Lecture 6: Synthetic unit hydrograph
Synthetic Unit Hydrograph In India, only a small number of streams are gauged (i.e., stream flows due to single and multiple storms, are measured) There are many drainage basins (catchments) for which no stream flow records are available and unit hydrographs may be required for such basins In such cases, hydrographs may be synthesized directly from other catchments, which are hydrologically and meteorologically homogeneous, or indirectly from other catchments through the application of empirical relationship Methods for synthesizing hydrographs for ungauged areas have been developed from time to time by Bernard, Clark, McCarthy and Snyder. The best known approach is due to Snyder (1938)
Synthetic unit hydrograph Snyder s method Snyder (1938) was the to develop a synthetic UH based on a study of watersheds in the Appalachian Highlands. In basins ranging from 10 10,000 mi. 2 Snyder relations are t p = C t (LL C ) 0.3 where t p = basin lag (hr) L= length of the main stream from the outlet to the divide (mi) L c = length along the main stream to a point nearest the watershed centroid (mi) C t = Coefficient usually ranging from 1.8 to 2.2
Synthetic unit hydrograph Snyder s method Q p = 640 C p A/t p where Qp = peak discharge of the UH (cfs) A = Drainage area (mi2) C p = storage coefficient ranging from 0.4 to 0.8, where larger values of cp are associated with smaller values of Ct T b = 3+t p /8 where Tb is the time base of hydrograph Note: For small watershed the above eq. should be replaced by multiplying t p by the value varies from 3-5 The above 3 equations define points for a UH produced by an excess rainfall of duration D= t p /5.5 Snyder s hydrograph parameter
Example Problem 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? C t = 1.8, C p = 0.6, L= 18mi, L c = 10mi Calculate t p t p = C t (LL C ) 0.3 = 1.8(18 10) 0.3 hr, = 8.6 hr Calculate Qp Q p = 640(c p )(A)/t p = 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= t p /5.5 hr = 8.6/5.5 hr = 1.6 hr
Example Problem 5000 4000 Q p W 75 = 440(Q P /A) -1.08 W 50 = 770(Q P /A) -1.08 (widths are distributed 1/3 before Q p and 2/3 after) Q (cfs) 3000 2000 W 75 W 50 Area drawn to represent 1 in. of runoff over the watershed 1000 0 0 5 10 15 20 25 30 35 40 Time (hr)
SCS (Soil Conservation Service) Unit Hydrograph Unit = 1 inch of runoff (not rainfall) in 1 hour t p Can be scaled to other depths and times Based on unit hydrographs from many watersheds The earliest method assumed a hydrograph as a simple triangle, with rainfall duration D, time of rise T (hr), time of fall B. and peak flow Q p (cfs). T Q p B SCS triangular UH
SCS Unit Hydrograph The volume of direct runoff is QpT Vol = 2 where B is given by + QpB 2 or Q p vol = 2 T + B B =1. 67T Therefore runoff eq. becomes, for 1 in. of rainfall excess, Q Q p p = = 0.75vol T 484A T where A= area of basin (sq mi) T = time of rise (hr) = Q p = 0.75(640) A(1.008) T
SCS Unit Hydrograph Time of rise T is given by D T = + 2 t p where D= rainfall duration (hr) t p = lag time from centroid of rainfall to Q P Lag time is given by t p = 0.8 L 1000 9 CN 0.5 19000y 0.7 where L= length to divide (ft) Y= average watershed slope (in present) CN= curve number for various soil/land use
SCS Unit Hydrograph unoff curve number for different land use (source: Woo-Sung et al.,1998)
Example Problem Use the SCS method to develop a UH for the area of 10 mi 2 described below. Use rainfall duration of D = 2 hr C t = 1.8, C p = 0.6, L= 5mi, L c = 2mi The watershed consist CN = 78 and the average slope in the watershed is 100 ft/mi. Sketch the resulting SCS triangular hydrograph. Solution Find t p by the eq. t p 0.8 L 1000 9 CN 0.5 19000y Convert L= 5mi, or (5*5280 ft/mi) = 26400 ft. = Slope is 100 ft/mi, so y = (100ft/mi) (1mi/5280 ft)(100%) = 1.9% Substituting these values in eq. of t p, we get t p = 3.36 hr 0.7
Example Problem Find T using eq. D T = + 2 t p Given rainfall duration is 2 hr, T = 4.36 hr, the rise of the hydrograph Then find Q p using the eq, given A= 10 mi 2 Q p = 484A T. Hence Q p = 1.110 cfs To complete the graph, it is also necessary to know the time of fall B. The volume is known to be 1 in. of direct runoff over the watershed. So, Vol. = (10mi 2 ) (5280ft/mi) 2 (ac/43560ft 2 ) (1 in.) = 6400 ac-in Hence from eq. Vol = Q T p 2 B = 7.17 hr + Q p 2 B
Example Problem 1200 1000 800 Q (cfs) 600 400 Q p = 1110 (cfs) 200 0 0 2 4 6 8 10 12 14 T =4.36 (hr) Time (hr) B=7.17 (hr)
Exercise problems 1. The stream flows due to three successive storms of 2.9, 4.9 and 3.9 cm of 6 hours duration each on a basin are given below. The area of the basin is 118.8 km 2. Assuming a constant base flow of 20 cumec, derive a 6-hour unit hydrograph for the basin. An average storm loss of 0.15 cm/hr can be assumed (Hint :- Use UH convolution method) Time (hr) 0 3 6 9 12 15 18 21 24 27 30 33 Flow (cumec) 20 50 92 140 199 202 204 144 84 45 29 20
Exercise problems 2. The ordinates of a 4-hour unit hydrograph for a particular basin are given below. Derive the ordinates of (i) the S-curve hydrograph, and (ii) the 2-hour unit hydrograph, and plot them, area of the basin is 630 km 2 Time (hr) Discharge (cumec) 0 0 2 25 4 100 6 160 8 190 10 170 12 110 Time (hr) Discharge (cumec) 14 70 16 30 18 20 20 6 22 1.5 24 0
Exercise problems 3. The following are the ordinates of the 9-hour unit hydrograph for the entire catchment of the river Damodar up to Tenughat dam site: and the catchment characteristics are, A = 4480 km 2, L = 318 km, L ca = 198 km. Derive a 3-hour unit hydrograph for the catchment area of river Damodar up to the head of Tenughat reservoir, given the catchment characteristics as, A = 3780km 2, L = 284 km, L ca = 184km. Use Snyder s approach with necessary modifications for the shape of the hydrograph. Time (hr) 0 9 18 27 36 45 54 63 72 81 90 Flow (cumec) 0 69 1000 210 118 74 46 26 13 4 0
Highlights in the Module This module presents the concept of ainfall-unoff analysis, or the conversion of precipitation to runoff or streamflow, which is a central problem of engineering hydrology. Gross rainfall must be adjusted for losses to infiltration, evaporation and depression storage to obtain rainfall excess, which equals Direct unoff (DO). The concept of the Unit hydrograph allows for the conversion of rainfall excess into a basin hydrograph, through lagging procedure called hydrograph convolution. The concept of synthetic hydrograph allows the construction of hydrograph, where no streamflow data are available for the particular catchment.