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

2. 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.

3. 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.

4. 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.

5. 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.

6. 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:

7. 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.

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

9. 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 .

10. 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

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

14. 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.

15. 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 =1

16. 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

17. 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

18. 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

19. 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

20. 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

21. 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 m3/s. Base width is increased from 20 to 24. So, peak discharge reduces and base widens