2. 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. It can also refer to a graph showing the volume of water reaching a particular outfall.

3. graphs are commonly used in the design of sewerage, more specifically, the design of surface water sewerage systems and combined sewers.

4. COMPONENTS OF A HYDROGRAPH Rising limb: The rising limb of hydro graph, also known as concentration curve, reflects a prolonged increase in discharge from a catchment area, typically in response to a rainfall event Recession (or falling) limb: The recession limb extends from the peak flow rate onward. The end of stormflow (aka quickflow or direct runoff) and the return to groundwater-derived flow (base flow) is often taken as the point of inflection of the recession limb. The recession limb represents the withdrawal of water from the storage built up in the basin during the earlier phases of the hydrograph.

5. Peak discharge: the highest point on the hydro graph when the rate of discharge is greatest Lag time: the time interval from the center of mass of rainfall excess to the peak of the resulting hydrograph Time to peak: time interval from the start of the resulting hydro graph Discharge: the rate of flow (volume per unit time) passing a specific location in a river or other channel

7. UNIT HYDROGRAPH

8. Unit Hydrograph Hydrograph usually consists of a fairly regular lower portion that changes slowly throughout the year and a rapidly fluctuating component that represents the immediate response to rainfall. The lower, slowly changing portion is termed base flow. The rapidly fluctuating component is called direct runoff. D = duration of excess rain Look at hyetograph, here 2 hours

9. UNIT HYDROGRAPH The amount of run-off resulting from 1 unit (1cm, 1mm, 1ft, etc.) of rainfall excess. is essentially a tool for determining the direct runoff response to rainfall. Once you know the watershed’s response to one storm, you can predict what its response for another will look like.

10. UH Basic Assumptions of UH 1. The effective rainfall is uniformly distributed within its duration 2. The effective rainfall is uniformly distributed over the whole drainage basin 3. The base duration of direct runoff hydrograph due to an effective rainfall of unit duration is constant. 4. The ordinates of DRH are directly proportional to the total amount of DR of each hydrograph 5. For a given basin, the runoff hydrograph due to a given period of rainfall reflects all the combined physical characteristics of basin (time-invariant)

11. Remove the base flow amount from the hydrograph. Calculate the net rainfall by removing the infiltration and retention storage from the hyetograph. Scale the new hydrograph units to yield a unit hydrograph for, say, 1 inch for one hour, or 1/2 in/hr for 2 hours, or 1/3 in/hr for 3 hr, etc. How to get a unit hydrograph, UH

12. net hydrograph net hyetograph A Unit Hydrograph has 1.0 inches of Direct Runoff for the storm duration. Here we start with 2 inches for 2 hours net rainfall. That is, 1 inch per hour for 2 hours. To get a 2-hour UH, we want 1/2 inches/hour for two hours. So we have to divide the hydrograph ordinates by 2.

13. We divided each hydrograph ordinate by two, resulting in a 2– hour Unit Hydrograph, i.e. One inch of direct runoff total from the 2 hour storm make a 1 hour UH. T b is the time base Define “ordinate” Notice hydrograph is not as tall

14. Convert the direct runoff hydrograph below to a 2-HR UH. In the hyetograph  = 0.5 in/hr In the hydrograph, base flow = 100 cfs EXAMPLE

15. 1 in/hr x 2 hours = 2 inches, so 2 inches for a 2 hour storm. This is twice too big. We want 1 inch total for the storm, so we must divide each NET hydrograph ordinate by 2 Draw the Net Hyetograph, and calculate the total direct runoff, in inches, over the watershed. Net Hyetograph Original Hyetograph

16. We removed the base flow from the gross hydrograph, then divided each ordinate by two, to get a unit hydrograph for a 2-hour storm. We have characterized our watershed; now we know how it will behave in a storm.

17. SYNTHETIC UNIT HYDROGRAPH

18. Synthetic Unit Hydrograph Synthetic hydrographs are derived by – Relating hydrograph characteristics such as peak flow, base time etc. with watershed characteristics such as area and time of concentration. – Using dimensionless unit hydrograph – Based on watershed storage

19. Need for synthetic UH UH is applicable only for gauged watershed and for the point on the stream where data are measured For other locations on the stream in the same watershed or for nearby (ungauged) watersheds, synthetic procedures are used.

20. Snyder’s Method Snyder’s method allows the computations of (a) lag time (t L ); (b) UH duration (t r ); (c) UH peak discharge (q p ); (d) Hydrograph time width at 50% and 75% (W 50, W 75 ) of peak flow UH

21. Snyder’s Method 1. Lag time (t L ): time from the center of rainfall – excess to the UH peak t L = C 1 C t (LL c ) 0.3 where t L = Time [hrs]; C 1 = 0.75 for SI unit; 1.0 for English unit; C t = Coefficient which is a function of watershed slope and shape, 1.8~2.2 (for steeper slope, C t is smaller); L = length of the main channel [mi, km]; L c = length along the main channel to the point nearest to the watershed centroid UH

22. Snyder’s Method 2. UH Duration (t r ): t r = t L / 5.5 where t r and t L are in [hrs]. If the duration of UH is other than t r, then the lag time needs to be adjusted as t pL = t L + 0.25 (t R - t r ) where t LR = adjusted lag time; t R = desired UH duration. 3. UH Peak Discharge (q p ): or where C 2 = 2.75 for SI unit; 640 for English unit; C p = coefficient accounting for flood wave and storage condition, 0.4 ~ 0.8; q p = specific discharge, [m 3 /s/km 2 ] or [ft 3 /s/mi 2 ] To compute actual discharge, Q p = A  q p UH

23. Snyder’s Method 4. Time Base (t b ): Assuming triangular UH, t b = C 3 / q p where t b – [hrs]; C 3 = 5.56 for SI unit, 1290 for English unit. 5. UH Widths:or where C W, 75 = 1.22 for SI unit; 440 for English unit; C W, 50 = 2.14 for SI unit; 770 for English unit;. W 50, W 75 are in hours; Usually, 1/3 of the width is distributed before UH peak and 2/3 after the peak Remember to check that the volume of UH is close to 1 cm or 1 inch UH

24. Use Snyder’s method to develop a UH for the area of 100mi2 described below. Sketch the appropriate shape. What duration rainfall does this correspond to? Ct = 1.8, L= 18mi, Cp = 0.6, Lc= 10mi EXAMPLE:

25. Calculate tp tp = tl= Ct(LLC)^0.3 = 1.8(18·10) 0.3 hr, = 8.6 Calculate Qp Qp= 640(cp)(A)/tp = 640(0.6)(100)/8.6 = 4465 cfs since this is a small watershed, Tb ≈4tp = 4(8.6) = 34.4 hr Duration of rainfall D= tp/5.5 hr = 8.6/5.5 hr = 1.6 hr SOLUTION