To calculate an oxygen removal rate pick a time point on the

-To calculate an oxygen removal rate pick a time point on the graph and see what the head space concentration is in the data provided. Since you know the initial headspace concentration at time zero you can calculate a removal rate. This will initially be in %/time but can be converted to mass per time by the density of air.

-After you find the oxygen removal rate make use of the mass ratio to find a HC removal rate.

-In part b I neglected to tell you that the depth of the contamination is 2m.

12) (20 Points) Soil is contaminated with ethylbenzene (C8Hio) at an industrial facility. You would like to determine if bioventing can be used to clean up the site. An in-situ test gives the following (graphed) results 25 Assume: e Soil bulk density: 1.9 kg/L 15 Porosity: 27 % e Density of oxygen: 1.4 g/L 10 Biodegradation is only the catabolic reaction (only CO25 and H20 produced) Density of air: 1.25 g/L R2 = 0.9946 e 10 20 30 Time (hours) 40 50 60 a) What is the estimated aerobic biodegradation rate (mg ethylbenzene per kg soil per day)? Some hints Use dimensional analysis to work toward your answer (mg ethylbenzene/kg soil-day) is the end point Assume the hydrocarbon (HC) degrades at the rate of oxygen consumption Write out the stoichiometric requirements of O2 and You can make use of the mass ratio= 106 mg HC/366 mg 02 b) Assume each bioventing well has a radius of influence (ROI) of 5m. At what rate (m3/day) should air be added to each well to provide enough oxygen to aerobic bacteria within the ROI? Some hints Consider the mass of soil in each ROI You can make use of the mass ratio 366 mg O2/106 mg HC c) If the soil is contaminated with 1000 mg/kg of ethylbenzene, how long w it take to clean up using bioventing?

Solution

a) WKT the consumption of oxygen is following the equation:

y = -0.361 x + 20.64

Therefore the per day oxygen consumption is,

y = -0.361 * 24 * 20.64 = 11.98%

The removal rate is therefore, 11.98%

The removal rate is therefore, 1.4 * 0.1198 = 0.177 g/L-day = 177 mg/L-day

The mass ratio = 106 mg of HC / 366 mg of O₂,

Therefore the anaerobic biodegradation rate = (106/ 366) * 177 mg/L-day = 51.26 mg/L-day

Since the soil bulk density is = 1.9 kg/L,

The anaerobic biodegradation rate = 51.26 / 1.9 = 26.98 mg/Kg-day = 26.98 mg ethylbenzene per Kg soil per day

b) The depth of contamination is 2m and the ROI is 5m.

Therefore, volume of soil = r² * d = 5² * 2 = 157.08 m³

In each ROI, the mass of soil would be = 157.08 m³ * 1900 kg/m³ = 298452 Kg

We know the anaerobic biodegradation rate 26.98 mg/Kg-day from the previous section,

Therefore, the mass of HC to be removed = 26.98 * 298452 = 8052235 mg of HC / day

The oxygen required would be =  366 mg of O₂ / 106 mg of HC * 8052235 mg of HC / day = 2947117995.36 mg of O₂ / day = 2947117.99 g of O₂ / day

Since, the density of O₂ is 1.4 g/L, the volume of oxygen required = 2947117.99 g/ day / 1.4 g/L = 2105084.28 L/day = 2105 m³/day

Since oxygen is only 20.95% of total air, the total amount of air need to be added is = 2105 / 0.2095 = 10048 m³/day

c) If the soil is contaminated with 1000mg/Kg of ethylbenzene then,

we know, the anaerobic biodegradation rate 26.98 mg/Kg-day

Therefore, time required = 1000 / 26.98 = ~37 days