Page 217 - Materials Chemistry, Second Edition

200 Practical Design Calculations for Groundwater and Soil Remediation

Therefore, 91,500 mg CO /m is equivalent to:
3
2
(91,500)(0.322) = 29,460 mg diesel/m 3

(d) Percentage removed by biodegradation
= (29,460) ÷ (29,460 + 3,534) = 89.3%

Percentage removed by volatilization = (1 − 89.3%) = 10.7%

Discussion:
1. The extracted vapor concentration data suggest that biodegrada-
tion accounts for ≈90% of the diesel removal.
2. The air-emission rate of diesel will determine if an off-gas treat-
ment system is needed.

Example 5.27: Determine the Rate of Biodegradation from Bioventing
For the bioventing project mentioned in Example 5.26, the air extraction is
operated on an intermittent basis. The extraction blower is only on for a con-
secutive 24-h period every 7 days. As mentioned, the average concentrations
of TPH and CO in the recently extracted air samples are 500 ppmV and 5%,
2
respectively. The air extraction rate is equal to 1.0 m /min. Estimate the rate
3
of biodegradation occurring in the subsurface and the emission rate of TPH
into the atmosphere.
Solution:
(a) As calculated in Example 5.26, 5% CO in the extracted air is
2
equivalent to 91,500 mg CO /m 3
2
Rate of CO emission = (Q)(G)
2
= (1.0 m /min)(91,500 mg CO /m )
3
3
2
= 9,150 mg/min = 91.50 g/min
(b) Total mass of CO emitted during the entire 1,440-min period
2
= (91.5)(1440)
= 131,760 g = 132 kg CO 2

(c) Total mass of TPH biodegraded over the 7-day period
= (132 kg CO ) × (0.322 kg TPH/kg CO )
2
2
= 42.5 kg TPH
(d) The rate of biodegradation over the 7-day cycle
= 42.5 kg ÷ 7 days = 6.1 kg/day = 13.4 lb/day