This memo style report analyzes an approach to the reduction of ground toxins in an urban environment using counter-current immiscible extraction technology. It emphasizes the applications of chemical engineering separations in real world situations. This report was created as part of a course observing advanced chemical separation processes at Worcester Polytechnic Institute.

1. Hazardous Waste Site Remediation Project
Beyond Separations Limited
CHE 2014
Submitted to the vice presidents of Stromboli Environmental
March 2, 2018
__________Zachary Powers____________
Zach Powers
Ryan Bowe _
Ryan Bowe
___________Jeffrey Page___________
Jeff Page

2. Introduction
The goal of this project is to design a process for extracting Tetrachloroethane (TCE) and
Dichloroethane (DCE) from the soil in the town of Euclid Ohio. TCE and DCE are harmful
pollutants which get released into the atmosphere during the metal cleaning process. Both TCE
and DCE are considered volatile organic compounds which can have negative effects on the
environment in high concentrations.1,2
TCE and DCE are both harmful when humans are
exposed by either inhalation or ingestion.1,2
Additionally, there is reason to believe that TCE and
DCE are both carcinogenic.1,2
It is feared that the presence of TCE and DCE in the soil could
lead to severe contamination of the town’s drinking water. Therefore, it would be very beneficial
for the long term health of the area both financially and environmentally for the TCE and DCE to
be removed from the project site in order for further development to be done. The design of the
system will use water as a solvent to extract TCE and DCE from the soil before the contaminated
water is processed at a different site. The system will be a countercurrent extractor in which
water with no solute flows in the opposite direction of contaminated soil. The system is
considered immiscible which means that none of the soil remains in the exiting solvent stream
and a small but negligible amount water remains in the soil. The system will produce two
streams one of which is a clean soil with a limited quantity of TCE and DCE which can be
reused at the same site it was taken from and the other is a contaminated water stream which
requires further processing.
Methodology
The first step to addressing this problem is identifying known flow rates and compositions of the
diluent, solvent, and solute. In this case, the diluent is the soil, the solvent is water, and the
solutes are DCE and TCE. The feed rate of the contaminated soil solution is 30 tons/hr.
Additionally, the feed composition is 50 ppm TCE and 15 ppm DCE. The solute is pure water,
meaning both contaminant compositions in this are 0. The treated stream compositions have a
maximum parameter of 8 ppm for TCE and 5 ppm for DCE. Since this is an immiscible
extraction, there is essentially no water in the returning soil. As mentioned in the introduction,
this process is completed using counter-current immiscible extraction, diagrammed below:
Fig. 1: Countercurrent Immiscible Extraction

3. The equations 𝑦 = 0.03𝑥2
+ 0.4𝑥 for DCE and 𝑦 = 0.02𝑥2
+ 0.3𝑥 for TCE are used to create
equilibrium lines. On these graphs, different operating lines were established using the slope of
FD/FS to determine the final solute composition in water and to then step stages to find total
stages required, where FD is the flow rate of the diluent and FS is the flowrate of the solvent. This
value will provide us with the means to calculate F1, which will be defined below. Additionally,
the point on the x-axis where the final stage ends will provide us with the value to calculate F3,
also defined below.
To calculate cost, the total leaching cost will be the sum of the cost of stages (F1), the cost of
solvent (F2), and the cost of the contaminants returning to the ground (F3). This whole equation
can be expressed:
𝐿𝐶 = 6000(# 𝑜𝑓 𝑆𝑡𝑎𝑔𝑒𝑠) + (𝐹𝑠)(2000)(1.1) + 500(𝑓𝑖𝑛𝑎𝑙 𝑥 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛)
This can provide us with an overall leaching cost, which then is multiplied by 3 to factor in labor
cost, and then 2.5 for site preparation and maintenance costs. Finally, this new number is added
to 200,000 which covers overhead and administrative costs to provide a final cost for this
operation.
Finally, to calculate the time required to conduct this entire operation, the volume of the soil on
site must first be established. Once done, this can be used to find the total mass of the soil using
the specific gravity of the soil. This can then be converted into tons and then divided by the rate
of soil processed to find the total time required.
Results and Discussion
The most cost effective solvent flow rate was determined to be 35 tons/hr of water. In a DCE
extraction, there are 4 stages required to reach the target 5 ppm. In a TCE extraction, it takes 7
stages to reach the target 8 ppm. Since we are doing both extractions within the same system, the
theoretical number of stages the system will have is 7. The stage stepping with the resulting
concentrations from each step are shown in figures 2 and 3. The overall process requires 7 stages
even though the DCE equilibrium data results 4 stages for the required separation because both
separation processes are happening in the same unit. The leaching cost using this flow rate was
determined to be $129,900. Therefore, the total cost of the operation will be $1,005,350. Finally,
the operation will take 200 days if the process runs for 24 hours a day. A more reasonable time
estimation is 600 days were the system operates 8 hours a day which is still well within the 2
year goal, making it a feasible operation.

4. Fig. 2: Stages required to reach target levels for DCE Fig. 3: Stages required to reach target levels for TCE
DCE TCE
Flow Rate
(tons/hr)
Stages Final DCE level
(ppm)
Flow Rate
(tons/hr)
Stages Final TCE level
(ppm)
30 9 3.6 30 12 6.8
35 4 2.0 35 7 7.8
40 4 3.4 40 6 3.5
45 4 0.5 45 4 6.6
60 3 1.0 60 4 5.0
Table 1: Results for Stages and Final Contaminant Levels
Flow Rate (tons/hr) F1 ($) F2, ($) F3 ($) Leaching Cost ($)
30 72,000 66,000 5,200 143,200
35 42,000 77,000 4,900 123,900
40 36,000 88,000 3,450 127,450
45 24,000 99,000 3,550 126,550
60 24,000 132,000 3,000 159,000
Table 2: Costs of Stages, Solvent, and Returning Contaminant

5. Conclusion:
With a TCE and DCE extraction where 30 tons of soil are being treated per hour with 35
tons of solvent per hour, all parameters specified in the project are met or exceeded. This process
ensures that DCE and TCE levels in returning soil are well below the target concentrations,
which, when diluted by groundwater, will be safe for the people and animals in the immediate
vicinity. Additionally, the project completes 4 months before the 2-year deadline, satisfying the
time concerns of the community. Overall, it appears this process will sufficiently provide a
solution to the current pollution problem in Euclid, Ohio.

6. References:
“Tetrachloroethane.” Pollutant Fact Sheet,
apps.sepa.org.uk/spripa/Pages/SubstanceInformation.aspx?pid=87.
1,2-Dichloroethane MSDS; 107-06-2; Science Lab: Houston Texas, 10/10/2005.
http://www.sciencelab.com/msds.php?msdsId=9927153 (Accessed 3/29/18)

7. Appendix A: Cost Calculations
DCE Guess 1: Fs= 30 tons/hr
FD/FS = 1.0
Stage # = 9
X1 = 3.6 ppm
F1 = 6000*9 = 54000
F2 = 30*2000*1.1 = 66000
F3 = 3.6*500 = 1800
LC = 121800
DCE Guess 4: FS = 40 tons/hr
FD/FS = 0.75
Stage # = 4
X1 = 3.4 ppm
F1 = 6000*4 = 24000
F2 = 40*2000*1.1 = 88000
F3 = 500*3.4 = 1700
LC = 113700
TCE Guess 1: FS = 30 tons/hr
FD/FS = 1.0
Stage # = 12 (More than
DCE)
X1 = 6.8 ppm
F1 = 6000*30 = 72000
F2 = 30*2000*1.1 = 66000
F3 = 6.8*500 = 3400
LC = 141400
TCE Guess 4: FS = 40 tons/hr
FD/FS = 0.75
Stage # = 6 (More than DCE)
X1 = 6.6 ppm
F1 = 6000*6 = 36000
F2 = 40*2000*1.1 = 88000
F3 = 500*6.6 =1750
LC = 125750
DCE Guess 2: FS = 60 tons/hr
FD/FS = 0.5
Stage # = 3
X1 =1.0 ppm
F1 = 6000*3 = 18000
F2 = 60*2000*1.1 = 132000
F3 = 500
LC = 150500
DCE Guess 5: FS = 35 tons/hr
FD/FS = 0.857
Stage # = 4
X1 = 2.0 ppm
F1 = 6000*4 = 24000
F2 = 35*2000*1.1 = 77000
F3 = 2.0*500 = 1000
LC = 102,000 (BEST
CHOICE)
TCE Guess 2: FS = 60 tons/hr
FD/FS = 0.5
Stage # = 4 (More than DCE)
X1 = 5.0
F1 = 6000*4 = 24000
F2 = 60*2000*1.1 = 132000
F3 = 500*5 = 2500
LC = 136900
TCE Guess 5: FS = 35 tons/hr
FD/FS = 0.857
Stage # = 7 (More than DCE)
X1 = 7.8
F1 = 7*6000 = 42000
F2 = 35*2000*1.1 = 77000
F3 = 500*7.8 = 3900
LC = 122900 (BEST
CHOICE)
DCE Guess 3: FS = 45 tons/hr
FD/FS = 0.67
Stage # = 4
X1 = 0.5 ppm
F1 = 6000*4 = 24000
F2 = 45*2000*1.1 = 99000
F3 = 500*0.5 = 250
LC = 123250
DCE Guess 6: FS = 90 tons/hr
FD/FS = 0.33
Stage # = 2
X1 = 2.0
F1 = 6000*2 = 12000
F2 = 90*2000*1.1 = 198000
F3 = 500*2 = 1000
LC = 211000
TCE Guess 3: FS = 45 tons/hr
FD/FS = 0.67
Stage # = 4 (Same as DCE)
X1 = 6.6
F1 = 6000*4 = 24000
F2 = 45*2000*1.1 = 99000
F3 = 6.6*500 = 3300
LC = 126300

8. Determining total cost: LCTCE + F3, DCE = total LC = 123900
Labor costs = 124150*3 = 371700
Maintenance costs = 124150*2.5 = 309750
LC = 123900
Overhead costs = 200000
Total cost 1,005,350
Appendix B: Time Calculations
5 acres = 217800 ft2
Site volume = 217800 ft2
* 15 ft = 3267000 ft3
Mass of soil = specific gravity*density of water*site volume
Mass of soil = 1.4*62.427 lbs/ft3
* 3267000 ft3
= 285528612.6 lbs = 142764 tons
Operation time = mass of soil/soil processing rate
Operation time = 142764 tons / 30 tons/hr = 4758.81 hours = 198.28 days at continuous operation
Realistic operation time = 198.28 days * 3 = 594.85 days operating for 8 hrs/day