My WebLink
|
Help
|
About
|
Sign Out
Home
Browse
Search
SITE INFORMATION AND CORRESPONDENCE
EnvironmentalHealth
>
EHD Program Facility Records by Street Name
>
S
>
SUTTENFIELD
>
24876
>
3000 – Underground Injection Control Program
>
PR0519201
>
SITE INFORMATION AND CORRESPONDENCE
Metadata
Thumbnails
Annotations
Entry Properties
Last modified
5/21/2020 11:20:46 AM
Creation date
5/21/2020 11:10:22 AM
Metadata
Fields
Template:
EHD - Public
ProgramCode
3000 – Underground Injection Control Program
File Section
SITE INFORMATION AND CORRESPONDENCE
RECORD_ID
PR0519201
PE
3030
FACILITY_ID
FA0014356
FACILITY_NAME
MILLER RES UIC DRUG LAB
STREET_NUMBER
24876
Direction
N
STREET_NAME
SUTTENFIELD
STREET_TYPE
RD
City
ACAMPO
Zip
95220
CURRENT_STATUS
02
SITE_LOCATION
24876 N SUTTENFIELD RD
P_LOCATION
99
P_DISTRICT
004
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.
/
38
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
F I <br /> 2 <br /> Assumptions, Formulae, Conversions, Computations <br /> 1. Septic Tank volume 1,600 gallons = 6,056 liters (1,600 allons){3.785 liters <br /> ) = 6,056 liters <br /> gallon <br /> 2. Assume that the Septic Tank is completely full of liquid, i.e.,this tank contains 6,056 liters of liquid <br /> 3. 1.0 A = 1.0 b mass — 1.0 µgrams -6 grams _3 liters of water, 9 grams <br /> µg pp ( ) ( )( 1.0 x 10 )(1 x 10 } = 1 x 10- = 1.0 ppb(mass) <br /> liter of water µgram gram billion grams <br /> 4. (1.0 µgrams)(1.0 X 10-3 milligrams)(tankvolume in liters) = milligrams of chemical in the tank = mg of chemical in the tank <br /> liter µgrams <br /> 5. 1.0 milliliters = 1.0 ml = 1.0 cm3 = 1.0 cc <br /> (mg of chemical)(1x10-3 grams) <br /> 6. density in grams <br /> mg = milliliters of chemical in the tank = nil of chemical in the tank <br /> cc <br /> 7. 1 A ml = 0.203 teaspoons <br /> CONCLUSIONS: <br /> 1. The quantity of vehicle gasoline in this septic tank is so insignificantly small (a total of 1.222 equivalent teaspoons of gasoline)that I <br /> do not believe anything additional needs to be done. I certainly do not believe that the tank needs to be pumped,nor do I believer that <br /> monitoring wells need to be installed in the septic tank's leach field. With regard to the total organics in this tank(gasoline+acetone), <br /> the tank contains an equivalent of only 1.372 teaspoons of organics. Expressed as a percentage of the tank's total volume,this <br /> amounts to only 0.000112% of the total tank volume. <br /> 2. In the event that the tank is found to be less than 100%full of liquid--i.e., if it contains less than 6,056 liters of liquid–then the <br /> equivalent amount of organics will be proportionally less by the fractional level to which the tank is actually filled. For example,if <br /> there were only 2,000 liters of liquid in this tank(approximately 33% of its capacity)then the equivalent amount of organics in it would <br /> be this fraction multiplied by the 1.372 teaspoons. On the assumption that the tank actually does contain only 2,000 liters of liquid, <br /> then the total volume of organics in it would be 0.453 teaspoons! <br /> In the event you have any questions on any of these calculations,determinations, or conclusions please do not hesitate to call me. <br /> Edward W. Finucane, PE, CSP, QEP,CIH <br /> �ti , <br />
The URL can be used to link to this page
Your browser does not support the video tag.