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Vegetated Filter Strip System Design Manual
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| Author: | Robert P. Stone, P. Eng., Engineer, Soil/OMAFRA |
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| Creation Date: | 4 July 2005 |
| Last Reviewed: | 16 August 2007 |
For Design #2, utilizing an existing external storage basin, the following modifications are required for the existing system:
In addition, the following components of the system must be installed:
Following is an outline of the decision process used to complete the system design.
| Step | Description | Method | Calculation |
|---|---|---|---|
| 1.1 | Establish extent of area contributing runoff | Identify on a map the drainage patterns around proposed collection area; define all areas contributing surface runoff to the collection area; eliminate all clean water sources, diverting clean flow and other waste flow. | |
| 1.2 | Define and measure extent of runoff collection area | Measure the area contributing water to the runoff collection area. | Width = 20 m Length = 50 m Area = 20 m × 50 m = 1,000 m2 |
| 1.3 | Select runoff coefficient (see Section 3.1.3) | The surface of the runoff collection area must be non-porous. A concrete surface with a runoff coefficient of 0.95 is assumed. | C = 0.95 |
| 1.4 | Determine storage/settling volume and peak discharge rate | ||
| 1.4.1(a) | Option 1 — Select design-storm event (see Section 3.1.4) for calculating the runoff storage/settling basin volume requirements using the conservative method) | Establish closest centre to farmstead from Table 6.1 and determine storage/settling volume required. | Oshawa WPCP For an area of runoff collection A = 1,000 m2, the storage/settling volume Vmax = 69.1 m3 |
| OR | |||
| 1.4.1(b) | Option 2 — Use IDF Tables and Equation 3.1 to determine the storage/settling volume | From IDF Tables determine 25-year/24-hour storm event for closest centre (Oshawa WPCP) and calculate storage/settling volume. | For an area of runoff
collection A = 1,000 m2, the 25-yr/24-hr storm event for Oshawa WPCP = 72.7 mm, the storage/settling volume Vmax = (0.95)(72.7 × 10-3)(1,000 m2) = 69.1 m3 |
| 1.4.2 (a) | Option 1 — Select minimum storage volume based on peak discharge rate of a 25-year storm with 5-minute duration and 15-minute holding time (see Section 3.1.4 for calculating the storage/settling basin volume requirements using the rational method) | Establish closest centre to farmstead from Table 6.2 and determine volume of runoff generated. | Oshawa WPCP For an area of runoff collection A = 1,000 m2, the minimum storage settling volume Vmin = 31.7 m3 |
| Calculate the peak discharge by dividing the volume obtained in Table 6.2 by the holding time of 900 seconds. | qp = Vmin/900
= 31.7/900 = 35 × 10-3 m3/s |
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| OR | |||
| 1.4.2(b) | Option 2 —
Use IDF Tables and Equations 3.2 and 3.3 to determine peak discharge
rate and minimum storage/settling volume Calculate minimum storage/settling volume |
Use Equation 3.2 Rainfall intensity of 25-year/5-minute storm event for Oshawa WPCP is 137.4 mm/hr. |
For an area of runoff
collection A= 1,000 m2, the peak discharge qp = (0.0027)(0.95)(137.4 mm/hr)(0.1 ha) = 35 × 10-3 m3/s |
| Use Equation 3.3 to determine minimum storage volume. | The minimum storage/settling volume Vmin = htm qp = (900 sec)(35 × 10-3 m3/s) = 31.5 m3 for a 15-minute settling time |
| Step | Description | Method | Calculation |
|---|---|---|---|
| 2.1 | Select storage type (see Section 3.2) | Determine preferred storage method: integrated using containment wall or dedicated external basin. | Dedicated external basin |
| 2.3 | Dedicated external storage/settling basin development | ||
| 2.3.1(a) | Determine maximum storage volume required | Volume is equal to the maximum storage volume calculated in Step 1.4.1 (a) or Step 1.4.1 (b). | Vmax = 69.1 m3 |
| OR | |||
| 2.3.1(b) | Determine minimum storage volume required | Minimum storage volume required is equal to volume calculated in Step 1.4.2(a) or Step 1.4.2(b). | Vmin = 31.7 m3 |
| 2.3.2 | Establish storage volume of existing facility | ||
| 2.3.2.1 | Establish volume capacity of existing storage facility | Subtract freeboard height and emergency overflow height before calculating maximum storage volume. | Existing storage basin dimensions:
depth = 1.5 m, diameter = 7 m V = [3.14(7)2/4] × (1.5 - 0.3 - 0.15) = 40.4 m3 Since volume capacity is greater than minimum storage volume (31.7 m3) required, the existing basin can be used to accommodate runoff water; however, existing storage does not have capacity to hold 25-yr/24-hr storm (69.1 m3). |
| 2.3.2.2 | Establish operating parameters for existing storage facility | Establish elevation of maximum storage volume. | Elev = 1.50 m |
| Establish elevation
of maximum storage volume minus freeboard and emergency overflow height. |
Elev = 1.05 m | ||
| Establish elevation of minimum storage volume. | Elev = 0.82 m (assumed) | ||
| 2.3.4 | Calculate discharge rate from external storage/settling basin | Use routing equation (Equation 3.6) since storage volume available is less than storage volume calculated in Step 2.3.1(a) but more than minimum storage volume from 25-year/5-minute storm event over a 15-minute period from Step 2.3.1(b). | qp = 35 × 10-3
m3/s V = 40.4 m3 = 4.04 × 10-3 ha-m R = 72.7 mm (IDF Table - Oshawa) A = 1,000 m2 = 0.1 ha qo = 35 × 10-3 [1.25 - (1500 × 4.04 × 10-3/72.7 × 0.1 + 0.06)0.5] = 10.7 × 10-3 m3/s |
Runoff Discharge System Collection Bay
| Step | Description | Method | Calculation |
|---|---|---|---|
| 3.1 | Establish collection point where runoff exits collection area to external basin | Select screening device to prevent coarse materials from entering external basin. | Wood picket fence screening device to have ¾" spacing between 2" × 6" vertical pickets |
| 3.4 | Calculate orifice plate opening based on orifice discharge capacity | Size the orifice opening to accommodate the maximum discharge rate of the external storage/settling basin for head equal to depth of liquid on orifice plate (Use Table 3.1 or Equation 3.7). Use discharge rate from Step 2.3.4. | Q = 10.7 × 10-3 m3/s h = 1.05 - 0.82 = 0.23 m Using Table 3.1 the orifice opening diameter D = 0.11 m Using Equation 3.7 Orifice area A = 10.7 × 10-3 m3/s/[(0.61)(2 × 9.8 × 0.23)0.5] = 8.26 × 10-3 m2 and orifice diameter D = (4 × 8.26 × 10-3/3.14)0.5 = 0.10 m |
| 3.5 | Calculate diameter of drainpipe | Select drainpipe to accommodate two times the discharge
rate of the external storage/settling basin (Use Table 3.1 or Equation
3.7). Use discharge rate from Step 2.3.4. |
Q = 10.7 × 10-3 m3/s ×
2 = 21.4 × 10-3 m3/s h = 1.05 - 0.82 = 0.23 m Using Table 3.1 the orifice opening diameter D = 0.14 m Using Equation 3.7 A = 21.4 × 10-3 m3/s /[(0.61)(2 × 9.8 × 0.23)0.5] = 16.5 × 10-3 m2 and orifice diameter D = (4 × 16.5 × 10-3/3.14)0.5 = 0.14 m |
| 3.6 | Determine minimum open area in perforated riser pipe | Use Equation 3.7 to determine the size of the open slot area. Increase the open area by 25%. | A = 10.7 × 10-3/[(0.61)(2
× 9.8 × 0.23)0.5] = 8.26 × 10-3 × 1.25 = 10.3 × 10-3 m2 = 103 cm2 To get 103 cm2 with 2 × 2 cm slots insert 26 slots |
| 3.7 | Select sump | Select appropriate size and type of sump. Maximum water depth in sump should be below frost line. Total sump depth = frost line depth + depth to accommodate volume required by pump or siphon device. Contact sump and siphon device or pump manufacturer for sump volume recommendations. | |
| 3.8 | Select siphon device | Contact a siphon device manufacturer to select the most appropriate mechanism for gravity flow (e.g., Flout™)1 |
1 The use of trademarks does not imply endorsement by OMAFRA and MOE.
| Step | Description | Method | Calculation |
|---|---|---|---|
| 4.1 | Define target flow rate | Target conveyance pipe flow rate is 10% greater than storage/settling basin discharge rate from Step 2.3.4. | Q = 10.7 × 10-3× 1.10 = 11.8 × 10-3 m3/s |
| 4.2 | Establish minimum pipe slope | Use Manning's Equation (Equation 3.9), assuming minimum velocity of 0.6 m/s and Manning coefficient = 0.009. | A = Q/V = 11.8 ×
10-3/0.6 = 0.020 m2 D = (4 × 0.020/3.14)0.5 = 0.16 m; select 200 mm pipe R = D/4 = 0.2/4 = 0.05 m S = [0.6 × 0.009/(0.05)2/3]2 = 0.0016 m/m |
| 4.3 | Establish design variables for evaluation of conveyance system | Establish inlet elevation of the conveyance pipe from the sump. | Inlet elevation = 253 m |
| Establish conveyance pipe run length from sump to top of infiltration area. | Conveyance pipe run length = 260 m | ||
| Establish the elevation of the existing grade at top end of the candidate infiltration area. | Elevation at top end of infiltration area = 270 m | ||
| Calculate outlet elevation of conveyance pipe. | Conveyance pipe outlet
elevation = 253 m - (0.0016 m/m × 260 m) = 252.6 m |
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| Compare conveyance pipe outlet elevation with elevation of existing grade at top end of infiltration area (see Step 4.4). | Conveyance pipe outlet elevation of 252.6 m vs. elevation of existing grade at top end of infiltration area of 270 m (see Step 4.4) | ||
| Calculate elevation change between inlet elevation of conveyance pipe from sump/drainpipe and elevation of existing grade at top end of infiltration area. | Elevation change = 253 m - 270 m = -17 m, thus a rise in elevation of 17 m for pump system |
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| 4.4 | Determine if gravity or pump system | If the existing grade at the top end of the infiltration area is higher than the conveyance pipe outlet elevation calculated, then gravity flow to the top end of the infiltration area is not possible. | See Step 4.3 (a pump system must be used) |
| 4.5 | Define target flow rate | See Step 4.1. | Q = 10.7 × 10-3× 1.10 = 11.8 × 10-3 m3/s |
| 4.6 | Determine total head losses between pump inlet and distribution pipe discharge | Use Darcy-Weisbach
Equation (Equation 3.8) with f = 0.020. For this case study it was
assumed localized friction losses were negligible. Add the pressure head of 0.9 m at the distribution pipe. |
Head differential
(difference in elevation) = 17 m Friction losses = f (L/D) (V2/2g) = 0.020 (260/0.15) (0.6)2/2 × 9.81) = 0.64 m Distribution pipe pressure head = 0.9 m Total head losses = 17 + 0.64 + 0.9 = 18.54 m |
| 4.7 | Select type and size of pump | Select submersible sewage pump (screw-induced flow preferred). | Obtain pump curves
from manufacturer and select pump to accommodate target conveyance
pipe flow rate and total head losses, i.e., 11.8 × 10-3 m3/s at total head losses of 18.54 m |
| 4.8 | Determine pipe size required | Recalculate pipe size based on pump selection and check that velocity does not exceed 1.5 m/s and friction losses are acceptable. | |
| 4.9 | Determine requirements for power and controls | Estimate distance from power source and effort required to power and automatically control pump. | Contact consultant or electrical contractor for recommendations |
| Step | Description | Method | Calculation |
|---|---|---|---|
| 5.1 | Determine distribution pipe length | Determine length of distribution pipe based on infiltration area width (calculated in Step 6.7). | Pipe length = 78.5 m (Step 6.7) |
| 5.2 | Select diameter of distribution pipe required | Use Manning's Equation (Equation 3.9) with the assumption that the pipe slope is between 0.1% and 0.3% and the pipe will run full with minimum velocity of 0.6 m/s and n = 0.009. | Assume slope S = 0.3% = 0.3 × 0.01
= 0.003 m/m D = 4[((Vn)/S0.5)1.5] = 4[((0.6 × 0.009)/0.0030.5)1.5] = 0.124 m Select 150 mm pipe |
| 5.3 | Determine capacity of orifice opening in distribution pipe | Use Equation 3.7 and assume the following: h = 0.9 m C = 0.61 Orifice diameter = 20 mm |
Qorifice = 0.61 × 3.14(0.02)2/4
× (2 × 9.8 × 0.9)0.5 = 0.8 × 10-3 m3/s |
| 5.4 | Determine number of orifices to meet flow capacity being delivered | Use a 1.25 target distribution volume. Divide the target discharge rate for the distribution pipe by the orifice discharge rate for each orifice opening as determined in Step 5.3. |
Q = 11.8 × 10-3× 1.25
= 0.015 m3/s No. of orifices = 0.015/0.8 × 10-3 = 19 orifices Select 20 orifices Orifice spacing = distribution pipe length (Step 5.1)/no. of orifices = 78.5 m/20 = 3.93 m Space orifices 78.5 m/20 = 3.93 m on centres |
| Step | Description | Method | Calculation |
|---|---|---|---|
| 6.1 | Determine minimum infiltration area required based on saturated hydraulic conductivity (Stage 1) | An in-situ determination of saturated hydraulic conductivity was performed and a value of 0.3168 m/day was obtained. | Q = (10.7 ×
10-3) × 1.10 = 11.8 × 10-3 m3/s = 1,020 m3/day A = 1,020/0.3168 = 3,220 m2 |
| 6.2 | Determine minimum infiltration area required based on liquid loading (Stage 2) | Use Table 3.3 to obtain the highest normal monthly precipitation for the closest station. | Highest normal monthly precipitation for closest centre (Trenton Airport) = 91.2 mm = 0.0912 m/month |
| Calculate highest normal weekly precipitation (highest normal monthly precipitation divided by 4). | Highest normal weekly
precipitation = 0.0912/4 = 22.8 × 10-3 m/week |
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| Calculate outside paved yard weekly runoff volume. Paved yard has area of 1,000 m2. | Paved yard weekly runoff
volume = 1,000 m2× (22.8 × 10-3) = 22.8 m3 |
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| Calculating limiting precipitation amount per week on VFS. Maximum allowable liquid loading is 0.05 m per week. | Limiting precipitation
amount/week on VFS = 0.05 m - (22.8 × 10-3) = 27.2 × 10-3 m2 |
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| Calculate minimum VFS area required, based on paved yard weekly runoff and limiting amount of precipitation per week on VFS area. | Minimum VFS area required
= 22.8 m3/27.2 × 10-3 m = 838 m2 |
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| 6.3 | Determine minimum infiltration area from Stage 1 (Step 6.1) and Stage 2 (Step 6.2) | See Step 6.1 and Step
6.2. Select larger infiltration area from Step 6.1 and Step 6.2. |
Minimum infiltration area from Stage 1 (Step 6.1) is 3,220 m2 |
| 6.4 | Determine minimum length of infiltration area | VFS slope is 4% (see Table 3.4). | For slope of 4% the length is 41 m |
| 6.5 | Determine minimum width of infiltration area | Use Equation 3.10 (see Table 3.4). | Flow depth 1.27 ×
10-2 m Velocity 4.54 × 10-2 m/s W = 11.8 × 10-3 m3/s/(1.27 × 10-2 m × 4.54 × 10-2 m/s) = 20.5 m |
| 6.6 | Calculate actual dimensions | Select largest infiltration
area determined from Stage 1 (Step 6.1), hydraulic conductivity measurement
or Stage 2 (Step 6.2), liquid loading limit. Use minimum length (Step 6.4) in determining final dimensions of infiltration area. Ensure that final width exceeds minimum width (Step 6.5). |
Largest infiltration
area results from Step 6.1 of 3,220 m2; therefore, A = 3,220 m2 (Step 6.1) L = 41 m (Step 6.4) W = A/L = 3,220 m2/41 m = 78.5 m > 20.5 m |
| 6.7 | Determine final dimensions of infiltration area | Final infiltration area dimensions. | Length = 41 m Width = 78.5 m Area = 3,220 m2 |
Should the farmer decide that a VFS system is not suitable for his or her operation, alternatives for managing the runoff could include:
If land application of runoff is limited by the nutrient content of the runoff, then the following alternatives may be considered for treatment:
These two alternatives rely on soil and plant properties for filtering nutrients and other contaminants from the runoff water. In the VIB system tile drains collect the infiltrated water and deliver the treated sub-surface discharge to a storage tank or to an additional treatment unit (e.g., constructed wetland or VFS system).
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