Runoff Hydrograph and Flow Routing

Runoff Hydrograph and Flow Routing

03/02/2006 Unit Hydrograph Reading: Sections 7.1-7.3, 7.5, 7.7, Hydrologic Analysis Change in storage w.r.t. time = inflow - outflow In the case of a linear reservoir, S = kQ Transfer function for a linear system (S = kQ).

Proportionality and superposition Linear system (k is constant in S = kQ) Proportionality If I1 Q1 then C*I2 C*Q2 Superposition If I1 Q1 and I2 Q2, then I1 +I2 Q1 + Q2 Impulse response function Impulse input: an input applied instantaneously (spike) at time and zero everywhere else An unit impulse at produces as unit impulse response function u(t-)

Principle of proportionality and superposition Convolution integral For an unit impulse, the response of the system is given by the unit impulse response function u(t-) An impulse of 3 units produces the 3u(t-) If I() is the precipitation intensity occurring for a time period of d, the response of the system (direct runoff) is I()u(t-)d The complete response due to the input function I() is givent by convolution integral Q(t ) I ( )u (t )d 0

Response of a linear system is the sum (convolution) of the responses to inputs that have happened in the past. Step and pulse inputs A unit step input is an input that goes from 0 to 1 at time 0 and continues indefinitely thereafter A unit pulse is an input of unit amount occurring in duration t and 0 elsewhere. Precipitation is a series of pulse

Unit Hydrograph Theory Direct runoff hydrograph resulting from a unit depth of excess rainfall occurring uniformly on a watershed at a constant rate for a specified duration. Unit pulse response function of a linear hydrologic system Can be used to derive runoff from any excess rainfall on the watershed. Unit hydrograph assumptions Assumptions Excess rainfall has constant intensity during duration Excess rainfall is uniformly distributed on watershed Base time of runoff is constant

Ordinates of unit hydrograph are proportional to total runoff (linearity) Unit hydrograph represents all characteristics of watershed (lumped parameter) and is time invariant (stationarity) Discrete Convolution t Continuo us Q(t ) I ( )u (t )d 0 n M

Discrete Qn PmU n m 1 m 1 Q is flow, P is precipitation and U is unit hydrograph M is the number of precipitation pulses, n is the number of flow rate intervals The unit hydrograph has N-M+1 pulses Application of convolution to the output from a linear system

Time Area Relationship Isochrone of Equal time to outlet A3 A4 A2 10hr 5hr A3 R2

R1 Area Excess Rainfall 15hr A1 R3 A4 A2

A1 Qn Ri A1 Ri 1 A2 ... R1 A j Time, t 0 5 10 Time, t 15 20

Application of UH Once a UH is derived, it can be used/applied to find direct runoff and stream flow hydrograph from other storm events. Given: Ex. 7.5.1 P1 = 2 in, P2 = 3 in and P3 = 1 in, baseflow = 500 cfs and watershed area is 7.03 mi2. Given the Unit Hydrograph below, determine the streamflow hydrograph 7.5.1 solution (contd)

See another example at: Gauged and ungauged watersheds Gauged watersheds Watersheds where data on precipitation, streamflow, and other variables are available Ungauged watersheds Watersheds with no data on precipitation, streamflow and other variables. 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. Synthetic UH 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 SCS dimensionless hydrograph

Synthetic UH in which the discharge is expressed by the ratio of q to qp and time by the ratio of t to Tp If peak discharge and lag time are known, UH can be estimated. Tc: time of concentration t p 0.6Tc

tb 2.67T p C = 2.08 (483.4 in English system) t Tp r t p 2 qp A: drainage area in km2 (mi2) CA Tp

Ex. 7.7.3 Construct a 10-min SCS UH. A = 3.0 km2 and Tc = 1.25 h t r 10 min 0.166 h t p 0.6Tc 0.6 1.25 0.75 h 0.833 h t Tp r t p 2 0.166 Tp 0.75 0.833 h 2 qp

q 7.49 m3/ CA 2.08 3 7.49 m3 / Tp 0.833 Multiply y-axis of SCS hydrograph by qp and x-axis by Tp to get the required UH, or construct a triangular UH

2.22 h t

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