CIV E 445 - APPLIED HYDROLOGY
SPRING 2009
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 = 55 mm/h • [ (0.3 • 150 • 30/100) + (0.5 • 150 • 40/100) + (0.9 • 150 • 30/100) ] ha • (10,000 m2/ha • 0.001 m/mm) / (3600 s/h) = Qp = 12.83 m3/s.  ANSWER.

Problem 5-2

Several rainfall durations are tried, as shown in the following

tr
(min)		I
(mm/h)		Subarea A
(C = 0.9)
(ha)		Subarea B
(C = 0.3)
(ha)		Σ(CA)Qp
(m3/s)
20147.76651.3374.830.68
30123.9667782.528.4
40107.566102.6790.226.95
5095.566128.3397.925.96
6086.166154105.625.26

The fraction of subarea B contributing to peak runoff increases linearly with rainfall duration.

Therefore: Qp = I Σ(CA), in m3/s.

The 50-y peak runoff is the maximum value, corresponding to a 60-min duration: 30.68 m3/s.  ANSWER.

Problem 5-3

Using Eq. 4-19, the equilibrium outflow is:

qe = iL/3600 = (40 mm/h • 100 m • 0.001 m/mm • 1000 L/m3) / (3600 s/h) =

qe = 1.11 L/s/m = 1.11 • 10-3 m3/s/m = 0.00111 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 • 0.015) / (3 • 1.0 • 10-6 m2/s) = 49,050 m-1s-1.

In Eq. 4-25, for laminar flow, b = CL, and m = 3. Therefore:

he = (qe / CL)1/3 = (0.00111/49,050)1/3 = 0.00283 m = 2.83 mm.  ANSWER.

For T = 30°C, ν = 0.801 • 10-6 m2/s.

Using Eq. 4-27: CL = 61,236 m-1s-1.

Therefore, with Eq. 4-25: he = 0.00263 m = 2.63 mm.  ANSWER.

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.