TR-62

TECHNICAL REPORT

A METHOD FOR ESTIMATING THE FLUSHING
TIME OF
ESTUARIES AND EMBAYMENTS

BLAIR W. GIBSON

Oceanographic Publications Branch
Division of Oceanography

JULY 1959

ua U.S. NAVY HYDROGRAPHIC OFFICE
743

dy) WASHINGTON, D.C.
M.- TR “bd ‘ Price 30 cents

ABSTRACT

A relationship is derived from the circulation pattern of a
piedmont-type estuary which expresses the time required
for a contaminated estuarine volume containing dissolved or
suspended matter to be replaced by tidal action and river
flow. The volume of such an estuary consists of an upper
layer of mixed water with a net seaward movement, overlying
a layer of sea water with a net landward movement. These
volumes are separated by a surface of no net horizontal
motion through which there is a net vertical movement of
sea water into the mixed layer. Hence, the lower layer is
renewed by tidal action alone while the mixed layer is re-
newed by tidal action and river runoff.

Since the net seaward flow of mixed water is equal in volume
to the net landward flow of sea water plus the river flow, the
balance should be indicated by tidal current data. Thus, it
is possible to compute the flushing time without first com-
puting the river flow.

In embayments and/or estuaries during a dry season, a rela-
aa pues Se

the segments.
oncentration of
1 the exchange

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FOREWORD

The equations and formulas developed through a purely theoretical
approach to oceanography sometimes are difficult to apply directly
to a specific naval operation or problem, At times slight modifications
in the theoretical approach make these formulas more usable for
practical problems; at other times a totally new or different approach
is required. In the flushing problem, a combination of theoretical
considerations has provided better methods of calculation for the
“general”

estuary. This report describes such a method currently
being used for determining the flushing characteristics of estuaries.

Ce

H. G. MUNSON
Captain, U. S. Navy
Hydrographer

0041

o 0301

lii

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ASTIA

JOHNS HOPKINS UNIVERSITY (CBI)
UNIVERSITY OF WASHINGTON

UNIVERSITY OF NORTH CAROLINA
UNIVERSITY OF MIAMI (MARINE LAB)
UNIVERSITY OF DELAWARE

UNIVERSITY OF SOUTHERN CALIFORNIA
UNIVERSITY OF TEXAS

UNIVERSITY OF FLORIDA

UNIVERSITY OF HAWAII

UNIVERSITY OF RHODE ISLAND (NML)
LAMONT GEOLOGICAL LAB

NEW YORK UNIVERSITY

YALE UNIVERSITY (BINGHAM OC. INST.)
RUTGERS UNIVERSITY

FLORIDA STATE UNIVERSITY

OREGON STATE COLLEGE

TEXAS A & M COLLEGE

MASSACHUSETTS INSTITUTE OF TECHNOLOGY (Meteor. Dept.)

iv

TABLE OF CONTENTS

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LIST OF FIGURES

Figure

1 Cross Section of a Piedmont-Type Estuary
Showing the Net Circulation Pattern

©, (0; (0; 6) Je; (e, je! “e) je) Je) (0: <0: ie

2 Schematic Movement of a Typical Water Particle

asa Result of Tidal Action): Sin ccs imeusee oO custleney isis
3 GeneralehlushinpiGurvesin caw ncmenenct tn lnc ononen =
4 Location of Observing Station in the James

Rie JRSUMENA, 556 0 6 (B10 0 610-6, 066900 Stu.5 soi OcyG. tn woRo ©

5 Mean Velocity - Depth Curves for Current
Sablon IaNl obs a0co bso conor ons 000s dag 540%

6 Mean Salinity - Depth Curve at 3 Stations in
themiamesuRawer iis tual yan em i eon wee mr-mr ail ei rn=inmtcMlcneyironteit=

v Flushing Time Curves for the James River
IDOE, 4 Gb Oooo BOO Oo DC Op FOO OOOOH OOO KOC

8 Diagrams Showing Redistribution of Contaminant
wala AGIA, JNSIGIN G6 Goo ooo obo ade ood oD oO O

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PART I
Estuaries

The river water entering an estuary mixes with salt water and
continues its movement to the sea. When the temperature of the
river water does not fall below that of the sea water, a net seaward
movement of mixed water occurs in the upper layers of the estuary.
When the volume of river water received by the estuary in a tidal
cycle is constant over a period of time, the volume of river water
reaching the sea becomes equal to that received from the land ina
tidal cycle.

Over a period of time the salinity at a point in the estuary will
remain practically constant. Since salt is constantly being lost from
the upper levels of the estuary, because of the net seaward volume
transport from these levels, there must be a net volume transport
of salt into the lower levels of the estuary to balance this loss. The
situation just described is illustrated in Figure 1.

From the standpoint of the net volume transport pattern of Figure l,

an estuary can be divided into two volume sections, A and B, as in
Figure 1. Here the net flow in A is toward the sea, while the net flow

RIVER ESTUARY SEA

_—o

FIGURE | CROSS SECTION OF PIEDMONT- TYPE ESTUARY SHOW-
ING NET CIRCULATION PATTERN

in B is toward the land. Because of the opposing directions of the
net flow velocities, a surface cd must exist where the net horizontal
transport of salt is equal to zero. Moreover, there must exist through
the surface cd a net volume transport of salt per tidal cycle from
B into A, equal to that entering the seaward boundary of B in a tidal
cycle. If, now, the volume of river water received from the land in
a tidal cycle is designated by R, and Qp represents the net volume
transport of sea water in B, the net volume transport of mixed water
in A will be represented by:

Q, =R+Q,; (1)

The hatched zone in Figure 1 is assumed to contain water which has
very little motion, and hence this zone does not relate to the problem
at hand. Also, the water in the shaded zone of Figure 1 will be con-
sidered to be dead, because the sea water arriving at point g is
expected to pass from f rather than uphill from some point e.

When an estuary is shallow, some river water will flow toward the
land in B, but when the depth is relatively great, the flow in B will

consist chiefly of sea water.

It is now assumed that a contaminant will be dispersed uniformly

WS
@

7

y
47 ”U

FIGURE 2 SCHEMATIC MOVEMENT OF A TYPICAL WATER PARTICLE
AS A RESULT OF TIDAL ACTION

through the entire estuary of Figure 1 at mean high water. Consider
now the motion of a water particle from position 1 through position 18
in Figure 2. Obviously, it is only the net movement of the particle
which contributes to its passage through the estuary. It is, therefore,
required to find this net movement in order to solve the problem at hand.

One method for accomplishing this would be todetermine the average
flood and ebb velocities through the seaward boundaries of A and B.
Then

Q, =hd (U,t, — U-t,) (2)

where, hd is the mean area at the seaward boundary of A, Up; and
Ur refer to mean ebb and flood, respectively, and t is the duration of
flood and ebb flow. Again, if the river flow R is computed inde-
pendently,

Q,=ed (U-t, — Uct,) (3)
from which (2) can be determined since Qa = R + Qp.

The case for which only river flow data is available for the solution
of the problem will be discussed later on.

When Qa and Qp can be determined, as above, the following state-
ments can be made:

(a) Except for the first several tidal cycles following the initial
contamination of the estuary, the contaminant in B must leave the
estuary via cd and the opening hd (Fig. 3).

(b) Any contaminant leaving the estuary in a tidal cycle will
be contained in a volume Qa.

The following assumptions are made to supplement (a) and (b):
The contaminant remaining in the estuary at any time becomes uni-
formly distributed at high tide; and if the removal of the contaminant
is excessive or deficient during a time interval, the rate of removal
will compensate during a following time interval.

We can now write exchange ratios for sections A and B, as

aes

r
aS A

(4)

and ~e Q. 5)

respectively. These ratios express the fraction of volume A which
is lost to the sea in a tidal cycle, and the fraction of volume B which
passes into section A in a tidal cycle. The actual volume of mixed
water containing contaminant, which is removed from the estuary with
each ebb of the tide, is;

TmA=Q~ (6)
Consider now the high water situation in the estuary at the end of
the first tidal cycle following initial contamination. The volume of

contaminated water removed with the first ebb tide is given by (6);
the contaminant remaining in section A would, therefore, be given by

A(1 — ra) (7)
had not the quantity
rgB (8)

entered section A from section B, The remaining volume of contaminant
in B,is:

BC a tp), (9)
at the end of the first tidal cycle.

The contaminant lost from section A during the second tidal cycle
is, from (7) and (8),

ray =ta[ AG —t) + re B) |, (10)

and that remaining in section A at the time of the second high water
is
C2 =[Ad —m)+reB ] (1—m) +reB(1—rg),
(11)
where, as before, the quantity r,B(1—r,)

entered A along with the salt necessary to replenish that lost to the
sea with volume Qa.

Similarly, for the third tidal cycle,

C3 = {faa —ta) + 1B] (1 — ta) + re BCL -rafca ST ERB (era)a > (12)
represents the remaining contaminant in section A.

The terms in (12) can be rearrangedto give the equivalent expression
Ca Aa =)? +reB[C ERG! Sip Oh Neal —r5)'| (13)

In (13) the sum involving the terms (l-r,) and (1-r,) is, except for
other constant terms, similar to the expanded form of

[1 —ra) + —08)]"”.
We can, therefore, write for the contaminant remaining in A at the
end of n tidal cycles

n

(n—1) (n — 2)
Cn=A(1 ra)" + reB[(1 — 1) AO =p)) Mane A

ca 14
lt Ol sts Faony-?]. es

This last expression is too complicated to be of practical use. If, in
the second member on the right of (14), ra and rp are replaced by

ea lfy SU js vee: ratte , (15)

we get

(n—1)
) c

Cp=A( —m)" +nrB(l —r (16)

On dividing and multiplying the second member on the right of (16)
by (l-r), we get

Coe NG Si) se gan B(1—r)", (17)

n being the number of tidal cycles.
Now, the numerical values of ra and rp in practice are expected

to lie between 1/10 and 1/50. Won substituted into a relation such
as

(l—nf *
1 \10
these values give =(1 ==) =0.3487,
Oa nin— = = a71R d = 0.367, where e is the base of the natural logarithms.
r=o e 2.718

and =(1- —)"= 0.3643.

For all practical purposes, therefore, we can write

1 hog
1 B a ?
Gi 4 SOS) array OTe Ts (18)
Tag) t
or its equivalent
Cy 1 =0.35A+ 0.358 , (19)
ih ( —r)

where 0.35A is the portion of the original contaminated volume A
which remains in section A at the end of

ckiecli
I= TA
tidal cycles, and
0.35B
(1 —r)
is that portion of the total contaminant arriving from B in
1

tidal cycles which would remain in A at the end of this latter time.
If the curves
m
Vi = (0:35) Ay m—sl 2setc: (20)

and

n

y, = (0.35)

ate , n=1, 2, etc. (21)
are plotted for the same coordinates, but with unit lengths on the
abscissa corresponding to the reciprocals of their r-values, the
sum of (20) and (21) will be a curve expressing the remaining con-
taminant in section A after any desired number of tidal cycles have
occurred as illustrated in Figure 3.

For example, let
A equal 10 units
B equal 14 units
Qa equal 0.4 unit
Qp equal 0.3 unit

The exchange ratios are rA equal 0.04, rp equal 0.0214, andr equal

0.0307, and unit lengths for the curves y); and y, are

=25
and

t>=32.5,

respectively. In Figure 3 the curves yj, Yo and y, + Y> are shown
for values of m and n equal to 1, 2, 3, etc.

In practice, sometimes very little information is available upon
which to base the preceding computations. We may, therefore, be
required to adopt the following procedure in order to obtain a first
order approximation of the flushing time.

The mean high tide volume is computed and divided into two parts,
giving A = B. Also, we can put ry, = rp = r*. When substituted into
(16) these give

Ch = AG SrieanrAGian) wane

ie sa nrA _ AW 22)
= A(1 —r) AGl=py r) (

=Aa "f+ ar]

1
=a _ yn 1 2k
Ca. py=AG 08 |

from which

1
=2A(1—-nr)r.,
approximately, or C1=(A+B)(0.35).
r

The only curve to be plotted in this situation is, therefore,

—(A+B)(0.35)", n=1, 2, 3, ete.
y=(A+B)( )n e (23)

The value r to be employed in order to determine the number of tidal
cycles corresponding to n, = I, n, = 2, etc. still remains to be found.
(For an example see Figure 3.) Now the total volume of sea water

* Also, the value mm can be retained in the first member on the right
of (20), and = © 2%) substituted in the second member on
the right of (20). This procedure is probably to be preferred to the
above.

**Here for simplicity (l-r) is assumed to be close to unity. For
example, when

po SH Ma HaOOBR
A 30

which enters the estuary on the flooding tide is, except for the river
water contributed in a half tidal cycle, equal to the tidal prism volume
P. The average excursion E of the seawater near the seaward boundary
a of the estuary is, therefore, approximately equal to:

male
E = (24)
where a = hd + ed (see Fig. 2).

Now, for the James River estuary, a relation existed between the
maximum current velocity and the net salt transport, which will
be assumed to hold in general. This is to the effect that the net volume
transport velocity is approximately equal to one-fifth the maximum
current velocity. Multiplying (24) by \'/2 gives the maximum excursion

GP
and multiplying (25) by one-fifth gives
7 P
Enet = 40a (26)

The net volume of sea water entering the estuary is

1 lw
Unet 2=— Umax @=— — Va= qo * 2Q8, or (27)
Q T P**

Therefore, ese 20

AOE Qe (28)
Gi mms} 1/2 (A+B
and

Ta =R+Q, ° (29)

*Up =UmaxSiN@ ,where @ varies between o andwduring the flooding tide
interval te . The area under the velocity profile is ST Umaxsind d 0=
=U —1—1]=2Umax- If U is the average flood velocity,U7=2Umax or
Umax = 5 U- IMGO Wrigley Cbs

** It is assumed that actual current velocities are not known.

The gross method just employed should not cause as great an
error in the flushing time as might at first be expected. For, when
A is taken too large or too small, a compensating effect results
from B being too small or too large. A serious deviation from a
reasonable estimate of the flushing time would occur when the net
salt transport velocity differed radically from one-fifth the value
of the maximum current velocity.

Example: Determine the flushing time of the James River estuary.

The navigation chart used for this example was U.S. Coast and
Geodetic Survey Chart No. 78. Ageneralized chart showing the locations
of the observing stations is shown in Figure 4. The area flushed
extended from a line connecting Pike Point and Newport News to
Shirley near the head of the estuary. Velocity-depth curves for
current station J-17 were used in place of river flow data and are
shown in Figure 5. The depth of the surface of no net motion at station
J-17 is about 9.5 feet and roughly corresponds to the inflection points
of the salinity-depth curves of Figure 6. The surface of no net motion
rises gradually from the head towards the mouth of the estuary,
and is estimated to be about 9 feet below the mean water level near
the boundary shown in Figure 4. The surface of no net motion will
be considered to be horizontal in this example.

In Figure 4, the net landward transport of sea water through the
boundary ab into volume B is estimated by multiplying the average
velocity found at station J-17 by the sectional area of B.

The dead water volumes involved in these considerations are
neglected when computing volume B (see Fig. 1). The average net
velocity for sections A and B at station J-17 can be obtained from
Figure 5 by means of a planimeter (i.e., by finding the centroid of
the area between the mean flood and ebb velocity curves). The velocities
thus obtained were:

Average net ebb velocity in A, , .0.30 knot
Average net flood velocity in B, ,0.27 knot

The effective areas corresponding to the volume sections A and B at
the boundary ab were computed to be:

Grossisection for Avis) 5 see 181,065 sq. ft.
Grossusection for Bl, 4. ao 180,087 sq. ft.,

and the corresponding volumes, A and,B, were:

Volume of section A ,...... 2,389x10! cu. ft.
Volume of sectionB,...... 2,254x10! cu. ft.

These and the above net flow velocities give:*

OQ, = 198xl0l cu. ft.
Qp = 177x10" cu. ft.

The exchange ratios are: fr =

and

y, = 2389 x 107(0.35)™

Yo = 2254 x 107(0.35)".
The plot of these curves and their sum (y; + yz) are shown in Figure 7.

PART II
Embayments (R = 0)

When the river water entering a bay is nil or insufficient to cause
the salinity of the bay to differ from that of the adjacent sea water,
then the circulation pattern of Figure 1 does not exist. In Figure 8A
the salinity of a bay at low water is supposed to be equal to that of
the adjacent sea volume, P.

Let- the volume P be equal to the tidal prism volume of the bay.
Suppose that the volume of the bay is contaminated,at high water.
Then Figure 8A represents the situation at the following low water.
It is assumed that the contaminant which is ejected into the sea
during any ebb tide is carried away from the bay entrance before the
following flood tide begins.

Now let volume P push bodily into the bay displacing the volumes

Pj} = P2 = P3, etc., all of which are numerically equal to P. We now
have a second high tide situation, Figure 8B. In Figure 8A the small

*Flood and ebb tide intervals were taken as 6 hours each.

10

numbered rectangles within Pj, Po, etc. represent uniform distri-
bution of the contaminant in the bay at low tide.

At high water, Figure 8B, some of the particles 1, which were in
volume Pp) at the previous low water, would still be expected to be
found near the entrance boundary of the bay. Similarly at high water
some of the particles 4 in Figure 8A would be expected to be found
at an excursion length farther into the bay as in Figure 8B. The
maximum span of the contaminant on the flooding tide is thus twice
that of the particle excursion distance. Thus, the contaminant in
P,} becomes distributed in the volume P, + P= 2P. The volume
of water containing contaminant which is lost during the next ebb
tide is equal to P, and therefore the exchange ratio for volume Pj
isr, =1/2.

Figure 8C shows what the situation would be at low tide if an adjust-
ment of the remaining contaminant did not now occur because of
diffusion. Since the volumes being considered are small, we assume
that the process of diffusion is complete before the next flood tide
begins. The effect of diffusion is to disperse the remaining contaminant
uniformly (Fig. 8D). Now the total contaminant remaining in P, at
the end of the first ebb tide is Py (l-r,). If P is taken equal to unity,
then, since rj, = 1/2, the contaminant remaining in P] is r}. On this
basis the contaminant in Pz, Figure 8C, is 2r), giving for the adjusted
value in each of the volumes P, and P, the amount 3 = 3/4, The con-
taminant lost from Pz is thus 1/4 = TSE This is the exchange ratio
for volume P2. Similarly, the contaminant in 12a and Po is now 2ry
L Be and for each of the volumes Pp and P3 this becomes 1/2
(2r] +34) = 7/8. Hence, the exchange ratio for P3 is r,>. In general
we can write for the exchange ratio of the nth segment P_,

ae
Now suppose that the bay volume at high water is equal to
Mm S IP ap IS) 485 9 0 6 0 P

The remaining contaminant near the head of the bay after

t=(+)

tidal cycles is approximately =

* Where e=is base of natural logarithm.

11

Obviously, if n is large, t will be too great to warrant consideration
of the flushing problem, when the region near the head of the bay
is considered alone. We will, therefore, consider the average situation
in the bay as a whole.

Let S be the sum of all the exchange ratios. Then

Nyy=S=rtr2+...... 49%,

Multiplying by r gives,
AS (SSP aS a5 go oc +
from which S(l-r) =r—r(n+Hor S = r/(l-r), sincer™+” is very small.

Since n segments were involved in the sum S the average exchange
ratio for the bay is

egal
Ve Tn
where P is the tidal prism volume and V, the high tide volume.

12

UNITS OF VOLUME

80

60
14,-B=14
13
= Mar
12 50
11
10--A=10 °
\ 40
9
8
30
7/ n
_ 140.35)
We BS Gai)
6 nN
5 : 20
4 e
3
we m
y,=10 (0.35.4 10
2 e
@
oo
. ie a
Mal ell GN es el Halaeate MWS) A
(0) \ a —j-m=4 == 0)
0 10 20 30 40 50 60 70 80 90 NOOR RO 120 140

TIDAL CYCLES

FIGURE 3 GENERAL FLUSHING CURVES

ils)

ORIGINAL CONTAMINATED VOLUME REMAINING (PERCENT )

| IN.212,097 FT.
|O-FT. DEPTH CONTOUR

FIGURE 4 LOCATION OF OBSERVING STATIONS IN THE JAMES RIVER
ESTUARY

14

DEPTH (feet)

SURFACE OF NO NET
MOTION 9.5 FEET

CURVES REPRESENT
WEIGHTED MEAN
VELOCITIES FOR
THREE PERIODS

FIGURE 5 MEAN VELOCITY-DEPTH CURVES FOR CURRENT
STATION J-I7

15

DEPTH (feet)

SURFACE OF NO NET
HORIZONAL MOTION

4 5 6 US) 14 15 16 17 18 nS
SALINITY (°/c0)

FIGURE 6 MEAN SALINITY-DEPTH CURVE AT 3 STATIONS
IN THE JAMES RIVER ESTUARY

16

jo)
ice)

70

(LNS0Y¥3d) AWNIOA G3SLVNINVLNOD SNINIVW3Y IWNIDINO

jo} je) fo)
© rs t o

SWNIOA JO SLINN

fe)
N

10

TIDAL CYCLES

FIGURE 7 FLUSHING TIME CURVES FOR THE JAMES RIVER ESTUARY

17

BAY

OCEANIC
p

i2fiijiofo|s{7te]s}4}3{2]i|

B. BAY AT SECOND HIGH WATER

2fiijiofo{s{7te{s] [4] [3|

C. BAY AT LOW WATER BEFORE ADJUSTMENT

OCEANIC
P

i3}i2}iifiofo{ 8 | 7] {5} fat [at |

D. LOW WATER AFTER ADJUSTMENT OF CONTAMINANT

FIGURE 8 DIAGRAM SHOWING REDISTRIBUTION OF CONTAMINANT
WITH TIDAL ACTION

18

BIBLIOGRAPHY

. KETCHUM, B.H. The exchanges of fresh and salt waters in tidal
estuaries, Journal of Marine Research, vol. 10, no. l, p. 18-38,
1951. Also in Woods Hole Oceanographic Institution Collected
reprints, 1951, Contribution No. 516, July 1952.

. PRITCHARD, D. W. A review of our present knowledge of the
dynamics and flushing of estuaries, Chesapeake Bay Institute
Technical Report No. 4, Reference 52-7. 45 p. 1952.

. - - - The physical structure, circulation, and mixing in a coastal
plain estuary, Chesapeake Bay Institute Technical Report No. 3,
Reference 52-2. 55 p. 1952.

. - - - A study of flushing in the Delaware Model, Chesapeake Bay
Institute Technical Report No. 7, Reference 54-4, 143 p. 1954.

19

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UOTJE[NIATD By} WIOIZ paatioap stdrysuoyjejlar yw

Aydeasoriqrg

‘(Z29-UL ‘O *H)
“8813 g ‘d 61 “6S61 Ate ‘uosqtn “mM atelq Aq
SLNAWAVAWNG GNV SAIUVALSA JO AWIL ONI

“HSNT4A AHL ONILWNILSA YOU GOHLAW V
201j3O dtydearZ01pAy AaeN ‘Ss ‘fp

"MOT IAATI pue
uotj0e Tepty Aq paoejdar aq 0} 1033eu1 papuad
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UOTJEMIAT 9y} UIOIJ paatiap stdiysuoyjelar vy

Aydeadoryqrg

‘(Z29-UL °O *H)
“8813 g ‘d 61 “6S61 Ammer ‘uosqrn “mM atelq Aq
SLINAWAVEWNG ANV SAIUVALSA JO ANIL DNI

“HSNT4 AHL ONILVWILSA YO GOHLAW WV
ao1jO s1ydeadoi1pAy AaeN *S “fp

29-UL “OH
“M IteTq ‘uosqry = :z0yyne
squeurAequigq pue
Satienjsq jo auity, Surysnyq ay3
Buryeurtjsq 10; poyyww Vy ‘21013

uotzoVy [TePLL
jo Sutysny ‘satrenjsq

29-UL ‘OH

"M Atelg ‘uosqrn ‘:r10yyne
szuaurAequigq pue

Satienjsg jo auity, Surysn[q 243
Sutyeurtjsqy I0y pouww VW 321913

uotzV [ePLL
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-xa yorym Arenjsa ad4j-juourpatd e jo uszazjed
UOTJEMIAT oy} WIOTJ paatiap stdrysuoryejlar yw

Aydeadorrqrg

‘(Z9-UL “O *H)

“8813 8 ‘d 61 “6s6T Ame ‘uosqrn “mM stetq Aq

SINAWAVGNG GNV SAIYVNLSA AO AWIL DONI
-HSN'T4 AHL ONILVWILSA YOA GOHLAN V

201730 DtyderdoapAy AneN ‘Ss ‘“

“MOTJ IDATI pue
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-xe yotym Arenjysa odAj-yuourpatd e jo uazezed
UOTJEMIATS ay} WIOAJ paatszap stdrysuoyelear yw

Aydeasortqrg

(Z9-UL ‘O “H)

“s313g “d 61 “6561 Ame ‘uosqtn “m steta Aq

SINAWAVGNG GNV SAIYUVNLSA AO AWIL ONI
-HSNT4 AHL ONILVWILSA YOU GOHLIW ‘V

a21jO dtyderd0a1pAH AaeN ‘SN

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29-UL “O°H “Ht
"M ATeT_ ‘uosqiy izoyyne “tt
sjusurAequigq pue
Satienjsy jo ouiry, Burysn{q au3
Butzeuitjsqy OF pouW VY :2131} ‘tT

uotzoV TePIL “Z
jo Burysny ‘sorrenjsq “|

29-UL “OH ‘Ht
“M ATeT@ ‘uosqry i104 Ne “IT
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uotzoVy [ePIL “2
jo Surysny ‘sotrenjsgq “[

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-xo yotym Arenjsa adfj-yuourpatd e jo uaazjed
UOTIE[NIATD Jy} WOT} paatiop stdrysuorjelar y

Aydeasoryqrg

(29-UL “O ‘H)

“8813 g “d 6T “6661 Amr ‘uosqtn “mM atetq Aq

SLNIWNAVGNG ANV SSIUVALSA 40 ANIL ONI
-HSN'1T4 AHL ONILVWILSA YOU GOHLAW VW

a01jO otydeadoi1pAy AACN ‘SQ

*MOTJ IATA pue
uot1j9e Tepty Aq paoeider aq 03 1233eur papued
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payeururejyuod e x0y paambaa aut} ay} sessaid
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UOTJE[MIATO ay} UIOIJ paatiap stdrysuorjelar yw

Aydessorqra

“(29-UL “O “H)
“8813 g “d 61 “6S61 Ame ‘uosqrn “mM steta Aq
SINAWAVENGA GNV SSIUVALSA JO ANIL ONI

“HSN TA AHL ONILWNILSA YOA GOHLAN V
ao1jO o1ydeaZ01pAy AaeN ‘Sf

29-UL “OCH “Tt
“M 2TeTA ‘uosqry :10yyne “Ir
squeurAequigq pue
Satienjsq jo awry surysn,q ay3
Buryeurt3sq Oy poyyaW VY :211) ‘t

uoroy TePIL “2
jo Surysnyy ‘satienqjsy “{

29-UL °O°H “Ht
"M 2TeTq ‘uosqry :104yNe “TT
syusurAequig pue
Satienjsgy jo awuity Butysntq ay}
Sutzeurtjsqy OJ pouW VW :21113 “t

uotIoy [EPIL “2
jo durysnyy ‘satienjsy “[

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uote Tepty Aq paoetdar aq 03 123;eur papuad
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-xa yotym Arenjsa adAj-yuourpatd e& jo uazajyzed
UOTJE[MIATS ay} UOT} paatiap stdtysuoyjejler wv

Aydeazorrqrg

(29-UL °O ‘H)

“8313. g ‘d 61 “6561 Amr ‘uosqtn “mM atetq Aq

SLNANAVENG GNV SAINVNLSA AO ANIL ONI
-HSN'T4 AHL ONILVWILSA YOU GOHLAN V

aoujO otydeaso0apAy Aaen ‘Sp

"“MOTZ IAATI pue
uorz0e yep; Aq padetdar aq 03 1933eUr papued
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-xa yotym Arenjysa ad4j-yuourpetd e jo uazazed
UOTJeINIATD yj UIOIF paatiap stdrysuojelar Ww

Aydeasorrqra

(Z29-UL “O “H)

88139 ‘d 61 “6561 Ame ‘uosqrd “mM atela Aq

SINANAVGANG GNV SAIYUVALSA 40 ANIL ONI
-HSN'1T4 AHL ONILVWILSA YO GOHLAW ‘V

201jO otydeas0apAy AAeN ‘S ‘f

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