Drainage Design Manual, Hydrology - Flood Control District of ...
Drainage Design Manual, Hydrology - Flood Control District of ... Drainage Design Manual, Hydrology - Flood Control District of ...
Drainage Design Manual for Maricopa County Hydrology: Application 4. Compute the Time of Concentration (T c ) and Intensity (i) for each subbasin using Equation (3.2) and the data from Table 9.5 based on the 100 year event. Use a log interpolation to compute i. Subbasin S1: T c 11.4L 0.5 0.52 – 0.31 = K b S i – 0.38 c = 11.4( 0.729) 0.5 ( 0.096) 0.52 ( 473.0) 0.31 T c = 0.426i – 0.38 Start with an initial estimate for T c of 15 minutes. – i – 0.38 - From Table 9.5, i = 5.26 inches/hour for a 15-minute duration. T c = 0.426( 5.26) – 0.38 = 0.227 hours = 13.6 min Recompute i for T c of 13.6 minutes. - From Table 9.5, i = 5.26 inches/hour for 15 minutes, and i = 6.37 inches/hour for 10-minutes ( ( ( 13.6 – 10) ⁄ ( 15 – 10 )) ( log 105.26 – log 10 6.37) + log 10 6.37) i = 10 = 5.55 inches/hour T c = 0.426( 5.55) – 0.38 = 0.222 hours = 13.3 min Recompute i for T c of 13.3 minutes. ( ( ( 13.3 – 10) ⁄ ( 15 – 10 )) ( log 105.26 – log 10 6.37) + log 10 6.37) i = 10 = 5.61 inches/hour T c = 0.426( 5.61) – 0.38 = 0.221 hours = 13.3 min Recompute i for T c of 13.3 minutes. ( ( ( 13.3 – 10) ⁄ ( 15 – 10 )) ( log 105.26 – log 10 6.37) + log 10 6.37) i = 10 = 5.61 inches/hour T c = 0.426( 5.61) – 0.38 = 0.221 hours = 13.3 min Difference is less than 2%. Round T c to nearest minute and recomput i. Use T c = 13 min, and i = 5.68 inches/hour 9-20 August 15, 2013
Drainage Design Manual for Maricopa County Hydrology: Application NOTE: There may be slight differences in results when DDMSW is used to perform these calculations due to numerical rounding. Using the above procedure, the T c and i for each subbasin are: Subbasin T c , min i, in/hr S1 13 5.68 S2 10 6.37 S3 12 5.90 S4 10 6.37 5. Compute the peak discharge for each subbasin using Equation (3.1): Subbasin S1: Subbasin S2: Subbasin S3: Subbasin S4: Q = ( 0.66) ( 5.68) ( 65.99) = 247 cfs Q = ( 0.50) ( 6.37) ( 12.60) = 40 cfs Q = ( 0.69) ( 5.90) ( 21.18) = 86 cfs Q = ( 0.65) ( 6.37) ( 27.80) = 115 cfs 6. Compute the peak discharge for concentration point C1 using the Combined Watershed Method. The combined area of subbasins S1 and S2 is 78.59 acres. The area-weighted C coefficient is: C1: ( 0.66) ( 65.99) + ( 0.50) ( 12.60) C w = ------------------------------------------------------------------------- = ( 78.59) 0.63 The area-weighted K b is: C1: = ------------------------------------------------------------------------------- ( 0.096) ( 65.99) + ( 0.065) ( 12.60) = ( 78.59) 0.091 Use the length, L, and slope, S, from subbasin S1 since both subbasins S1 and S2 join at C1 and subbasin S1 has the longer T c flow path. C1: L = 0.729 miles C1: S = 473.0 feet/mile August 15, 2013 9-21
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<strong>Drainage</strong> <strong>Design</strong> <strong>Manual</strong> for Maricopa County<br />
<strong>Hydrology</strong>: Application<br />
4. Compute the Time <strong>of</strong> Concentration (T c ) and Intensity (i) for each subbasin using Equation<br />
(3.2) and the data from Table 9.5 based on the 100 year event. Use a log interpolation to compute<br />
i.<br />
Subbasin S1:<br />
T c<br />
11.4L 0.5 0.52 – 0.31<br />
= K b S<br />
i – 0.38<br />
c<br />
= 11.4( 0.729) 0.5 ( 0.096) 0.52 ( 473.0) 0.31<br />
T c<br />
= 0.426i – 0.38<br />
Start with an initial estimate for T c <strong>of</strong> 15 minutes.<br />
– i – 0.38<br />
- From Table 9.5, i = 5.26 inches/hour for a 15-minute duration.<br />
T c<br />
= 0.426( 5.26) – 0.38 = 0.227 hours = 13.6 min<br />
Recompute i for T c <strong>of</strong> 13.6 minutes.<br />
- From Table 9.5, i = 5.26 inches/hour for 15 minutes, and i = 6.37 inches/hour for 10-minutes<br />
( ( ( 13.6 – 10) ⁄ ( 15 – 10 ))<br />
( log 105.26 – log 10 6.37) + log 10 6.37)<br />
i = 10 =<br />
5.55 inches/hour<br />
T c<br />
= 0.426( 5.55) – 0.38 = 0.222 hours = 13.3 min<br />
Recompute i for T c <strong>of</strong> 13.3 minutes.<br />
( ( ( 13.3 – 10) ⁄ ( 15 – 10 ))<br />
( log 105.26 – log 10 6.37) + log 10 6.37)<br />
i = 10 =<br />
5.61 inches/hour<br />
T c<br />
= 0.426( 5.61) – 0.38 = 0.221 hours = 13.3 min<br />
Recompute i for T c <strong>of</strong> 13.3 minutes.<br />
( ( ( 13.3 – 10) ⁄ ( 15 – 10 ))<br />
( log 105.26 – log 10 6.37) + log 10 6.37)<br />
i = 10 =<br />
5.61 inches/hour<br />
T c<br />
= 0.426( 5.61) – 0.38 = 0.221 hours = 13.3 min<br />
Difference is less than 2%. Round T c to nearest minute and recomput i. Use T c = 13 min, and<br />
i = 5.68 inches/hour<br />
9-20 August 15, 2013