In This Section |
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 |
The first step in the design of a VFS system for this farm was to confirm that the site conditions are suitable, with reference to Table 2.1 and Figure 2.2. A brief summary of the work required was prepared, to ensure that the project met the farmer's expectations.
For this case study, the following modifications are required to the existing systems:
In addition, the following components of the system must to be installed:
Following is 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 maximum 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 maximum storage/settling volume Vmax = 69.1 m3 |
| OR | |||
| 1.4.1(b) | Option 2 — Use IDF Tables and Equation 3.1 to determine the maximum storage/settling volume | From IDF Tables determine 25-year/24-hour storm event for closest centre (Oshawa WPCP) and calculate maximum storage/settling volume. | For an area of runoff
collection A = 1,000 m2, the 25-year/24-hour storm event for Oshawa WPCP = 72.7 mm, the maximum storage/settling volume Vmax = (0.95)(72.7 × 10-3 m)(1000 m2) = 69.1m3 |
| 1.4.2 (a) | Option 1— Select
minimum storage volume based on peak discharge rate of a 25-year/5-minute
storm event over a 15-minute holding time Calculate peak discharge rate |
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 peak discharge by dividing volume obtained in Table 6.2 by holding time of 900 seconds. | qp = Vmin/900
= 31.7/900 = 35 × 10-3 m3/s |
||
| 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 | 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 rate qp = (0.0027)(0.95)(137.4 mm/hr)(0.1 ha) = 35 × 10-3 m3/s |
| Calculate minimum storage/settling volume | Use Equation 3.3 to determine storage/settling 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. | Integrated storage/settling basin |
| 2.2 | Develop integrated storage/settling basin | Use if runoff collection is suitable for development of a watertight basin and sufficient volume can be generated using a containment wall around all or portion of runoff containment area. | |
| 2.2.1(a) | Determine maximum storage volume required | Volume required is equal to the conservative maximum storage volume calculated in Step 1.4.1 (a) or Step 1.4.1 (b). | Vmax = 69.1 m3 |
| OR | |||
| 2.2.1(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 |
| 2.2.2 | Establish containment wall height | ||
| 2.2.2.1 | Establish containment wall height assuming consistent slope perpendicular to runoff flow to a low side | Calculate containment wall height using Equation 3.4, assuming a consistent slope of 0.01 m/m and runoff collection area length perpendicular to the runoff flow. | V = 69.1 m3 L = 50 m S = 0.01 m/m h = SQRT[(2 × 69.1 × 0.01/50)] = 0.17 m |
| 2.2.3 | Establish final height required to accommodate runoff storage + freeboard + emergency spillway | ht = h + 0.3 + 0.15 | ht = 0.17 + 0.3 + 0.15 = 0.62 m (An integrated storage/settling basin can be accommodated for this application) |
| 2.2.4 | Calculate discharge rate from the runoff collection area for integrated storage/settling basin with a holding time ranging from 4 to 10 hours (see Section 3.3.1) | Establish closest centre to farmstead from Table 6.1 and determine discharge rate for a retention period of 4 hours. | Oshawa WPCP For an area of runoff collection A = 1,000 m2 and a retention period of 4 hours, qmax = 4.8 × 10-3 m3/s |
| OR | |||
| Using storage volume (Vmax) from Step 2.2.1(a), determine discharge rate for a retention period of 4 hours. | Discharge rate qmax = 69.1m3/4 × 60 × 60 sec = 4.8 × 10-3 m3/s |
Runoff Discharge System Collection Bay
| Step | Description | Method | Calculation |
|---|---|---|---|
| 3.1 | Establish collection/discharge bay dimension to accommodate screens and drainpipe area for integrated storage basin | Minimum width of collection bay opening
= 1.0 m Length of collection/discharge bay opening Collection/discharge bay height equal to containment wall height |
Width = 1.0 m
Length = 2.0 m Height = 0.62 m |
| 3.2 | Establish screen configuration | Determine size of screens based on the opening measurements of the collection/discharge bay. | Width = 1.0 m Height = 0.62 m |
| 3.3 | Select screen material | Material: galvanized metal screen Coarse-vertical spacing of approx. 25 mm Medium-vertical spacing of approx. 10 mm Fine-vertical spacing of approx. 3 mm |
Installation of screens to be vertical or at an angle of 60 degrees above the horizontal and sloped away from the flow |
| 3.4 | Calculate orifice plate opening based on orifice discharge capacity | Size the orifice opening to accommodate the maximum discharge rate of the integrated storage/settling basin (empty time of 4 hours) for the head equal to final containment wall height (use Table 3.1 or Equation 3.7). Use discharge rate from Step 2.2.4. | Q = 4.8 × 10-3 m3/s h = 0.62 m Using Table 3.1 the orifice opening diameter D = 0.05 m Using Equation 3.7 orifice area A = 4.8 × 10-3 m3/s/[(0.61)(2 × 9.8 × 0.62)05] = 2.25 × 10-3 m2 and orifice diameter D = (4 × 2.25 × 10-3/3.14)0.5 = 0.054 m |
| 3.5 | Calculate diameter of drainpipe | Select drainpipe to accommodate two times
the maximum flow rate through the orifice plate opening of the integrated
storage/settling basin (empty time of 4 hours) for the head equal
to final containment wall height (use Table 3.1 or Equation 3.7).
Use discharge rate from Step 2.3.4. |
Q = 4.8 × 10-3 m3/s
× 2 = 9.6 × 10-3 m3/s h = 0.62 m Using Table 3.1 the orifice opening diameter D = 0.08 m Using Equation 3.7 A = 9.6 × 10-3/3.14[(0.61)(2 × 9.8 × 0.62)0.5] = 4.5 × 10-3m2 and orifice diameter D = (4 × 4.5 × 10-3/3.14)0.5 = 0.076 m Since the orifice diameter required is 0.08 m (approx), the drainpipe diameter will have to be 0.10 m diameter |
| 3.6 | Determine minimum open area in the perforated riser pipe | Use Equation 3.7 to determine the size of the open slot area. Increase the open area by 25%. | A = 4.8 × 10-3/[(0.61)(2
× 9.8 × 0.62)0.5] = 2.25 × 10-3× 1.25 = 2.8 × 10-3 m2 = 28 cm2 To get 28 cm2 with 2 × 2 cm slots, it will be necessary to insert 7 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. Contact sump and pump manufacturer for sump volume recommendations. |
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 2.2.4. | Q = 4.8 × 10-3× 1.10 = 5.3 × 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's n coefficient = 0.009. | A = Q/V = 5.3 ×
10-3/0.6 = 8.8 × 10-3 m2 D = (4A/3.14)0.5 = (4 × 8.8 × 10-3/3.14)0.5 = 0.11 m; Select 150 mm pipe R = D/4 = 0.15/4 = 0.0375 m S = (Vn/R2/3)2 = [0.6 × 0.009/(0.0375)2/3]2 = 0.0023 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 end 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 - (0.0023 m/m × 260 m) = 252.4 m | ||
| 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 252.4 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 |
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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 = 4.8 × 10-3×
1.10 = 5.3 × 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 | 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., 5.3 × 10-3 m3/s @ 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 = 35 m (see 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 = 10 mm |
Qorifice = CA(2gh)0.5 = 0.61 × 3.14(0.01)2/4 × (2 × 9.81 × 0.9)0.5 = 0.2 × 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 = 4.8 × 10-3× 1.10
× 1.25 = 6.6 × 10-3 m3/s No. of orifices = 6.6 × 10-3/0.2 × 10-3 = 33 orifices Select 34 orifices Orifice spacing = distribution pipe length (Step 5.1)/no of orifices = 35 m/34 = 1.03 m Space orifices on 1.03 m 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. Table 6.3 can also be used after soil texture has been determined. |
Q = 4.8 × 10-3×
1.10 = 5.3 × 10-3 m3/s = 458 m3/day A = 458 m3/day/0.3168 m/day = 1,445 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/4). | Highest normal weekly precipitation = 0.0912/4 = 22.8 × 10-3 m/week | ||
| 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 |
||
| Calculate limiting precipitation amount per week on VFS. Maximum allowable liquid loading is 0.05 m. | Limiting precipitation amount/week on VFS = 0.05 m - (22.8 × 10-3) = 27.2 × 10-3m | ||
| Calculate minimum VFS area required based on paved yard weekly runoff and limiting amount of precipitation/week on VFS area. | Minimum VFS area required = 22.8 m3/27.2 × 10-3m = 838 m2 |
||
| 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 1,445 m2 |
| 6.4 | Determine minimum length of infiltration area | VFS slope is 4%. See Table 3.4 |
For slope of 4% the VFS length is 41 m |
| 6.5 | Determine minimum width of infiltration area | Use Equation 3.10.
See Table 3.4. |
Flow depth 1.27 ×
10-2m Velocity 4.54 × 10-2 m/s W = 5.3 × 10-3 m3/s/(1.27 × 10-2m × 4.54 × 10-2m/s) = 9.2 m |
| 6.6 | Calculate actual infiltration area 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 (see Step 6.5). | Largest area results
from Step 6.1 of 1,445 m2, therefore, A = 1,445 m2 Step 6.1 L = 41 m Step 6.4 W = A/L = 1,445 m2/41 m = 35.2 m > 9.2 m |
| 6.7 | Determine final dimensions of infiltration area | Final infiltration area dimensions. | Length = 41 m Width = 35.2 m Area = 1,445 m2 |
5.1.2 Design #2 — Design of VFS System (External Storage)
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