Module 5: Channel and Slope Protection Example Assignments A) Example Project Assignment on Slope and Swale Design North America Green Software Example (Erosion Control Materials Design Software) The following is excerpted from a homework assignment prepared by Heather Hill, a student at the University of Alabama at Birmingham, as part of the Construction Site Erosion and Sediment Control Class taken during the summer of 2005. The assignment was given as follows: 1) Identify several different slope categories on your construction evaluation site and propose suitable control practices for each type. Justify your selections with appropriate calculations. The following is an example output screen from the North American Green software to assist in the selection of turf reinforcement mats for slopes. NAG SC150 Temporary Cover 100 slope 2) Design an appropriate diversion swale for your evaluation site. Using the previously calculated flow rates, select a suitable channel lining, including the consideration of check dams. Justify your selections with appropriate calculations. 1
There is not a diversion swale at this construction site. The creek that runs through the site will actually be rereouted through a 15 culvert pipe and covered to level out the site. In order to reroute the stream and install the culvert, an impermeable diversion dam was installed where the stream entered the site. A bypass pump was set up at this location to pump the water to the end of the site where it naturally releases. The average daily flow of this stream is approximately 2600 gpm. The following is an example of applying the North American Green software to evaluate a channel lining material to this channel, assuming that an open channel was an optional method. B) Example Project Assignment on Slope and Swale Design Example Hand Calculations The following is modified from a homework assignment prepared by Regan Johnson, a student at the University of Alabama, as part of the Construction Site Erosion and Sediment Control Class taken during the summer of 2005. 1) Identify several different slope categories on your construction evaluation site and propose suitable control practices for each type. Justify your selections with appropriate calculations. The construction site described here is located along Kicker Road in Tuscaloosa, AL. The City of Tuscaloosa has granted permission to develop St. Charles Place, a townhouse residential development located. The following is the detailed site topographic map donated by the site developer, and a USGS aerial photograph of the surrounding area. 2
This site was separated into the following slope categories: 0-2%, 2-5%, 5-20%, and 20% and greater. One critical slope is examined in the following calculations for slope stability requirements, having the following characteristics: Slope = 0.5 = 50% Width of slope = 175ft Manning s n = 0.02 The modified Manning s equation was used to calculate the nominal depth for sheetflow on this particular slope, assuming a previously calculated peak flow rate of 6.4 ft 3 /sec. Where: y is the flow depth (in feet), q is the unit width flow rate (Q/W) n is the sheet flow roughness coefficient for the slope surface s is the slope (as a fraction) q= Q/W= 6.39/175= 0.0365 cfs n= 0.02 for sandy loam soils s= 0.50 ft/ft y= ((0.0365)(0.02) /(1.49/ 0.50 0.5 )) 3/5 = 0.051 ft, or about a half an inch 3
Therefore, the basic shear stress equation can be used to calculate the maximum shear stress expected on a slope: where: γ = specific weight of water (62.4 lbs/ft 3 ) y = flow depth (ft) S = slope (ft/ft) Thus, τ o = (62.4)(0.051)(0.50)= 1.58 lb/ft 2 Since the allowable shear stress for the soils on this hillside is only 0.15 lb/ft 2, a vegetated mat will therefore be needed. The next step is to check the shear stress under the mat. We can solve for the needed roughness factor n of the mat to find a mat that will work given the following equation: τ e = 0.15 = 1.58(1-0)[0.02/n mat ] 2 n mat = 0.065 Therefore, a mat is needed having an n value of at least 0.065 to provide proper soil protection. Additionally, the mat also needs a C factor to meet the maximum allowable erosion rate on the slope (0.25 inches, or less). Using RUSLE: R = 350/yr (Tuscaloosa) k = 0.21 LS = 12.75 (for 150 ft slope at 50%) The base (unprotected) erosion rate is therefore: (350)(0.21)(12.75) = 643 tons/acre/year This corresponds to 643(0.00595) = 3.8 inches per year. With a maximum allowable erosion loss of 0.25 to 0.5 inches per year, the C factor for the mat should therefore be: 0.5 to 3.8 = 0.13; or 0.25/3.8 = 0.065, or less A NAG P300 mat has a C of 0.09 (intermediate in the above range) and an n of 0.02 for this slope and unvegetated condition. The shear stress is too large, as the mat n is only 0.02, and not the desired 0.065. Therefore, the only reasonable solution for this steep and long slope is to use terraces to divide the slope into several segments, and to use diversion down-slope drains to collect the water from each terrace bench and safely carry it to the bottom of the slope. If the slope was divided into 50 ft lengths, 1/3 of the original slope length, the Q would also be 1/3, or 2.1 ft 3 /sec, and the q would be 0.012 cfs. The resulting flow depth would therefore be: y= ((0.012)(0.02) /(1.49/ 0.50 0.5 )) 3/5 = 0.0125 ft, or 0.15 inch The resulting shear stress is therefore: τ o = (62.4)(0.0125)(0.50)= 0.39 lb/ft 2 4
The needed value for n (unvegetated) is therefore: τ e = 0.15 = 0.39(1-0)[0.02/n mat ] 2 least n mat = 0.032 at The NAG P300 still is not rough enough. If the slope was divided into 25 ft lengths, 1/6 of the original slope length, the Q would also be 1/6, or 1.1 ft 3 /sec, and the q would be 0.006 cfs. The resulting flow depth would therefore be: y= ((0.006)(0.02) /(1.49/ 0.50 0.5 )) 3/5 = 0.0043 ft, or 0.052 inch The resulting shear stress is therefore: τ o = (62.4)(0.0043)(0.50)= 0.13 lb/ft 2 The needed value for n (unvegetated) is therefore: τ e = 0.15 = 0.13(1-0)[0.02/n mat ] 2 least n mat = 0.019 at Therefore, this slope length is suitable, as the n for the mat is 0.02. As an alternative, it may be suitable to re-examine the slope itself and consider reducing it from 50% to 40%, and with terraces at 50 ft spacing: y= ((0.012)(0.02) /(1.49/ 0.40 0.5 )) 3/5 = 0.007 ft, or 0.08 inch The resulting shear stress is therefore: τ o = (62.4)(0.007)(0.40)= 0.17 lb/ft 2 The needed value for n (unvegetated) is therefore: τ e = 0.15 = 0.17(1-0)[0.02/n mat ] 2 least n mat = 0.021 at This is close to the available n of 0.2 and is also a likely a suitable solution. Either of these solutions to modify the slope would also reduce the resulting erosion rate. 2) Design an appropriate diversion swale for your evaluation site. Using the previously calculated flow rates, select a suitable channel lining, including the consideration of check dams. Justify your selections with appropriate calculations. This site consists of one channel that diverts water from the upper portion of the watershed. This channel is located in the back of the site and it will be designed to handle the flow rates that were calculated through an earlier analysis. The cross section of the channel will be a trapezoidal in shape. Other factors such as slope, Manning s n, and soil type also affect the channel s performance as well. There are two important parameters involved when designing a diversion swale (1) Allowable velocity (V 0 ) and (2) Allowable shear stress (τ 0 ). The first step of the design is to determine the applicable values associated with site specific soil conditions. The site soil is sandy loam. The following parameters should therefore be met for this design: 5
Maximum permissible velocity (V 0 ): 2.5 ft/sec Allowable shear stress (τ 0 ): 0.075 lb/ft 2 For this particular swale design, the Manning s equation for open channel flow will be used with the Chow shape factor: Where: Q= 16.3 cfs S= 0.055 ft/ft n= 0.02 It is therefore possible to calculate the nominal depth of channel flow within the swale for different swale cross sections, using an Excel spreadsheet. The spreadsheet allowed an examination of various base widths (b) and side slopes (z) of the channel. The selected alternative for the channel dimension is one with a 3ft base and 2:1 side slope. The resulting shear stress and channel velocity are also shown on this table. Channel Design Option b (ft) z (ft) Top (ft) AR 2/3 b 8/3 AR 2/3 /b 8/3 y n /b (From Chow s figure) y n τ (lb/ft 2 ) R V (ft/s) 3 2 15 8.01 18.72 0.43 0.35 1.05 3.60 0.70 2.50 4 2 20 8.01 40.32 0.20 0.3 1.20 4.12 0.83 1.94 5 2 25 8.01 73.10 0.11 0.27 1.35 4.63 0.96 1.48 6 2 30 8.01 118.87 0.07 0.153 0.92 3.15 0.71 1.91 3 3 21 8.01 18.72 0.43 0.32 0.96 3.29 0.62 2.43 4 3 28 8.01 40.32 0.20 0.27 1.08 3.71 0.73 1.85 5 3 35 8.01 73.10 0.11 0.24 1.20 4.12 0.84 1.46 6 3 42 8.01 118.87 0.07 0.142 0.85 2.92 0.63 1.83 Note: Highlighted areas indicate the best option for channel design. Since the shear stress is higher than permissible, the channel will be fitted with a liner or vegetation mat. Installing a channel liner will cause the effective shear stress to decrease, thus, reducing the potential of excessive sediment erosion. Moreover, the vegetation mat will provide adequate support for the channel s exposed sediment surface. The North American Green website provides a list of suitable of potential mats to be used for erosion control for construction sites. For this channel, a P300 polypropylene fiber erosion control blanket was selected. The following calculations show that this liner meets the permissible shear stress criteria. 6
Where: τ e = effective shear stress exerted on soil beneath vegetation γ = 62.4 lbs/ft 3 D = the maximum flow depth in the cross section=1.05 ft S = hydraulic slope = 0.055 ft/ft C f = vegetation cover factor =0.90 (Bermuda grass) n s = roughness coefficient of underlying soil = 0.02 n = roughness coefficient of erosion control blanket = 0.44 τ e = 62.4*1.05 *0.055(1-0.90)[0.02/0.044]2 = 0.074 Therefore, τ e < τ o and the NAG P300 mat will be an acceptable solution to the for this the channel. 7