Module 9 Storm Sewer Design PDF

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Iloilo Science and Technology University

EC Lazarte

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storm sewer design hydrologic principles rational method engineering

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This document details the design principles for storm sewers, covering topics like hydrologic principles, the rational method for predicting peak flows, and calculations for determining pipe diameters. It includes various examples and tables.

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MODULE 9 STORM SEWER DESIGN Hydrologic Principles c l Rational Method Hydrology The Rational Method is an intensity based rain...

MODULE 9 STORM SEWER DESIGN Hydrologic Principles c l Rational Method Hydrology The Rational Method is an intensity based rainfall prediction method, meaning that it can be used to predict peak flows based on the characteristics of the watershed and a rainfall event. e Storm Sewer – design to carry flow from a rainfall Q = CiA event away from areas where it is unwanted such as parking lots and Where: roadways. Q = is the design flow rate (ft3/sec or m3/sec) C = is the weighted rational coefficient for the Watershed – it is an area that drains to a common drainage area point of discharge. i = is the rainfall intensity (in/hr or mm/hr) l l A = is the drainage area (sq. meters, sq. ft, acres or Time of Concentration – it is the amount of time hectares) that it takes for the first drop of 1 acre = 43, 560 sq. ft. water that land at any watershed to 1 hectare = 10, 000 sq. m. work its way to the discharge point. c c The Rational Coefficient, C Basis of calculating time of concentration 1. Type of ground cover The rational coefficient, C, is the parameter that is 2. Slope of the land most open to engineering judgment. It is a value 3. Distance along the flow path from 0 to 1 that relates the rate of rainfall over a watershed to the ratio of discharge from that e e Minimum time of concentration recommended watershed. The coefficient is highly dependent on 5 to 10 minutes (small watershed, like a the land use and slope. section of a parking lot draining to a storm sewer) Area Type Description C Values Business Downtown 0.70-0.95 Neighborhood 0.50-0.70 Return Period – it represents, on average, the Residential Single Family 0.30-0.50 length of time between rainfall Multiunit Detached 0.40-0.60 l events of a specific magnitude. Multiunit Attached 0.60-0.75 Suburban Resident 0.25-0.40 Frequency – it is a measure of how often a specific Apartment 0.50-0.70 1.2 acre lots or more 0.30-0.45 rainfall event will be equaled or Lawns, Heavy Flat 2% 0.13-0.17 exceeded. Soils Average, 2-7% 0.18-0.22 c Steep> 7% 0.25-0.35 Rainfall intensity – it is a measure of the rate of Industrial Light 0.50-0.80 rainfall, usually measured in Heavy 0.60-0.90 inches/hr or mm/hr. The higher the Lawns, Sandy Flat, 2% 0.05-0.10 Soils Average, 2-7% 0.10-0.15 intensity, the “harder” it is raining. Steep> 7% 0.15-0.20 e Pavement Asphalt/Concrete 0.70-0.95 Duration – it is indicative of how long the rainfall Brick 0.70-0.85 will fall with that intensity. Drives and Walks 0.75-0.85 Miscellaneous Parks and Cemeteries 0.10-0.25 Playgrounds 0.20-0.40 Railroads Yards 0.10-0.30 Unimproved 0.20-0.40 EC Lazarte 29 Sample Problem 1. A single watershed contains 0.50 acres of pavement, 0.25 acres of steep lawn, and 0.90 acres of trees and bushes. Calculate the weighted CA value for this watershed. l Talbot’s Formula (Rainfall Intensity) c 1. for heaviest storm ί = 360 / (t + 30) 2. for ordinary storm e ί = 105 / (t + 30) Use C = 0.90 for pavement C = 0.30 for steep lawns with heavy soils where: ί = rainfall intensity (in/hr) C = 0.20 for unimproved areas with tress and t = time of concentration (min) brushes = 15 minutes for conservative design Solution: Rational Method l l Total CA = 0.50 acres (0.90) + 0.25 acres (0.30) + 0.90 acres (0.20) Q=CίA Total CA = 0.70 acres where: Q = storm-water runoff rate (m3/sec) C = coefficient of runoff c c ί = rainfall intensity (m/hr) Manning’s Formula A = area to be drained (m2) V = 1/n R2/3S1/2 Conversion Factors e e Cross-sectional Area = пD2/4 1 m = 39.37 in Wetted Perimeter = пD 1 in = 25.40 mm 1 m = 3.28 ft where: V = flow velocity (m/sec) 1 hectare = 10,000 m2 n = pipe roughness = 0.013 (most frequently used value) For Urban Drainage System R = pipe hydraulic radius = Cross-sectional Area / Wetted max. manhole interval = 50 m (for circular l Perimeter conduits) R = 0.25 D S = pipe slope min. invert slope = 0.005 D = pipe diameter min. velocity to avoid silting = 3 m/sec maintenance hole interval = 100 m to 150 m c Chezy Formula Q = AV e where: Q = discharge (m3/sec) A = area of flow = п D2 / 4 V = flow velocity (m/sec) EC Lazarte 30 Sample Problem: 2. Design the drainage system of the given area shown in the figure using the following data and specifications indicated on the plan. c l 18 m e 9m Elev= 62.000 m 9m Q1 9m 9m 9m P1 Q2 9m Elev= 61.999 m l l R O A D P2 Q3 18 m c c R O A D R O A D P3 e e 18 m Q4 12 m 12 m 12 m R O A D P4 LEGEND: SPECIFICAIONS: Elev= 61.998 m l – manhole Road Width = 8 meters – outlet C = 0.32 – pipe n = 0.013 t = 15 min c Solution: Rainfall Intensity (use heaviest storm) Pipe Slope ί = 360 / (t + 30) For P1 e ί = 360 = 8 in/hr x 1m x 1hr S = 62.000 − 61.999 = 18.52x10-6 15 + 30 39.37in 3600 sec 9(6) ί = 56.44x10-6 m/sec For P2, P3, and P4 S = 61.999 − 61.998 = 19.23x10-6 8(2) + 18(2) EC Lazarte 31 Discharge Q= CiA For Q1 A = 3 [9x18] = 486 m2 Q = AV l Pipe Diameter c where: A = п D2 / 4 V = 1/n R2/3S1/2 e Q1 = 0.32 (56.44x10-6 m/sec) (486) Q1 = 8.78x10-3 m3/sec For P1 (use Q1): 8.78x10-3 m3/sec = D x 2 1 -6 1/2 For Q2 (0.25D)2 / 3 (18.52x10 ) 4 0.013 A = 3 [9x18] = 486 m2 Q2 = 0.32 (56.44x10-6 m/sec) (486) = 8.78x10-3 = 23.98 D8/3 (18.52x10-6)1/2 Q2 = 8.78x10-3 m3/sec D1 = 0.39690 m D1 = 396.90 mm say 400mm l l For Q3 A = 3 [12x18] = 648 m2 For P2 (use Q1 and Q2): Q3 = 0.32 (56.44x10-6 m/sec) (648) 8.78x10-3 + 8.78x10-3 = 23.98 D8/3 (19.23x10-6)1/2 Q3 = 11.70x10-3 m3/sec D2 =0.51110 m D2 = 511.10 mm say 525mm c c For Q4 A = 3 [12x18] = 648 m2 For P3 (use Q1, Q2, and Q3): Q4 = 0.32 (56.44x10-6 m/sec) (648) 8.78x10-3 + 8.78x10-3 + 11.70x10-3 = 23.98 D8/3 (19.23x10-6)1/2 Q4 = 11.70x10-3 m3/sec D3 = 0.61896 m D3 = 618.96 mm say 625 mm e e For P4 (use Q1, Q2, Q3, and Q4): 8.78x10-3 + 8.78x10-3 + 11.70x10-3 + 11.70x10-3 = 23.98 D8/3 (19.23x10-6)1/2 D4 = 0.70218 m D4 = 702.18 mm say 725 mm l Tabulated Results Pipe No: Total Discharge Slope Computed Diameter Final Diameter P1 8.78x10-3 m3/sec 18.52x10-6 396.90 mm 400 mm P2 17.56x10-3 19.23x10-6 511.10 525 c P3 29.26x10-3 19.23x10-6 618.96 625 P4 40.96x10-3 19.23x10-6 702.18 725 EC Lazarte e 32

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