Interpretation of Laboratory Analysis of Biosolids Samples
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 nicetoknow facts.
Table 1
Information necessary for a complete laboratory report of the analysis of a biosolids sample
General information  

 
Analytical information  
All parameters, except percent of solids, are measured on a dry matter basis. The preferred units of expression are milligrams constituent per kilogram dry biosolids (milligrams per kilogram).  
Required  
Additional tests required if TKN is more than 50,000 milligrams per kilogram and/or application rate of biosolids is more than 2 dry tons per acre:
 
Useful

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
_ Weight of dry solids _ Weight of initial sample  x 100  = percent total dry solids 
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)
Constituent  Concentration (milligrams per kilogram, dry weight) 

Total solids  49,000 (4.9 percent) 
Arsenic (As)  <5 
Cadmium (Cd)  5 
Chromium (Cr)  63 
Copper (Cu)  709 
Lead (Pb)  147 
Mercury (Hg)  8 
Molybdenum (Mo)  14 
Nickel (Ni)  50 
Selenium (Se)  4 
Zinc (Zn)  1,252 
Total Kjeldahl Nitrogen  45,226 
Organic N  25,470 
Ammonium N  19,757 
Nitrate N (NO_{3} – N)  28 
Total P  19,033 
Total K  3,789 
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
ppm x  ___1___ 10,000  = percent 
 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
ppm x 0.002 =  __pounds___ dry ton 
Some laboratories report results on an asreceived 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 asreceived to the dry basis or viceversa is easy if you know the percent total dry solids in the sample (Equation 6 and 6a):
Equation 6
percent wet basis percent solids/100  = percent dry basis 
Equation 6a
percent dry basis x  percent solids 100  = percent wet basis 
For example, 1 ppm of an element on the asreceived (wet) basis in a sample with 5 percent total solids is:
1 ppm ÷  __5__ 100  = 20 ppm dry basis 
The wettodry 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 P_{2}O_{5} and K_{2}O — the units used by the fertilizer industry to sell phosphorus and potassium fertilizers.
You may convert P_{2}O_{5} percentage and K_{2}O percentage to the preferred elemental basis (P percentage and K percentage) as follows:
Percentage P_{2}O_{5} x 0.44 = percentage P
Percentage K_{2}O 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 ammoniumN (NH_{4}^{+}N), but not nitrateN (NO_{3}^{}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} (NH_{4}N) + NO_{3}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 (NH_{4}^{+}N)
A portion of N in biosolids may be ammonium nitrogen (NH_{4}^{+}N). When biosolids containing NH_{4}^{+}N are applied to soil or crop residues, some of the NH_{4}^{+}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 surfaceapplied biosolids. If the biosolids are injected there is no loss, and the availability factor equals 1.
Nitrate nitrogen (NO_{3}^{}N)
The portion of total N in the nitrate (NO_{3}^{}N) form is soluble and not subject to direct volatilization. All NO_{3}^{}N in biosolids is considered plant available.
Organic nitrogen
Organic nitrogen is the portion of total N in biosolids not present as NH_{4}^{+}N or NO_{3}^{}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:
2 dry tons per acre x  2,000 pounds tons  x  __5__ 100  = 200 pounds TKN per acre per year 
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  NH_{4}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} (NH_{4}N) + NO_{3}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):
Nutrient milligrams per kilogram dry basis x  percent total solids 100  = nutrient milligrams per kilogram wet basis 
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):
1,000,000 pounds biosolids  = 120,482 gallons biosolids 
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 NH_{4}N per kilogram (Table 2)
19,757 x  _4.9_ 100  x 0.0083 = 8.0 pounds per 1,000 gallons 
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 NH_{4}^{+}N (19,757 milligrams per kilogram dry basis):
Remember milligrams per kilogram is equal to ppm.
mg NH_{4}N per kilogram wet = 19,757 x  _25_ 100  = 4,939 milligrams per kilogram 
__4,939 pounds__  x 62.4 pounds = 0.308 pounds NH_{4}^{+}N per cubic foot 
Or, NH_{4}N per wet (as applied) ton could be calculated as follows:
19,757 ppm NH_{4}N x 0.002 = 39.5 pounds NH_{4}^{+}N per dry ton
39.5 pounds NH_{4}N per dry ton x 0.75 = 29.6 pounds NH_{4}^{+}N per wet ton