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Design Methodology for Permeable Reactive Barriers Combined with
Monitored Natural Attenuation
Amine Hafsi
The Charles E. Via Department of Civil Engineering
(ABSTRACT)
Permeable reactive barrier (PRB) technology is increasingly considered for in situ treatment of
contaminated groundwater; however, current design formulas for PRBs are limited and do not
properly account for all major physical and attenuation processes driving remediation. This study
focused on developing a simple methodology to design PRBs that is easy to implement while
improving accuracy and being more conservative than the available design methodologies. An
empirical design equation and a simple analytical design equation were obtained to calculate the
thickness of a PRB capable of degrading a contaminant from a source contaminant concentration
sC to a maximum contaminant level MCLC at a Point of compliance POC . Both equations integrate
the fundamental components that drive the natural attenuation process of the aquifer and the
reactive capacity of the PRB. The empirical design equation was derived from a dataset of random
hypothetical cases that used the solutions of the PRB conceptual model (Solution I). The analytical
design equation was derived from particular solutions of the model (Solution II) which the study
showed fit the complex solutions of the model well. Using the hypothetical cases, the analytical
equation has shown that it gives an estimated thickness of the PRB just 15 % lower or higher than
the real thickness of the PRB 95 percent of the time. To calculate the design thickness of a PRB,
Natural attenuation capacity of the aquifer can be estimated from the observed contaminant
concentration changes along aquifer flowpaths prior to the installation of a PRB. Benchscale or
pilot testing can provide good estimates of the required residence times (Gavaskar et al. 2000),
which will provide the reactive capacity of the PRB needed for the calculation. The results of this
study suggest also that the installation location downgradient from the source of contaminant is
flexible. If a PRB is installed in two different locations, it will achieve the same remediation goals.
This important finding gives engineers and scientists the choice to adjust the location of their PRBs
so that the overall project can be the most feasible and cost effective.
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AB C
................................................................................................................................................................ 1
INTRODUCTION................................................................................................................................................................
1.1 B ..........................................................................................................................................................
1.2 A ......................................................................................................................................
1.3 .............................................................................................................................................
1.4 B ....................................................................................................................
1.5 B ............................................................................................
REFERENCES....................................................................................................................................................................
............................................................................................................................................................. 15
MATHEMATICALMODEL ANDSOLUTIONS........................................................................................................................1
2.1 .............................................................................................................................................................
2.2 .......................................................................................................................................
2.3 B C ......................................................................................................................................
2.4 .............................................................................................................................................................
2.4.1 ................................................................................................................................................
2.4.2 .............................................................................................................................................
2.5 ..................................................................................................................................
2.5.1 C ..................................................................
2.5.2 B( ) A L ..........26
2.5.3 B( ) B L ..................................................................................29
2.5.4 AC B .....................................................................................................
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.............................................................................................................................................................. 33
DESIGNMETHODOLOGY...................................................................................................................................................
3.1 .....................................................................................................................................
3.1.1 .............................................................................................
3.1.1.1 A ( A) ............................................
3.1.1.2 .................................................................................................................
3.1.2 ....................................................................................................................................
3.1.3 ...............................................................................
3.1.4 ....................................................................................................
3.2 ....................................................................................................................................
3.3 ............................................................................................................................................
............................................................................................................................................................. 47
CASESTUDY....................................................................................................................................................................
4 1. B ........................................................................................................................................
4 2. ...................................................................................................................................
4 2 1 A NAC .................................49
4 2 2 B .......................................52
4 2 3 B B NAC ................................................52
4 2 4 B ...............................................................................................................
REFERENCES......................................................................................................................................................................
............................................................................................................................................................. 54
ENGINEERINGSIGNIFICANCE...........................................................................................................................................
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........................................................................................................................................................... 56
........................................................................................................................................................... 77
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F 2 1: C
.......................................................................................................................................................
F 2 2 : C S I S II
E ( ) ..........................................................................................................................................
F 2 2 : C S I S II
E ( ) ..........................................................................................................................................
F 2 3: S PRB
........................................................................................................................................................
F 2 4 : C 3 A L
E ( ) ..........................................................................................................................................
F 2 4 : C 3 A L
E ( ) ..........................................................................................................................................
F 2 5 : E ( ) C C ,
PRB M I M II,
. ........................................................................................................................................
F 2 5 : E ( ) C C , PRB M I M II,
. .......................................................................................................................................
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F 2 6: E ( ) C C ,
PRB M I, nd B NAC ...........31
F 3 1 : PRB D C sstd
Awall C C vs Dk A ,. ...................................................................................37
F 3 2 : D C F A ................................................................................................38
F 3 3 : D C F A ................................................................................................38
F 3 4: P VA A .............................................................................................................................................
F 3 5: P PO B ...........................................................................................................................................
F 3 6: E D E ................................................................................
F 3 7: A D E ............................................................................
F 4 1: M S F (B ., 1998).........................................................
F 4 2 : M F (B ., 1998) ....................................................................
F 4 3 : TCE (IT C ., 1991).....................................................................
F 4 4: L M B M W PR
2005) ...............................................................................................................................................................
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AB
T 1 1: S (U.S. EPA R
T F S , EPA/600/F 97/008) .......................................................................................
T 3 1: ( )( )1
1
001.0,,400001.0,,
===
d V m LC
d V LC
A APOC A
A APOC A
..................................................................................................39
T 3 2: ( ) ( )( )
===
==
1
11
001.0,,400001.0,,ln001.0, / 1.0,ln
d V m LC d V LC
C d d mV LC
A APOC A
A APOC Astd
A
A APOC A
......40
T 3 3:( )
std A
A APOC A
C
d V m LC 1001.0,,400 == . AV ........................................................................................42
T 4 1: B T R D (B ., 1998) .................
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Chapter 1
1.1 B
G . I U
56% (V L
H
. C
(K
2001). F , A
(D , 1999).
T U.S E P A (EPA)
217,000 U.S $ 190
(EPA, 1996).C
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2
. H ,
, ,
(M , 2003). C ,
. C
(PCE), (TCE), (TCA), N
( ; )
.
G
,
(T .1990; G .1992; N R C , 1994). T
. H ,
P R B (PRB )
(P . 199
P ,
, TCE
(S . 2000), . C , PRB
,
, (P
T
. S ,
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3
(P . 1998).
T PRB . I
. H ,
. F PRB
;
. I ,
,
T , PRB
(MNA) .
1.2 A
W , ,
, , (U.S. EPA, 199
N O H S P
(NCP)
EPA, 1990 ). U , (U.S.
H , , ,
. T EPA MNA
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4
MNA
,
, ,
(U.S. EPA, 1997, .1).
T , , MNA
(L . 1997). S
( . ., )
MNA . C
U.S. A
P G 6 . O
B C MNA
( , , , ,
) F . (2001). T .
1.3
I ,
. I ,
. A G
,
F V G N P 1912 (P
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1998). F , 1980 , S
1982 G
C (VI) H
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, ,
(I S R T S , EPA, 1995).
T 15
(U.S. EPA, 2002). T EPA
(U.S. EPA, 1997). T EPA 20%
. T 1
. 1997).
R US
. PRB
. T , F O,
PCE TCE,
R ,
(B . 1996). T
,
. O
, .
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7
1 1: S (U.S. EPA RT F S , EPA/600/F 97/008)
,
C
,B ,
,C ,
C ,
C , C
,,
C
S . 1995 D . 1995 J . 1996 J 1995 O . 1996
C & . D
2 /L DCE 300 /LTCE
400 /LTCE
10 /L TCE10 /L
C (VI)
700 /L TCE & DCE15 /L VC
E & RV
F &G
CT
F &M G
N A S W SB
S
NA
S JS P
N A 100 + 100 490 +490
NA
1040
. N A 1 R 1 NA
4
F F F F F
5 16 11 A . 23 10 15
44 NA 20 150 40 (4 40 = 160)
4.5 16 3 2 G , = 2 = 6
90 15 70 450 NI
D
15 20 18 40 17 28 3 26 10 15 20 25
44 A . 200 1000 150 1200
& .
HDPE F
H2O F
W H2O
,
T .. C
F
L . G
.
C N I $375 K $400 K $500 K NI
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M 26 PRB US
.(U.S. EPA , RTDF, 2001). A
,
. H ,
.
I J 1996, , 46
, 7.3 0.6 U.S. C G S C
(USCG) E C , N C
(TCE) (B . 1999).
I S 1997, F C U ,
10 ,
( VI) ,
(AFO) , . A 1 ( VI)
(AFO)
99.9% (U.S. G S , 1997).
I 1999, C C A F S F ,
(0.01%) 20,000 DNAPL, TCE
3 /L TCE
1,2 (DCE). A 12.2 , 12.2
1.2 E 12.2 1.2
.% VI, 79 .% 5 .%
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9
TCE
M 1999 MCL
. 2000).
I A 1999, P A F B N H PRB
VOC 1,000 /L
TCE . A 487 359
+50 U.S. ) 473 3 . I
, 20
6 PVC . T
(C . 2000).
I 1998, PRB C T S S
J 915 TCE
6,000 8,000 /L 50 /L . T
TCE P R
1950 1984
TCE. T 3,000
7.6 , 46 27.4 250
. T
, 50% TCE ,
7,000 /L 3,500 /L. T
PRB
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H
.
B
, PRM
.
1.5 D B
T PRB
. T E A UK (N G & C
L C , 2002)
:
F Bres B S V t L **= . (1 1)
( )k
C C t T res
0ln= . (1 2)
T C = ( / ); 0C = ( / ); rest =
( ); BV = ( / );k = (1); F S =
.
T
. T
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N , PRB ,
(R .2005). T
. A ,
.
R . I
PRB N 2005 . A
, ,
PRB. A ,
PRB
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A , W.L.; S , E.L. I . E . C .1977, 16(2), 158 162
B , R.C. ;R.D. N ; R.E. H ; R. B ; P.L. M C ; L. S ; J.T. W ; C ; M. R ; E.J. B ; R.C. B ; T.M. V ; J.M. T ; C.H. W . NB H C G W . B , LP , B R , FL. 1994.
B D.W.; C.J. P ; J.L. J . I C (VI) : L .. . . 1997. 31(12), 3348 3357.
B D. W. . A . V 2, P M , . . A
A/600/ 99/095 ( . . A, , , 1999).
B , B. T., C , S. W. S , D. J. A. & M , D. M. A BTEX .
.1996
C , J. I P R W P A F B U BT : P RTDF P B A T M. 16 17, 2000,
.
C , F.H., P.M. B , C.C. C . 2005. B F . 25, . 2:
131 141
D , F. P; J. N. J . M N A C S .A.1999
F ,M.G.; M.A.W ; J.C.L .B C M N A . .. 2001.
F. L , F. L. T , D. K. T , T W E (L , C , MI, 1990).
G , R. W.; S. F. O H . E D H A V I . . . 32, . 6, 958 967.1994
G , R.W; D.R.B . R . , C ,
24, , . 1992
H , G. V H F E PRB . B A , A 15 16, 1998, B , .
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13
I , H.; T. K ; J.J. L . T R C C UP . A C . 1997.
K , P.; T. L ; P. K . R TCE C P RB V I . 2001.1
L , K.; P. F . P S N A CO G W .A.1997.
M ,D.K.;P.C.J ;C.B . B MTBE P RT . C .
N R C . A . A ,, .C., 315 . 1994
P , R. W. S P R D , NRMRL, ORD, U.S. EPA
P , R. W.; R. M. P ; D. W. B ; J. L. V ; D. S ; T. S ; P. D. P ; R. LP R B T F C R .. . A .1998
P ,R.M. P & A S S , L V , NV
R , A.J.; R. S ; J. R. C . A M D I BR B . 2005.
R , D.R.; M.B. C ; C.A. C ; J. Q . NASA PRB I DM :P RTDF P B A T M . 16 17,
2000, , . S F.; T. M . R
C & , . 8, 2, 103 1(2000)
S , J. N G &C L C , E A
S , K. H.; J. R. F . R D H P .1972. . . .3,640,821.
T , C.C., D , C.B. 1990. C S
E S T . 24:1464 1466.
U.S. E P A . F A I S R TP R B . 2002
U.S. E P A . R T D F. . / / /PRBSUMMS. A 2001
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U.S. E P A . U M N A S , . A 1999
U.S. E P A . O S W E R 9200. U M N A S , RCRA C AU S T .1997.
U.S.E P A . C U N W S : M TT . A 542 96 005A.1997.
U.S. E P A . N O H S PC P : F (NCP).. .55(46):8733 8734. 1990
U.S. G S ,U.S. B L M ,U.S. D E D O P R B T R D U F GF C , U . . 1 2.1997
W , T.H.; M.A., S ; D.E.; M ; E.K G ; J.T W ; B.H W ; D.H KJ.E. H ; P. H ; F.H. C .. AC . 1 5. A C A B , A , . 1997
W , P.; W H.; K A.; H S.; S K. G (B /W , G 2002
V , J. P B R P C T S . B A , A 15 16, 1998, B , . 1998
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Chapter 2
2.1
T PRB
.
F :
1- S 1)
PRB ,
, 2
, ,
,
PRB .
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2- C 7,500
PRB
sC MCLC P
POC
3- I
PRB
PRB
PRB 1
PRB ,
4-
PRB .
2.2
T , ,
N
. I PRB,
. D PRB,
.
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T 1 D (
,S C , ) ,
: (F
B , , Bv ,
B B A A nV nV = . (2 1)
An = L3/ L 3 ; Bn = L3/ L 3 ;
AV = .
CA , A CB , B CA , A
CS
2 1: C .
LALB
LPOC
V A
Source
A A B
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C ,
,
022
=
B B B
B B
B C xC
V xC
D . (2 2)
BC M/L3
B A A L L x L + ; B D L2/T ; B
T1
. S
, , ,
022
=
A A A
A A
A C xC
V xC
D . (2 3)
AC M/L3
A L x 0
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( ) 0= xC A .(2 4 )
:
B A C C = A L x = .(2 5)
B A C C = B A L L x += .(2 6)
A A A A
A A B B B B
B B C nV xC
n DC nV xC
n D +
=+
A L x = .(2 7)
A A A A
A A B B B B
B B C nV x
C n DC nV
x
C n D +
=+
B A L L x += .(2 8)
A L
B L
.
E 2 7
E 2 8
2.4
2.4.1
T
AU L
x X = .(2 9 )
B
A B L
L x X = ..(2 9 )
B APOC
B A D L L L
L L x X
= .(2 9 )
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U )10( U ;
B )10( B ; D
< D0 ; POC L
POC. T POC L
.
T E 2 2, 2 3 (A A)
:
1. F 10 = AU L
x X
)exp()1()exp()( 22 U nd
AU nd
AU R A X NAC C X NAC C X C += .(2 10)
2. F 10
= B
A B L
L x X
)exp(*)exp()exp(*)exp()( 13 Bnd
B R B
nd B B
nd B
R B
nd B B
R B X NAC L NAC C X NAC L NAC C X C += ...
.(2 11)
3. F 10
=
B APOC
B A D L L L
L L x X
( ) )exp(*)1(exp)( 4 D RPOC nd A R Bnd A D R A X L NAC L NAC C X C += .(2 12)
= R AC s A C C = = R BC
s B C C = . E 2 10, 2 11, 2 12
=nd A NAC [ ] Ae A Ae Ae P Dk PP 4)(21 2 + = NAC .(2 13)
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=nd A NAC [ ] Ae A Ae Ae P Dk PP 4)(2
1 2 ++ = NAC .(2 14)
A
A A A
e D
v LP = = P ;
A
A A
A V
L Dk
= = D ,
I
=nd B NAC [ ] Be B Be Be P Dk PP 4)(21 2 + = NAC PRB ..(2 15)
=nd B NAC [ ] Be B Be Be P Dk PP 4)(2
1 2 ++ = NAC PRB (2 16)
B
B B Be D
v LP = = P ;
B
B Bwall V L
Dk = = D
, I
A
B R B L
L L = .(2 17)
A
B APOC RPOC L
L L L L
= .(2 18)
+= A
e
A A P
Dk 411
21
++= A
e
A A P
Dk 411
21 .(2 19)
+= B
e
B B P
Dk 411
21
++= B
e
B B P
Dk 411
21 .(2 20)
:
++
= Be B Be R B
B A
B A P Dk P L 4) / 11(exp*C 21 .(2 21)
sC C C '''
'CC 2122 += .(2 22 )
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W1''
1'C
22
=C
.(2 22 )
+
+
+
++=
Ae A
Ae
Be B
Be
A B
A B
A B
A B
Be B
Be
P Dk PP Dk P
P Dk PC
4exp*4exp*
4exp1''
22
2
2
.(2 22 )
( )
( )
+
+
++
++
=
Ae A
Ae
B
A B
B
A B R B
Be B
Be
Ae A
Ae
B
A
B
A
R B
Be B A
e A
R B
Be B
Be
P Dk P L
P Dk PC
P Dk P
L
PP
L
P Dk PC
C
4)(exp*4
exp
4)(exp11
*exp
4exp1
'''
21
2
1
2
2
2
..
.(2 22 )
( ) ( )
+++
+
+++
+=
R B
B
e B
B
e
B
e R B
B
e B
B
e
B
e
Ae A
Ae
Ae
Ae A
Ae
Ae
L
P Dk PPC
L
P Dk PP
P Dk PPC P Dk PPC C
2
4)(exp
2
4)(exp
4)(21
exp)1(4)(21
exp
2
1
2
22
22
3 .(2 23)
( ) ( )
( )
++
++++
++
=)1(*4)(
21
exp
11*4)(
21
exp1
1*4)(21
exp
2
21
23
4 R B
Ae A
Ae
Ae
R B
Be B
Be
Be R
B
Be B
Be
Be
LP Dk PP
LP Dk PPC
LP Dk PPC
C
....(2 24)
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2.5 D
2.5.1 C
I S I S II,
( ) ( ) . B
MCLC lg / 5
POCs S I . U lg / 5 MCLC , S
II
, , PRB. F 2 2 F 2 2
.
2 2 : C S I S
E ( )
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2 2 : C S I S
E ( )
T S I S II
D ,
. T
0.2
8.2 ( ). T
. T S II
S I.
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2.5.2 B( ) A L
2 3: S PRB S 1 PRB )1( A L .
S 2 PRB )2( A L .
F 2 3 PRB )1( A L
MCLC P POC PRB )2( A L
. T PRB placed away from the
C A
Source
C s
Plume
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T S I ( ) ( )
. T :
1. W PRB
A L m2
2. W PRB POC
A L m40
3. W PRB 10 ( ) m50 POC
A L m90
2 4 : C 3 A L
- Example (c)-
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2 4 : C 3 A L
E ( )
T F 2 4 F 2 4 PRB
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2.5.3 B( ) B L
2 5 :C C ,
PRB M I M II, PRB
E ( )
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2 5 : C C ,
PRB M I M II, PRB
E ( )
S I
PRB
( ) ( ). T F 2 5 2 5 ,
PRB N
.
A , PRB
.
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2.5.4 AC B
2 6: C C , PRB M I, nd B NAC
-E ( )
S I
PRB
( ). B
B L Bv B PRB, nd
B NAC 13.8, 10.5
5.6. T F 2 6,
PRB N .
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A , PRB nd B NAC ,
. A , F 2 6 PRB
nd
A NAC .
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Chapter 3
D
3.1 D
3.1.1
W PRB B L
A L ,
PRB. T
PRB .
T POC
:
),,,,,,,,,,,()( POC s B A B A B A B A B APOC A LC nn L LV V D D f L xC == .(3 1)
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P L ( ). A
, (NAC)
X NAC C C As A = )ln()ln( ), A
A A A A A D
DV V NAC 2
42 ++=
T , P As L NAC C PCL = )ln()ln( PCL ( . .,lg / 2 ).
T
A
sP NAC
C L
)2ln(= .(3 5)
S A L A V D = 2 34 2 35:
A
A
A A A A
s A V
D
DV V
C D
414.2
2
2
4
)2ln(log83.0
++=
.(3 6)
A
A
A A A A
s A A V
D
DV V
C D D f
414.2
2
2
4
)2ln(log83.0)(
++=
.(3 7)
A D E (3 7).
. . .
A 2.5.2, POC
PRB A L E 3 1. A
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3.1.1.1 E 3 2 POC PRB
:
),,,,,,()( POC s B B A B APOC A LC LV V f L xC == .(3 7)
AV : F L/T .
BV : T L/T .
A : A T1 .
B : F T1 .
B L : T L .
POC L : D L .
sC : I M/L3 .
3.1.2
U S I 7,500
s
A APOC A
C V L xC ),,( =
POC
. T wall Dk . wall Dk .
s
A APOC A
C
V L xC ),,( =
(>89%) (F 3 1).
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3 1 :PRB D C sstd
Awall C C vs Dk A ,.
I , sstd
Awall C C vs Dk A ,. wall Dk .
s
A APOC A
C
V L xC ),,( = POC L m400 AV d m / 1.0
3.1.3
A F 3 1wall Dk :
A A
A
s
std A
wall C C
Dk
=
,
* .(3 8)
A A A . T
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:
A
Ae
1088*248.0= .. .(3 9)
113.0*34.46 += A A .(3 10)
3 2 : D C F A
3 3 : D C F A
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3 2:( ) ( )
( )
===
==
1
11
001.0,,400
001.0,,ln
001.0, / 1.0,ln
d V m LC
d V LC
C
d d mV LC
A APOC A
A APOC Astd
A
A APOC A
C ( )
A ( / ) 25 75 200 600 800 1000
0.02 0.230043 0.230042891 0.2300428 0.229392 0.229392 0.229392
0.04 0.432424 0.43242356 0.4324236 0.432424 0.432423 0.432422
0.07 0.721675 0.721675027 0.7216751 0.721675 0.721675 0.721675
0.1 1 1 1 1 1 1
0.2 1.921336 1.921335624 1.9213356 1.921336 1.921336 1.921336
0.6 5.576004 5.576003942 5.5760035 5.576005 5.576004 5.576003
0.8 7.397108 7.397107277 7.3971074 7.397109 7.397108 7.397106
1 9.209606 9.209612596 9.2095838 9.209656 9.209622 9.209636
F T 3, VA A :
( ) ( )
( ) VA
A APOC A
A APOC Astd
A
A APOC A Ad V m LC
d V LC
C
d d mV LC =
==
=
==
1
11
001.0,,400
001.0,,ln
001.0, / 1.0,ln
..(3 11)
A , d mV A / 1.0= ( T 3 1), POC B
:
( ) )exp(001.0, / 1.0, 1 POC std A
A APOC A BC
d d mV LC === .(3 12)
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C , (3 11) (3 12):
( )( )
====
1
1
001.0,,400001.0,,ln*
d V m LC d V LC A B
A APOC A
A APOC AVAPOC
( )( )
=
===
VA
POC
A APOC A
A APOC A
A
B
d V m LC
d V LC exp
001.0,,400001.0,,
1
1
( )
====
VA
POC A APOC A A APOC A A
Bd V m LC d V LC exp*001.0,,400)001.0,,( 11 .. (3 13)
3 4:P VA A
072.0*151.9 += AVA V A .(3 14)
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)*009.0(*0275.01
POC POC e B =
.(3 15)
3 5:P PO B
594.3*009.0 += POC POC L B .(3 15)
1001.0,,400 == d V m LC A APOC A std AC
AV .( T 3 3)
3 3:( )
std A
A APOC A
C
d V m LC 1001.0,,400 == . AV
AV ) / ( d m
( )std
A
A APOC A
C d V m LC 1001.0,,400 ==
0.02 7.85296E 06
0.04 0.009319187
0.06 0.1182682740.08 0.443341395
0.1 1
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AV ) / ( d m
( )std
A
A APOC A
C d V m LC 1001.0,,400 ==
0.2 5.398266307
0.3 9.685354635
0.6 17.64225402
0.8 20.55409107
1 22.54114178
F T ,( )
VAstd
A
A APOC A DC
d V m LC =
== 1001.0,,400 ..(3 16)
:
+=
+=
1V0.1when3.7-V*49.5V*23.7-
1.0V04.0when0.47V*22.69-V*279.82
AA2A
AA2A
VA
VA
D
D
.(3 17)
F , T 2, 3 4
A 10012.0 d , 10015.0 d , 10017.0 d ,
1002.0 d , 10025.0 d :
U E 3 13 3 16:
( ) A
A A A
VA
POC VA
std A A APOC A A
B DC V LC
001.0
,,001.0
exp**),,(
= .. (3 18)
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:
0639.0*5.3439**588.8, ++= APOC APOC L B A .(3 19)
3.1.4 D
F E 3 8 3 18 wall Dk
MCLC . T PRB B L ,
PRB MCL
C POC,
Bwall
B B Dk
V L
*= ..(3 20)
113.034.46001.0
,exp
**
1*)1088exp(*248.0
+
=
A
A
VA
APOC
MCL
S VA
Awall
A B
C
C D Dk
.
.(3 21)
3.2 D
U ( 2.4.2),
PRB MCLC POC L (S A B):
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A B
POC A MCL
s
B NAC NAC
L NAC C
C
L
=*ln
.(3 22)
A NAC = = A
A A A A A D
DV V NAC
2
42 ++= ,
B NAC = PRB = B
B B B B B D
DV V NAC
2
42 ++=
3.3
U ,
PRB 7,500
. F
F 3 7
.
E S
. I , 95 ,
PRB 15 % 45
.
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3 6:E D E
-Model I-
3 7: A D E
M II
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Chapter 4
C
4 1. B
A PRB ( VI) A 199
F M V C (F 4 1)
(CVOC ) ,
1,2 (1,2 DCE), (PCE),
5,000 (F 4 2), N /N E , TCE
5,000 / (F 4 3).
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4 1: M S F (B ., 1998)
4 2 :M F (B ., 1998)
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4 3 : TCE (IT C ., 1991)
4 2.
4 2 1 A NAC
M M F P
(F 4 4). M WCI 1 WCI 3
PRB. A
J 1997, PRB. U
CVOC ( T 4 1) J 1997,
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lg / 3180 WIC 1. A lg / 3160
WIC 3. U 4 4, T feet 28 . A
PRB ,
A NAC 100023.0 m . H A NAC
(F 4 3). U ,
85 1) lg / 6200 , 2250
(W09 10) lg / 1600 A NAC
10006.0 m .
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4 4: L M B M W
(B ., 2005)
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4 2 2 B
A M F
PRB B (T 4 2). T
PRB B E 4 1. F
17.4 PRB B 199.3 d .
21)2ln(
T B = .(4 1)
4 1: B D (B ., 1998)
4 2 3 B B NAC
T 0.2 2
An 0.30 PRB Bn 0.33 (B ., 1998). I
PRB
2 / PRB
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W B B D d m E 20560.8
B NAC 119.7 m .
4 2 4 B
U F 4 3, 200 PRB
. B CVOC W9 35,
PRB ,
lg / 8000 . T
sC .
T E 3.22 PRB
(TCE, PCE, DCE, ) lgC s / 8000 = lgC MCL / 2 =
WIC 3 225 :
feet L design B 7.30006.019.7
225*0006.02
8000ln
=
=
T 95% , 15 %
PRB . T M PRB
feet 3.4 PRB 6 . T
PRB .
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Chapter 5
T
PRB . A
PRB
sC
MCLC P POC . B
PR
T 7,500
PRB (S I). T
(S II) . T
. U 7,500 ,
PRB 15 %
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PRB 95 . A ,
15
95
T PRB, N
PRB. B
( G . 2000) , P
.
A PRB
. T
. I PRB
. T
PRB .
G , A. R., G , N., S , B., F , T., J , R. J., H . J. T. D R P
US AIR FORCE, B , C , O . 2000
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MODEL I
POC L
A L
B L
B A L L +
A A A
A A A
B B B
nV
nV
n
V
Source
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MODEL I
Equations (ODEs, BC, Intermediate conditions)
:0 A L x 011
21
2
=
C xC
V xC
D A A A (1)
: B A A L L x L + 022
22
2
=
C xC
V x
C D B B B (2)
: B A L L x + 033
23
2
=
C xC
V x
C D A A A (3)
sC xC == )0(1 (4)
0)(3 + x xC (5)
)()( 21 A A L xC L xC === (6)
)()( 32 B A B A L L xC L L xC +==+= (7)
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11
22 C nV
xC
n DC nV xC
n D A A A A B B B B +
=+
when 1 L x = (8)
33
22 C nV
x
C n DC nV
xC
n D A A A A B B B B +
=+
when B A L L x += (9)
B
A A B B B A A B A n
nV V nV nV qq === (10)
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SOLUTION I
:0 A L x 011
21
2
=
C xC
V x
C D A A A (1)
The characteristic equation is:
04
02
2
>+=
=
A A A A
A A A
DV
xV x D
1a and 2a are the solutions to the equation above:
02
4
2
02
4
22
2
2
1
>++
=+
=
+=
=
B B B B
B B B
DV
xV x D
1b and 2b are the solutions to the equation above:
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02
4
2
0
2
4
22
2
2
1
>++
=+
=
+=
=
A A A A
A A A
DV
xV x D
1a and 2a are the solutions to the equation above:
02
4
2
02
42
2
2
2
1
>++
=+
=
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Using Boundary conditions
s
C xC == )0(1
(4)
Using xa xa
A Be Ae xC L x 21)(0 1 +=
=+== saa C Be Ae xC 0*0*1
21)0( )( I
C B A s=+
0)(3
+ x xC (5)
Using xa xa
B A HeGe xC L L x21
3 )( +=+ where :
0
2
4
2
02
4
22
2
2
1
>++
=+
=