I. Units and Conversions

1. A) The decimal equivalent (DE) of the percent total solids equals the percentage divided by 100.

   % TS	=	DE of Total Solids
   100

   Example:	5% TS	=	0.05 DE
   100

Laboratory analyses are reported on either a wet weight ("as-received") basis or on a dry weight basis. The units for a wet weight basis are millgrams per liter (mg/l – weight per volume). The units for a dry weight basis are milligrams per kilogram of solids (mg/kg – weight per weight). Sludge represents a material in which most of the solid matter is undissolved and the dissolved fraction is of minor importance. It is generally simpler to perform sludge calculations by using dry weight units. Furthermore, some calculations must ultimately be expressed in dry weight units to be correct. For these reasons all the sample calculations are worked on a dry weight basis. However, since may laboratories report results on a wet weight basis, conversion relationships are provided below. Be aware that some laboratories report results in mg/kg on an "as-received" basis. You should consult the laboratory to confirm this and then convert the units to a dry weight basis. Finally, note that the specific gravity of liquid and most dry sludges can be assumed to be 1.0 (equal to water) and sufficient accuracy in the calculations will be obtained.

B) Wet Weight to Dry Weight Basis

1. C) Dry Weight to Wet Weight Basis

   (mg/kg)

   x

   (DE)

   =

   mg/l (wet wt. basis)

   Example:

   (30,000 mg/kg TKN)

   x

   (0.05)

   =

   1500 mg/1 TKN (wwb)

2. D) Other useful conversions

   1)	ppm (wet.)	x	100	=	ppm (dry)
   %TS

3. 2) ppm x 0.002 = lb/ton
4. 3) 10,000 ppm (or mg/l or mg/kg) = 1%

5. 4) 1 acre = 43,560 square feet

   5)	(dry tons sludge)	x	100	=	wet tons sludge
   %TS

   6)	Wet tons	x	2000	=	wet tons sludge
   acre	8.345

6. 7) 1 cubic yard of drying bed sludge is approximately equal to 0.45 dry tons.
7. 8) 1 mg/kg = 0.002 lb/ton

8. II. Assumptions for Sludge Calculations

   1. A) Sludge will be surface applied with incorporation by disking or chisel plowing.
   2. B) Soils are non-sandy at the application sites.
   3. C) The sludge has been well stabilized by heat anaerobic digestion.

   4. D) Laboratory analysis of sludge (dwb):

      5% TS

      Total Kjeldahl Nitrogen (TKN) = 30,000 mg/kg

      Ammonia Nitrogen = 10,000 mg/kg

      Phosphorus = 8,000 mg/kg

      Potassium = 3,500 mg/kg

      Cadmium = 30 mg/kg

      Copper = 2000 mg/kg

      Manganese = 1000 mg/kg

      Nickel = 400 mg/kg

      Lead = 1000 mg/kg

      Zinc = 4000 mg/kg

   5. E) Corn for grain is grown and the average yield is 110 bushels per acre per year.

9. III. Calculating Agronomic Nitrogen Application Rates of Sludge

   1. A) Determine the availability of nitrogen forms by referring to Section 391.411.

2. Organic nitrogen plant availability is 20% for the first year and decreases as shown in Table I.

1. B) First Year Application Rate

1) Organic nitrogen is not a laboratory test. It is a calculated value as shown below.

Organic N = Total Kjeldahl N - Ammonia N

Organic N = 30,000 - 10,000 = 20,000 mg/kg

2) Calculate the plant available nitrogen (PAN) in the sludge as follows:

Ammonia Nitrogen: 10,000 x 0.8 = 8,000 mg/kg

Organic Nitrogen: 20,000 x 0.2 = 4,000 mg/kg

PAN = 8,000 + 4,000 = 12,000 mg/kg

This means that each dry ton of sludge solids will have 24 pounds of nitrogen available for utilization by plants when the sludge has been disked into the soil. Note that if the sludge has been injected into the soil there would have been 28 pounds of plant available nitrogen provided. If the sludge had been surface applied without incorporation there would have been only 18 pounds of plant available nitrogen provided.

3) Calculate the agronomic nitrogen requirement for the corn grain crop using the yield and the values from Appendix B – Table IV:

This means that each acre of corn with the stated yield requires 143 pounds of plant available nitrogen for proper growth.

4) Calculate the sludge application rate needed to provide the required plant available nitrogen.

5) For convenience during the actual sludge application it is usually helpful to convert the application rate into gallons per acre.

1. C) Second Year Application Rate

2) Determine the amount of organic nitrogen (O.N.) applied during the first year that is remaining for plant uptake during the second year by using Table I values and the first year application of organic nitrogen.

(20,000 mg/kg O.N.) x (0.20) = 4,000 mg/kg O.N. used up first year

20,000-4,000 = 16,000 mg/kg Organic N remains for second year use

(16,000 mg/kg) x (10%) = 1600 mg/kg O.N. available during second year

This represents:

Therefore, the second year crop will need:

3) Calculate the second year sludge application rate.

1. D) Third year application rate

a) From the first year application:

(16,000 mg/kg O.N.) x (0.10) = 1600 mg/kg O.N. used the second year

16,000-1600 = 14,400 mg/kg O.N. remaining from the first year sludge application.

(14,400 mg/kg O.N.) x (5%) = 720 mg/kg O.N. is available during the third year

This represents:

b) From the second year application:

1. c) Sum the available organic nitrogen from previous years application of sludge and subtract that sum from the crop nitrogen requirements.

   143-(8.58 + 16.5)	=	118	lb. PAN	of additional nitrogen is needed.
   acre

2. d) Calculate the third year application rate.

   118	=	4.92	dry tons sludge
   24	acre

   4.92	dT	x	2000	x	1	=	23,580	gallons
   acre	8.345	0.05 DE	acre

(20,000 mg/kg O.N.) x (0.20) = 4,000 mg/kg O.N. used

20,000 - 4,000 = 16,000 mg/kg O.N. remains

(16,000 mg/kg) x (0.10) = 1600 mg/kg O.N. available

E) Fourth year application rate

1. 1) In the fourth year some organic nitrogen applied during the previous three years will become available for plant utilization; 2.5%, 5%, and 10% respectively from Table I.
2. 2) Determine the amount of organic nitrogen applied during previous years that is available to plants during the fourth year.

3. a) From the first year application:

   14,400-mg/kg O.N.) x (0.05) = 720 mg/kg used during the third year.

   14,400-720 = 13,680 mg/kg O.N. remaining

   (13,680 mg/kg) x (2.5%) = 342 mg/kg O.N. available

   (342 mg/kg) x (0.002) x	(	5.96	dT	)	= 4	lb. O.N.	available
   acre	acre

4. b) From the second year application:

   (16,000 mg/kg O.N.) x (0.10) = 1600 mg/kg O.N. used

   16,000-1600 = 14,400 mg/kg O.N. remains

   (14,400 mg/kg) x (0.05) 720 mg/kg O.N. available

   (720 mg/kg) x (0.002) x	(	5.17	dT	)	= 7.4	lb. O.N.	available
   acre	acre

5. c) From the third year application:

   1. d) Sum the available organic nitrogen from previous years application of sludge and subtract that sum from the crop nitrogen requirements.

      143-(4 + 7.4 + 15.7)	=	116	lb. PAN	of additional nitrogen is needed.
      acre

   (20,000 mg/kg O.N.) x (0.20) = 4,000 mg/kg O.N. used

   20,000-4,000 = 16,000 mg/kg O.N. remains

   (16,000 mg/kg) x (0.10) = 1600 mg/kg O.N. available

   (1600 mg/kg) x (0.002) x	(	4.92	dT	)	= 15.7	lb. O.N.	available
   acre	acre

6. e) Calculate the fourth year application rate.

   116	=	4.83	dry tons sludge
   24	acre

   4.83	dT	x	2000	x	1	=	23,150	gallons
   acre	8.345	0.05 DE	acre

F. Fifth year application rate

1. 1) In the fifth year some of the organic nitrogen applied during the previous four years will become available for plant utilization; 1.25%, 2.5%, 5%, and 10% respectively from Table I.
2. 2) Determine the amount of organic nitrogen applied during previous years that is available to plants during the fifth year.

3. a) From the first year application:

   (13,680 mg/kg) x (0.025) = 342 mg/kg used during the fourth year.

   13,680-342 = 13,338 mg/kg O.N. remaining

   (13,338 mg/kg) x (0.0125) = 167 mg/kg O.N. available

   (167 mg/kg) x (0.002) x	(	5.96	dT	)	= 2	lb. O.N.	available
   acre	acre

4. b) From the second year application:

   (14,400 mg/kg) x (0.05) = 720 mg/kg O.N. used

   14,400-720 = 13,680 mg/kg O.N. remaining

   (13,680 mg/kg) x (0.025) = 342 mg/kg O.N. available

   (342 mg/kg) x (0.002) x	(	5.17	dT	)	= 3.5	lb. O.N.	available
   acre	acre

5. c) From the third year application:

   (16,000 mg/kg) x (0.10) = 1600 mg/kg O.N. used

   16,000-1600 = 14,400 mg/kg O.N. remaining

   (14,400 mg/kg) x (0.05) = 720 mg/kg O.N. available

   (720 mg/kg) x (0.002) x	(	4.92	dT	)	= 7	lb. O.N.	available
   acre	acre

6. d) From the fourth year application:

   1. e) Sum the available organic nitrogen from previous years application of sludge and subtract that sum from the crop nitrogen requirements.

      143-(2 + 3.5 +7 + 15.5)	=	115	lb. PAN	of additional nitrogen is needed.
      acre

   (20,000 mg/kg) x (0.20) = 4,000 mg/kg O.N. used

   20,000-4,000 = 16,000 mg/kg O.N. remaining

   (16,000 mg/kg) x (0.010) = 1600 mg/kg O.N. available

   (1600 mg/kg) x (0.002) x	(	4.83	dT	)	= 15.5	lb. O.N.	available
   acre	acre

7. f) Calculate the fifth year application rate.

   115	=	4.79	dry tons sludge
   24	acre

   4.79	dT	x	2000	x	1	=	22,960	gallons
   acre	8.345	0.05 DE	acre

8. G) During and after the sixth year there will not be any significant amount of organic nitrogen available to the crop from sludge applied during the first year. An equilibrium situation was reached in this sample problem during the fourth year since there is practically no difference between the fourth and fifth year application rates. A conservative operating plan would be to apply sludge at approximately 4.8 dry tons per acre per year. Note, however, that this would mean that additional nitrogen would be needed from fertilizers during the first, second, and third years of sludge application to meet the nitrogen requirement for the crop grown.

IV. Metal Loading Rate Calculations

1. A) It is useful to estimate the lifetime of a site receiving sludge on the basis of metal loadings. The method for performing this estimation is given below assuming the sludge chemical analysis provided the following quantities.

   Percent Solids = 5% (0.05 decimal equivalent)
   Cadmium	30 mg/kg
   Copper	2000 mg/kg
   Nickel	400 mg/kg
   Lead	1000 mg/kg
   Zinc	4000 mg/kg
   Manganese	1000 mg/kg

   Application rate 4.8 dry tons/acre/year

2. B) Set up and complete a table as follows:

   Metal	Sludge Analysis	Loading Factor	Annual Metal Loading (lb/acre/year)	Maximum Metal Loading (lb/acre)	Site Life (Years)
   Cadmium	30 mg/kg	0.0096	0.29	10	34
   Nickel	400 mg/kg	0.0096	3.84	100	26
   Copper	2000 mg/kg	0.0096	19.2	250	13
   Zinc	4000 mg/kg	0.0096	38.4	500	13
   Manganese	2000 mg/kg	0.0096	19.2	900	46
   Lead	1000 mg/kg	0.0096	9.6	1000	104

3. C) The loading factor column will be the same for each metal and is calculated as follows:

   0.002 #/dry ton	x	(dry tons/acre/year) = 0.002 x 4.8 = 0.0096
   mg/kg

4. D) Multiply the metal analysis value by the loading factor value (0.0096) to obtain the Annual Metal Loading (lb/acre/year) column.

   Divide the maximum metal loading for each metal (from Table II, Section 391.420) by the annual metal loading rate (pounds/acre/year) to obtain the site lifetime.

5. E) This example indicates that the zinc and copper loading is the most restrictive to the site lifetime. Therefore a site should be used for no more than 13 years, however updated sludge analyses and different application rates may change these site lifetimes. Records should be kept on each site that sludge is applied to for each year's metal loading.