Example Problem

For an example, we will determine the 25-year 24 hour discharge for Swedetown Creek in Hancock, Michigan. For the purposes of this example, we will assume the following:

With this information and the topographic map of the region, it should be easy to find the maximum discharge for a 25-year 24 hour storm.

Estimate Runoff

Since the basin has Type B soil with thin-stand forest, we can use Table 1 to find a CN value of 66. Using Equation 1, the potential abstraction can be calculated:

S = 1,000/66 - 10 = 5.15 in

Now, the total runoff can be calculated with Equation 2. Remember Ia = 0.2 S = 1.03 in.

Q = (3.4 in - 1.03 in)2 * [(3.4 in + 1.03 in) + 5.15 in]-1 = 0.59 in

Time of Concentration

The routes taken to calculate the time of concentration are shown in this link. Frame 1 is for sheet flow, Frame 2 is for shallow concentrated flow, and Frame 3 is for channel flow. For this example, only one path is shown. In practice, multiple pathways should be chosen, with the longest being used as the time of concentration.

We can assume a sheet flow length of 100 ft. The topographical map shows that slope of the sheet flow segment has a slope of 0.05. From Table 2, the Manning's roughness coefficients for sheet flow is 0.4 for woods with light underbrush. This should be enough to use Equation 3.

Tt = 0.007 (0.4 * 100 ft)0.8 * (2.3 in0.5 * 0.050.4)-1 = 0.29 hr

The shallow concentrated flow time of concentration can be calculated by Equation 4 .

Tsc = 4,224 ft * (58,084.2 * 0.0470.5)-1 = 0.34 hr

Equation 6 requires a Manning's coefficient for the stream channel, which can be found in Table 3. For Equation 5, we'll assume a rectangular channel with a width of 5 ft and a depth of 2 ft. Thus, R = 1.11 ft.

Tch = 0.050 * 10,560 ft * (5,364 * 1.112/3 * 0.03031/2)-1 = 0.53 hr

Combine the results from the time computed above, and the resultant time of concentration can be found in Equation 6:

Tc = 0.29 hr + 0.34 hr + 0.53 hr = 1.16 hr

Graphical Discharge Method

Since we know that the basin has a SCS Type II rainfall distribution, and Ia/P = 1.03 in / 3.4 in = 0.30, we can find the coefficients for Equation 7 from Table 4.

log(qu) = 2.46732 + -0.62257 * log (1.16 hr) + -0.11657 * [log (1.16 hr)]2
qu = 267 ft2/s

From the topographic map, the total percentage of area that is covered by swamps and ponds is about 1%. From Table 5, that gives a Fp of about 0.87. We now have all the variables to complete Equation 8.

qp = 267 ft2/s * 0.87 * 0.59 in * 3.2 mi2 = 439 ft3/s

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