km2 the duration of rainfall is 3 hours. Derive the 3-hour unit hydrograph for the basin and plot the same.
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30. Solution
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31. Unit Hydrograph for Different Duration:
There are several ways derivation standard hydrograph which sustainability nD - an hour from a standard water scheme sustainability D - an hour, and most important of these ways:
Super Position Method .
S - Curve Method .
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32. Y-coordinates record hydrograph
Super Position Method .
If the availability of a standard hydrograph who sustainability D - an hour and was required is a derivative standard water scheme nD for - hour, where n is an integer,
Example (8) / information given is the Y-coordinates record hydrograph sustainability 4 - hour, derived the y-coordinate is 12 - hour water hydrograph record.
Solution : make table
Time hr 4 8 12 16 20 24 28 32 36 40 44 48 52
Y-coordinates record hydrograph 80
130
150 90 27 15 5 -
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34. Ex / Prepared solution of the previous example S curved way.
S - Curve Method :
This method is used if desired derivative record hydrograph sustainability mD where m non an integer.
Ex / Prepared solution of the previous example S curved way.
الوقت (ساعة)
إحداثيات UH-4 hr
منحني S
إحداثيات منحنيS (2+3)
منحني S متخلف بـ 12 ساعة
عمود 4 – عمود 5
العمود 6 ÷ (4/12) 1 2 3 4 5 6 7 - 20
6.7 8 80
100
33.3 12
130
230
76.7 16
150
380
360
120
510
410
136.7 24 90
600
370
123.3 28 52
652
272
90.7 32 27
679
169
56.3 36 15
694 94
31.3 40
699 47
15.7 44 48
1.7
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35. Ex/ Vertical coordinates of hydrograph 4 - hour shown below
Ex/ Vertical coordinates of hydrograph 4 - hour shown below. Use these coordinates and derived hydrograph coordinates sustainability hydrograph 2 - hour for the same basin by S - Curve Method .
الوقت (ساعة)
إحداثيات UH-4 hr
منحني S
إحداثيات منحنيS (2+3)
منحني S متخلف بـ 2 ساعة
عمود 4 – عمود 5
العمود 6 ÷ (4/2)
( UH – 2hr) 1 2 3 4 5 6 7 - 8 16 12 20 24 31 51 62 49
100 98 10 61
161
122 69
230
138 14 77
307
154 73
380
146 18
119
449
510 22
561
102 39
600 78 26
631 28 21
652 42 30 17
669 34 32
679
689 36
694 38
699
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36. Distribution percentages:
The distribution shows the percentages of total unit hydrograph, which occur during successive uniform time increments.
The procedure of deriving the distribution graph is first to separate the base flow from the total runoff .
Example / Analysis of the DRO for a one day unit storm over a basin for the following data: 18,96,120,82,47,25,12,and 2m3. Determine the distribution graph percentages.
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37. Distribution of ER (cm)
Example /
Basin area of 200 hectares, the rainfall in three consecutive days and the depths of the rain was the 7.5, 2 and 5 cm, respectively. the index rate Ф 2.5 cm / day, the distribution percentage of surface runoff per day rainstorm that one day is 5,15, 40,25, 10,5. determine the hydrograph of runoff ?
السيح في يوم واحد = (200*104 / 24*60*60) = م2 / ثا
0.25 /100*(23.148) = m3/s
time (day)
Rainfall
(cm)
Ф (cm/day)
ER (cm)
distribution percentages %
Distribution of ER (cm)
runoff 5
2.5 cm
m3 / s
0 -1
7.5
0.25
0.0579*
1 -2 2 15
0.75
0.1736
2 – 3 40
0.125
2.125
0.4919
3 – 4 25
1.25
0.375
1.625
0.3762
4 – 5 10
0.5 1
1.5
0.3472
5 – 6
0.625
0.875
0.2025
6 – 7
0.0579
7 – 8
0.0289
8 - 9
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38. APPLICATION OF UNIT HYDROGRAPH:
The application of unit hydrograph consists of :
(i) From a unit hydrograph of a known duration to obtain a unit hydrograph of the demand duration, either by the S-curve method or by the principle of superposition.
(ii) From the unit hydrograph, so derived to obtain the flood hydrograph corresponding to a single storm or multiple storms. For design purposes, a design storm is assumed with the help of unit hydrograph, gives a design flood hydrograph.
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39. Example
The 3-hr unit hydrograph ordinates for a basin are given below. There was a storm, which commenced on July 15 at (6-hr ), which was followed by another storm on July 16 at (12-hr). the amount of rainfall on July 15 was 5.75 cm for first (3-hr ) and 3.75 cm for next (3-hr) , and on July 16, 4.45 cm for all (12-hr). Assuming an average loss of 0.25 cm/hr and 0.15 cm/hr for the two storms, respectively, and a constant base flow of 10 m3. Determine the stream flow hydrograph and peak flow .
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40. Solution
Since the duration of the UG is 3 hr, the 6-hr storm can be considered as 2-unit storm .
A net rain of – 0.25 × 3 = 5 cm in the first 3-hr period and a net rain of 3.75 – 0.25 × 3 = 3 cm in the next 3-hr period.
The unit hydrograph ordinates are multiplied by the net rain of each period by 3 hr. Similarly,
Another unit storm by 12 hr (next day) produces a net rain of 4.45 – 0.15 × 3 = 4 cm which is multiplied by the UGO .
Table show the rainfall excesses due to the three storms are added up to get the total direct surface discharge ordinates.
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42. Example /Storm rainfalls of 3. 2, 8. 2 and 5
Example /Storm rainfalls of 3.2, 8.2 and 5.2 cm occur during three successive hours over an area of 45 km2. The storm loss rate is 1.2 cm/hr. The distribution percentages of successive
hours are 5, 20, 40, 20, 10 and 5. Determine the stream flows for successive hours assuming a constant base flow of 10 m3.
Solution
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