CE322 Basic Hydrology - Final Exam - Solution
Jorge A. Ramirez

Questions are in boldface.

Problem 1 (50 points) The U.S. Army Corps of Engineers is worried about the safety implications of one of their reservoirs. If the dam fails, a sharp- crested hydrograph of the following form will be released into a channel below the dam:

Time (h) Q (m3/s)
6 50
12 400
18 300
24 200
30 100

The channel below the dam has routing parameters Co = 0.1 and C1 = 0.4. This channel connects the upstream reservoir with another reservoir downstream. The downstream reservoir has vertical walls and a bottom area of 20 106 m2. The level-pool ( storage-indication) equation of the downstream reservoir is,

(3)

where Q is in m3/s, S is in m3, and Dt is 21600 s.. The maximum capacity of the spillway of the downstream reservoir is 200 m3/s.

a) Will the spillway of the downstream reservoir be able to handle the flood wave resulting from the failure of the upstream dam? The initial storage in the reservoir is 600 106 m3, and the initial discharge from the reservoir is zero. In obtaining an answer, determine the inflow hydrograph to the downstream reservoir and the outflow hydrograph from the downstream reservoir.

A) Route hydrograph from upstream reservoir to inlet of downstream reservoir. Use Muskingum method:

(4)

From the given values,

C2 = 1- Co - C1 = 1 - 0.1 - 0.4 = 0.5.
Time (h) Channel Inflow (m3/s) Co x Ii+1 C1 x Ii C2 x Oi Channel Outflow (m3/s)
6 50       0
12 400 40 20 0 60
18 300 30 160 30 220
24 200 20 120 110 250
30 100 10 80 125 215
36   0 40 107.5 147.5
42   0 0 73.75 73.75
48   0 0 36.875 36.875

B) Route Channel Outflow hydrograph through downstream reservoir. Use Storage-Indication method:

Time (h) Inflow (m3/s) Ii + Ii+1 2Si/Dt-Oi 2Si+1/Dt + Oi+1 Outflow (m3/s)
6 0        
12 60 60 55555.56 55615.56 4.619658
18 220 280 55606.32 55886.32 25.4474
24 250 470 55835.42 56305.42 57.68626
30 215 465 56190.05 56655.05 84.58069
36 147.5 362.5 56485.89 56848.39 99.45289
40 73.75 221.25 56649.48 56870.73 101.1717
42 36.875 110.625 56668.39 56779.01 94.11642

Use equation 3 above to obtain the outflow from the reservoir. The peak flow at the reservoir outlet (i.e., at the spillway) is 101.17 m3/s, which is less than the spillway capacity of 200 m3/s.

b) What are the values of the Muskingum parameters k and x associated with this problem? What do parameters k and x represent?
 
 

(6)

(7)

(8)

Using these equations obtain:

Adding Eqs 6 & 7:

Dt = 6 h = 0.5 C3. Thus, C3 = 12 h.

Adding Eqs 7 & 8:

k = 0.9 C3 = (0.9) (12 h) = 10.8 h

From Eq. 6:

x = (0.5 (6 h) - 0.1 (12 h))/10.8 h = 0.167

k: average travel time of a wave through the channel reach.

x: relative influence of the inflow in determining the S vs. Q relationship.

Problem 2 (20  points) The 25-min UH for a given watershed is tabulated below.

Time (min) 25 50 75 100 125 150
UH (m3/s/cm) 90 150 240 180 90 45
  1. Use a 75-min UH to determine the DRH that would be observed from the following ERH.
    2. Time (min) 0-75 75-150 150-225
    ERH (cm/h) 0.5 2.0 1.5

    P1 = (0.5 cm/h) (75 min/60 (min/h)) = 0.625 cm

    P2 = (2.0 cm/h) (75 min/60 (min/h)) = 2.5 cm

    P3 = (1.5 cm/h) (75 min/60 (min/h)) = 1.875 cm

    Time (h)
    UH

    (m3/s/cm)
    S-Hyd

    (m3/s/cm)
    Lagged

    S-Hydr

    (m3/s/cm)
    S1 -S2

    (m3/s/cm)
    D'-UH

    (m3/s/cm)
    P1*UH

    (m3/s)
    P2*UH

    (m3/s)
    P3*UH

    (m3/s)
    DRH

    (m3/s)
    1
    0
    0
    		  
    0
    0
    0
    		    
    0
    2
    		 90 90   90
    30
    		 18.75     18.75
    3
    150
    		 240  
    240
    80
    		 50     50
    4
    240
    		 480 0 480
    160
    		 100
    0
    		   100
    5
    		 180 660 90 570 190 118.75 75   193.75
    6
    		 90 750 240 510 170 106.25 200   306.25
    7
    		 45 795 480 315 105 65.625 400
    0
    		 465.625
    8
    0
    		 795 660 135 45 28.125 475 56.25 559.375
    9
    0
    		 795 750 45
    15
    		 9.375 425 150 584.375
    10
    0
    		 795 795
    0
    0
    0
    		 262.5 300 562.5
    11
    		   795 795
    0
    0
    0
    		 112.5 356.25 468.75
    12
    		   795 795
    0
    0
    0
    		 37.5 318.75 356.25
    13
    		   795 795
    0
    0
    0
    0
    		 196.875 196.875
    14
    		   795 795
    0
    0
    0
    0
    		 84.375 84.375
        795 795      
    0
    		 28.125 28.125
        795 795        
    0
    0


     


     
  2. Describe the procedure to derive UH's from concurrent observations of precipitation and streamflow.


    3. A) Separate baseflow from total streamflow hydrograph to obtain the Direct Runoff Hydrograph.

    B) Determine the total volume of Direct Runoff.

    C) Normalize the DRH hydrograph to obtain a Unit Hydrograph.

    D) Determine the duration of the effective rainfall hyetograph, D.

  3. Define lag-time, tl, and time base, tb of UH's.


    4. Time Lag: time interval between the centroid of the ERH and the centroid of the DRH. In some synthetic unit hydrograph procedures this definition is modified as the time interval between the centroid of the ERH and the peak of the DRH.

    Time Base: is the duration of the Direct Runoff associated with the Unit Hydrograph.

  4. Define Synthetic Unit Hydrographs; Define and describe the synthetic unit hydrograph procedures presented in class.
Synthetic Unit Hydrographs: when concurrent observations of precipitation and streamflow are not available so that a UH can not be derived, synthetic UH can be developed. These are UH's which relate the main characteristics of a UH, namely the peak flow, the time to peak, and the time base to geomorphological characteristics of the basin. In class, we presented several Synthetic UH procedures: the Nash model or cascade of linear reservoirs model, the Geomorphologic Instantaneous Unit Hydrograph, the Snyder's synthetic unit hydrograph, and the Soil Conservation Service Dimensionless Unit Hydrograph.

Problem 3 (30 points) In defining the design specifications for a flood warning system, insurance company investigators are trying to determine the magnitude of a flood that destroyed a bridge located 150 km downstream of a reservoir. If the flood wave behaves as a kinematic wave and the observed travel time of the flood wave between the reservoir and the bridge was 4.35 h, estimate the magnitude of the flood that washed away the bridge. Assume a wide rectangular channel with a width of 150 m, slope of 0.01, and Manning roughness of 0.040.

Kinematic Wave Celerity for a wide rectangular channel:

(1)

For the conditions of our problem, the wave traveled 150 km in 4.35 h, thus,

ck = L/Dt = 150000 m/(4.35 x 3600)s = 9.578 m/s

Substituting ck = 9.578 m/s for ck in Eq (1) and solving for y:

(2)

obtain

y = 3.486 m.

Finally, the magnitude of the discharge associated with this wave is,

Q = VA = (3/5) (9.578 m/s) (150 m) (3.486 m) = 3005. 0 m3/s

b) When is the kinematic wave an appropriate distributed flow routing solution?

The kinematic wave is an appropriate distributed flow routing solution when the local and convective acceleration terms (inertial terms), the pressure force term, and the lateral inflow term of the momentum equation are negligible when compared to the gravitational and frictional terms; or when the kinematic parameter is large (i.e., greater that 20).

Kinematic waves do not attenuate; kinematic wave conditions will lead to steepening of the rising limbs of hydrographs, and flattening of the recession limbs of hydrographs.