University of Missouri Extension

WQ326, New August 1995

Settling Basins and Terraces for Dairy Waste

Charles D. Fulhage and Donald L. Pfost
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.

Schematic representation of four-zone settling 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

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.

A typical concrete settling basin designed for mechanical removal of solids Figure 2
A typical concrete settling basin designed for mechanical removal of solids (from MWPS18.)
 

A typical earthen basin designed for mechanical removal of solids Figure 3
A typical earthen settling basin designed for mechanical removal of solids (from MWPS18)
 

Duplex concrete settling basin 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.

Design of concrete settling basins

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

Example 1

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 251 (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 202 2.8 5.1
1One year per 10 is sufficient in most design situations. However, if a settling terrace is adjacent to a stream, use one year per 25.
2For areas greater than 20 acres, refer to MU publication G1518, Table 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

Q = peak runoff rate in cubic feet per second per acre (Table 1)
L = location factor (Figure 5)
T = topographic factor (Table 2)
QT = inflow rate into settling basin, cubic feet per hour

QT = Q x L x T x square feet lot area x 0.0411 = cubic feet per hour

QT = 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 QT from the size of the sewer pipe in Table 3.

QT = 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

QT = estimated gpm x 8 = cubic feet per hour

Location factors, L 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
1Obtained from MU publication G1518, Estimating Peak Rates of Runoff from Small Watersheds.

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) = QT (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

SA = basin surface area, square feet (from item 3)
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 = QT/1,250. If a riser pipe is used as an overflow device instead of a rectangular weir, riser pipe diameter, inches = QT/274. Do not use a riser pipe smaller than 6 inches in diameter.

Minimum rectangular weir length = QT/1,250 = 3,993 /1,250 = 3.2 feet

Minimum riser pipe diameter = QT/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
*Multiply by 1 for mechanical removal or by 2 for pumping.

Basin depth, BDC (concrete lot or confinement building)

BDC = 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
*Multiply by 1 for mechanical removal or by 2 for pumping.

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.

Basin outlets

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

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.

Perforated pipe outlets

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 second2; 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

Example

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.

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

Design of earthen settling basins

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.

Design of settling channels

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.

Example 2

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

Q = peak runoff rate in cubic feet per second per acre (Table 1)
L = location factor (Figure 5)
Y = topographic factor (Table 2)
QT = inflow rate into settling basin, cubic feet per hour
QT = Q x L x T x square feet lot area x 0.0411 = cubic feet per hour
QT = 5.5 cubic feet per second per acre x 0.96 (L) x 0.92 (T) x 20,000 square feet x 0.0411 = 3,993 cubic feet per hour = 1.1 cfs

Your design

QT = __ 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 QT from the size of the sewer pipe in Table 3.

QT = __ 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

QT = __ 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

Worksheet for concrete settling basins

Open lot area draining into the basin

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

2. Inflow rate into settling basin

Q = peak runoff rate in cubic feet per second per acre (Table 1)
L = location factor (Figure 5)
T = topographic factor (Table 2)
QT = inflow rate into settling basin, cubic feet per hour
QT = Q x L x T x square feet lot area x 0.0411 = cubic feet per hour

QT = __ 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 QT from the size of the sewer pipe in Table 3.

QT = __ 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

QT = __ estimated gpm x 8 = __ cubic feet per hour

Surface area (SA) of settling basin

SA (square feet) = QT (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 = QT/1,250. If a riser pipe is used as an overflow device instead of a rectangular weir, riser pipe diameter, inches = QT/274. Do not use a riser pipe smaller than 6-inches diameter.

Minimum rectangular weir length = QT/1,250 = __ /1,250 = __ feet

Minimum riser pipe diameter = QT/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
*Multiply by 1 for mechanical removal or by 2 for pumping.
Basin depth (concrete lot or confinement building) = 0.5 x ____ feet3 per day manure prod. (Table 4) x _ days storage x (1* or 2*)/ __ square feet surface area (item 3) = __feet
*Multiply by 1 for mechanical removal or by 2 for pumping.

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.

References

WQ326, new August 1995

WQ326 Settling Basins and Terraces for Dairy Waste | University of Missouri Extension

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