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इंटरनेट मानक

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“The Right to Information, The Right to Live”

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“Knowledge is such a treasure which cannot be stolen”

“Invent a New India Using Knowledge”

है”ह”ह

IS 5477-2 (1994): Fixing the capacities of reservoirs -Methods, Part 2: Dead storage [WRD 10: Reservoirs andLakes]

IS 6477 ( Part 2 ) : 198'+

~ qr;rcp

\if~TWll1 CflT efqcrT f;ralf~a Cf)~;f Cf)T ~follf

1fT1T 2 atst~ \i11~"

( ~.m ~'fU&fUT )

Indian Standard

FIXING THE CAPACITIES OFRESERVOIRS- METHODS

PART 2 DEAD STORAGE

( First Revision)

UDC 627-815-6

e SIS 1994

BUREAU OF. INDIAN STANDARDSMANAK BRAVAN, 9 BAHADUR SHAH ZAPAR MARO,

NEW DBLHI 110002

D,cember 1994

REAFFIRMED

. JUL200~

Price Group S

Reservoirs Sectional Committee, RVD 4

FOREWORD

This Indian Standard ( First Revision) was adopted by the Bureau of Indian Standards, after the draftfinalized by the Reservoirs Sectional Committee had been approved by the River Valley DivisionCouncil.

By providing extra storage volume in the reservoir for sediment accumulation. in addition to the livestorage, it is ensured that the Jive storage, although it contains sediment, will function at full efficiencyfor an assigned number of years. This volume of storage (in the fixation of which the minimum drawdown level is also a major criterion in case of power projects) is referred to as the dead storage andis equivalent to the volume of sediment expected to be deposited in the reservoir during the designedlife of the structure.

The distribution pattern of sediments in the entire depth of a reservoir depends on many factors, suchas slope of the valley, length of reservoir, constriction in the reservoir, particle size of the suspendedsediment and capacity inflow ratio; but the reservoir operation has an important control over otherfactors. However, a knowledge of this pattern is essential. especially, in developing areas, in order tohave an idea about the formation of delta and the recreational spots and the consequent increase inback water levels after the reservoir comes into operation.

This standard ( Part 2 ) was first publisbed in 1969. The present revision has been prepared to incorpo­rate the latest knowledge in this field in this revision an additional figure for determining the type ofreservoir has been incorporated in addition to modifying Fig. 1 and 2 and some tables.

This standard consists of four parts, Part 1 covers general requirements, Part 3 covers live storage andPart 4 covers flood storage.

For the purpose of deciding whether a particular requirement of this standard is complied with.the final value, observed or calculated, expressing the result of a test or analysis, shall berounded off in accordance with IS 2: 1960 -Rules for rounding off numerical values (revised )'. Thenumber of significant places retained in the rounded off value should be the same as that of thespecified value in this standard.

IS 5477 ( Part 2) : 1994

Indian Standard

FIXING THE CAPACITIES OFRESERVOIRS - METHODS

PART 2 DEAD STORAGE

I ( First Revision)

3 TERMIN()IJ()(;Y

1 S(~OPE

For the purpose of tbis standard, tbe definitious givenin IS 4410 ( Part 6 ) : 1983sball apply.

4 MEASlJREMENT OF SEDIMENT YIELDS

This standard (part 2) covers the methods for comput­ing the sediment yield and for predicting the probablesediment distribution in tbe reservoir below normal(full) reservoir level (F.R.L.).

2 REFEREN(~ES

record covered by the survey will then be equal to thetotal weigbt of tbe sediment deposited in the reservoirplus that which bas passed out of the reservoir based onthe trap efficiency. In this way, reliable records may bereadily and economically obtained on long-term basis,

4.%.2 The density ofdeposited sediment varies with thecomposition of the deposits, location of the depositwithin the reservoir, tbe flocculation characteristics ofclay content and water, tbe age of deposit, ell". Forcoarse material (0.0625 mm and above) variation ofdensity with location and age may be unimportant.Nonnallya time and space average density ofdepositedmaterials applicable for tbe period under study isrequired for finding tbe overall volume of deposits. Forthis purpose the trapped sediment for the period understudy would have to he classified in different fractions,Most of the sediment escape from getting depositedinto the reservoir should be from tbe silt and clayfractions. In some special cases local estimates of den­sities at points in the reservoir may be required insteadof average density over the whole reservoir.

4.Z.3 The trap efficiency mainly depends upon thecapacity-in-Dow ratio but may vary with location ofoutlets and reservoir operating procedure. Computationof reservoir trap efficiency may be made using trapefficiency curves, sucb as tbose developed by Bruneand by Churchill (see IS 12182 : 1987).

4.1.4 The sedimentation rates observed in adjacentreservoirs also serve as guide while designing deadstorage capacity for a new reservoir, the rate ofsedimentation observed in similar reservoirs and/oradjacent basin should be suitably modified keeping inview the density of deposited material, trap efficiencyand sediment yield from the catchment,

4.3 Sediment Load Measurements

Periodic samples from the stream should be taken atvarious discharges along with tbe stream gaugingobservations and tbe suspended sediment concentra­lion should be measured as detailed in IS 4890 : 1968.A sediment rating curve which is a plot of sedimeutconcentration against the discharge is then preparedand is used ill conjunction with stage duration curve (ornow duntion) based on uniformly spaced daily orshorter time units 1ata in case ofsmaller river basins toassess sediment load. For convenience, the correlationbetween sediment concentration against discharge OIlY

TitleGlossary of tenus rela ting to rivervalley projects : Part 6 Reser­voirs (first revision)Methods of measurement of sus­pended sediment ill open channels

Guidelines for determination ofeffects of sedimentation in plan­ning and performance of reservoirs

121M2: 1987

4890: 1968

4.1 The sediment yield in a reservoir maybe estimatedby anyone of the following two methods:

a) Sedimentation surveys of reservoirs withsimilar catchment characteristics, or

b) Sediment load measurements of the stream,

4.1 Reservoir Sedimentation Survey

4.%.1 The sediment yield from the catchment is deter­mined by measuring the accumulated sediment in areservoir for a known period, by means of echosounders and other electronic devices since the normalsounding operations give erroneous results in largedepths. The vol ume ofsediment accumulated in a reser­voir is computed as the difference between the presentreservoir l'apal'ity and the original capacity after thecompletion of the dam, The unit weight of deposit isdetermined in tbe laboratory trom the representativeundisturbed samples or by field determination using acalibrated density probe developed for tbis purpose.The total sediment volume is then converted to dry­weight of sediment Oil the basis of average unit weightof deposits. The total sediment yield for the period of

The following Indian Standards are necessary adjunctsto this standard:

IS No.

4410(Part 6) :1~83

t

IS S477 (Part Z) : 1994

be altered to the relation of sediment load againstrun-off for calculating sediment yield. Where observedstage/Dow data is available for only shorter periods,these have to be. suitably extended with the help oflonger data on rainfall. The sediment discharge ratingcurves may also be prepared from hydraulic considera­tions using sediment load formula, that is, modifiedEinstein's procedure.

4.3.1 The bed load measurement is preferable. How­ever, where it is 110t possible, it may be estimated usinganalytical methods based 011 sampled data or as a per­centage of suspended load (generally ranging from 10to 20 percent). This should be added to the suspendedload to get the total sediment load.

5 PREDICTIN(; SEDIMENT DISTRIBUTION

5.1 The sediment entering into a storage reservoir getsdeposited progressively with the passage of time andthereby reduces the dead as well as live storage capacityof the reservoir, This causes the bed level near the damto rise and the raised bed level is termed as new zeroelevation. It is, tberefore, necessary to assess therevised areas and capacities at various reservoir eleva­tions thatwould be available ill future and could be usedill simulation studies to test the reservoir performanceand also tbe new zero-elevation.

The following procedure may be adopted for fixing thedead storage level and sill levels of the outlets:

a) The distribution of the estimated sediment loadfor the feasible service time of the reservoirshould be carried out and new zero-elevationssbould be determined, and

b) The minimum drawdown level is fixed a littleabove the new zero-elevation computed in (a)above. When other considerations like com­mand area elevation, providing extra head forpower generation, etc, prevail, this elevation isfixed higher than one of these.

S.z Several methods are ill use for predicting sedimentdistribution in reservoirs for design purposes. Either theempirical area reduction method or the area incrementmethod maybe used.S.1.1 Empirical AreaReductionMethod

This method isbased on tbe analysis ofdata ofsedimentdistribution. In this method, reservoirs are classifiedinto four types, namely, (8) gorge, (b) hill, (c) Ooodplain-foot bill, and (d) lake, based on the ratio of tbereservoir capacity to the reservoir depth plotted on alog-log scale (see Fig. 1). Figures 2 and 3 give thesediment distribution-area design curves for each typeof these reservoirs. The equation for the design curveused is:

...... (1)

where

Ap • a non-dimensional relative area at relativedis..nee 'p' above the stream bed, and

C, m and =non-dimensional constants which have beenn fixeddepending on the type of reservoir.

5.1.1.1 These curves are used to work out the probablesediment deposition in the reservoir at different depths.This method is more reliable than the area incrementmethod. An example of the usage of this method isgiven in Annex A.

NoI.: m= x S !y

~ 7 m II io ro1.5.oT fFlE:1 v(GOA::r....JI~ ..

l/l r-,~

V :~ ~~ V m• 1.5to2.51YPE III (Hlll)

V x L--~

'~ I.......~

~

L......-". I

II ~ 1

- -I ~ ~m. 25to3.5TVFtEIi (flO D[ I~N FOOT HILL~ --........- - I.' ~ II""" L....o ,.., m. 3.5to4.5 TYPE I (lAKE)II /" ......

~ ~..- _I--"" ~ t.- IIIII~

III ~ ~~ I- .....-~

~ I

!/lI' ~ I-~~~

~1-~

~~l..oIII'" 1..00---

~~~ ~~ "..I-'~ ~

~ ....CAPACln(C)

FIG. 1 C~IFICA11ON OFREsERVOJRpFYTH VERSUSCAPACTY RaAnONSHIP

2

IS !477 (Part %) : 1994

,where

VI - the sediment volume to be distributed in thereservoir in hectare metres,

Ao - tbe area correction factor in hectares whichis original reservoir area at the new zeroelevation of the reservoir,

H • tbe reservoir depth below full reservoirlevel (F.R.L.) in metres,

ho • tbe depth in metres to which the reservoiris completely filled with sediment, and

Vo • the sediment volume below new zeroelevation in hectare metres,

5.1.2.1 In other words, the equation mathematicallyexpresses that the total sediment volume Vs consists oftwo parts, namely, (a) tbe protion which is uniformlydistributed vertically over the height H - 110 with an

!.2.2 Area Increment MethodThe basic assumption in this method is that tbe sedi­ment deposition in tbe reservoir may be ap­proximated by reducing the reservoir area at eachreservoir elevation by a fixed amount. Successiveapproximations are made. Average end area (or pris­moidal formula) is used to compute the reservoircapacities on the basis of reduced surface areas untiltbe total reservoir capacity below the full reservoirlevel is the same as the predetermined capacity ob­tained by subtracting the sediment accumulationwith time from tbe original capacity.

The basic equation in this method is:

VI • Ao (H - ho) + Vo ••••••••(2)

t----tt-......-IV Ap;: 1.486P -0.25 ( 1 _ p) 1.34_

III Ap. 16.967p 1.15 (1-p) 2.32

t--.......---4~ II Ap =2.487 P 0.57 (1 _ p) 0.41

I Ap = 5.074 P 1.85 (1 _ p) 0.36

0.0 ....--'-.--.I""'--.....&-.--............-......-......._~~~o 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

RELATIve DEPTH ~

,(MEASURED FROM BOTTOM)

FlO. 2SED~ DISTRIBUTION - AREA DESIGN CURVES(BT ON REsERVOIR STORAGE CuRVES )

2.8

2.8

2.4

2.2

2.0

!1.8

L5 1.6ex:-c 1.4

~ 1.2

IX1.0

0.8

0.6

0.4

'00~~

TYPE1""\ / r-' JTYPE 11--;\ ~V

~

/" j/ r\ V / ~

/" V ~VI/ / Y \ V/ ~

I / /~ V LTYPE III

I / ~

/ /v: ,

K'V ./

II~/ /~ \ -TYPE I'

~ t

,L~~

•

o 20 40 60 80 100 120

PERCENT SEDIMENT DEPOSITED

FIG. 3 TYPE CuRVES OF PERCENTSmlMen' DEPOSITED V,.ASUS PER(~ENT

REsERVOIR DEPlH BASFD ON Acrue, Oa:uRRENCES •

3

where

f (P) • a function of the relative depth of reservoirfor one of tbe four types of theoreticaldesign curves,

V (p). relative volume at a given elevation,

(, (P). relative area at a given elevation,

f' V}) - a function of tbe relative depth of reservoir

IS 5477 ( Part 1 ) : 1994

area equal to Ao and (b) the portion Vo below the newzero elevation of the reservoir.

S.1.1.1 Anexampleof theusageof this method isgivenin Annex B.

NOTE-lbe applicability of this metboddecreaseswith. . . sediment deposit

the I nerease I n the rano of . . . If thereservoi r capacityhundred years sediment, accumulation exceeds 15 per­cent of the original capacity, a more exact method shouldbe applied.

S.l.J Moody:~Metlwd to Find New Zero Elevation

This method is used to determine the newzero elevation0, directly without trial and error process. Twoparametersf (p) andf' (P) as explained below are madeuse of:

1(P). 1 - V(p~a (P)

f' (P). S - V (pH)HA(pH)

....(3)

....(4)

fora particular reservoirand itsanticipatedsediment storage,

S • total sediment inthereservoir inbeda~metres,V (pH)• reservoir capacity at a given elevation in

hectare metres,H. the total depth ofreservoir for nonnal water

surface in metres, andA (pH) • reservoir area at a given eleva tion itt hec-

tares.S.1.3.1 Table 1gives the values of the functionj' (p) fortbe four types ofreservoirs (see 5.2.1) and Fig. 4 sbowsthe plotting of!Cp)against relative reservoir depth, P,for the four types of reservoirs of tbe empirical areamethod (see 5.1.1) and also for the area incrementmethod (see 5.1.1).S.1.J.l To determine the new zero eleva tion, f (P)should equal!' (p). This is done graphically by plottingthe values of I' (P) and superposing this over therelevant ICp)curve. The intersection gives the relativedeptb of (Po) reservoir at new zero elevation aftersedimentation. New zero-elevation may be computedby adding the product Po- H to tbe original stream bedelevation. After arriving at the new zero elevation,either empirical area method (see S.1.1) or the areaincrement method (see S.%.1) is used.5.1.3.3 An example to lind out the new zero eleva lionis given in Annex C.

1 000-0

500-0

100·0

50-0

'"~.... 10-0

0·1

~05

0-01

..

I I

\ rl'VPE I

,\,

~

" " ,a, "-~~ ~TlPEII '-~,~

~

'" ....... 1

~ 1......... 11oo.:: - .........I~ 10......- r---- -. ~ 1

/ .... I I - ~---- ~~ '-~ :rVPE IV V ""'lIIIIIiiII~ .......... """~~J -,...;:::~ '-0..

-. VPE III ..--. -.. --

, ARE A INC ~EME~T"'-~" .""

I I ~ vI ~

0-' 0-2 &3 ~4 ()& 0-6 0-7 0-8 0-' 1-0RELATIVE 'DEPTH (p)

FIG. 4 CURVES TODETERMlW THE DEPTH OF SFDIMENT IN mE REsERVOIR

4

IS 5477 ( Part Z) : J9,..

llIble 1 Values of the FUDdioD/(P) Cor the Four1)'pes o(Reservoirs(Clause 5.2.3.1)

p 1)pe

II III IV

(1) (2) (3) (4) (5)

0

0.01 996.7 10.568 12.03 0.2023

0.02 277.5 3.758 5.544 O.23Q 0

0.05 51.49 2.233 2.057 0.2796

0.1 14.53 1.495 1.013 0.2911

0.15 6.971 1.169 0.682 1 0.2932

0.2 4.145 0.9706 0.5180 0.2878

0.25. 2.766 0.8299 0.4178 0.278 1

0.3 1.900 0.7212 0.348 6 0.2556

0.35 1.495 0.6323 0.296 8 0.251 8

0.4 1.109 0.5565 0.2333 0.236 5

0.45 0.9076 0.4900 0.221 2 O.21Q7

0.5 0.726 7 0.4303 0.1917 0.2010

0.55 0.5860 0.376 R 0.168 7 0.1826

0.6 0.4732 0.3253 0.1422 0.1637

0.65 0.3805 0.2780 0.120 7 0.1443

0.7 0.3026 0.2333 0.100 8 0.1245

0.75 0.2359 0.1907 0.08204 0.1044

0.8 0.1777 0.1500 0.064 28 0.083 Q7

0.85 0.1202 O.IIO? 0.04731 0.06330

iJ.9 0.080 11 0.07276 0.031 OJ 0.04239

0.95 0.05830 0.02698 0.01527 0.021 23

0.98 0.01494 0.01425 0.006057 0.008 534

0.99 0.007411 0.007 109 0.003020 0.002470

1.0 0.00 0.00 0.00 O.(X}

5

IS 5477 (Part 2) : 1994

ANNEXA(Clause 5.2.1.1 )

EMPIRICAL AREA REDUCTION MEmOD

A-Z.l.l Column 6 is K multiplied Ap values at eachsucceeding increment, which is sediment area.

tions on a log-log paper. A line is drawn through tbeplotted points. Reciprocal of the slope of the line willgive thevalue of m by which reservoir type is selected(see Fig. 1). in Ibis case, it is Type II (see Fig. 5). Theplot may sometimes indicate a curve baving differentslopes in different parts. III sucb cases, tbe reservoirmaybe classified into the type ill whicb major portionof the sediment would deposit.

A-Z.z Ap in col 5 is obtained from the relevant curveof Fig. 2. Zero elevation is assumed wbich in this caseis 1 277.00 0). Surface area corresponding to this eleva­tion is 121.40 bectares. Ap at this elevation is 1.10 (seeFig. 2). Find out K, tbe ra tio ofo~iginalarea a t assumednew zero elevation ( col 2 ) to the corresponding Apvalue ( col 5 ).

A-I DA1:4

A-I.I The data given are as follows:

a) Origiual capacity curve,

b) Annual sediment inOow =37.00bectare metres,c) Period of sedimentation 100 years (total sedi­

ment for the period =3 700 hectare metres),d) Bed elevation =1 265.00 Ill,

e) Normal water surface elevation =1 302.50 m,and

f) Spillway crest =1 3025001.

A-2 PR()CEDlrRE

A-Z.l Referring to Table 2, the data given itl coli, 2and 3 are from original area capacity curves. The rela­tive depth ratio lor different levels to the total depthfrom spillway crest (or FRL) to tbe stream bed isentered ill ("01 4. Reservoir depth is plotted as ordinateagainst reservoir capacity as abscissa It different eleva-

K. 1~.li: - 110.36 ......(5)

100.080.0

70.0

60.0

30.0enw~ 20.0w:i

~

,IE 10.0

I~ 8.0

cr: 6.0

~w 4.0(J)wa: 3.0

2.0

0

~.

.-J--'---,- -- '-' ,-_. '--f---

I I

--~~

---_. -,--_.~

~lI"'"

~- ~~ f------- ._~-~--.. '-, -- ~----,~

~ I y=3.5

~~v I

~. - .»: x =10.5~

~~,~-- -------. ..... ....... L

-~~------

V""""~t/

x -= 10,5 Y=3,5--

~/'

--~- ~-- ~- ~-~--,.,10.5

m=15 :;3.0t--

RESERVOIR CAPACITY IN HECTARE METRES

FIG. S CURVE 10 DETBMINE VALUE OF 'm'

6

IS 5477 ( Pal11 ) : 1994

Table Z Sedlmeat Deposition Computation by Empirical Area Reduction Method(Clause A-2.1 )

Elevation Original Relative Ap Sediment AcculDu- Reylsed Revisedm Depth (lype III) lated Area C.apaclty

Area Capacity Area \blume SedimentVolume

Hectare Hectare Hectare Hectare Hectare Hectares HectareMetres Metres Metres Metres

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)

1 302.5 1882.00 19330.00 1.00 0.000 0.00 0.0 3708.16 1 882.00 15621.84

1 301.0 165Q.OO 16460.00 0.96 0.65 71.73 53.8 3654.36 1 587.27 12805.64

1 298.0 1 295.00 11 960.00 0.88 0.97 107.05 268.17 3386.19 1 187.95 8 173.81

1 295.0 991.50 8480.00 0.80 1.13 124.70 347.63 3036.54 866.8 5441.44

1292.0 700.10 S 889.00 0.72 1.22 134.64 389.01 2649.55 573.46 3239.45

1289.0 505.80 4039.00 0.64 1.26 139.05 410.54 2239.01 36'.75 17Q9.9Q

1 286.0 364.20 2713.0()' 0.56 1.28 141.26 420.47 1 818.54 222.94 894.46

1283.0 263.00 1 758.00 0.48 1.25 137.95 418.82 1 399.72 125.05 358.28

1280.0 182.10 1079.00 0.40 1.19 131.33 403.92 995.8 50.67 83.20

1 277.0 121.40 616.70 0.32 121.40 379.10 616.7 0 0

1274.0 80.93 3OJ.30 0.24 80.93 308.4 308.3 0 0

1 271.0 40.47 123.30 0.16 40.47 185.0 123.3 0 0

1 268.0 20.23 30.83 0.08 20.23 92.47 30.83 0 0

1 265.0 0 0 0 0 0 30.83 0 0 0

3708.16

A·2.1.1 CoIunm 7 is the inaement of sediment volume A·Z.2.3 Sum of col 7 gives the sediment volume ofcomputed by the average end area (onnula, that isequal to: 3 708.16 which is nearly equal to the computed sedi-

h ...... (6)ment volume of 3 700.00 hectare metres. Hence, the

2(Al +A2)· V zero elevation assumed is correct If the values do nottally, further trials have to be made till tbe sediment

where values differ by not more than one percent. Column 8

h- the height of the segment, gives the sediment accumulation volume in hectare

AllndA2 • the areas at the end of the segment, and metres. Revised areas in 001 9 are obtained by subtract-ing values in col 6 from col 2. Revised capacity in colv. the volume of the segment 10 is obtained by subtncting values in 0018 from col 3.

7

IS 5477 ( Part 2 ) : 1994

ANNEXB( Clause 5.2.2.2)

AREAINCREMENTMETHOD

B-1 DATA

8-1.1 Data given are the sameas in A-I.I.

B-1 PROCEDURE

8-%.1 Table 3.gives typical calculations of an examplefor working out the revised capacity. The procedure isIS given below:Step 1 - Assume lao and corresponding to this ho read

Ao IDd Vo from the original area capacitycurve. Substitute the values in the basicequation Vs - Ao (H - ha) + Vo• In this caseIto - 12.0 metres, Ao - 121.40 hectares,Vo - 616.70 hectare metres andVs • 3 712.4 hectare metres which is nearlyequal to the 3 700.00 hectare metres of thetotal sediment load within one percent,

Step 2 - Compute tbe cumulative volume of sedi-

ment (col 5)by applying the area correctionfactorwhich is incol4 ofTlble 2andget thevolume by average end area formula. Com­puted volume should then be within onepercent of the predetermined sedimentvalue.

Step 3 - Revisedareasincol 6 are obtained by reduc­ing the original area at each increment in col2 by the Irea correction factor in col 4.

Step 4 - The revised capacity is determined byreducing the original capacityat each incre­ment by the sediment accumulation (col 7 =col 3 - col 5).

B-Z.Z The result obtained should be compared withactual resurvey curve. After verifying, the probablesediment deposition in the reservoir at different depthsmay be worked out for the sediment volume to bedistnbuted in the periodequal to the life of the reservoir.

18ble 3 Area Increment Method

(Clause 8-2.1)

Elevation Original Area Originallit Capacity

Hectares HectareMetres

( 1) (2) (3)

1 302.5 1882.00 19330.00

1 301.0 1659.00 16460.00

1 298.0 1 295.00 11 960.00

1 295.0 991.50 8480.00

1 292.0 708.10 5889.00

1 289.0 505.80 4039.00

1286.0 364.20 2 713.00

J 283.0 263.00 1 758.00

1 280.0 182.10 1 079.00

1 277.0 121.40 616.70

1274.0 80.93 308.30

1 271.0 40.47 123.30

1 268.0 20.23 30.83

1 265.0 0 0

Hectares

(4)

121.40

121.40

121.40

121.40

121.40

121.40

121.40

121.40

121.40

121.40

SO.Q3

40.47

20.23

o

S~~lment RevisedVo,ume

,a,

Area Capacity

~ectare Hectares HectareMetres Metres

(5) (6) (7)

3712.40 176.60 1 5617.60

3530.3 1 537.60 1292Q.70

3 166.10 1 173.60 87Q3.Q{)

2801.90 870.10 5678.1

2437.7 586.70 3451.3

2073.5 384.40 1 <)85.5

17CR.30 242.80 1 003.7

1 345.10 141.40 412.QO

980.90 60.70 <)8.10

616.70 0 o

~.30 0 0

123.30 0 0

30.83 0 0

0 0 0

8

IS 5477 (Part Z ) : 1994

ANNEXC( Clause 5.2.3.3 )

MOODY'S METHOD TO FIND NEW ZERO ELEVATION

c-i DATA

e-l.l The data given are as follows:

a) Reservoir Type II,

b) Bottom elevation = 1 265.00 m,c) Total depth (H) = 37.5 m, and

d) Total sediment in reservoir (5) = 3 700 hectaremetres,

e-l.% Referring to Table 4, the data given in col 1 and3 are taken from tbe known original c-apacity versusdeptb curve. COIUID11 2 is worked out as the result of

division of each depth corresponding to the elevationsgiven in col 1 by the total depth (H). Col U11111 5 isworked out from the known original area versus depthcurve and it is the product of the area at a specifiedelevation and the total depth. Column 4 is obtained bysubtracting col 3 from '5'. Column 6 is obtained fromFig. 4, and column 7 is worked out by equation (4).

C-l.2.1 In Fig. 6, f(P) and I' (P) curves are drawnagainst relative reservoir depth (P) and their intersec­lion corresponds to Po of 0.32. Therefore,PcII- 12.00 m and the new zero elevation is1 265oCx) + 12.00 :I: 1 277.00 Ill.

Table 4 Moody's Method for Determination of New Zero Elevation

( Clause C-l.2 )

(1) (2)

1 268.00 0.08

1 271.00 0.16

1 274.00 0.24

1 277.00 0.32

1280.00 0.40

1 283.00 0.48

1 286.00 0.56

(6) (7)

1.80 4.83

1.22 2.36

0.85 1.12

0.68 0.68

O~56 0.38

0.46 0.20

0.39 0.07

Elevationm

p V (pm S - VCpH) HA (PH)

Hectare Hectare HectareMetres Metres Metres

(3) (4) (5)

30.83 3669 759

123.30 3577 1 518

308.30 3392 3035

616.70 3083 4553

1 079.00 2621 6829

1 758.00 1942 9863

2 713.00 987 13650

9

l(p) fromFig. 4

I' (p) fromEq4

If' (p).

S-V(pH)

~ - HA(pH)

"- ,~ ,~ !o~

" ""(~"' ~"

'~,~, PoS! 0.32

'~o ,. PoH =: 0.32 X'f" 5 =: 12.00M

·1 IElEVAllON OF SEDIMENT DEPOSITED AT DAM =1277.00M

IS 5477 ( Part Z) : 1994

2.0

1.0

~..0.5-i

~-u.0enw::>

~ 0.2

0.1o 0.1 0.2 0.3 0.4 0.5 0.6 0.7

RElATIVE DEPTH (p)

FIG. 6 EXAMPLE FOR DIRECT DETERMINATION OF NEW ZERO ELEVATION

10

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AmeDdmeDts IUBed SiDce Publlc.do.

Amend No. Date of Issue

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