Reviewed by Zhiqiang Hu
Professor of Civil and Environmental Engineering

Biosolids are useful for land application. Examples of beneficial use include application to agricultural land and reclamation sites (e.g., mining sites).

Although, it contains organic matter and plant nutrients, such as nitrogen (N) and phosphorus (P), biosolids material from municipal waste treatment plants does not have a uniform composition. Waste characteristics and treatment technologies are different for each community. Therefore, it is required to determine the composition of the biosolids to meet nutrient management guidelines.

General characteristics

The biosolids dry matter content (total solids content) for land application ranges from a low of 0.5 percent to more than 50 percent. Fortunately, biosolids material from a specific plant has a narrow range in solids content over time. Each plant is designed to yield certain amounts of biosolids and effluent.

The sewage treatment facility that generates biosolids from wastewater must meet certain sampling requirements. A daily sample determines the percentage of total dry solids. In addition, each facility must sample the biosolids for chemical content. The Missouri Department of Natural Resources (DNR) dictates the sampling schedule by the volume of biosolids generated,

The facility submits the biosolids samples to an approved laboratory. The lab uses standard U.S. Environmental Protection Agency (EPA) and/or Missouri Department of Natural Resources (DNR) guidelines. The lab reports the analysis to the treatment facility. It is this report that is the subject of this guide.

The laboratory report

All reports include certain information, but the format may vary. Table 1 lists the essentials found in a good report, as well as some nice-to-know facts.

Table 1
Information necessary for a complete laboratory report of the analysis of a biosolids sample

Units

The report's measuring unit should be weight of measured material per unit of total dry solids. The lab takes a volume of biosolids, weighs it, dries it and then weighs the dry material. The following is a calculation of percent of total solids (Equation 1):

Equation 1

The lab results are reported in a total dry solids measurement in one of three units — parts per million (ppm), milligrams per kilogram (mg/kg) or percent (Table 2). These same units can also refer to an "as is" or wet weight analysis, so be sure that the lab report says "Dry Weight." Milligrams of a constituent per kilogram dry biosolids (milligrams per kilogram) is the preferred unit to use. PPM and milligrams per kilogram are numerically equal as shown in Equation 2.

Table 2
Example of analytical information found in a laboratory report made on biosolids from a publicly owned treatment works (POTW)

Source: Columbia — Perche Creek Facility annual report, 1994.

Other conversions are necessary, since other units may be used, (Equations 2, 3 and 4):

• of ppm to milligrams per kilogram:

Equation 2

1 ppm = 1 milligram per kilogram

Note
1 milligram (mg) = 0.001 gram and 1 kilogram (kg) = 1,000 grams

• of ppm to percent:

Equation 2

• of percentage to ppm:

Equation 4

percent x 10,000 = ppm

Another useful calculation is conversion of ppm or milligrams per kilogram to pounds per dry ton. A dry ton is 2,000 pounds.

Equation 5

Some laboratories report results on an as-received basis (wet basis). In such cases, the lab should also report the percentage of total solids in the sample. If the report does not specify wet or dry, contact the lab. Percentage moisture or water may be reported by some laboratories,

Percent water + percent solids = 100 percent

Percent solids = 100 percent – percent water

Conversion from the as-received to the dry basis or vice-versa is easy if you know the percent total dry solids in the sample (Equation 6 and 6a):

Equation 6

Equation 6a

For example, 1 ppm of an element on the as-received (wet) basis in a sample with 5 percent total solids is:

The wet-to-dry conversion and its reverse calculation are important. The biosolids are applied wet, so you must calculate the quantity of wet biosolids per acre.

Terminology

Table 1 contains the general information found in laboratory reports. There are labs that report in terms of phosphate and potash. In this case, you are not quite sure what is being reported. Likely the lab means P₂O₅ and K₂O — the units used by the fertilizer industry to sell phosphorus and potassium fertilizers.

You may convert P₂O₅ percentage and K₂O percentage to the preferred elemental basis (P percentage and K percentage) as follows:

Percentage P₂O₅ x 0.44 = percentage P

Percentage K₂O x 0.83 = percentage K

Table 2 contains a biosolids analysis. The following list explains some terms and acronyms used either on lab reports or used to modify the analytical results.

Total Kjeldahl Nitrogen (TKN)
This term refers to certain forms of N in the sample. Kjeldahl first developed this standard procedure. TKN usually includes organic nitrogen and ammonium-N (NH₄⁺-N), but not nitrate-N (NO₃^--N).

Plant Available Nitrogen (PAN)
PAN is an estimate of the fraction of applied nitrogen which is available to plants. The calculated value includes a percentage of the organic nitrogen and ammonium nitrogen, plus the nitrate nitrogen

Equation 7

PAN = f_o (organic N) + f_a (NH₄-N) + NO₃-N

The availability factors, f_o and f_a, are necessary because all of the organic N is not plant available the first year and some of the ammonium nitrogen may volatilize.

Ammonium nitrogen (NH₄⁺-N)
A portion of N in biosolids may be ammonium nitrogen (NH₄⁺-N). When biosolids containing NH₄⁺-N are applied to soil or crop residues, some of the NH₄⁺-N may be lost by volatilization. The amount of volatilization depends upon temperature, pH of the soil and rainfall or snow, and the time period until a tillage or incorporation operation is performed. An availability factor f_a (0.7) is used for surface-applied biosolids. If the biosolids are injected there is no loss, and the availability factor equals 1.

Nitrate nitrogen (NO₃^--N)
The portion of total N in the nitrate (NO₃^--N) form is soluble and not subject to direct volatilization. All NO₃^--N in biosolids is considered plant available.

Organic nitrogen
Organic nitrogen is the portion of total N in biosolids not present as NH₄⁺-N or NO₃^--N. Typically, 20 percent of the organic nitrogen is plant available the year the biosolids are land applied, i.e., f_o = 0.2. The nitrogen becomes available through mineralization by soil microbes.

Calculations

How can the biosolids data in Table 2 relate to the plant needs on an application site? After all, nutrient management plans require a connection to plant needs. This prevents increasing soil nutrients to levels that will adversely affect ground and surface waters.

Nitrogen

The first limit to land application of biosolids deals with nitrogen.

DNR rules state, "Annual biosolids application rates shall not exceed agronomic rates for plant nitrogen needs." The rules set up two different cases for documenting N applications.

Case 1

When the biosolids N content does not exceed 50,000 milligrams per kilogram (5 percent) and the biosolids application rate does not exceed 2 dry tons per acre, PAN reporting is not required.

This guideline means that up to 200 pounds of TKN per acre may be applied without further calculations:

Case 2

In situations not covered by Case 1, report nitrogen compounds in the PAN calculations. Use MU Extension fertility guidelines to determine PAN loading and crop removal to determine agronomic rates.

For example, assume the biosolids in Table 2 are surface applied at 3 dry tons per acre.

Step 1
Calculate organic N concentration:

Organic N = TKN - NH₄-N

Organic N = 45,226 milligrams per kilogram -19,757 milligrams per kilogram

Organic N = 25,469 milligrams per kilogram

Step 2
Calculate PAN:

PAN milligrams per kilogram = f_o (organic N) + f_a (NH₄-N) + NO₃-N

PAN milligrams per kilogram = (0.2) (25,469) + 0.7 (1,975) + 28

PAN milligrams per kilogram = 18,952 milligrams per kilogram

PAN pounds per dry ton = 0.002 x 18,952 (Equation 5)

PAN pounds per dry ton = 38 pounds PAN per dry ton at 3 dry tons per acre

Total PAN pounds per acre = 3 tons x 38 pounds per ton = 114 pounds per acre

The next question that arises concerns the agronomic requirement of the crop. Refer to application requirements.

Other nutrients

Currently, N is the key rate to applying biosolids. Application limits depend upon PAN in the biosolids and crop removal. It is helpful to calculate the amount of other nutrients delivered by the biosolids, also.

There is no availability factor used with any elements listed in the lab report, except N. If you apply 3 dry tons per acre of the biosolids represented in Table 2, the quantity of P and K is 114 and 23, respectively:

19,033 milligrams per kilogram x 0.002 x 3 tons per acre = 114 pounds P per acre

3,789 milligrams per kilogram x 0.002 x 3 tons per acre = 23 pounds K per acre

Pollutants

Check the metal pollutants in the biosolids to determine if any limits are exceeded. The calculation is the same as for P and K.

An example is zinc and mercury when 3 dry tons per acre of biosolids are applied (Table 2):

1,252 milligrams Zn per kilogram x 0.002 x 3 dry tons = 7.5 pounds Zn per acre

8 milligrams Hg per kilogram x 0.002 x 3 dry tons = 0.048 pounds Hg per acre

Application rates in gallons

The sample of biosolids represented in Table 2 has only 4.9 percent solids. It is handled as fluid from the treatment facility to the land application site. In this case, the units of measure may be gallons.

You may prefer to calculate the biosolids per 1,000 gallons, so the hauler can more easily determine the rate to apply per acre.

In the following calculations, nutrient refers to TKN, PAN, P, K and metal.

Step 1
Converting the concentration dry basis to wet basis (Equation 6a):

Step 2
Converting a weight of wet biosolids to gallons:

Assume the weight of a gallon of water (8.3 pounds) is the same as a gallon of biosolids (specific gravity of biosolids is assumed to equal 1):

This means that if a nutrient concentration in wet biosolids is 1 ppm (1 pound per 1,000,000 pounds biosolids), the concentration is 1 pound per 120,482 gallons or 0.0083 pound per 1,000 gallons. An example is:

19,757 milligrams NH₄-N per kilogram (Table 2)

Application rate in cubic feet

Biosolids with more than 20 percent total dry solids may likely be handled in cubic feet or in cubic yards. It is assumed that the specific gravity is one. Thus, one cubic foot of biosolids weighs 62.4 pounds.

For example, assume the percent solids in Table 2 is 25 percent and that we are dealing with NH₄⁺-N (19,757 milligrams per kilogram dry basis):

Remember milligrams per kilogram is equal to ppm.

Or, NH₄-N per wet (as applied) ton could be calculated as follows:

19,757 ppm NH₄-N x 0.002 = 39.5 pounds NH₄⁺-N per dry ton

39.5 pounds NH₄-N per dry ton x 0.75 = 29.6 pounds NH₄⁺-N per wet ton

Original authors: James R. Brown, School of Natural Resources; Dennis Sievers, Department of Agricultural Engineering; Robert Magai, Environmental Specialist, Missouri Department of Natural Resources