Case study for sewage sludge application

by Ted Sammis

Introduction

Sewage material for the city of Las Cruces New Mexico passes through a sewage treatment plant and the output of the final digester is sewage sludge. For the last 15 years this material has been land applied to 39 acres of land. The Forman at the site stated that the sewage sludge distribution truck applies 4000 gal a load and twenty loads a day are applied to the land 3 times a week . The depth to the water table is three hundred feet and the monitoring well located in the south east corner of the land has not detected any nitrogen increase from the background levels. Soil samples indicated that the nitrogen level was 400 mg/l at the 4 ft depth and about 140 mg/l at the 20 ft depth. EPA has indicated that the site should be closed in the next two years.

Question

The question is how long before the nitrates in the system reaches the water table if the system continues to operate and how long will the front take to reach the water table if the disposal system is shut down. Should the disposal system be shut down.

Solution

The location of the site is 1/2 mile south of highway 10 on the west mesa out side of Las Cruces New Mexico at  T23S R1W.  Map 48 on the Dona Ana soil survey.
The soil is classified at a blue point loam sand 1-15 percent slope.

Calculation of sewage sludge application per year.

Gallons of water application = 4000gal/trip*20trip/day*3day/week*52weeks/year = 12.48 *10^6 gal /year

Depth = volume/area                                                                                                                                        (1)

volume 12.48 *10^6 gal/year = 44.6 ac-ft
see http://www.speckdesign.com/Tvolume.html for conversion table.

Area = 39 ac

Depth = 44.6 ac/ft/ 39 ac = 1.14 ft of water applied each year

Rainfall in the area  on a long term average at State University Las Cruces is 9.29 inches =0.77 ft

See http://www.wrcc.dri.edu/cgi-bin/cliMAIN.pl?nmstat For climate data

Total depth  of rainfall plus sewage sledge assuming all the rainfall is effective rainfall which would be the worse case scenario is:
Depth = 1.14 ft + 0.77 ft = 1.9 ft.

Calculation of depth of sewage sludge front.

Assumption

Assume piston flow with the center of the dispersion front equal to:  depth of water applied/ water content of the soil profile.

Depth of front= depth of water applied / water content soil profile                                                                                             (2)

Darcy's Law controls water movement. The flow of liquid through a porous media is in the direction of and at a rate proportional to the driving force acting on the liquid (dH/dx,                 hydraulic gradient) and the property of the conducting media to transmit the liquid conductivity.

q= K (q )* dH/dx                                                                                                                                                                                       (3)

where:
q is the flux with units L/t,
K(q ) is the unsaturated conductivity with units L/t
H and x have units Length (L)

Q= q*a where Q has units of v/t and a is area  normally a unit area.

At below 5-10 ft the gradient approaches a unit gradient where dH/dx = 1

Then : q= K(q )

The water content will reach a level that under unit gradient  will have a conductivity of 1.9 ft /year.

Based on Darcy's law the value of K(q ) can be calculated from Campbell's hydraulic conductivity equation for unsaturated flow.

K(q )= Ksat (q /q s)^(2b+3)                                                                                                                                                                     (4)

where:
Ksat is the saturated conductivity
q is the moisture content of the soil cm^3 H2 O/ cm^3 soil
q s is the saturated moisture content of the soil cm^3 H2 O/ cm^3 soil
b is a constant depending on soil type.

The values for Ksat and b and the other soil parameters can be gotten on the internet at: http://weather.nmsu.edu/irrdoc/soilfc.html

For a loamy sand the values are:
Ksat= 2.44 inch/hr
b= 2.856
q s= 1-Bd/Pd
Bd is bulk density = 1.54 g/cm
Pd= particle density- 2.65 g/cm
q s= 0.42 cm^3 H20/cm^3 soil
K(q ) is 1.9 ft/ year = 0.0026 inch/hr

Solving  for q in eq. 4 results in  q = 0.19 cm^3 H2 O/ cm^3 soil

Check this answer against field capacity . The value for field capacity is 0.183 cm^3 H2O/ cm^3 soil
The numbers should be similar or the number should be lower than field capacity. However, over a 15 year period the soil will reach an equilibrium moisture content that will                conduct 1.9 ft a year of sewage sledge.

To check the calculations go to the site and measure the moisture content.

Returning to the problem,  solve eq. 2 for the depth of the sewage sledge front:
Depth = 1.9 ft/year/ 0.19 cm^3 H2 0/ cm^3 soil = 10 ft /year

Over a 15 year period the sewage sledge will reach 150 ft or half the distance to the water table. Consequently, the need to close the sewage sledge application facilities in the next two years is not critical. However, if the system is keep opened , the front will continue to move at 10 ft / year and in 15 more years will reach the water table. If the sewage sledge application facilities is closed the front will move very slowly toward the ground water as the moisture content in the 150 depth decreases and is redistributed over deeper depths. The rate of movement will depend on the moisture content and unsaturated conductivity at that moisture content over the wetting front profile. The rate of movement will decrease with time after the facilities are shut down. .

Depth of Chemical front

If the sledge contained chemical material then the chemical could move slower then the water.The equation controlling the depth of the chemical front is:

Depth= q/ q R
R = retardation factor has no units
R= 1+ rKd/q
r  =
density of soil g/cm^3
q is the moisture content of the soil cm^3 H2 O/ cm^3 soil

Kd is the linear sorption coefficient or the partition coefficient of the chemical in the soil   cm^3 H2 O/g soil or g H2 O/g soil because  the density of water is 1 g/cm^3
Kd= Koc OC
Koc= the linear sorption coefficient normalized by the organic carbon content (OC) of the soil.

The chemical can also be degraded by microorganisms resulting in a half life for the chemical in the upper portion of the soil profile-top 3 ft. Below 3 ft microorganisms are small in number. Pesticides have half   life's range from  1-500 days and  Koc range from  1- 12,400 ml/g OC

Nitrogen will move with the water but any phosphorus will be adsorbed with a Kd value of 60.    P in sewage sludge is 0.6% of moist sludge and the moisture content of sledge is 45%

Calculation of phosphorus front
R= 487     because phosphorus does not move in the soil profile
The depth of the phosphorus  front is:
Depth 1.9ft/year  / 487 *0.19  *15 years =  0.3 ft

The sledge also contains heavy metals that will have different retardation factors. The sledge will contain other chemicals depending on the waste stream coming into the sewage treatment plant.  A calculation for the depth of the heavy metals of other chemicals can be calculate as was done for the phosphorus front.

References

Campbell, Gaylon S. Soil physics with BASIC : transport models for soil-plant systems / Gaylon S. Campbell. PUBLISHER: Amsterdam ; New York : Elsevier, 1985.