Page 197 - Materials Chemistry, Second Edition

180 Practical Design Calculations for Groundwater and Soil Remediation

Determine the required cleanup time by using Equation (5.10):

T = M removal ÷ R removal = (403 kg) ÷ 5.58 kg/day = 72.2 day
2
(i) In the second interval, the amount of the TPH mass to be removed
is the same as that of the first interval, 403 kg. The initial theoreti-
cal vapor concentration is 117 mg/L (corresponding to 1,314 mg/
kg), and the final theoretical vapor concentration (corresponding
to 100 mg/kg) can be easily found as:

G final = 117 × (100/1,314) = 8.9 mg/L

The geometric average of these two is used as the average con-
centration for this interval:

G = (117)(8.9) = 32.3 mg/L

Calculate the mass removal rate by using Equation (5.9):

R removal = [(η)(G)](Q)
= [(0.11)(32.3 g/m )](0.216 m /min)
3
3
= 0.77 g/min = 1.1 kg/day
Determine the required cleanup time by using Equation (5.10):

T = M removal ÷ R removal = (403 kg) ÷ 1.1 kg/day = 365 days
3

The Entire Project
The total cleanup time required = T + T + T 3
2
1
= 61.2 + 72.2 + 365
= 498 days

Discussion:
1. For the three intervals, the average mass removal rates drop sig-
nificantly from 18.85 kg/day in the first interval to 5.6 kg/day,
then to 1.1 kg/day in the third interval.
2. For the two intervals during the absence of free product, the sec-
ond interval takes 365 days and the first interval takes only 72
days to remove the same amount of TPH.
3. The cleanup time of 498 days is not acceptable in most project
applications. One may consider increasing the extraction flow
rate or adding more wells.