1000 Hydraulics Chapter - Ministry of Transportation
1000 Hydraulics Chapter - Ministry of Transportation
1000 Hydraulics Chapter - Ministry of Transportation
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SUPPLEMENT TO TAC GEOMETRIC DESIGN GUIDE<br />
MoT Section 1020 TAC Section Not Applicable<br />
BC MoT<br />
1020.07 HYDROLOGY EXAMPLE<br />
Background<br />
Thames Creek is located on the east<br />
side <strong>of</strong> Vancouver Island near<br />
Denman Island.<br />
Problem<br />
Since the highway crosses Thames<br />
Creek, a bridge or culvert will be<br />
required. Estimate the 200-year<br />
(Q 200 ) flow.<br />
Solution<br />
Step 1 - Determine Basin Size and Creek Length<br />
From the 1:50000 scale mapping, the following<br />
dimensions were measured:<br />
A = 6.6 km 2 = 660 ha<br />
L = 8.2 km<br />
Step 2 - Determine Basin Slope<br />
A pr<strong>of</strong>ile <strong>of</strong> the main channel was plotted. Since the<br />
upper portion <strong>of</strong> basin is steep, the basin slope was<br />
estimated using the Equivalent Slope Method.<br />
s = 0.051 m/m = 5.1%<br />
Step 3 - Determine Land Characteristics<br />
Design flows are estimated assuming worst case<br />
conditions. Considerations include basin slope, type <strong>of</strong><br />
vegetation, recurrence intervals, snowmelt, antecedent<br />
moisture condition (AMC) etc. Since the Thames Creek<br />
basin is relatively low with light forest cover, the<br />
following land characteristic values were selected:<br />
r = 0.60, deciduous timber land<br />
CN = 85, forest land with good cover, Hydrologic<br />
Soil Group C, AMC III<br />
C = 0.40, flat, forested<br />
Step 4 - Determine Time <strong>of</strong> Concentration<br />
There are numerous ways <strong>of</strong> estimating the time <strong>of</strong><br />
concentration (t c ). A few different methods will be used<br />
and an “average” value will be selected.<br />
Method 1 - BC Rational Formula Method<br />
A = 66 . km = 2.6km<br />
t c = 3.6hr (interpolated)<br />
2<br />
Method 2 - Hathaway Formula<br />
t = (rL) 0.467 0.467<br />
(( 0. 60)(8.<br />
2km))<br />
c<br />
= = 2.6hr<br />
0234 .<br />
0.234<br />
165 . s 1. 65( 0.<br />
051m/m)<br />
Method 3 - SCS Curve Number Method<br />
S = 254( 100<br />
CN − 1) = 254( 100 − 1)<br />
= 44.8<br />
CN<br />
85<br />
0.8<br />
0.7<br />
T = L ( 0.<br />
039SCN<br />
+1)<br />
L<br />
0.5<br />
735 s<br />
0.8<br />
( 8200m)<br />
( 0. 039( 44.8) + 1)<br />
=<br />
05 .<br />
735(.<br />
51%)<br />
t c = 1.7T L = 1.7(1.7hr) = 2.8hr<br />
07 .<br />
= 17 . hr<br />
Method 4 - Bransby Williams Formula<br />
t = 0.605L 0605 . ( 82 . km)<br />
c = = 30 . hr<br />
0.2 0.1 0.2 2 01 .<br />
s A (5.1%) ( 66 . km )<br />
Taking an “average”, it is assumed that t c =3 hours.<br />
Page 1020-6 June, 2007
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