WQ326, New August 1995

Department of Agricultural Engineering

Solids settle out of a liquid flow when velocity is reduced. Often, settling basins are designed to limit flow velocities to 1.5 feet per second, or less. Settling may remove 35 percent to 60 percent of the solids from dairy slurry, with as little as 10 minutes detention time (30 to 60 minutes detention time is common). Settled solids, submerged in water, typically are about 15 percent dry matter. After dewatering, the solids are usually 15 percent to 25 percent dry matter and can be used as fertilizer or composted. Figure 1 is a schematic of a typical settling basin.

**Figure 1**

Schematic representation of four-zone settling.

Liquids separated from animal waste must be contained in a lagoon, storage basin or settling basin until they are land-applied to a vegetative filter or soil-plant filter. The liquid should be applied to a vegetative or soil-plant filter large enough to use the nitrogen content.

Settling basins add to the overall cost of a waste management system. They may require two handling systems: one for solids and one for liquids; and they may require additional labor and management.

Settling basins may be used to reduce the nutrient-loading on a lagoon from a gutter-flushing system, but are used more commonly to reduce the nutrient-loading on a vegetative filter strip from lot runoff. Adding a settling basin to remove a portion of the solids may decrease the required lagoon volume for a new facility, or it may let you increase the animal units served by an existing lagoon. Settling may be a good way to remove undesirable material, such as hay and straw, from the waste flow to a lagoon. It may reduce odors and/or prevent a crust or mat from forming on the lagoon surface. You may need a baffle to retain floating solids (such as straw) in a settling basin. By removing the larger solids, you may reduce plugging of liquid handling equipment, such as pumps and irrigation sprinkler nozzles.

There are two types of settling basins, based on the method of removing solids. With one type, the solids are removed mechanically (after the free water has drained away), usually with a front-end or skid-steer loader. The depth of accumulated solids should not exceed 1.5 feet. The other type uses hydraulic (pump) removal of the solids. Typically, pumping is initiated when the basin is half full of solids and the remainder is water. Vigorous agitation is needed to mix the liquid and the solids, preferably by propeller-type agitators or pumps with agitation nozzles.

Settling basins may be either concrete or earthen structures. For concrete basins, a common recommendation is a minimum depth of 2 feet plus the depth required for solids storage. Figure 2 shows a typical concrete settling basin. Earthen structures may be compact basins, settling terraces, settling diversion terraces or settling channels. Figure 3 shows a typical earthen settling basin. Earthen basins to be cleaned with loaders are usually designed to be shallow (not more than 3 feet deep) and to cover a large area. Earthen settling basins should have a concrete entrance ramp and a concrete runway on the bottom to allow entry of equipment for solids removal. Figure 4 shows a duplex concrete settling basin that allows one side to receive effluent while the other side is thoroughly dewatered and the solids removed.

**Figure 2**

A typical concrete settling basin designed for mechanical removal of solids (from MWPS18.)

**Figure 3**

A typical earthen settling basin designed for mechanical removal of solids (from MWPS18)

**Figure 4**

Duplex concrete settling basin

A settling terrace, a settling diversion terrace or a settling channel is a wide, shallow, gently sloping, flat-bottomed waterway in which runoff solids settle due to low velocity. The channel is sometimes grassed to improve settling and reduce erosion. Grass may not survive in the channel and may make cleaning more difficult. Grass should be maintained on the sideslopes, if possible. Solids should be removed annually, or more often if required, to maintain capacity. Berm tops should be at least 2 feet wide to maintain the design height and at least 12 feet wide for vehicle traffic.

In Missouri's humid climate, inadequate drying of the solids and the channel may limit the usefulness of earthen settling terraces and channels.

An example will illustrate the design of a concrete settling basin. A blank form is included for your calculations.

Design a concrete settling basin for a dirt lot, 100 feet by 200 feet, on a 6-percent slope. The solids will be pumped out at 6-month intervals. The basin length is to be four times the basin width. The location is in the northeast corner of Missouri. Design using peak runoff rate for a 1-year per 10-year storm from Table 1.

**Table 1**

Peak rates of runoff to be expected from watersheds in Missouri (Q)

Acres | One year per 10 (cubic feet per second per acre) | One year per 25^{1} (cubic feet per second per acre) |
---|---|---|

0 to 1 | 5.5 | 0.2 |

1 to 2 | 4.6 | 6.0 |

2 to 3 | 4.2 | 5.5 |

3 to 4 | 3.0 | 5.2 |

4 to 20^{2} |
2.8 | 5.1 |

**Open lot area draining into the basin**

Lot length 200 feet x lot width 100 feet = lot area 20,000 square feet

**Inflow rate into settling basin**

L = location factor (Figure 5)

T = topographic factor (Table 2)

Q

Q_{T} = Q x L x T x square feet lot area x 0.0411 = cubic feet per hour

Q_{T} = 5.5 cubic feet per second per acre (Q) x 0.96 (L) x 0.92 (T) x 20,000 square feet x 0.0411 = 3,993 cubic feet per hour

If inflow arrives at settling basin via a sewer pipe, estimate Q_{T} from the size of the sewer pipe in Table 3.

Q_{T} = gpm (Table 3) x 8 = cubic feet per hour

If inflow arrives at settling basin by other means, explain (i.e., a dairy flush alley discharging into settling basin).

**Estimate inflow rate**

Q_{T} = estimated gpm x 8 = cubic feet per hour

**Figure 5**

Location factors (L) from G1518, *Estimating Peak Rates of Runoff From Small Watersheds.*

**Table 2**

Topographic factors for different land slopes.^{1}

Average land slope | Topographic factor (T) |
---|---|

1 percent | 0.65 |

2 percent | 0.72 |

3 percent | 0.78 |

4 percent | 0.83 |

5 percent | 0.88 |

6 percent | 0.92 |

7 percent | 0.96 |

8 percent | 1.00 |

9 percent | 1.04 |

10 percent | 1.07 |

12 percent | 1.14 |

14 percent | 1.20 |

**Table 3**

Diameter, slope and expected flow rates for sewer lines

Diameter | Slope | Expected flow rate |
---|---|---|

4 inches | 0.02 foot per foot | 115 GPM |

6 inches | 0.013 foot per foot | 255 GPM |

8 inches | 0.009 foot per foot | 460 GPM |

10 inches | 0.007 foot per foot | 750 GPM |

**Surface area (SA) of settling basin**

SA (square feet) | = | Q_{T} (cubic feet per hour)/ 4 cubic feet per hour per square feet |
= | 3,993 cubic feet per hour/ 4 cubic feet per hour per square feet | = | 998 square feet |

**Basin dimensions**

Design basin for length = 3 to 5 times basin width.

Basin width = [SA/R]^{0.5}

R = length-width ratio = 4

width = [998 square feet (SA) ÷ 4 (R)]^{0.5} = 15.8 feet

length = 16 feet (width) x 4 (R) = 64 feet

**Basin overflow**

Provide a rectangular overflow weir at downstream end of basin. Weir height = 6 inches. Maximum weir length = width of settling basin. Minimum weir length, feet = Q_{T}/1,250. If a riser pipe is used as an overflow device instead of a rectangular weir, riser pipe diameter, inches = Q_{T}/274. Do not use a riser pipe smaller than 6 inches in diameter.

Minimum rectangular weir length = Q_{T}/1,250 = 3,993 /1,250 = 3.2 feet

Minimum riser pipe diameter = Q_{T}/274 = 3,993 /274 = 14.6 inches

**Basin depth**

Use the following as a guide for volume of solids and calculate depth required for desired storage period.

Dirt lots = 2,800 cubic feet per acre-yr

Concrete lots and confinement buildings = 0.5 times manure production volume (Table 4).

**Table 4**

Approximate daily manure production

Animal | Weight | Cubic feet per day (liquid + solid) |
---|---|---|

Dairy | 1,000 | 1.32 |

Beef (a) Concentration ration (b) Corn silage (c) 25 percent haylage (d) 50 percent haylage |
1,000 1,000 1,000 1,000 |
0.35 0.45 0.34 0.43 |

Swine, finish | 1,000 | 1.10 |

Sow + litter | 1,000 | 1.44 |

Gestation sow | 1,000 | 0.55 |

Boar | 1,000 | 0.55 |

Sheep | 1,000 | 0.62 |

Poultry (a) Layers (b) Broilers |
1,000 1,000 1,000 |
0.88 1.20 |

Horse | 1,000 | 1.50 |

If solids are to be removed from basin by pumping, design basin to hold an equal volume of water above settled solids. Solids must be diluted and agitated for pumping. If solids are to be removed mechanically (i.e., front-end loader), provide concrete entrance to settling basin with at least 10:1 slope (20:1 slope preferred). Additional dewatering via a hardware cloth dam and/or a perforated riser pipe is desirable for mechanical removal of settled solids.

Indicate desired storage period, days = 182

(Lot acres = 100 feet x 200 feet/43,560 = 0.46 acres)

Basin depth (dirt lot) = | 2,800 x 0.46 acres lot x 182 days storage x 2*365 days per yr x 998 square feet surface area (item 3) |
= 1.3 feet |

Basin depth, BD_{C} (concrete lot or confinement building)

BD_{C} = |
0.5 x __ cubic feet per day manure prod (Table 4) x __ days storage x (1* or 2*)________ square feet surface area (item 3) |
= _______ feet |

**Note**

When settling basin discharges into a lagoon, the size of the lagoon may be reduced as follows:

Design volume (with settling basin) = Design volume (without settling basin) x 0.5

Manure storage volume (with settling basin) = Manure storage volume (without settling basin) x 0.5

Minimum design storage period is 90 days when the lagoon design volume is reduced by 50 percent as noted above. Less storage may be used if the lagoon design volume is based on 100 percent loading.

Various types of basin outlets are used to drain liquids from the full depth of basins and allow the solids to dewater. The porous plank dam (Figure 4, Sec. C-C) ahead of either a perforated or a slotted riser pipe are frequently used outlets. Manure tends to plug even large openings in outlets. Unplugging is required frequently. A hoe may be used to scrape solids off of openings. Also, a slanted expanded metal or quarry screen with 1-inch to 1.5-inch openings may be used around the outlet to increase the screening area and reduce clogging.

Porous dams may be made of welded wire fabric, expanded metal mesh or spaced boards. Porous dams may be used to dewater settling basins or to remove large solids that tend to cause excessive clogging of the openings in perforated pipe outlets. Dams constructed with spaced boards usually have 0.75-inch spaces between the boards. The boards usually range from 2-by-6s to 2-by-12s. Expanded metal and welded wire fabric have even greater open areas. Due to the large open area in a porous dam, little design is required. As a general rule, the open area in a porous dam should be twice the area of the perforations in the riser pipe it precedes. As a practical matter, a porous dam 4 feet long or more, should suffice for the common sizes of outlet pipes. In some applications, there is no outlet pipe and the porous dam forms one wall of the settling/storage basin.

Material for perforated pipe is usually PVC plastic, galvanized steel or concrete. Perforations can be 5/8-inch to 1-inch diameter holes or 1-inch by 4-inch slots. The outlet is sized to match the anticipated flow rates in order to assure adequate detention time. Flow rate is controlled by the amount of open area (slots or holes) in the pipe. Table 5 gives opening requirements for perforated pipes.

**Table 5**

Riser pipe outlet design for settling basins (from MWPS18)

Based on Q = (C)(A)(2gh) ^{0.5}, where Q = flow rate in cfs; C = slot constant, assumed to be 0.61; A = open slot area in square feet; g = 32.174 feet per second^{2}; h = head on openings in feet. The pipe height is divided into 0.5-foot increments. The head on all the slots in the first 0.5-foot increment is assumed to be 0.25-foot. The head on subsequent 0.5-foot increments increase at 0.5-foot increments. |
||||||||

Open slot area per foot of pipe height, in ^{2} per foot |
Head, feet | |||||||
---|---|---|---|---|---|---|---|---|

0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 | |

cfs | ||||||||

4 | 0.034 | 0.093 | 0.169 | 0.259 | 0.361 | 0.473 | 0.596 | 0.728 |

6 | 0.051 | 0.139 | 0.253 | 0.388 | 0.541 | 0.710 | 0.894 | 1.091 |

8 | 0.068 | 0.186 | 0.338 | 0.518 | 0.721 | 0.947 | 1.192 | 1.455 |

10 | 0.085 | 0.232 | 0.422 | 0.647 | 0.902 | 1.183 | 1.480 | 1.819 |

12 | 0.102 | 0.279 | 0.507 | 0.776 | 1.082 | 1.420 | 1.788 | 2.183 |

14 | 0.119 | 0.325 | 0.591 | 0.906 | 1.262 | 1.657 | 2.086 | 2.546 |

16 | 0.136 | 0.371 | 0.675 | 1.035 | 1.443 | 1.894 | 2.384 | 2.910 |

18 | 0.153 | 0.418 | 0.760 | 1.164 | 1.623 | 2.130 | 2.682 | 3.274 |

20 | 0.170 | 0.464 | 0.844 | 1.294 | 1.803 | 2.367 | 3.890 | 3.638 |

22 | 0.187 | 0.511 | 0.929 | 1.423 | 1.984 | 2.604 | 3.277 | 4.001 |

24 | 0.204 | 0.557 | 1.013 | 1.542 | 2.164 | 2.840 | 3.575 | 4.365 |

26 | 0.221 | 0.603 | 1.097 | 1.682 | 2.344 | 3.077 | 3.873 | 4.729 |

28 | 0.238 | 0.650 | 1.182 | 1.811 | 2.525 | 3.314 | 4.171 | 5.093 |

30 | 0.255 | 0.696 | 1.266 | 1.940 | 2.705 | 3.550 | 4.469 | 5.456 |

32 | 0.272 | 0.743 | 1.351 | 2.070 | 2.885 | 3.787 | 4.767 | 5.820 |

34 | 0.289 | 0.789 | 1.435 | 2.199 | 3.066 | 4.024 | 5.065 | 6.184 |

36 | 0.306 | 0.836 | 1.519 | 2.329 | 3.246 | 4.260 | 5.363 | 6.548 |

38 | 0.323 | 0.882 | 1.604 | 2.458 | 3.426 | 4.497 | 5.661 | 6.911 |

40 | 0.340 | 0.928 | 1.688 | 2.587 | 3.607 | 4.734 | 5.959 | 7.275 |

Design a basin outlet to allow outflow to equal peak flow rate off the lot in Example 1 when the basin is full. The inflow rate in Example 1 is 3,993 cubic feet per hour and the depth is 1.3 feet.

- For a perforated pipe riser, determine the required open area per foot of pipe height from Table 5. Outflow = 3,993 cubic feet per hour = 1.1 cubic feet per second. By interpolation in Table 5, we find that with 1.3 feet of head the required opening area in the riser pipe is 32 in
^{2}per foot of pipe height. - Size the outlet pipe from the data in Table 3. Outflow in gpm = 1.1 cubic feet per second x 450 gpm per cfs = 495 gpm. From Table 3, an 8-inch pipe at 0.009 slope will carry 460 gpm and a 10-inch pipe at 0.007 slope will carry 750 gpm. Depending on slope, use an 8-inch or 10-inch pipe (an 8-inch pipe will carry 1.1 cfs at slopes greater than 1 percent, Ref. Figure 4.5b in MWPS18).

Check on the time to withdraw the 9,167 cubic feet of runoff from the 25-year, 24-hour storm at 1.1 cfs.

Time in hours = 9,167 cubic feet ÷ (1.1 cubic feet per second x 3,600 second per hour) = 2.3 hours at maximum head

To meet approval by the Missouri Department of Natural Resources, earthen basins must be built as follows. Berms shall have minimum slopes of 3:1. If solids are to be removed using mechanical equipment, a concrete pad shall be installed in the bottom of the basin and a concrete access ramp with a maximum slope of 10 percent shall be provided. If the settled solids are to be removed by pumping, the basin must be designed to contain an equal volume of water above the solids to allow for agitation and dilution of the solids. Access points for the mixing equipment must be indicated on the construction drawings.

For more information on the design of earthen storage basins/lagoons, consult your local NRCS engineer or your MU Extension regional agricultural engineering specialist. You may get their names from your local NRCS office or MU Extension center. The engineers have a computer program available for design of earthen storage basins.

A settling diversion terrace, settling terrace or a settling channel is a wide, shallow, gently sloping, flat-bottomed channel, in which suspended solids contained in runoff water are settled out. A settling channel may be either of earthen or concrete construction. The settling channel may be grassed to improve settling and reduce erosion. Runoff water from the channel is stored in a lagoon or storage pond. Wastes settled from the runoff are allowed to dry before removal with mechanical equipment, typically with a tractor and front-end loader. Usually, solids are removed from the channel once per year or when accumulated solids reduce the settling ability of the channel.

Sideslopes for settling channels usually range from 3:1 to 4:1, depending on soil properties. The bottom slope of the channel should be between 0.1 percent and 0.3 percent to maintain low velocities and rapid settling.

An example design shows how to design a settling terrace or channel. Blanks are provided for your specific design.

Design a settling terrace (or channel) for the dirt lot 100 feet by 200 feet on a 6-percent slope in Example 1. The location is in the northeast corner of Missouri. Design using peak runoff rate for a 1-year per 10-year storm from Table 1. Assume 3:1 sideslopes and a 0.1 percent bottom slope. Design for a maximum velocity of 1 fps and a detention time of one hour.

**Open lot area draining into the basin**

Lot length __200 __feet x lot width __100__ feet = lot area 20,000 square feet = 0.46 acre

**Your design**

Lot length __ feet x lot width __ feet = lot area __ square feet = __ acre

**Inflow rate into settling basin**

L = location factor (Figure 5)

Y = topographic factor (Table 2)

Q

Q

Q

**Your design**

Q_{T} = __ cubic feet per second per acre x __ (L) x __ (T) x __ square feet x 0.0411 = __ cubic feet per hour

__ cubic feet per hour ÷ 3,600 second per hour = __ cubic feet per second (cfs)

If inflow arrives at settling terrace/channel via a sewer pipe, estimate Q_{T} from the size of the sewer pipe in Table 3.

Q_{T} = __ gpm (Table 3) x 8 = __ cubic feet per hour = __ cubic feet per second

If inflow arrives at settling terrace/channel by other means, explain (i.e., a dairy flush alley discharging into terrace/channel basin).

**Estimate inflow rate**

Q_{T} = __ estimated gpm x 8 = __ cubic feet per hour = __ cubic feet per second

**From Table 6**

Find the width and depth for limit velocity to 1 fps. A channel 20 feet wide and 0.5 foot deep on a 0.2 percent slope will limit velocity to 1.0 fps and will drain better than a flatter slope.

Your width: __ feet; __ depth: __ feet; __ slope: __ percent

**Table 6**

Trapezoidal settling channel design

These low-slope channels settle solids out of runoff. The lower the water velocity, fps, the higher the settling rate. Construct the channel at least 0.5 foot deeper than the table value to allow for solids storage. Bottom slope as shown, sideslopes = 3:1 and n = 0.04. | ||||||||

Bottom slope | Depth (feet) | Bottom width (feet) | ||||||
---|---|---|---|---|---|---|---|---|

10 | 15 | 20 | 25 | 30 | 35 | |||

0.1 percent | 0.5 | fps | 0.6 | 0.7 | 0.7 | 0.7 | 0.7 | 0.7 |

cfs | 3.7 | 5.5 | 7.4 | 9.2 | 11.1 | 12.9 | ||

1.0 | fps | 0.9 | 1.0 | 1.0 | 1.0 | 1.1 | 1.1 | |

cfs | 12.2 | 17.9 | 23.6 | 29.4 | 35.2 | 41.0 | ||

1.5 | fps | 1.2 | 1.2 | 1.3 | 1.3 | 1.3 | 1.4 | |

cfs | 25.1 | 36.0 | 47.1 | 58.4 | 69.7 | 81.1 | ||

2.0 | fps | 1.3 | 1.4 | 1.5 | 1.5 | 1.6 | 1.6 | |

cfs | 42.8 | 60.0 | 77.7 | 95.6 | 113.7 | 131.9 | ||

0.2 percent | 0.5 | fps | 0.9 | 0.9 | 1.0 | 1.0 | 1.0 | 1.0 |

cfs | 5.2 | 7.8 | 10.4 | 13.0 | 15.6 | 18.2 | ||

1.0 | fps | 1.3 | 1.4 | 1.5 | 1.5 | 1.5 | 1.5 | |

cfs | 17.3 | 25.3 | 33.4 | 41.6 | 49.8 | 58.0 | ||

0.3 percent | 0.5 | fps | 1.1 | 1.2 | 1.2 | 1.2 | 1.2 | 1.2 |

cfs | 6.4 | 9.6 | 12.8 | 15.9 | 19.1 | 22.3 |

**Determine the minimum channel volume required for a 1-hour detention time**

Minimum channel volume cubic feet = (channel capacity, 1.1 cfs) x (detention time, 1 hour) x 3,600 second per hour = 3,960 cubic feet

**Your design**

__Channel volume__ cubic feet = (channel capacity __ cfs) x (detention time __ hour) x 3,600 second per hour = __ cubic feet

**Divide the minimum channel volume by the volume per foot of channel length values (in Table 7) to get the minimum channel length based on volume required for detention time**

From Table 7 for the 0.5-foot deep channel, 20-feet wide, as selected in Step 3, volume per foot is 10.75 cubic feet.

Channel volume __3,960 __cubic feet ÷ volume per foot __10.75 __cubic feet = minimum length of channel __368 __feet

**Your design**

Channel volume __ cubic feet ÷ volume per foot __ cubic feet = minimum length of channel __ feet

**Table 7**

Settling channel volumes

Assumptions 3:1 sideslopes, vertical ends, bottom nearly flat, volume = LD(W + 3D), use LD(W + 4D) for 4:1 sideslopes. |
||||||

Channel width | Channel depth (feet) | |||||
---|---|---|---|---|---|---|

0.5 | 1.0 | 1.5 | 2.0 | 3.0 | 4.0 | |

Cubic feet per foot of channel length | ||||||

10 feet | 5.75 | 13.0 | 21.75 | 32 | 57 | 88 |

15 feet | 8.25 | 18 | 29.25 | 42 | 72 | 108 |

20 feet | 10.75 | 23 | 36.75 | 52 | 87 | 128 |

25 feet | 13.25 | 28 | 44.25 | 62 | 102 | 148 |

30 feet | 15.75 | 33 | 51.75 | 72 | 117 | 168 |

35 feet | 18.25 | 38 | 59.25 | 82 | 132 | 188 |

40 feet | 20.75 | 43 | 66.75 | 92 | 147 | 208 |

50 feet | 25.75 | 53 | 81.75 | 112 | 177 | 248 |

60 feet | 30.75 | 63 | 96.75 | 132 | 207 | 288 |

**Determine the depth of channel required to store the accumulated solids volume assuming once per year solids removal**

Annual solids volume (cubic feet) = 2,800 cubic feet per acre-year x 0.46 acre = 1,288 cubic feet

If a 6-inch depth will contain 3,960 cubic feet in 368 feet of channel, then the required depth for 1,228 cubic feet of solids can be approximated as follows:

6 inches x 1,228 ÷ 3,960 = 1.95 inches

**Your design**

Liquid depth __ inches x solids volume __ cubic feet ÷ liquid volume cubic feet = solids depth __ inches

**Determine total channel depth for both solids and liquid**

6 inches for liquid + 1.95 inches for solids = 7.95 inches total depth

Construct channel at least 1 foot deep to allow for freeboard.

**Your design**

__ inches for liquid + __ inches for solids = __ inches total depth

**Open lot area draining into the basin**

Lot length __ feet x lot width __ feet = lot area __ square feet

**2. Inflow rate into settling basin**

L = location factor (Figure 5)

T = topographic factor (Table 2)

Q

Q

Q_{T} = __ cubic feet per second per acre x __ (L) x __ (T) x __ square feet x 0.0411 = __ cubic feet per hour

If inflow arrives at settling basin via a sewer pipe, estimate Q_{T} from the size of the sewer pipe in Table 3.

Q_{T} = __ gpm (Table 2) x 8 = __ cubic feet per hour

If inflow arrives at settling basin by other means, explain (i.e., a dairy flush alley discharging into settling basin).

**Estimate inflow rate**

Q_{T} = __ estimated gpm x 8 = __ cubic feet per hour

**Surface area (SA) of settling basin**

SA (square feet) = | Q_{T} (cubic feet per hour)/ 4 cubic feet per hour per square feet |
= cubic feet per hour/ 4 cubic feet per hour per square feet | =__ square feet |

**Basin dimensions**

Design basin for length = 3 to 5 times basin width.

Basin width = [SA/R]^{0.5}

SA = basin surface area, square feet (item 3)

R = length-width ratio

width = [ __ square feet (SA) ÷ __ (R)]^{0.5} = __ feet

length = __ feet (width) x __ (R) = __ feet

**Basin overflow**

Provide a rectangular overflow weir at downstream end of basin. Weir height = 6 inches. Maximum weir length = width of settling basin. Minimum weir length, feet = Q_{T}/1,250. If a riser pipe is used as an overflow device instead of a rectangular weir, riser pipe diameter, inches = Q_{T}/274. Do not use a riser pipe smaller than 6-inches diameter.

Minimum rectangular weir length = Q_{T}/1,250 = __ /1,250 = __ feet

Minimum riser pipe diameter = Q_{T}/274 = __ /274 = __ inches

**Basin depth**

Use the following as a guide for volume of solids and calculate depth required for desired storage period.

Dirt lots = 2,800 cubic feet per acre-yr

Concrete lots and confinement buildings = 0.5 x manure production volume (Table 4)

If solids are to be removed from basin by pumping, design the basin to hold an equal volume of water above settled solids. Solids must be diluted and agitated for pumping. If solids are to be removed mechanically (i.e., front end loader), provide concrete entrance to settling basin with at least 10:1 slope. Additional dewatering via a hardware cloth dam or perforated riser pipe is desirable for mechanical removal of settled solids.

Indicate desired storage period, days = __

Basin depth (dirt lot) = | 2,800 x __ acres lot x __ days storage x (1* or 2*)/ 365 days per year x __ square feet surface area (item 3) | = __ feet |

Basin depth (concrete lot or confinement building) = | 0.5 x ____ feet^{3} per day manure prod. (Table 4) x _ days storage x (1* or 2*)/ __ square feet surface area (item 3) |
= __feet |

**Note
**When settling basin discharges into a lagoon, the size of the lagoon may be reduced as follows:

Design volume (with settling basin) = Design volume (without settling basin) x 0.5

Manure storage volume (with settling basin) = Manure storage volume (without settling basin) x 0.5

Minimum design storage period is 90 days when the lagoon design volume is reduced by 50 percent, as noted above. Less storage may be used if the lagoon design volume is based on 100 percent loading.

- Missouri Manual 121,
*Design Guidelines for Animal Waste Management for Concentrated Animal Feeding Operations,*Second Edition, July 1989. Missouri Department of Natural Resources - Water Pollution Control Program, P.O. Box 176, Jefferson City, Mo. 65102. - Schneider, John H.; Susan B. Harrison and Paul B. Freeze. 1993.
*No Discharge Gated Pipe Distribution of Feedlot Runoff.*ASAE Paper number 93-4566. St. Joseph, Mich. 49085-9659.

WQ326, new August 1995

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