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ARCHIVED REPORTS_XR0005026
EnvironmentalHealth
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2900 - Site Mitigation Program
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PR0507952
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ARCHIVED REPORTS_XR0005026
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Last modified
9/16/2020 12:50:34 AM
Creation date
3/17/2020 2:04:46 PM
Metadata
Fields
Template:
EHD - Public
ProgramCode
2900 - Site Mitigation Program
File Section
ARCHIVED REPORTS
FileName_PostFix
XR0005026
RECORD_ID
PR0507952
PE
2950
FACILITY_ID
FA0007846
FACILITY_NAME
OLYMPIAN CFN
STREET_NUMBER
983
STREET_NAME
MOFFAT
STREET_TYPE
BLVD
City
MANTECA
Zip
95336
APN
22115006
CURRENT_STATUS
01
SITE_LOCATION
983 MOFFAT BLVD
P_LOCATION
04
P_DISTRICT
005
QC Status
Approved
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SJGOV\sballwahn
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EHD - Public
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t <br /> 1 <br /> 27 October 2004 <br /> AGE-NC Project No 96-0254 <br /> i <br /> ' Page 6 of 10 <br /> • The effective porosity of soil at the site is 35% (probably overestimated), and <br /> 1 • Data collected from the July 2004 ground water sampling event are representative of <br /> existing site conditions <br /> The area of an ellipse can be calculated by the formula <br /> Ae = nab, where a is the long radius of the ellipse and b is the short radius <br /> For the estimated ellipse a= 140 ft and b = 30 ft, the column depth c = 30 ft <br /> 1r <br /> The average MTBE concentration for the ellipse Is estimated to be 200 micrograms per liter(µg11), <br /> as an overestimate based on one impacted ground water well and the maximum distance from well <br /> MW-1 which has detectable contamination <br /> The area of the given Is ellipse b <br /> P g Y <br /> ' 0 <br /> Ae 'nab = n(140 ft) (30 ft) = 13,188 ftz <br /> The volume of the elliptical column is given by <br /> ' • <br /> Ve =Aexc= 13,188f�2 x30ft = 395,640ft' <br /> ' Water occupies the porosity in the soil,which Is estimated to be 35%of the soil volume,so the total <br /> volume of water in the saturated portion of the elliptical column is approximated by <br /> Ve = (0 35)(395,640 ft3) = 138,474 ft' <br /> ' One cubic foot of water comprises a volume of 7 48 gallons, so the volume of the water in the <br /> elliptical column is given by <br /> Ve =(138,474 ft3)(7 48 gal/ft) = 1,035,785 5 gallons <br /> One gallon of water is equal to 3 785 liter so the volume of water expressed as liters is <br /> ' Ve =(1,035,785 5 gallons)(3 785 l/gal) = 3,920,448 liters of water <br /> Multiplying the volume of water in the elliptical column by the average MTBE concentration <br /> (200 µg11) yields the approximate mass of MTBE remaining <br /> ' MMTBE = (3,920,448 liters of water)(200 µg11) = 784,089,623 µg of MTBE <br /> Advanced GeoEnvironmental,Inc <br />
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