SPRING 2011
SOLUTION TO HOMEWORK 5, CHAPTER 4
Problem 5-1
Since rainfall duration is greater than time of concentration,
the flow is superconcentrated and the entire catchment is
contributing. For subcatchments with different runoff
coefficients,
use a weighted formula for peak
runoff (see Eq. 4-14): Qp = I Σ(CA) =
Qp =
45 mm/h • [ (0.3 • 80 • 25/100) + (0.5 • 80 • 35/100) +
(0.7 • 80 • 40/100) ] ha • (10,000 m2/ha • 0.001 m/mm) / (3600 s/h) =
Problem 5-2
Several rainfall durations are tried, as shown in the following table.
The fraction of subarea B contributing to peak runoff increases linearly with rainfall duration.
Therefore: Qp = (10/3600) I Σ(CA) = 0.00278 I Σ(CA).
The 25-y peak runoff is the maximum value, corresponding to a 10-min duration:
Problem 5-3
Using Eq. 4-19, the equilibrium outflow is:
qe = iL/3600 = 75 mm/h • 144 m / 3600 = 3 L/s/m
qe = 3 L/s/m = 3 • 10-3 m3/s/m = 0.003 m2/s.
For T = 20°C, ν = 1.0 • 10-6 m2/s (Table A-1).
Using Eq. 4-27: CL = (9.81 m/s2 • 0.01) / (3 • 1.0 • 10-6 m2/s) = 32,700 m-1s-1.
In Eq. 4-25, for laminar flow, b = CL, and m = 3. Therefore:
For T = 25°C, ν = 0.893 • 10-6 m2/s. Using Eq. 4-27: CL = 36,618 m-1s-1. Therefore, with Eq. 4-25:
Problem 5-4
The rainfall excess in m/s is:
i = (25 mm/h • 0.001 m/mm) / (3600 s/h) = 6.94 •10-6 m/s.
qe = 6.94 •10-6 m/s • 80 m = 0.0005555 m3/s/m = 0.5555 L/s/m.
For 75% turbulent flow, m = 2. Therefore, in Eq. 4-29:
te = [ 2 • (0.05 • 80)1/2 ] / [(6.94 •10-6)1/2 • 0.011/4 ] = 4800 s.
Using Eq. 4-36, the rising limb of the overland flow hydrograph is calculated as shown in the following table. ANSWER.
tr
(min)I
(mm/h)Subarea A
(C = 0.7)
(ha)Subarea B
(C = 0.3)
(ha)Σ(CA)
(ha)Qp
(m3/s)
10 141.73 30 11.67 24.5 9.65
20 120.35 30 23.33 28.0 9.36
30 105.42 30 35 31.5 9.22
40 94.31 30 46.67 35.0 9.17
50 85.67 30 58.33 38.5 9.16
60 78.74 30 70 42.0 9.19
