Assoc.Prof. dr.tarkan erdik HYDROLOGY Lecture 11 IUH Assoc.Prof. dr.tarkan erdik

Instantaneous Unit Hydrograph: Instantaneous unit hydrograph (IUH) is obtained when the duration of the rainfall excess is infinitesimally small. Why do we need IUH? Because some basic assumptions below of Unit Hydrograph Theory is violated. 1-Large areas, areas more than 5000 km square unit hydrograph approach will not be valid. 2- Actual rainfall event may not follow a uniform intensity for a large a large period of time. In those cases above, rainfall will be uniform so IUH should be used.

i Unit Hydrograph What would happen if we reduce D to D/2 intensity will be increased to 2i What would happen if we reduce D to D/4 intensity will be increased to 4i What would happen if we reduce D to 0 intensity will be increased to infinity We can reduce the duration to 0 and that is why this is called an instantaneous unit hydrograph. The intensity is infinite of course, the duration is 0.

IUH is for instantaneous precipitation with total depth of rainfall = 1cm Intensity is not uniform So at any time τ, the intensity of rain fall is given by I (τ). I (τ)d(τ)= depth of rain fall, occurring over the catchment instantaneously dQ=I (τ)d(τ)u(t-τ) Q= 0 𝑡 I (τ)d(τ)u(t−τ) Our aim is to find out the DRH ordinate at the point t which we will call qt.

FLOOD ROUTING As the flood moves in the downstream direction, not only is the hydrograph lagged by the travel time but the shape of the hydrograph varies due to the effect of storage in the stream cross-section. Its peak discharge reduces and its base width increases.

Hydraulic methods: beyond the scope of the lecture A non-uniform and unsteady flow occurs in the stream as the flood wave moves. Two types of methods can be used in studying this phenomenon (flood routing): Hydraulic methods: beyond the scope of the lecture Hydrologic methods: These simpler methods based on the continuity equation and only give approximate results. Let x(t) be the hydrograph of the flow that enters a stream reach, y(t) the hydrograph of the flow that goes out, and S(t) is the volume , stored in the reach. The continuity equation can be written as:

In flood routing, x1, x2, y1, and S1 are known In flood routing, x1, x2, y1, and S1 are known. To solve the unknowns S2 and y2, it is necessary to express S2 in terms of other variables.

Computation of a storage volume in a reach Time variation of the storage

Plotting the volume S versus y, S is not a function of y only Plotting the volume S versus y, S is not a function of y only. The volume S during the rise of the water stage is larger than the volume during the fall of the stage for same value of y .

Plotting the volume S versus y, S is not a function of y only Plotting the volume S versus y, S is not a function of y only. The volume S during the rise of the water stage is larger than the volume during the fall of the stage for same value of y . S1 S2

If x>y ds/dt>0 stage increase If x<y ds/dt<0 stage decreases

to plot ax+(l-a)y with respect to S to plot ax+(l-a)y with respect to S. The value of a that minimizes the scatter of the points around a straight line is the desired value, and the inverse of the slope of this line is the value of K.

Example: The flood hydrograph observed at the entrance of the streamis given below. Please determine the outflow hydrograph at the exist, 18 km from the entrance, by Muskingum method. The flood wave propagates in the stream with a velocity of 2m/s (please asssume a=2, ∆x=6 km and ∆t=2hr ). K= ∆𝑥 𝑢 = 6000𝑚 2𝑚/𝑠 =3000𝑠 co=0.48, c1=0.75, c2=-0.23 0.48 +0.75 -0.23=1

x(km) t (hr) 10 2 18 4 50 6 107 8 147 150 12 105 14 59 16 33 17 20 22 24 26 28 Base flow

x(km) t (hr) 6 10 2 18 13.84 4 50 107 8 147 150 12 105 14 59 16 33 17 20 22 24 26 28 x1 x2 y1 y2 10 18 13.84 X1 y1 X2 y2 co=0.48, c1=0.75, c2=-0.23 Y2=COX2+C1X1+C2Y1 Y2=0.75×10+0.48×18-0.23×10

x(km) t (hr) 6 10 2 18 4 50 107 8 147 150 12 105 14 59 16 33 17 20 22 24 26 28 x1 x2 y1 y2 10 18 13.84 X1 y1 X2 y2

x(km) t (hr) 6 10 2 18 13.84 4 50 34.35 107 81.3 8 147 132.4 150 149.9 12 105 125.2 14 59 77.9 16 33 41.9 17 23.1 20 12.2 22 9.5 24 10.1 26 28 x1 x2 y1 y2 10 18 13.84

x(km) t (hr) 6 12 10 2 18 13.84 11.9 4 50 34.35 24.4 107 81.3 59.6 8 147 132.4 111.2 150 149.9 145.9 105 125.2 138.8 14 59 77.9 99 16 33 41.9 55.5 17 23.1 29.6 20 12.2 16.3 22 9.5 9.9 24 10.1 9.7 26 10.2 28

x(km) t (hr) 6 12 18 10 2 13.84 11.9 10.9 4 50 34.35 24.4 18.2 107 81.3 59.6 43 8 147 132.4 111.2 88.6 150 149.9 145.9 133.4 105 125.2 138.8 145.4 14 59 77.9 99 117.9 16 33 41.9 55.5 73.5 17 23.1 29.6 38.8 20 12.2 16.3 21 22 9.5 9.9 12.1 24 10.1 9.7 9.3 26 10.2 28 150 m3/s is reduced to 145.4 m3/s. Base width is increased from 20 to 24. So, peak discharge reduces and base widens