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1000 Hydraulics Chapter - Ministry of Transportation

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BC MoT<br />

SUPPLEMENT TO TAC GEOMETRIC DESIGN GUIDE<br />

MoT Section 1020 TAC Section 2.1.2<br />

Step 5 - Determine Rainfall Intensity<br />

The nearest rainfall gauging station is located at Comox<br />

Airport (El. 24 m). Since the basin elevation varies from<br />

El. 20 m to El. 760 m, a precipitation gradient is<br />

expected. The 10-year rainfall intensity corresponding to<br />

the time <strong>of</strong> concentration will be used due to the increased<br />

reliability <strong>of</strong> rainfall data over more frequent return<br />

periods (e.g. 2-year). A previous hydrological study<br />

estimated the average intensity over the basin will<br />

increase at a rate <strong>of</strong> 5% per 100 m rise in elevation.<br />

i = (9mm / hr)(( 740m )( 05 . )( 005 . ) + 1) = 107 . mm / hr<br />

100m<br />

Step 6 - Determine Design Flow<br />

There are numerous ways <strong>of</strong> estimating the design flow.<br />

A few different methods will be used.<br />

Method 1 - Rational Formula<br />

Since the basin is small and there is limited data, the<br />

Rational Formula will be used to determine the 10-year<br />

flow. The 10-year flow will then be converted to a 200-<br />

year flow. Studies have shown that the Q 200 /Q 10 ratio is<br />

approximately 1.7 for this region.<br />

Method 2 – SCS Peak Flow Method<br />

For this creek, a 24-hour, Type 1A rainfall distribution<br />

will be used in the analysis. The 10-year 24 hour total<br />

rainfall volume will be obtained from the Comox Airport<br />

IDF curve. The estimated 10-year flow will be converted<br />

to a 200-year flow.<br />

Total Rainfall = (3.2mm / hr)(24hr) = 76.8mm<br />

Q 10 = 7.4m 3 / s<br />

3 3<br />

Q 200 = 1.7Q 10 = 1.7(7.4m / s) = 12.6m / s<br />

Method 3 - Regional Frequency Analysis<br />

Hydrological studies have resulted in regional frequency<br />

curves for the area.<br />

Q<br />

200<br />

= (A)(unit run<strong>of</strong>f)(peaking factor)<br />

2 3 2 3<br />

= (6.6km )( 18 . m / skm )( 15 . ) = 17.8m / s<br />

Since the results do not vary significantly, an “average”<br />

will be taken. The 200-year flow is estimated to be 15<br />

m 3 /s.<br />

Q = CiA ( 040)(107 . . mm / hr)(660ha)<br />

10 =<br />

360<br />

360<br />

3<br />

= 7.9m<br />

/ s<br />

3 3<br />

Q 200 = 1.7Q 10 = 1.7(7.9m / s) = 13.4m / s<br />

June, 2007 Page 1020-7

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