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 5/23/2019

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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. 5/23/2019

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 5/23/2019

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 : 5/23/2019

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) 5/23/2019

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 5/23/2019

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. 5/23/2019

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). 5/23/2019

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 . 5/23/2019

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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 5/23/2019

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)]= 1.4904*106 m3 Depth of DRH = Runoff vol./ Area = 1.4904*106 / 27*106 = 5.52 cm. المطرالصافي))Pnet Total Rainfall = 2.8 + 3.8 = 6.6 cm. Time of Duration = 8 hr. Ф index = (6.6 – 5.52) / 8 = 0.135 cm/hr. 5/23/2019

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? 5/23/2019

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. 5/23/2019

The steps for the derivation of unit hydrograph can be formulated as follows : 5/23/2019

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 = 270- 20 =250 m3/s Peak for 3-UH = 250/5= 50 m3/s /cm 5/23/2019

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. 5/23/2019

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. 5/23/2019

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? 5/23/2019

Solution (i) The flood hydrograph due to the two consecutive hourly storms is computed in Table: 5/23/2019

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 5/23/2019

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) 5/23/2019

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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 5/23/2019

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 5/23/2019

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? 5/23/2019

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 5/23/2019

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. 5/23/2019

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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 . 5/23/2019

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 - 5/23/2019

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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 5/23/2019

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 5/23/2019

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. 5/23/2019

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) = 23.148 م2 / ثا 0.25 /100*(23.148) =0.0579 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 5/23/2019

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. 5/23/2019

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 . 5/23/2019

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 5.75 – 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. 5/23/2019

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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 5/23/2019

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