mathematical model of reservoir routing for...

http://www.iaeme.com/IJCIET/index.asp 341 [email protected]

International Journal of Civil Engineering and Technology (IJCIET)

Volume 10, Issue 02, February 2019, pp. 341–354, Article ID: IJCIET_10_02_037

Available online at http://www.iaeme.com/ijciet/issues.asp?JType=IJCIET&VType=10&IType=2

ISSN Print: 0976-6308 and ISSN Online: 0976-6316

©IAEME Publication Scopus Indexed

MATHEMATICAL MODEL OF RESERVOIR

ROUTING FOR SPILLWAY OF WADI HORAN

DAM

Alaa H. Alshami

Civil Engineering department, AL-FARABI University Coll. Eng., Baghdad Iraq

ABSTRACT

This research is to analyses hydraulic parameters of the spillway design for WADI

HORAN Dam. The spillway design of type Ogee overflow and the design based on

Water Experiment Station (W.E.S) of the U.S. Corp of Engineers. The inflow are

routed in order to decrease the maximum discharge passing on the spillway. The

maximum discharge passing are 1400 m3/sec with maximum head over the crest equal

to 3.4 m and 50 m3/sec Attenuation and 2 hrs Reservoir Lag. Froude number is

determined in order to select the type of stilling basin, the Froude number equal to 5

that can choose type II stilling Basin. The method used for routing is Inflow- Storage

Discharge ( I.S.D.) The profile of the downstream and upstream are calculated.

Key words: Spillway, Froude number, Reservoir Routing, Wadi Horan.

Cite this Article: Alaa H. Alshami, Mathematical Model of Reservoir Routing for

Spillway of Wadi Horan Dam, International Journal of Civil Engineering and

Technology (IJCIET) 10(2), 2019, pp. 341–354.

http://www.iaeme.com/IJCIET/issues.asp?JType=IJCIET&VType=10&IType=2

1. INTRODUCTION

Hydrology is a branch of Earth Science. The important of hydrology in the assessment

development, utilization and management of water resources, of any region is being

increasingly realized at all levels. It was in view of this that the United Nation proclaimed the

period of 1965-1974 as the International Hydrology decade during which, intensive efforts in

hydrologic education research, development of analytical techniques and collection of

hydrological information on a global basis, were promoted in University, Research Institution,

and Government Organizations (Raghunath, 2006).

Hydraulics is a technology and applied science using engineering, chemistry, and other

sciences involving the mechanical properties and use of liquids. In fluid power, hydraulics is

used for the generation, control, and transmission of power by these of pressurized liquids.

Hydraulic topics range through most science and engineering disciplines, and cover concepts

such as pipe flow, dam design, fluidics and fluid control circuitry, pumps, turbines,

hydropower, computational fluid dynamics, flow measurement, river channel behavior and

erosion (Raghunath, 2006).

Mathematical Model of Reservoir Routing for Spillway of Wadi Horan Dam


Hydraulic structures are structures submerged or partially submerged in any body of

water, which disrupts the natural flow of water. They can be used to divert, disrupt or

completely stop the flow. A hydraulic structure can be built in rivers, a sea, or any body of

water where there is a need for a change in the natural flow of water. An example of a

hydraulic structure would be a dam, which slows the normal flow rate of the river in order to

power turbines. Hydraulic design involves the application of flow theory to the design of

various water containment and control structures. The designer should ensure that the

structure and the systems will function effectively and economically under all foreseeable

service conditions. Hydraulic structures must be capable of withstanding flow conditions and

forces caused by both static and flowing water loading. They must resistant deterioration,

aging, and extemporaneous forces due to weather extremes and earthquakes (Icold, 2018)

spillways are important in dams and study site, Water normally flows over a spillway only

during flood periods when the reservoir cannot hold the excess of water entering the reservoir

over the amount used so Spillways insure that the water does not overflow and damage or

destroy the dam Usually, spillways are provided with gates, which provide a better control on

the discharges passing through If the spillway is gated, then the discharging water (Q) is

controlled by the gate opening and hence the relation of Q to reservoir water level would be

different from that of an ungated spillway (Kargpur, 2008). The capacity of a spillway is

usually worked out on the basis of a flood routing study, is seen to depend upon the following

major factors: The inflow floods, The volume of storage provided by the reservoir, Crest

height of the spillway, Gated or ungated. The aim of this study is to analysis hydrological data

of inflow discharge at Horan dam area at Iraq and find the maximum flow by reservoir

routing analysis. In addition determine the best hydraulic section of dam spillway.

2. DESCRIPTION OF THE STUDY AREA

Figure 1 Study area (Sayl et al., 2017)

https://en.wikipedia.org/wiki/Dam

https://en.wikipedia.org/wiki/Volumetric_flow_rate

https://en.wikipedia.org/wiki/Turbines

Alaa H. Alshami


Wadi Horan is the longest Wadi in Iraq. Located in Al Anbar Governorate west of Iraq, at

Latitude 33.0333°, Longitude 40.2500° Wadi Horan is roughly midway between Damascus

and Baghdad near the town of Rutba, it stretches for 350 kilometres (220 mi) from the Iraq-

Saudi border to the Euphrates river near Haditha.

Wadi Horan dam is located in the middle course of the (Khatsuria, 2004) Wadi Horan

channel serving the area of Block 7 area covered with Wadi Horan catchment on that profile

is 12,180 sq.km. The mean annual runoff is estimated at 19.3×106cu.m Figures (1) .The dam

location is selected having regard of wadi channel width and of ground topography far

positioning of spillway. The dam height at this location is limited by low-lying and rather thin

saddles, present on the right bank. The saddles are reducing the pool level to about 15 m. The

profile of the valley at the said elevation is about 350 m wide. At the elevation of 6 m above

the channel bottom the width is almost the same.

3. RESERVOIR CAPACITY

As said above the saddles on the right bank are practically limiting the normal pool level in

the reservoir on this profile and reduce it to 15 m. Higher normal pool level would impose

considerably greater scope of the works, both on the dam and in securing safety of its

abutments against detrimental effect of percolation. The reservoir capacity to be formed at

this pool level would amount to 25×106 cu.m with this capacity, the uncontrolled part of

Wadi Horan runoff due to spilling over would reach 42% of runoff, for the controlled release

of water of 3.0cm/sec. The average evaporation rate of the reservoir is estimated at 1.7x106

cu.m/year. In this way, the average net annual yield of the reservoir might be estimated at

9.5x106 Cu.m, or 49% of the average Wadi Horan runoff, under the assumption that other

reservoirs are not constructed upstream of that dam site.

4. DAM DESIGN

The dam will be 21.5 m high above the river bed. Its width at crest will be 7 m and the length

504 m. At an elevation of 13 m on the upstream side, a 4m wide berm will be built. Total

volume of the dam is estimated at 360,000 cu.m. The impervious part of the dam will have e

volume of 70,000 cu.m and the pervious fill and pitching 290,000 cu.m. At the abutments the

dam will be founded in dolomite and lime stones. The impervious zone of the dam will cut 2

m deep into the bed rock. In river bed and on terraces, the dam will be founded on alluvial

deposits and on bedrock. Pervious Zones of the dam will be founded on sediments, except in

the sections which first have to be cleared of the materials of poor bearing capacity or

geological-engineering characteristics. The watertight core will be founded on bedrock cutting

2 m deep into it. After this section is excavated and cleared of alluvium through the

foundation width. The terrace and recent sediments, in which the foundation trench is to be

excavated for the core, are expected to be 5-8 m deep. Alluvium will be investigated in the

dam site by means of excavating wells, drilling of holes and examination in situ of density

and permeability of the Alluvium. The dam will be filled with local materials from borrow

pits near the site. The watertight core will be built of clay available on nearby Plateaus 0.5 to

3 km far from the dam site. Pervious zones of the dam will be filled with gravel and send of

terrace and recent alluvium from borrow pits near the dam Site. These materials are estimated

to be available in adequate quantities and to be of good quality. While excavating terrace

material blasting might be required. Fitter zones will be filled with processed material,

because no suitable material has been found in the prospected terrain which could be used

enslaved. Rock for upstream pitching and dry stones drain and rock spalls for dam crest and

base to rock fill will be supplied from the excavations for water discharge structures of the

http://www.wikiwand.com/en/Wadi

http://www.wikiwand.com/en/Iraq

http://www.wikiwand.com/en/Al_Anbar_Governorate

http://www.wikiwand.com/en/Latitude

http://www.wikiwand.com/en/Longitude

http://www.wikiwand.com/en/Damascus

http://www.wikiwand.com/en/Baghdad

http://www.wikiwand.com/en/Rutba

http://www.wikiwand.com/en/Euphrates

http://www.wikiwand.com/en/Haditha



dam and from nearby quarry.The contact of watertight core and the bed rock will be protected

by a grout blanket 5 m deep through the dam length.

5. SPILLWAY

The spillway position is predisposed by the saddle present on the right dam abutment. The

Spillway is dimensioned at flood flow of 0.1 frequency, with Confidence interval of 50%

amounting to 3,960 cu.m/sec. The spillway consists of ungated concrete weir and chute

channel. The weir is to suit overflowing head of 4.5 m. The weir is proposed to be buried into

Sound limestone down to depth of 2 m. The chute shall be constructed as 200 m wide open

unlined channel. Such a solution is caused both by the distance from the dam and by the

expected geological condition of rock making up spillway and chute foundations. The type of

spillway to be chosen shall depend on:

Inflow flood, Availability of tail channel, its capacity and flow hydraulics, Power, house,

tail race and other structures downstream.

6. OVERFLOW (OGEE) SPILLWAY

The overflow type spillway is most commonly used as the integral overflow section of a

concrete or masonry dam Figure 2. It is most commonly used with gravity dams. However, it

is also used with earth and rock fill dams with a separate gravity structure; the ogee crest can

be used as control in almost all types of spillways; and it has got the advantage over other

spillways for its high discharging efficiency (Kargpur, 2008).

Figure 2 Ogee spillway

7. MATHEMATICAL MODEL

7.1. Area Elevation Capacity Curve

Curve measures the volume (capacity) of the reservoir below a certain elevation the area

capacity curve are necessary for defining the storage capacity of reservoir (Hussein, 2010).

An area – capacity curve is obtained by plan- metering enclosed within each contour line in

the reservoir area. The graphical plotting of area and capacity curve relates the surface area

and the storage capacity of the reservoir to the elevation of the water surface.

Alaa H. Alshami


7.2. Topographical Survey at proposed site at WADI HORAN

The result of the topographical survey at WADI HORAN site , the following reading of the

elevation ground level versus the area in Hectar, the Table (1) represent the data of area

versus Elevation. Table 1 Area, Elevation data

Area (Hectare) Elevation(m) Area(Hectare) Elevation (m)

172.9 526 3.8 519

233.8 527 4.2 520

329.1 528 5.6 521

330.6 529 17.6 522

381.7 035 42.5 523

431.2 035 65.3 524

108.6 525

7.3. Establishing Storage – Discharge Relationship

The area at reservoir site are surveyed in detail and contour map is prepared with contour

interval one meter (1.0 meter) . From the map the area enclosed by various contours are plan

metered and curve of elevation versus area is prepared. Hence, the incremental volume of

water stored between any two successive contours can be determined using one of the

following equations:-

1- Trapezoidal Formula

= h/2 (A1 + A2)

2- Cone Formula

= h/3 (A1 + A2√

3- prismoidal formula

Where A1, A2 are area corresponding to two successive contour values and (h) is the

difference between them ,Am is the area enclosed by contour line midway between the two

adjacent a contours .If these incremental volumes are accumulated up to any contour value,

then that sum represent the storage volume of reservoir up to elevation of that contour. This

information is used to prepare a curve of elevation versus storage.

7.4. Out flow from the reservoir

The out flow from reservoir corresponding to any elevation can be determined by using the

discharge equation, one for ogee spillway and the sluices way opening

1- Ogee (Overflow) spillway

For Ogee spillway

Qsp = C L ( H–Hsp)3/2

Where Qsp = the discharge from the spill way

C = Coefficient of discharge of the spillway = 2.25

L = effective length of spillway, assumed as in HORAN DAM equal to 100meters

2- Sluicegate water way

Qsl = n Cd A √

Where



Qsl = discharge from sluice gate

n= the number of sluice ( 2 circular sluice pipes with one meter diameter and are placed at

elevation 520 m

A = area of each sluice opening

Hsl = sill level of the sluice opening (520 m)

Cd = Coefficient of discharge of sluice = 0.8

Trash racks for the opening of the sluices 2

Trash rack is a structure, whichis provided at the entrance to the intakes and sluiceways to

prevent entrance to debris. These are made in the form of semi – polygonal grid of iron or

steel bars as shown in the Figure 3.

Figure 3. Trash Rack

7.5. Elevation – Storage Curve

The calculation to obtain the storage versus elevation curve is given in the Table 2. The

contour interval is uniform and equal to 1 meter; Trapezoidal formula has been used to

compute incremental volumes. The storage in Hectar-meter is converted in to m3/sec will be

usually convenient in the routing problems (A.Hussein and Alshami, 2018). The Elevation-

area curve and elevation – storage curve are prepared from the Table 2 as shown in the Figure

4 and Figure 5.

Figure 4. Elevation – area curve

518

520

522

524

526

528

530

532

0 50 100 150 200 250 300 350 400 450 500

Elev

atio

n

Area (hectar)

Alaa H. Alshami


Figure 5. Elevation – Storage curve

Table 2. Calculation Elevation area and storage curve

Storage

Cumecs-day

Cumulated

storage H-m = h(A1+A2)/2 A1 +A2

Area

(Hectar) Elevation (m)

6 5 4 3 2 1

0 0 3.8 519

0.463 4 4 4 4.2 520

1.030 8.9 4.9 4.9 5.6 521

1.91 16.5 11.6 11.6 17.6 522

4.82 41.65 30.05 30.05 42.5 523

9.72 83.95 53.9 53.9 65.3 524

16.30 140.85 86.95 86.95 108.6 525

26.35 227.70 140.75 140.75 172.9 526

42.88 368.55 203.35 203.35 233.8 527

56.11 484.80 281.45 281.45 329.1 538

70.76 611.35 329.90 329.90 330.7 529

79.41 686.10 356.20 356.20 381.7 530

88.27 762.65 406,45 406.45 431.2 531

8. CALCULATIONS AND RESULTS

8.1. Out flow Discharge

The discharge from the pipe sluices is given by

Qsl = n Cd A √ )

The data used are as follows

Number of pipe used are 2 with a diameter equal to one meter, Coefficient of discharge = 0.8

Area of the pipe = 2 (1)2𝜋 /4 = 1.571 m

2

Qsl = 2 0.8 1.571 √ √

Qsl= 11.13√ .. (1)

The discharge from Ogee (over flow) spillway

Qsp = C L ( H –Hsp)3/2

Qsp = Discharge from the spillway

C =coefficient of discharge = 2.25

L = effective length of the spillway = 100m

518

520

522

524

526

528

530

532

0 10 20 30 40 50 60 70 80 90 100

Ele

vati

on

Storge (cumecs day)



Hsp = Sill level of the spillway = 527m

Qsp = 2.25 100 ( H – 527)3/2

Qsp = 225 ( H–527)3/2

(2)

Total discharge = Discharge of sluice gate + discharge of Ogee spillway

Q = Qsl + Qsp

The discharge obtained from reservoir at different levels are calculated using the above

equation (1) and (2) and result are tabulated in Table 3 and elevation discharge curve is

presented in Figure 6.

Figure 6. Elevation – Discharge

Table 3. Calculation for Elevation – Discharge curve

Q = Qsl+Qsp Qsp (H-Hsp) Qsl (H-Hsl) Elevation m

6 5 4 3 2 1

0 0 0 0 0 519

0 0 0 0 0 520

11.13 0 0 11.13 1 521

15.74 0 0 15.74 2 522

19.28 0 0 19.28 3 523

22.26 0 0 22.26 4 524

24.89 0 0 24.89 5 525

27.26 0 0 27.26 6 526

29.45 0 0 29.45 7 527

256.48 225 5 31.48 8 528

669.78 636.39 2 33.39 9 529

1204.33 1169.13 3 35.20 10 530

1836.91 1800.0 4 36.91 11 531

The final result of the Elevation - Storage in Cumecs-day and discharge in m3/sec is

represented in the following Table 4.

Table 4. Elevation - storage - discharge

Discharge m3/s Storage cumecs day Elevation

0 0 519

0 0.463 520

11.13 1.030 521

15.74 1.91 522

19.28 4.82 523

22.26 9.72 524

518

520

522

524

526

528

530

532

0 200 400 600 800 1000 1200 1400 1600 1800 2000

Ele

vati

on

Discharge m^3/sec

Alaa H. Alshami


24.89 16.30 525

27.26 26.35 526

29.45 42.88 527

256.48 56.11 528

669.78 70.76 529

1204.33 79.41 530

1836.91 88.27 531

8.2. Inflow-Storage-Discharge Method (I.S.D. Method)

The inflow- storage-discharge method was developed by L.G. Puls of U.S.Army corps of

engineers. According to the method, the equation is arranged as follows:-

I – Q = ds/dt (3)

where:

I = inflow

Q = Out flow from the reservoir, and S = Storage in the reservoir. The suffixes 1 and 2 are

used to denote a given quantity at the beginning and the end of the time interval and if the

inflow and outflow have straight-line variable within time interval then equation (1) can be

written

(i1 + i2)/2) -(Q1+ Q2 /2) t = S2 – S1 (4)

Or

(I1 + I2)/2 + S1 –(Q1/2 =S2 +(Q2/2

In the above equation all the quantity on the left hand side are known and hence the

quantity (S2 + Q2/2

From the quantity, the value of Q2 can be find out from the storage discharge relationship

as explained below-

From the available storage discharge curve the curve of (S–Q/2 ) verus Q and (S+ Q /2

versus Q are developed such curve may called the ROUTING CURVE

From the known value of Q1 the value of (S1–Q1/2 t)is read from Q Vis ( S – Q/2 ) curve . This value is then added to(I1+ I2/2) t is given ( S2 + Q2/2 t) . Entering the graph

with the value of (S2 + Q2/2 t), the value of Q2 is read and from Q Vis (S + Q/2 t)curve the

value of Q2thus determined , becomes Q1 from next time interval.

The procedure is repeated for all the subsequent routing, period until the complete out

flow hydrograph is obtained.In preparing the routing curve, care should be taken to see that(S

+ Q/2 t) and ( S – Q/2 ) will have consistent unit for example if Q is in m3/s and tis in

day , S should be in cumecs- day. Actually the routing can be performed with only one

routing curve that is Q is( S+Q/2 t). This because (S-Q/2 t) for any Q can be obtained by

first reading (S+Q/2 t) from Q vis (S+ Q/2 t)curve and then subtracting Q from this curve.

If the inflow and out flow hydrograph are plotted on the same time scale it will be observed

that the peak flow of out flow hydrograph is less than the peak flow of the inflow hydrograph.

In other words the peak flow is reduced similarly the time to peak in the out flow hydrograph

is more than the time to peak in the inflow hydrograph. These are effect of reservoir storage

on the movement of flood wave through reservoir. The reduction in peak is known as the

ATTENUATION of the difference in time to peak is known as the reservoir LAG

8.3. Routing Curve by I.S.D. Method

The inflow data for Wadi Horan Reservoir at period of one hour shown in the table 5 and

Table 6 shows the calculation the Routing curve by using the I.S.D method. Figure 7

represent a curve between Discharge Q m3/s and (S+Q/2 t) and (S–Q/2 t).



Table 5 Inflow of Wadi Horan

Discharge m3/s Time (hr) Discharge m

3/s Time(hr)

1160 11 0 5

1024 12 70 1

800 13 250 2

560 14 890 3

400 15 1250 4

320 16 1400 5

256 17 1450 6

160 18 1390 7

96 19 1300 8

0 20 1300 9

1285 10

Table 6. Routing calculation by I.S.D. method

S – Q/2 t

Cumecs -day

S + Q/2 t

Cumecs -day

Q/2 t

cumecs- day Discharge Q m

3/s

Storagecumecs -

day

0 0 0 0 0

0.463 0.463 0 0 0,463

0.797 1.263 0.233 11.13 1.030

1.580 2.240 0.330 15.74 1.91

4.415 5.225 0.405 19,28 4.82

9.253 10.187 0.467 22.26 9.72

15.777 16.823 0.523 24.89 16.30

25.778 26.922 0.572 27.26 26.35

42.262 43.49 0.618 29.45 42.88

50.724 61.496 5.386 256.48 56.11

56.695 84.825 14.065 669.78 70.76

54.12 104.700 25.290 1204.33 79.41

49.695 126.845 38.575 1836.91 88,27

Figure 7. Curve between Discharge Q m3/s and (S + Q/2 t) and (S – Q/2 t).

8.4. Routing Through Reservoir By I.S.D. Method

The following Table 7 shows the routing through the reservoir. The Procedure for finding the

out flow can be represent, Column (1) and (2) of the given inflow hydrograph, Column (3)

=(I1+I2/ 2 ) t =( 0 + 70 /2 )1/24= 1.47, Column (4) = (S – Q1/2 t)obtained from Q vs ( S –Q/2

t)curve fig.(5-3) against the value in column (6) first one assumed equal to 150 m3/s then

0

200

400

600

800

1000

1200

1400

1600

1800

2000

0 20 40 60 80 100 120 140

Dis

char

ge m

^3

/se

c

(S+Qt/t), (S-Qt/2) cumecs - day

S+Qt/2

S-Qt/2

Alaa H. Alshami


Column (5) = ( S2 + Q2/2 t) = Col. 3 + col.4. Column (6) = Q obtained from Q Vs ( S + Q/2

t)curve in Fig( 3-5) against the value of (S2+Q2/2 t) in col.(5) and Column (7)=Elevation of

water surface in the reservoir is obtained from elevation discharge curve in Fig (3-4) against

the value of Q in col(6)

Table 7. Routing through the reservoir

Elevation H

(m)

7

Out flow Q

6 m3/s

S2+ Q2/2 t

5

S1-Q1/2 t

4

I1+I2/2

3

Inflow m3/s

2

Time hr

1

527.6 150 0 0

527.4 100 49.47 48 1.47 70 1

537.6 120 52.70 46 6.72 250 2

528.4 420 72.0 48 24.0 890 3

529.6 1090 99.0 54 45.0 1250 4

529.8 1160 110.5 55 55.7 1400 5

530.3 1350 113.5 54 59.55 1450 6

530.4 1400 111.0 52 59.6 1390 7

530.4 1400 110.5 53 57.54 1350 8

530.1 1320 109.0 54 55.65 1300 9

530.2 1300 108.0 54 54.30 1285 10

530.0 1200 104.0 53 51.34 1160 11

529.8 1120 101.0 56 45.80 1024 12

529.6 900 94.0 56 38.30 800 13

529.2 680 84.0 56 28.30 500 14

528.8 550 77.0 57 20.16 400 15

528.4 380 71.0 56 15.20 320 16

528.2 300 64.0 52 12.10 256 17

528.0 280 62.7 54 8.7 160 18

527.8 180 55.4 50 5.4 96 19

527.4 80 48.0 46 2.0 0 20

The inflow and out flow hydrograph and also the graph ofdischarge and storage are shown in

the Figure 8 and Figure 9 from which the above result are obtained from the calculation in the

Table 7, The Attenuation= 50 m3/sec and the reservoir Lag equal to 2hr

Figure 8. Inflow and out flow hydrograph

0

200

400

600

800

1000

1200

1400

1600

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Dis

char

ge

Time

Inflow

outflow



Figure 9. Discharge and storage

8.4. Ogee (Overflow) Profile Spillway and Stilling Basin

Preface the Ogee profile to be acceptable should provide possible hydraulic efficiency,

structural stability and economy and avoid the formation of object able sub- atmospheric

pressure. Different types of up-stream face of spillway, which are crest profile for vertical up-

stream face and crest profile for inclined up stream face . Figure 10 shows the different profile

of the upstream face shape

Figure 10. W.E.S Standard Spillway crest

0

10

20

30

40

50

60

70

80

90

100

0 200 400 600 800 1000 1200 1400 1600 1800 2000

Stor

ge H

ecta

r.m

Discharge m^3/sec

Alaa H. Alshami


In this study, type (a) is selected which called crest profile for vertical up-stream profile as

suggested by Water Way Experiment Station(W.E.S.), based on extensive experiment

conducted by U.S.B.R. Army Corps of Engineers.

The details of the downstream Crest profile for vertical upstream face by using the

following equation

X1.85

= 2 (Hd )0.85

y

Where x and y are co-ordinates of the crest profile measured from the apex of the crest

and Hd is design head excluding head due to velocity of approach. According to the latest

analytical studies of the U.S. Army the upstream curve of the Ogee shape has the following

equation

Y = 0.724 (x+0.27Hd)1.85

/Hd)0.85

+ 0.126Hd –0.4315(Hd)0.375

(x+ 0.27Hd)0.625

It should be noted here that the upstream curve at the crest should neither be made too

sharp no too broad. To determination downstream profile thee equations used is:

X1.85

= 2(Hd )0.85

y and

Q = C L(Hd)3/2

Where:

Hd = Head of water above the Crest, we can find from spillway equation

Q= discharge as maximum when routed = 1400 m3/s

C = 2.25

L = 100m

1400= 2.24 100 (Hd)3/2

Hd = 3.4 m which is equal to the maximum depth of water above the crest after routed

From equation above

Y = X1.85

/ 2( Hd)0.85

= X1.85

/ 2 ( 3.4)0.85

= X1.85

/ 5.66

Y = 0.1767 X1,85

……. (1)

Let us now determine the tangent point (T.P.) for the downstream slope of0.75(H) :1 (V)

Hence dy/dx = 1/0.75…..(2)

Differentiating equation (1)

Y = 0.1767 X1.85

dy/dx = 1.85 0.1767 ( X )0.85

dy/dx = 0.327X0.85

1/0.75 = 0.327 X 0.85

……(3)

1 = 0.75 0.327X0.85

X =( 4.077 )1.176

= 5.11

Y = 0.1767(5.11)1.85.

Y = 3.612 and X = 5.11

For vertical upstream face the upstream profile is given by the following equation

Y = 0.724(x + 0.27 Hd) 1.85

/Hd0.85

+ 0.126 Hd – 0.4316Hd0.375

( x + 0.27 Hd)0.625

The upstream curve extend up to

X = - 0.27 Hd= 0.27 3.4 = 0.918m

Y = 0.724(x+0.27Hd)1.85

/Hd0.85

+ 0.126Hd – 0.4316Hd0.375

(x + 0.27Hd)0.625

Y= 0.724(( x+ 0.27(3.4))1.85

/(3.4)0.85

+ 0.126 3.4 – 0.4316 (3.4)0.375

(x+ 0.27 3.4)0.625



Y = 0.256(x+0.918)1.85

+ o.428– 0.683(x +0.918)0.625

X = 0.96

Y = 0.526(0.96+ 0.918)1.85

+ 0.428 – 0.683 (0.96+ 0.918)0.625

Y = 1.103

The value of the coordinate X = 0.96, y = 1.103m

Stilling basin are used to dissipate the energy of water exiting spillway of a dam. The

purpose is to prevent scouring that occurs when high velocity water enters the downstream

reach of the dam. This scouring can damage the foundation of the dam, leading to overtopping

and also cause severe erosion downstream. The primary method of dissipating energy is to

generate a hydraulic jump to transition flow from supercritical to subcritical. The design are

selected based off of the FROUD NUMBER of the flow and the velocity. The velocity

approach = 12.68 m/s and the Froude Number (F)= V2/ √ =5 . The stilling basin type

which used in this case is Type II.

9. CONCLUSIONS

A spillway is the overflow portion of the dam over which surplus discharge flows from the

reservoir to the downstream. The most common type of spillway is the Ogee spillway or

Overflow spillway, which guided smoothly over the crest and the profile of the spillway so

that the overflow does not break contact with spillway surface. If this is not assured, a vacuum

may form at the point of separation and cavitations may occur. Inflow discharge from the

reservoir is routed through the spillway and two pipes sluices gated of one-meter diameter

placed at elevation of 520 m and find the maximum discharge is 140 0m3/sec with

Attenuation of 50 m3/sec and Reservoir Lag equal to 2 hrs. The method used for routing is

Inflow- Storage – Discharge ( I.S.D.)

The profile of the downstream and upstream are calculated depending on Water

Experiment Station (W.E.S.) belong to the U,S. Army corps of Engineers.

REFERENCES

[1] a.Hussein, H. & Alshami, A. H. 2018. Evaluation Of The Hydrological Behavior In The

Greater Zab River Basin. International Journal of Civil Engineering and Technology,

Volume 9, 240–258.

[2] HUSSEIN, H. A. 2010. Dependable discharges of the upper and middle diyala basins.

Journal of Engineering, 16, 4960-4969.

[3] ICOLD, C. 2018. Flood Evaluation and Dam Safety, Taylor & Francis Group.

[4] KARGPUR, I. 2008. Water Resources Engineering, IIT Karagpur, 2008: Water

Resources Engineering.

[5] KHATSURIA, R. M. 2004. Hydraulics of Spillways and Energy Dissipators, CRC Press.

[6] RAGHUNATH, H. M. 2006. Hydrology : Principles, Analysis And Design, New Age

International (P) Limited.

[7] Sayl, K. N., Afan, H. A., Muhammad, N. S. & Elshafie, A. 2017. Development of a

Spatial Hydrologic Soil Map Using Spectral Reflectance Band Recognition and a

Multiple-Output Artificial Neural Network Model. Hydrol. Earth Syst. Sci. Discuss.,

2017, 1-16.

mathematical model of reservoir routing for...

Documents