Page 196 - Materials Chemistry, Second Edition

Vadose Zone Soil Remediation 179

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

R removal = [(η)(G)](Q)
= [(0.11)(551 g/m )](0.216 m /min)
3
3
= 13.1 g/min = 18.85 kg/day

(f) Determine the required cleanup time by using data from (c) and
(e) and Equation (5.10):

T = M removal ÷ R removal = (1,153 kg) ÷ 18.85 kg/day = 61.2 day
1

Absence of the Free-Product Phase
(g) At the end of the free-product removal, the TPH concentration in
soil is 2,528 mg/kg, corresponding to a theoretical vapor concen-
tration of 226 mg/L. The cleanup level of soil for this project is 100
mg/kg. The average of 2,528 and 100 is equal to 1,314. To estimate
the required cleanup time, we divide it into two intervals. The
first interval is the time required to reduce the concentration from
2,528 to 1,314 mg/kg, and the other is from 1,314 to 100 mg/kg.
(h) Determine the amount of TPH to be removed in the first interval
by using Equation (5.13) as:
M removal = (X initial − X final )(M )
s
= (2,528 − 1,314 mg/kg)(332,000 kg)
= 4.03 × 10 mg = 403 kg
8
For this interval, the initial theoretical vapor concentration is 226
mg/L (corresponding to 2,528 mg/kg), and the final theoreti-
cal vapor concentration (corresponding to 1,314 mg/kg) can
be easily found as:

G final = 226 × (1,314/2,528) = 117 mg/L

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

G = (226)(117) = 163 mg/L

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

R removal = [(η)(G)](Q)
= [(0.11)(163 g/m )](0.216 m /min)
3
3
= 3.87 g/min = 5.58 kg/day