My WebLink
|
Help
|
About
|
Sign Out
Home
Browse
Search
ARCHIVED REPORTS_XR0008925
EnvironmentalHealth
>
EHD Program Facility Records by Street Name
>
T
>
26 (STATE ROUTE 26)
>
8203
>
3500 - Local Oversight Program
>
PR0545707
>
ARCHIVED REPORTS_XR0008925
Metadata
Thumbnails
Annotations
Entry Properties
Last modified
11/20/2024 8:48:28 AM
Creation date
5/13/2020 4:05:13 PM
Metadata
Fields
Template:
EHD - Public
ProgramCode
3500 - Local Oversight Program
File Section
ARCHIVED REPORTS
FileName_PostFix
XR0008925
RECORD_ID
PR0545707
PE
3528
FACILITY_ID
FA0003591
FACILITY_NAME
JOHN M RISHWAIN
STREET_NUMBER
8203
Direction
E
STREET_NAME
STATE ROUTE 26
City
STOCKTON
Zip
95215-9536
APN
10114021
CURRENT_STATUS
02
SITE_LOCATION
8203 E HWY 26
P_LOCATION
99
P_DISTRICT
002
QC Status
Approved
Scanner
LSauers
Tags
EHD - Public
There are no annotations on this page.
Document management portal powered by Laserfiche WebLink 9 © 1998-2015
Laserfiche.
All rights reserved.
/
58
PDF
Print
Pages to print
Enter page numbers and/or page ranges separated by commas. For example, 1,3,5-12.
After downloading, print the document using a PDF reader (e.g. Adobe Reader).
View images
View plain text
I <br /> I <br /> Soil-Vapor Extraction Volume-Mass Calculations <br /> Former Mel Bokides Petroleum - Linden <br /> 8203 East Highway 26, Stockton, California <br /> The hydrocarbon mass removed during the operating period can be calculated using the following <br /> equation M = C•Q•t <br /> where M = cumulative mass recovered (kg) <br /> C = vapor concentration (kg/m') <br /> Q = extraction flow rate (m'/hr) <br /> t = operational period (hrs) <br /> The calculations for the determination of volume and mass of hydrocarbons removed over the <br /> reporting period are provided below <br /> 11-17-04 to 12-21-04 <br /> using C,a,,,r= (500+650 µg/l) — 2 = 575 micrograms per liter <br /> converted to 0 000575 kg/m3 <br /> Q = 62 scfm (average) x 1 69 = 105 M3/hr <br /> t = 840 hours (sum of known operation) <br /> 0 000575 kg/m3 • 105 M3/hr • 840 hours = 50 715 kg gasoline <br /> 50 715 kg gasoline - 2 205 lbs/kg = 111 8 lbs gasoline <br /> to convert lbs gasoline to gallons gasoline, use 0 16 gal/lb <br /> 111 8 lbs • 0 16 gal/lb = 17 89 gallons of gasoline <br /> 12-21-04 to 01-21-05 <br /> Iusing C,aP,,, = (650+450 µg/1) — 2 = 550 micrograms per liter <br /> converted to 0 00055 kg/m' <br /> Q = 67 scfm (average) X 1 69 = 113 m'/hr <br /> t = 720 hours (sum of known operation) <br /> 0 000550 kg/m' - 113 M3/hr• 720 hours = 44 75 kg gasoline <br /> 44 75 kg gasoline - 2 205 lbs/kg= 98 67 lbs gasoline <br /> I to convert lbs gasoline to gallons gasoline, use 0 16 gal/lb <br /> 98 67 lbs - 0 16 gal/lb = 15 79 gallons of gasoline <br /> Advanced GeoEnvu onmental,Inc <br /> I <br />
The URL can be used to link to this page
Your browser does not support the video tag.