The unit peak discharge is computed with Equation 5.6 by interpolating c 0, c, and c Table 5.5 using a type II distribution. The peak discharge is also calculated as follows. from Variable SI Unit U Unit.5444 0.6587 log ( 0.6 ) - 0.598 [ log ( 0.6 )] q u = 0 = (0.00043) 0.85 = () 0.85 = 0.305 m 3 /s/km /mm = 708 ft 3 /s/mi /in q p = q u Q = 0.305 (0.76 km )(6 mm) = 3.3 m 3 /s = 708 (0.068 mi ) (.46 in) = 0 ft 3 /s 5.3 RTIONL METHOD One of the most commonly used equations for the calculation of peak discharges from small areas is the rational formula. The rational formula is given as: Q = i (5.8) Q = the peak flow, m³/s (ft 3 /s) i = the rainfall intensity for the design storm, mm/h (in/h) = the drainage area, ha (acres) = dimensionless runoff coefficient assumed to be a function of the cover of the watershed and often the frequency of the flood being estimated = unit conversion constant equal to 360 in SI units and in U units. 5.3. ssumptions The assumptions in the rational formula are as follows:. The drainage area should be smaller than 80 hectares (00 acres).. The peak discharge occurs when the entire watershed is contributing. 3. storm that has a duration equal to t c produces the highest peak discharge for this frequency. 4. The rainfall intensity is uniform over a storm time duration equal to the time of concentration, t c. The time of concentration is the time required for water to travel from the hydrologically most remote point of the basin to the outlet or point of interest. 5. The frequency of the computed peak flow is equal to the frequency of the rainfall intensity. In other words, the 0-year rainfall intensity, i, is assumed to produce the 0-year peak discharge. 5.3. Estimating Input Requirements The runoff coefficient,, is a function of ground cover. Some tables of provide for variation due to slope, soil, and the return period of the design discharge. ctually, is a volumetric coefficient that relates the peak discharge to the "theoretical peak" or 00 percent runoff, occurring when runoff matches the net rain rate. Hence is also a function of infiltration and other hydrologic abstractions. Some typical values of for the rational formula are given in 5-3
Table 5.7. Should the basin contain varying amounts of different covers, a weighted runoff coefficient for the entire basin can be determined as: i Weighted = i i = runoff coefficient for cover type i that covers area i = total area. (5.9) 5.3.3 heck for ritical Design ondition When the rational method is used to design multiple drainage elements (i.e. inlets and pipes), the design process proceeds from upstream to downstream. For each design element, a time of concentration is computed, the corresponding intensity determined, and the peak flow computed. For pipes that drain multiple flow paths, the longest time of concentration from all of the contributing areas must be determined. If upstream pipes exist, the travel times in these pipes must also be included in the calculation of time of concentration. In most cases, especially as computations proceed downstream, the contributing area with the longer time of concentration also contributes the greatest flow. Taking the case of two contributing areas, as shown in Figure 5.3a, the longest time of concentration of the two areas is used to determine the time of concentration for the combined area. When the rainfall intensity corresponding to this time of concentration is applied to the rational equation, as shown below, for the combined area and runoff coefficient, the appropriate design discharge, Q, results. Q = ( + (5.30) ) i However, it may be possible for the larger contributing flows to be generated from the contributing area with a shorter time of concentration. If this occurs, it is also possible that, if the longer time of concentration is applied to the combined drainage area, the resulting design flow would be an underestimate. Therefore, a check for a critical design condition must be made. t Q = ( + Q = design check discharge t = time of concentration for area t = time of concentration for area. ) i t (5.3) 5-3
Table 5.7. Runoff oefficients for Rational Formula (SE, 960) Type of Drainage rea Runoff oefficient Business: Downtown area 0.70-0.95 Neighborhood areas 0.50-0.70 Residential: Single-family areas 0.30-0.50 Multi-units, detached 0.40-0.60 Multi-units, attached 0.60-0.75 Suburban 0.5-0.40 partment dwelling areas 0.50-0.70 Industrial: Light areas 0.50-0.80 Heavy areas 0.60-0.90 Parks, cemeteries 0.0-0.5 Playgrounds 0.0-0.40 Railroad yard areas 0.0-0.40 Unimproved areas 0.0-0.30 Lawns: Sandy soil, flat, < % 0.05-0.0 Sandy soil, average, to 7% 0.0-0.5 Sandy soil, steep, > 7% 0.5-0.0 Heavy soil, flat, < % 0.3-0.7 Heavy soil, average to 7% 0.8-0. Heavy soil, steep, > 7% 0.5-0.35 Streets: sphalt 0.70-0.95 oncrete 0.80-0.95 Brick 0.70-0.85 Drives and walks 0.75-0.85 Roofs 0.75-0.95 5-33
If Q > Q, Q should be used for design; otherwise Q should be used. Equation 5.3 uses the rainfall intensity for the contributing area with the shorter time of concentration (area ) and reduces the contribution of area by the ratio of the times of concentration. This ratio approximates the fraction of the area that would contribute within the shorter duration. This is equivalent to reducing the contributing area as shown by the dashed line in Figure 5.4. Highway t t = 0 min t c = 5 min t c = 0 min B t t = min t c = 0 min Drainage area boundary Surface flow path Pipe To outfall Figure 5.4. Storm drain system schematic Example 5.5. flooding problem exists along a farm road near Memphis, Tennessee. lowwater crossing is to be replaced by a culvert installation to improve road safety during rainstorms. The drainage area above the crossing is 43.7 ha (08 acres). The return period of the design storm is to be 5 years as determined by local authorities. The engineer must determine the maximum discharge that the culvert must pass for the indicated design storm. The current land use consists of.8 ha (53.9 acres) of parkland,.5 ha (3.7 acres) of commercial property that is 00 percent impervious, and 0.4 ha (50.4 acres) of single-family residential housing. The principal flow path includes 90 m (95 ft) of short grass at percent slope, 300 m (985 ft) of grassed waterway at percent slope, and 650 m (,30 ft) of grassed waterway at percent slope. The following steps are used to compute the peak discharge with the rational method:. ompute a Weighted Runoff oefficient: The tabular summary below uses runoff coefficients from Table 5.7. The average value is used for the parkland and the residential areas, but the highest value is used for the commercial property because it is completely impervious. 5-34
Description Value SI Unit U Unit rea (ha) i i rea (acres) i i Park 0.0.8 4.36 53.9 0.8 ommercial (00% 0.95.5.43 3.7 3.5 impervious) Single-family 0.40 0.4 8.6 50.4 0. Total 43.7 3.95 08.0 34.5 Equation 5.9 is used to compute the weighted : Variable Value in SI Value in U i Weighted = i 3.95 34.5 = = 0.3 = = 0.3 43.7 08.0. Intensity: The 5-year intensity is taken from an intensity-duration-frequency curve for Memphis. To obtain the intensity, the time of concentration, t c, must first be estimated. In this example, the velocity method for t c is used to compute t c : SI Unit U Unit Flow Path Slope(%) Velocity Velocity Length (m) (m/s) Length (ft) (ft/s) Overland (Short grass) 90 0.30 95.0 Grassed waterway 300 0.64 985. Grassed waterway 650 0.46,30.5 The time of concentration is estimated as: T c Variable Value in SI Value in U L 90 m 300 m 650 m 95 ft 985 ft,30 ft = = + + = + + 0.3 m/s 0.64 m/s 0.46 m/s.0 ft/s.ft/s.5 ft/s V =,80 s = 36 min =,80 s = 36 min The intensity is obtained from the IDF curve for the locality using a storm duration equal to the time of concentration: i = 85 mm/h (3.35in/h) 3. rea (): Total area of drainage basin, = 43.7 ha (08 acres) 4. Peak Discharge (Q): Variable Value in SI Value in U ( 0.3 )( 85 )( 43.7 ) 3 ( 0.3 )( 3.35 )( 08 ) 3 Q = i = = 3.3 m /s = = 6 ft /s 360 5-35