My WebLink
|
Help
|
About
|
Sign Out
Home
Browse
Search
ARCHIVED REPORTS XR0003983
Environmental Health - Public
>
EHD Program Facility Records by Street Name
>
L
>
LODI
>
204
>
3500 - Local Oversight Program
>
PR0544962
>
ARCHIVED REPORTS XR0003983
Metadata
Thumbnails
Annotations
Entry Properties
Last modified
11/6/2019 10:09:05 AM
Creation date
11/6/2019 9:42:16 AM
Metadata
Fields
Template:
EHD - Public
ProgramCode
3500 - Local Oversight Program
File Section
ARCHIVED REPORTS
FileName_PostFix
XR0003983
RECORD_ID
PR0544962
PE
3528
FACILITY_ID
FA0003651
FACILITY_NAME
ARTS & ARTISTS
STREET_NUMBER
204
Direction
E
STREET_NAME
LODI
STREET_TYPE
AVE
City
LODI
Zip
95240
APN
04719102
CURRENT_STATUS
02
SITE_LOCATION
204 E LODI AVE
P_LOCATION
02
P_DISTRICT
004
QC Status
Approved
Scanner
SJGOV\wng
Tags
EHD - Public
There are no annotations on this page.
Document management portal powered by Laserfiche WebLink 9 © 1998-2015
Laserfiche.
All rights reserved.
/
93
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
TEST #3: MONITORING WELL MW-7 <br /> I = depth of well - depth to water = 63.10 ft - 59.65 ft = 3.45 ft <br />' d = oft <br />' = 35% (porosity of filter pack is assumed to be same as aquifer) <br /> m = 50 ft (assumed; no boring has penetrated more than 20 feet of aquifer) <br />' r,,, = 4" = .333 ft <br /> rc2 T (1 in)2 x (1-.35) + .35(4 in)2 = .65 int + .35(16 in) = 5.6 int = .466 ft, <br /> A = 1.9 from Figure 23.1 of Dawson and Istok (1991) <br /> 1 y <br /> B = .27 from Figure 23.1 of Dawson and Istok (1991) <br />' tL = 16.75 minutes from recovery curve MW-7 A <br /> or 15.08 minutes from recovery curve MW-7 B <br /> rW) <br /> In R/ = 1.1 + 19 + .35 x In 50 ft - 3.45 ft 1 <br /> ( <br /> ln(3.45 ft/.333 ft) (3.45 ft - 0 ft)/.333 ft = <br />' <br /> 11 + 1.9 + .35(3.84� -1 <br /> _ _ <br /> 2.33 10.36 - 1.472 + .3131-1 = 1.27 <br />' z <br /> K = .466 ft x 1.27 = .59 ftz = .0051 ft/min = 7.35 da <br /> 2(3.45 ft - 0 ft)16.75 min 115.57 ft-min <br /> or K = .466 ft2 x 1.27 = .59 ft' = .0056 ft/min = 8.16 Mdgy <br /> 2(3.45 ft - 0 ft)15.08 min 104.05 ft-min <br /> 1 <br /> A"P Wn-E-MA7 <br />
The URL can be used to link to this page
Your browser does not support the video tag.