Runoff Hydrographs. The Unit Hydrograph Approach

Transcription

1 Runoff Hydrographs The Unit Hydrograph Approach

2 Announcements HW#6 assigned

3 Storm Water Hydrographs Graphically represent runoff rates vs. time Peak runoff rates Volume of runoff Measured hydrographs are best But not often available Methods are available to develop a synthetic hydrograph Use a unit hydrograph (UHG)

4 Unit Hydrograph Widely used method of empirical storm flow analysis Def: Basin outflow resulting from 1 unit (in./mm/cm/etc) of direct runoff generated uniformly over the drainage area at a uniform rainfall rate during a specified period of rainfall intensity.

5 Unit Hydrographs Assumptions: rainfall intensity is not considered linear relationship between stormwater runoff and rainfall UHG is independent of antecedent conditions uniform rainfall distribution

6 Unit Hydrographs Derived from observed records extensive data requirements streamflow and rainfall record pairs generally not used for small catchments UHG represent direct surface runoff baseflow (ground water) must be removed For small catchments ==> synthetic UHG models models give ==> tp (time to peak), qp(peak flow rate and mathematical shape of curve

7 Duration of the Unit Hydrograph Unit hydrograph have a duration that is the same as the duration of the rainfall excess that produced it Conceptually it is possible to have an infinite # of hydrographs corresponding to different durations. Practically, unit hydrographs are limited to rainfall excesses up to 25% different than the duration of the unit hydrograph.

8 How is the unit hydrograph used? For a unit hydrograph of duration, D, the volume underneath the hydrograph is always 1, produced by 1 unit of excess rainfall. A hydrograph for a block of rainfall excess of any depth is obtained by multiplying the ordinates of the unit hydrograph by the depth of the rainfall excess block. The result are the ordinates of the runoff hydrograph.

9 Unit Hydrograph Development Approximate hydrograph using a triangle Need to find: Time to peak (t p ) Time lag (t l ) Time of the base (t b ) Peak flow (q p )

10 Lag time, t L SCS Equation to calculate time lag L = hydraulic length of watershed (feet) S = curve number parameter (inches) Y = average land slope of the watershed (%) t l = time lag (hours) t L L 0.8 (S 1) 1900Y Eq. 1

11 Unit Hydrograph Development Equation to calculate time to peak t p = t l + D/2, ( Eq. 2), where: t l = time lag D = time increment of rainfall excess SCS Equation to calculate peak flow (q p ) q p = 484A / t p, (Eq. 3), where: A = watershed area in mi 2 t p = time to peak in hours q p = peak flow in cfs per inch of runoff*** SCS Equation for time of base (t b ) t b = 2.67t p (Eq. 4)

12 Example 5.10 in Text Given: 1-hr storm = 2.5 in. P = 2.5 in. 500 ac watershed = 21,780,000 ft 2 Land use = commercial business Watershed soil HSG = D Average watershed slope = 1% Hydraulic length of watershed = 6,000 ft Required: Find the storm hydrograph Use the triangular unit hydrograph method

13 Example 5.10 in Text Solution: HSG = D / Commercial T. 5.1 CN = 95 S = 0.53 in. S = (1000 / CN) - 10 Assume AMC = II Q = 1.96 in. of runoff Using the SCS CN method Q = (P - 0.2S) 2 / (P + 0.8S) Step #1 Find points to develop the unit hydrograph t l = 0.75 hr (45 min) Use Eq. 1 t p = 1.25 hr (75 min) Use Eq. 2 t b = 3.33 hr (200 min) Use Eq. 4 q p = 302 cfs / 1 in. of runoff Use Eq. 3 Plot unit hydrograph Check area under the triangle 1 in.

14 Q (cfs) Volume under triangle = 1/2bh = ½(200 min x 60 sec/min)302.5 ft 3 /sec = 1,815,000 ft 3 Surface runoff depth = 1,815,000 ft 3 / 21,780,000 ft 2 = ft = 1.0 in. ok q p = cfs Watershed area (A) = 21,780,000 ft t p = t b = 200 T(min)

15 Example 5.10 in Text Step #2 q p 2.5 rain = cfs x 1.96 in. of SRO - SRO = 1.96 in. from the SCS CN Method q p 2.5 rain = cfs Plot storm hydrograph Check area under the triangle 1.96 in.

16 Q (cfs) 600 q p = cfs Volume under triangle = 3,557,400 ft 3 Surface runoff depth = 1.96 in. ok 3,557,400 ft 3 / 21,780,000 ft 2 = ft = 1.96 in t p = t b = 200 T(min)

17 Example 5.11 in Text Given: Same as Example 5.10 but this time more detailed rainfall information 1-hr storm = 2.5 in min = 0.5 in min = 1.0 in min = 0.75 in min = 0.25 in. 500 ac watershed Land use = commercial business Watershed soil HSG = D Average watershed slope = 1% Hydraulic length of watershed = 6,000 ft Required: Find the storm hydrograph

18 Example 5.11 in Text Solution: HSG = D / Commercial T. 5.1 CN = 95 S = 0.53 in. (same as Ex. 5.10, S = (1000 / CN) - 10) Assume AMC = II Q = 1.96 in. of runoff (same as Ex. 5.10, Q = (P - 0.2S) 2 / (P + 0.8S) Find points to develop the unit hydrograph t L = 0.75 hr (45 min) (same as Ex. 5.10, t L = [L 0.8 (S + 1) 0.7 ] / [1900 x Y 0.5 ] New since D = 15 min. t p = t L + D/2 = 0.88 hr (52.5 min) New since t p = 52.5 min q p = 484(A)/t p = 432 cfs / 1 in. of runoff

19 Example 5.11 in Text 0 15 min P = 0.5 in. (given) Q 1 = [ (.53))] 2 / [ (.53)] Q 1 = 0.17 in. of runoff q p1 = 432 cfs / 1 of SRO x 0.17 = cfs t p1 = t L + D/2 = /2 = 52.5 min t b1 = 2.67(t p ) = 2.67(52.5) = min

20 Example 5.11 in Text 0 30 min P = = 1.5 in. Q 1 + Q 2 = [ (.53))] 2 / [ (.53)] Q 1 + Q 2 = 1.01 in. of runoff Q 2 = 1.01 in. - Q 1 Q 2 = = 0.84 in. of runoff q p2 = 432 cfs / 1 of SRO x 0.84 = cfs t p2 = = 67.5 min Shift on the x-axis by 15 min t b2 = = min Shift on the x-axis by 15 min

21 Example 5.11 in Text 0 45 min P = = 2.25 in. Q 1 + Q 2 + Q 3 = [ (.53))] 2 / [ (.53)] Q 1 + Q 2 + Q 3 = 1.72 in. of runoff Q 3 = 1.72 in. - Q 1 - Q 2 Q 3 = = 0.71 in. of runoff q p3 = 432 cfs / 1 of SRO x 0.71 = cfs t p3 = = 82.5 min Shift on the x-axis by 30 min t b3 = = min Shift on the x-axis by 30 min

22 Example 5.11 in Text 0 60 min P = = 2.50 in. Q 1 + Q 2 + Q 3 + Q 4 = [ (.53))] 2 / [ (.53)] Q 1 + Q 2 + Q 3 + Q 4 = 1.96 in. of runoff Q 4 = 1.96 in. - Q 1 - Q 2 - Q 3 Q 4 = = 0.24 in. of runoff q p4 = 432 cfs / 1 of SRO x 0.24 = cfs t p4 = = 97.5 min Shift on the x-axis by 45 min t b4 = = min Shift on the x-axis by 45 min

23 Y = mx + b Y = -mx + b

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25 Y = mx + b m = (73.4 0) / ( ) m = Y = X + b b = Y X = (52.5) b = Y = X Incremental Hydrograph #1 Q (cfs) (52.5, 73.4) (140.2, 0.0) Time (min)

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