energy dissipator

7/29/2019 energy dissipator

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CHAPTER 2

ENERGY DISSIPATOR

2.1 Energy Dissipation Below Spillways

Water flowing over a spillway has a very high kinetic energy because of the

conversion of the entire potential energy to the kinetic energy. If the water flowing with such

a high velocity is discharged directly into the channel downstream, serious scour of the

channel bed may occur. If the scour is not properly controlled, it may extend backward and

may endanger the spillway and the dam. In order to protect the channel bed against scour, the

kinetic energy of the water should be dissipated before it is discharged into the d/s channel.

The energy-dissipating devices can be broadly classified into two types.

1. Devices using a hydraulic jump for the dissipation of energy.2. Devices using a bucket for the dissipation of energy.The choice of the energy-dissipating device at a particular spillway is governed by the

tail water depth and the characteristics of the hydraulic jump, if formed, at the toe. If the tail

water depth at the site is not approximately equal to that required for a perfect hydraulic

jump, a bucket-type energy dissipating device is usually provided.

The characteristics of the hydraulic jump are discussed in the following section. The

sequent depth (conjugate depth or post-jump depth) y2 is determined for different values of

the discharge, and a jump height curve (JHC) is plotted between the conjugate depth y2 as

ordinate and the discharge (Q) as abscissa. The tail water rating curve (TWRC) at the spillway

site is determined by stream gauging, (Q) as abscissa. As discussed later, the correct choice of

the energy-dissipating device is made after comparing the relative positions of the jump

height curve (JHC) and the tail water rating curve (TWRC). For the design of spillways, the

discharge per unit length (q) is usually taken as abscissa instead of Q. Different types of

stilling basins have been developed which are quite effective for the formation of a stable

hydraulic jumps and for confining the hydraulic jump. Stilling basins are commonly used for

spillways and other hydraulic structures, such as weir and barrages. In a stilling basin, chute

blocks, basin blocks (baffle blocks) and an end sill are usually provided. Chute blocks are

triangular blocks installed at the upstream end of the basin. An end sill is constructed at the

downstream end of the basin. It may be a solid sill or a dentated sill. Baffle blocks are

installed on the basin floor between the chute blocks and the end sill. These are also known as

baffle blocks or baffle piers.

2.2 Characteristics of A Hydraulic JumpHydraulic jump is a sudden and turbulent rise of water which occurs in an open

channel when the flow changes from the supercritical flow state to the subcritical state. It is

accompanied by the formation of extremely turbulent rollers and considerable dissipation of

energy. Thus a hydraulic jump is a very effective means of dissipation of energy belowspillways.


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Types of jumps The type of jump and its characteristics depend mainly upon the Froude

number of the incoming flow or the initial froude number (F1), given by

F1 = V1/ 1gy (2.1)

where V1 is the mean velocity of flow before the hydraulic jump, g is the acceleration due to

gravity andy1is the pre-jump depth (or the initial depth of flow).

For the formation of a hydraulic jump, the initial Froude number F1 should be greater

than unity. Different types of hydraulic jump are as follows:

1. Undular Jump An undular jump is formed when F1 = 1.0 to 1.70. In anundular jump, the water surface shows some undulation. The energy dissipation is about 5%.

2. Weak Jump When F1= 1.70 to 2.50, a weak hydraulic jump occurs. In thiscase, a series of small rollers develops on the surface of the jump, but the downstream water

surface remains quite smooth. The velocity is uniform throughout. The energy dissipation is

about 20%.

3. Oscillating Jump An oscillating hydraulic jump occurs when F1 = 2.50 to4.50. There is an oscillating jet entering the jump bottom to surface and back again without

any periodicity. The energy dissipation is between 20 to 40 %.4. Steady Jump A steady jump occurs when F1 = 4.50 to 9.0. The jump is quite

stable and balanced. This jump is not much sensitive to variations in the tail water depth. The

steady jump has very good performance, and most of the hydraulic structures utilize this type

of jump for the dissipation of energy. The energy dissipation is between 45 to 70 %.5. Strong Jump A strong jump occurs when F1> 9.0. The jump action is quite rough

but effective. It causes a rough water surface with strong surface waves downstream. The

energy dissipation is between 70 to 85 %. Because of rough action, a strong jump is avoidedin spillways, as far as possible.

Jump Heigh Curve (JHC) A hydraulic jump will occur in a rectangular open

channel if the following equation between the initial depth y1 and the sequent depth (post-

jump depth)y1 is satisfied (See any text of Fluid Mechanics).

[ ]1)/(81231

212 += gyqyy (2.2)

where q is the discharge intensity (i.e. discharge per unit length).

Eq. 2.2 is usually written in terms of the initial froude number (F1) as

[ ]1812

2

1

1

2 += Fy

y (2.3)

where3

11

11

gy

q

gy

VF == (2.4)


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The mean velocity V1 of the incoming flow for an ogee-shaped spillway can be

determined by applying the Bernoulli equation to points A and 1 (Fig. 2.1). Neglecting losses

and the velocity of approach,

Fig 2.1

gVyHP2

2

11 +=+ (2.5)

The mean velocity of flow V1 at the toe of spillway is equal to (q/y1). Therefore,

2g

1x)/( 211 yqyHP +=+ (2.6)

By substituting the values ofP,H, and q, the value ofy1 can be found from Eq. 2.6.

Thus the value ofy1 is determined for a given discharge intensity q over the spillway.

The corresponding value of the sequent depthy2 can be determined from Eq. 2.2. Likewise,

for different values of the discharge intensity, the values of the sequent depth y2 can be

computed. A plot is then made between the discharge intensity q as the abscissa and the

corresponding value of the sequent depth y2 as ordinate [Fig. 2.2 (a)]. The curve is known as

the jump height curve (JHC) or jump rating curve (JRC).

Fig 2.2

Tail water rating curve The tail water rating curve (TWRC) gives the relation

between the tail water depthy2'(i.e. the actual water depth in the river on the downstream) as

ordinate and the discharge intensity q as abscissa [Fig. 2.2 (b)]. The actual tail water depth

corresponding to any discharge intensity q depends upon the hydraulic characteristics of the


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river downstream. The values ofy2'corresponding to different values of q are obtained by

actual stream gauging.

If there is a suitable control somewhere downstream of the spillway where the depth

of water and discharge can be accurately measured, the tail water depthy2'at the spillway

can also be determined by backwater computation.While plotting the tail-water rating curve, an allowance for channel retrogression,

which is likely to occur, must be made.

2.3 Location of A Hydraulic JumpFor a given discharge intensity (q), the sequent depthy2 and the tail water depthy2'are

fixed. The location of hydraulic jump will depend upon the relative magnitudes ofy2 andy2',

and hence on the JHC and TWRC. There are five cases, depending upon the relative positions

of JHC and TWRC, as discussed below.Case-1 JHC and TWRC coincide throughout In this case, the JHC and TWRC

curves concide for all discharges [Fig. 2.3 (a)]. As the tail water depth y2'is exactly equal to

the sequent depth y2 required for the formation of hydraulic jump, a perfect jump is formed

just at the toe of the spillway as shown in Fig. 2.1. However, this case indicates a highly

idealised condition, which rarely occurs in practice.

Case-2 TWRC always lower than JHC In this case, the tail water rating curve

(TWRC) is below the jump height curve JHC for all discharges [FIG. 2.3 (b)]. Such a

condition occurs when the tail water is carried away quickly due to a rapid or a fall

somewhere on the downstream of the spillway. In this case, the jump will be located at apoint on the downstream of the toe of spillway. The high velocity jet would sweep down the

toe and scour the river bed. Therefore, severe erosion may occur in the portion of the river

between the spillway and the section where the hydraulic jump is formed.

Case-3 TWRC always higher than JHC In this case, the tail water rating curve is

above the jump height curve for all discharges [Fig. 2.3 (c)]. This condition usually occurs

when the river cross-section on the downstream of the spillway is narrow and therefore the

tail water backs up. The hydraulic jump in this case is located upstream of the toe on the

spillway face. The hydraulic jump is drowned or submerged, and the high velocity jet dives

under the tail water. The energy dissipation in a drowned hydraulic jump is not good.Case-4 TWRC lower than JHC at low discharges, but higher at high discharges

In this case, the tail water rating curve is lower than the jump height curve at low discharges,

but it becomes higher at a particular discharge and then remains higher than the jump height

curve [Fig. 2.3 (d)].

It is a combination of cases 2 and 3. The hydraulic jump is formed further

downstream of the toe at low discharge, as in the case 2; but at higher discharges, it is

drowned, as in the case 3.


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Case-5 TWRC higher than JHC at low discharges, but lower at high discharges.

It is also combination of cases 3 and 2. However, in this case, at low discharges, the jump is

drowned; whereas at high discharges, it is formed further downstream of the toe [Fig. 2.3

(e)].

Fig. 2.3

2.4 Measure Adopted For Dissipation of Energy

Various measures are adopted at or near the toe of the spillway so that a perfect jump

is formed for the dissipation of energy. The measures adopted will depend upon the relative

positions of the tail water rating curve (TWRC) and the jump height curve (JHC). Measures

are discussed separately for all the five cases discussed in the preceding section.

Case-1 In this case, the tail water rating curve and jump height curve concide for all

discharges. There is no need of any special measure for the formation of hydraulic jump, as a

perfect jump will always form at the toe.

A horizontal apron is however provided on the downstream of the toe for the

protection of the river bed (Fig. 2.4). The length of a horizontal apron is taken equal to the

maximum length of the hydraulic jump. Sometimes, baffle blocks are also constructed on the

horizontal apron for dissipation of energy. However, if the baffle blocks are placed too nearthe toe, they may be subjected to cavitation and abrasion.

Fig. 2.4


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If may be noted that the case-1 rarely occurs in practice. However, by suitably

choosing the length of the spillway, the TWRC and JHC may be made to coincide to some

extent.

Case 2 As the tail water rating curve is lower than the hydraulic jump curve, the hydraulicjump forms at a certain section downstream of the toe. The following measures are adopted.

(i) A depressed horizontal apron is formed by excavating the river bed on the

downstream of the toe of the spillway to increase the tail water depth [Fig. 2.5 (a)]. The

length and depth of the apron should be such that, for all discharges, the jump is confined to

the apron. Sometimes, the depressed apron is made sloping instead of horizontal.

Fig. 2.5

(ii) A low secondary weir (or dam) is constructed downstream of toe to raise the

tail water [Fig. 2.5(b)].

(iii) A stilling basin is formed on the downstream of toe and a sill or baffle wall is

provided at the end of the stilling basin. The length and depth of the stilling basin should be

sufficient to contain the hydraulic jump for all discharges (See sect 2.5).

(iv) If the river bed consist of solid rock, a ski jump bucket can be provided which

throws the water up so that it strikes the bed at a safe distance away from the toe (See sect.

2.6 for details).

Case-3 In this case, the tail water rating curve is higher than the jump height curve and the

hydraulic jump is drowned, the following measures are adopted.

(i) A sloping apron is constructed above the river bed level extending from the

spillway surface to the toe [Fig. 2.6 (a)]. The sloping apron raises the level of the point where

the hydraulic jump is formed. The slope of the apron should be such that a perfect jump will

form somewhere on the sloping apron for all discharges. A large quantity of concrete is

however required for the construction of the sloping apron.


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Fig. 2.6

(ii) The river bed may be excavated to provide a drop in the river bed to lower the

tail water [Fig. 2.6(b)].(iii) A roller bucket is provided near the toe, which forms rollers for the dissipation

energy (Sect 2.6).

Case-4 In this case, the tail water rating curve is lower than the jump height curve at low

discharges but higher at high discharges. Thus at low discharges, the hydraulic jump is

shifted to a downstream point; but for high discharge, it is shifted upstream of the toe and the

jump is drowned. The following measures are adopted.

(i) A sloping apron is provided which lies partly above and partly below the river bed

level so that a perfect jump will form in the lower portion of the apron at low discharges andin the higher portion of the apron at high discharges (Fig. 2.7).

Fig. 2.7

(ii) A low secondary dam (or a sill) with a stilling basin is provided downstream of

the toe to raise the tail water level at low discharges. This arrangement is combined with a

sloping apron at a higher level for developing a jump at high discharges (Fig. 2.8). It is found

in practice that the low secondary dam has negligible effect at high discharges.


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(iii) If the velocity is not greater than 15 m/s, baffle blocks or dentated sills may be

constructed to break up the jet and raise tail water level at low discharges to assist jump

formation. At high discharges, the high velocity jet dives under the tail water and breaks up

and the energy is dissipated in internal turbulence, though jump is not formed.

Case-5 In this case, the tail water depth is higher than jump height curve at low discharges,

but lower at higher discharges. The case is similar to case 4 but the range of discharge is

different. The following measures are usually adopted.

(i) A sloping apron is provided which is partly above the river bed level and

partly below the river bed level, as in Fig. 2.7. In this case, the jump will form in the upper

portion of the apron at low discharges, and in the lower portion, at high discharges.

(ii) A low secondary dam (or a sill) with a stilling basin is provided to increase thedepth at high discharges as in Fig. 2.8. However, at low discharges, this arrangement will

further increase the tail water depth, which is already quite high. Therefore, at low

discharges, the jump will be more drowned and consequently, there will be less dissipation of

energy. If this arrangement is not likely to cause much scour, if may be acceptable.

Fig. 2.8

2.5 Stilling Basins

A stilling basin is a basin-like structure in which all or a part of the energy is

dissipated. In a stilling basin, the kinetic energy causes turbulence and it is ultimately lost as

heat energy. The stilling basins commonly used for spillways are of the hydraulic jump type,

in which dissipation of energy is accomplished by a hydraulic jump.

A hydraulic jump can be stabilised in stilling basin by using appurtenances

(or accessories such as chute blocks, basin blocks and end sill.


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(a) Chute blocks These are triangular blocks with their top surfaces horizontal.

These are installed at the toe of the spillway just at upstream end of the stilling basin. They

act as a serrated device at the entrance to the stilling basin. They furrow the incoming jet and

lift a portion of it above the floor. These blocks stabilise the jump and thus improve its

performance. These also decrease the length of the hydraulic jump.(b) Basin blocks (or baffle blocks or baffle piers) These are installed on the

stilling basin floor between chute blocks and the end sill. These blocks also stabilise the

formation of the jump. Moreover, they increase the turbulence and assist in the dissipation of

energy. For low flows, baffle blocks also help compensate a slight deficiency of the tail water

depth, and for high flows, they help deflect the flow away from the river bed. However,

baffle blocks are prone to cavitations on the downstream face, and are not recommended

when the velocity is greater than 15m/s.

(c) End sill It is constructed at the downstream end of the stilling basin. It may be

solid or dentated. Its function is to reduce the length of the hydraulic jump and to controlscour. For large basins designed for high incoming velocities, the sill is usually dentated to

perform an additional function of diffusing the residual portion of the high velocity jet that

may reach the end of the basin. In a dentated sill, there are teeth with small gaps which

diffuse the jet. (These gaps and the projections between them look like human teeth).

Types of stilling basins There are various types of stilling basins. The type of stilling

basin most suitable at a particular location mainly depends upon the initial Froude number

(F1) and the velocity V1 of the incoming flow. The stilling basins are usually rectangular in

plan. However, sometimes these are flared. These are made of concrete. The length of the

basin, measured in the direction of flow, depends upon the sequent depth y2 and the initialFroude No. F1. It is different for different type of basins.

The following types of basins are commonly used in practice.

A. U.S.B.R Stilling basins

1. Type I basin 2. Type II basin

3. Type III basin

B. Indian Standards basins

1. Horizontal floor-Type I 2. Horizontal floor Type II

3. Sloping apron Type III 4. Sloping apron Type IV

A.U.S.B.R Stilling basins

No special stilling basin is required to still flow if F1 is less than 1.70. However, the

channel length beyond the point from where the water depth starts increasing, should not be

less than 4.00y2, wherey2 is the sequent depth.

For F1 between 1.7 to 2.5, there is not much turbulence, only a horizontal apron is

provided. However, the apron should be sufficiently long to contain the jump. A length of 5.0

y2 is usually provided. No accessories such as baffles or sills are provided.


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1. U.S.B.R. Type I basin for Froude number F1 between 2.5 to 4.5 For this

range ofF1, type I basin has proved to be quite effective for dissipating most of the energy

(Fig. 2.9). However, it is not able to dampen the oscillating flow entirely. The water depth in

the basin should be about 1.10 y2 to check the tendency of the jump to sweep out and to

suppress wave action. The basin is provided with chute blocks of the size, spacing andlocation as shown in the figure. All the dimensions are in terms of the initial depthy1.

Fig. 2.9

The length L of the stilling basin varies from 4.3y2 to 6 y2, depending upon the value

ofF1, as given in Table 2.1F1 2 3 4 6

L/y2 4.3 5.3 5.8 6.0

2. USBR Type II basin for Froude number F1 greater than 4.5 and V1 less

than 15 m/s. For F1 greater than 4.5 and V1 less than 15 m/s, type II basin shown in Fig. 2.10

is provided.

Fig. 2.10


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The basin is provided with chute blocks, baffle blocks (baffle piers) and end sill, as

shown. The size, spacing and location of the chute and baffle blocks are as shown in Fig.

2.10. The lengthL of the stilling basin, the height h3 of the baffle block and the height h4 of

the end sill are obtained from Table 2.2

Table 2.2

F1 5 6 8 10 12 14 16

L/y2

h3/y1

h4/y1

2.3

1.5

1.2

2.5

1.7

1.3

2.6

2.0

1.5

2.7

2.3

1.6

2.8

2.7

1.7

2.8

3.0

1.8

2.8

3.3

1.9

The height of the chute blockh1 is equal toy1, wherey1is the initial depth.

The length of the stilling basin is considerably less than that in Type I because thedissipation of energy occurs by the impinging action of the incoming flow against the baffle

blocks. However, the baffle blocks are subjected to large impact forces. They should be

designed for the extra force due to dynamic action against the upstream face, given by

F= 2 w A (y1+ h) (2.7)

Where w is the specific weight of water, A is the area of upstream face of the block,

and (y1+ hv) is the specific energy of flow entering the basin. Moreover, on the d/s face of

the baffle blocks, suction (or negative) pressure usually develops, which further increase the

force. Hence, the baffle blocks should be properly anchored at the base.

3. U.S.B.R. Type III basin for Froude number F1 greater than 4.5 and V1 greater

than 15m/s. When F1 is greater than 4.5 and V1 is greater than 15 m/s. Type III basin is

provided. In this case, baffle blocks are not provided, because of the following reasons:

(i) The block would be subjected to very high impact forces due to high velocity

V1 of incoming flow.

(ii) There is a possibility of cavitations on the downstream faces of the blocks.

The stilling basin therefore consists of only chute blocks and a dentated sill. As the

dissipation of energy occurs mainly by hydraulic jump, the length of basin is greater than that

in Type II basin. The size, spacing and location of the chute blocks are the same as in Type II

basin. The length of the basin is obtained from Table 2.3

Table 2.3

F1 5 6 8 10 12 14

L/y2 3.85 4.0 4.2 4.3 4.3 4.3

To check the tendency of the jump to sweep out of the basin, the water depth in the

basin should be about 5 percent greater than the computed value ofy2.


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B. Indian standards stilling basins

IS : 4997 1968 recommends 4 types of stilling basinso, two types are with

horizontal apron, and two types are with sloping apron.

(a) Stilling basins with a horizontal apron Stilling basins with a horizontal floor

may be provided when the jump height curve (JHC) and the tail water rating cure (TWRC)are not much different from each other i.e one curve may be slightly above or below the

other. In this case, the required condition for the development of the hydraulic jump are

obtained on a horizontal apron or near the bed level, and hence there is no necessity of a

sloping apron.

Depending upon the froude number F1, there are two types of basins with a horizontal

apron.

(i) Type I for F1 < 4.5

(ii) Type II for F1 >,. 4.5

1. I.S Type I basin for F1 < 4.5. This type of the basin is provided when F1 is

less than 4.5. Such a case usually occurs on weirs, barrages and low dams. The basin is

provided with chute blocks, basin blocks (baffle blocks) and a dentated sill (Fig. 2.11).

However, the basin blocks should not be used if the velocity of flow exceeds 15m/s. The

water depthy2'should normally not exceed 1.10y2wherey2 is the conjugate depth.

Fig. 2.11

The chute blocks should be kept at a height equal to 2y2 at the end of the glacis slope.

The top length should be also equal to 2y1 and a space equaly1/2 should be left at the ends.

The width and spacing of the basin blocks should be equal to their heights. The height

(hb) of the blocks is obtained from Table 2.4. These blocks should be placed at a distance of

0.8y2 from the d/s face of the chute blocks.


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Table 2.4

F1 2 3 4

hb/y1 0.5 0.9 1.2

These values of (hb/y1) in Table 2.4 are interpolated form the chart given in the code.

The height of the dentated sill is 0.2 y2, the width of dents is 0.15 y2 and the spacing is also

0.15 y2. The top thickness of the dentated sill is 0.02y2.

The lengthLb of the basin is obtained for different values ofF1, as given in Table 2.5.

Table 2.5

F1 2 3 4 4.5

Lb/y2 3.15 4.0 4.75 5.0

This basin is shorter than U.S.B.R. Type I basin because of basin blocks.

2. IS Type II basin for F1 4.5 This type of basin is used when F1 is equal to or greater

than 4.5. This basin is usually required in the case of spillways of the medium and high dams.

The stilling basin is provided with chute blocks, basin blocks (or baffle blocks) and a

dentated sill (Fig. 2.12)

Fig. 2.12

The height, width and spacing of the chute blocks, each should be kept equal to y1.

The width and spacing may be slightly varied to eliminate fractional blocks. A space equal to

y1/2 is preferable at ends.

Table 2.6

F1 5 6 8 10 12 14 16

hb/y1 1.5 1.7 2.0 2.3 2.7 3.0 3.3


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The height of the basin blocks (hb) is interpolated from table 2.6. The width and

spacing of the basin blocks is 0.75 times the height. These blocks should be placed at a

distance of 0.8y2 downstream from the chute blocks.

The end sill is the same as for IS Type I basin.

The lengthLb of the stilling basin may be obtained from Table 2.7.

Table 2.7

F1 5 6 8 10 12

Lb/y2 2.9 3.2 3.7 3.9 4.0

Modified IS Type II basin If F1 > 4.5 and the inflow velocity V1 is greater than 15 m/s,

Modified IS Type II basin is used. In this case, basin blocks are not provided. Moreover, the

floor of the basin is kept at a depth equal toy2 below the tail water level.

(b) Stilling basins with a sloping apron floor (IS Type III basin and type IV basins)

A sloping apron is provided when the tail water depth y2' is very large as compared to the

sequent depthy2. In that case, a drowned jump would develop if no sloping apron is provided.

With a sloping apron, an efficient hydraulic jump is formed at a suitable level on the sloping

apron. There are two types of basins: (i) IS Type III basin [Fig. 2.13 (a)] (ii) IS Type IV basin

[Fig. 2.13 (b)].

Fig. 2.13

IS basin Type III is recommended where the tail water rating curve (TWRC) is higher than

the jump height curve (JHC) at all discharges. IS Basin Type IV is suitable where the tail

water depth y2'at the maximum discharge exceeds y2 considerably but is equal to or slightly

greater thany2 at lower discharges.

The design criteria for sloping aprons have not been standarised to the same extent as

in the case of the horizontal apron. The slope and overall shape are determined from

economic considerations. The length of the basin is fixed, depending upon the type and

soundness of the river bed.


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IS Type III basin is usually provided with a sloping apron for the entire length,

whereas IS Type IV basin is provided with a partly sloping and a partly horizontal apron. For

both these basins, only a solid or dentated sill is provided No other accessory is required.

IS Type III Basin

The following procedure may be adopted for the design:

(a) Assume a certain level at which the front of the jump will form for the maximum tail

water depth and discharge.

(b) Determine y1 from the known upstream total energy line by applying Bernoulli's

theorem and calculate F1.

(c) Assume a certain slope and determine the conjugate depthy2 and length of the jump

for the above Froude number from figure 2.14 and figure 2.15 respectively. The length of the

apron should be kept equal to 60 percent of the jump height.

(d) Test whether the available tail water depth at the end of the apron matches the

conjugated depthy2. If not, change the slope or the level of the upstream end of the apron or

both. Several trails may be required before the slope and the location of the apron are

compatible with the hydraulic requirement.

Fig. 2.14 Conjugate depth on sloping floor


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Fig. 2.15 Length of jump in terms of conjugate depth D2 sloping aprons

Fig. 2.16 Values of k

(e) The apron designed for maximum discharge may then be tested at lower discharge say

, and . It the tail water depth is sufficient or in excess of the conjugate depth for the

intermediate discharge, the design is acceptable. If not, a flatter slope at lower apron level

should be tried or Basin IV may be adopted.

(f) The basin should be supplemented by a solid or dentated end sill of height equal to

0.05 to 0.2 with an upstream slope of 2:1 to 3:1.

y'2 (depth conjugate to d for sloping apron or partly sloping may be determined from the

relation,


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+

= 11

tan21

cos8

cos2

1

3

1

'

2'

2

K

F

y

y(2.8)

The value of K can be determined from figure 2.16. However, d'2 may also be determined

from figures 2.14 and 2.17

IS Type IV Basin

In the design of basin IV, the following procedure may be adopted.

(a) Determine the discharge at which the tail water depth is most deficient.

(b) For the above discharge, determine the level and length of the apron (as for basin I

and II)

(c) Assume a certain level at which the front of the jump will form for the maximum tailwater depth and discharge.

(d) Determine d1, from the known upstream energy line and calculate F1 and findy2.

(e) Determine a suitable slope (by trail and error) so that the available tail water depth

matches the required conjugate depthy2 determined from figure 2.17

(f) Determine the length of the jump for the above slope from figure 2.14. If the sum of

the length of the inclined portions and horizontal portion is equal to about 60 percent

of the jump length, the design is accepable. If not, fresh trials may be done by

changing the level of the upstream end of the jump formation.

(g) The basin should be supplemented by a solid or dentated end sill of height 0.05 to 0.2y2 and upstream slope of 2:1 to 3:1.

Fig. 2.17 Conjugate depths for sloping apron


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Evaluation of Prejump Velocity and Depth of Flow

The theoretical value of velocity at the bucket invert or at the start of hydraulic jump

can be calculated by the following relation:

)5.0(2 dT HHgV =

where

VT = velocity ( theoretical ) in m/sec.

H = Difference of u.s reservoir elevation and bucket invert in m.

Hd = Head over the spillway crest in m.

g = Acceleration due to gravity in m/sec2.

= 9.81 m/sec2.

Fig 2.18 Relation of actual and theoretical velocities

However because of surface friction etc., the actual velocity VA is slightly less than

VT. Figure 3.17 gives the ratio of actual to theoretical velocity at entrance to stilling basin for

different values ofHandHd.


19/41

52

Design of rectangular stilling basin

Fig. 2.19

Consider the rectangular stilling basin, it is assumed that Hd, Vo, Y3, elevation A,

elevation C and the discharge are known.

ElevationB, Y1, V1, Y2, V2 andH2 are to be determined.

The critical depth Yc for a rectangular channel at a given discharge can be given by

y

q

gc =

2

3

(2.9)

where yc = critical depth

q = unit discharge of spillway = Q/b

g = gravitational force

Applying Bernoullis equations, the energy loss from point A to point C is determined

by Elevation A + P + Hd +V

g

O

2

2= Elevation C + y3 +

V

g

3

2

2+ HL

HL = ( Elev. A + P + Hd + Vg

O

2

2) - (Elev. C + y3 + V

g

3

2

2) (2.10)

The dimensionless relationship involving the head loss in rectangular channels by the

hydraulic jump are

[ ]Hy

y y

y y

L

1

2 1

2 1

1

4=

-( / )

( / )(2.11)


20/41

53

and

3/1

1

212

1

1/

2

+

=

y

yyy

yc

y(2.12)

The product of these two equations becomes

[ ]

3

1

2

1

212

3

12

11

2

)/(4

1)/(

+

=

y

y

y

yyy

yy

y

HL (2.13)

The values ofH

y

L

C

can be calculated by equation 2.9 and 2.10 and corresponding

value of (y2/ y1) can be found from equation 2.13 by trial and error solution. After knowing(y2/ y1), equation 2.11 or 2.12 can be used to calculate y1.y2 can be found from the known

valuey2/ y1. Thus, V1 and V2 can be found.

By applying Bernoullis equation, the elevation of the pool floor can be calculated.

Elev.A + P +Hd+V

g

O

2

2= Elev. B + y1 +

V

g

O

2

2+ hf , assume hf0

Elev.B = ( Elev. A + P + Hd +V

g

O

2

2) - (y1 +

V

g

1

2

2)

Example 2.1

Find the elevation and design the rectangular stilling basin for the spillway as shown

in the figure.

Fig.

Given data:

Design discharge = 9000 cusec

R.L of river bed at weir site = 664.0 ft

R.L of tail water = 667.0 ft

R.L at weir crest = 671.5 ft

Total design head = 4.39 ft

Assume base width of stilling basin = 250 ftProvide USBR type stilling basin and sketch the appurtenances in detail.


21/41

54

Fig.

Tail water depth = 667 - 664 = 3.0

Weir height = 671.5 - 664 = 7.5

Design discharge per unit width q =9000 cu sec

ft t250

= 36 cusec/ unit width

approach Velocity Vo =36

7 5 4 39. .+= 3.03 ft/sec

V

g

O

2

2= 0.14 ft

V3 =

36

3 =12ft/sec,

V

g

3

2

2 =2.24 ft

Applying Bernoullis equation, from pt. A to pt.C

Elev. A + P + Hd +V

g

O

2

2= Elev. C +y3 +

V

g

3

2

2+ HL

7.5 + 4.39 + 0.14 = 3 + 2.24 +HL (Elev. A = Elev. C)

HL =6.79 ft

critical depthyc =q

g

2 2

3336

32 2=

.=3.427 ft

L. H. S =H

y

L

C

= =6 79

3 427198

.

..

Assumey

y

2

1

= 6.1, R. H. S =

1

2

3

1

2

4

1

y

y

y

y

3/1

1

2

1

2 1

2

+

y

y

y

y

R. H. S = 1.95 L . H . S (O.K.)

Take yy

2

1

= 6.1


22/41

55

3/1

1

2

1

2

1

1

2

+

=

y

y

y

yy

y

c

=

3./1

)11.6(1.6

2

+

= 0.36

y1 = 0.36 yc = 0.36 x 3.427 = 1.234 ft

y2 = 6.1 y1 = 6.1 x 1.234 = 7.527 ft

incoming velo. V1 =q

y1= 29.17 ft/sec < 50 ft/sec

V2 =q

y 2=4.78 ft/sec,

V

q

2

2

2= 0.3552 ft

Applying Bernoullis eq :n,

Elev. A + P + Hd +V

qO

2

2= Elev . B + y2 +

V

g2

2

2+ HL

664 + 7.5 + 4.39 + 0.14 = Elev. B + 7.527 + 0.3552 + 6.79

Elev. B = 661.35 ft

Fr1 =V

g y x

1 2917

32 2 12341=

.

. .= 4.6275 > 4.5

V1 < 50 ft/sec (15 m/s)Choose USBR Type II stilling basin

L/y2= 2.3 , L = 2.3 x y2 = 2.3 x 7.527 = 17.3121 ft= 17.5 ft

h3/y1 = 1.5, h3 = 1.5 x y1 = 1.5 x 1.234 = 1.851 fth4 /y1 = 1.2, h4 = 1.2 x y1 = 1.2 x 1.234 = 1.48 ft

Fig.


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56

2.6 Bucket Type Energy Dissipators

Bucket type energy dissipators are commonly used for the dissipation of energy below

the overflow (ogee-shaped) spillways. The dissipator consists of an upturned backet

( or curved apron) provided at the toe of the spillway in continuation of its downstream face.Bucket type energy dissipators are usually of small size and are more economical than

conventional hydraulic jump stilling basins. These are especially useful when the Froude

number F1 exceeds 10, because in that case, the difference between the initial depth and the

sequent depth is quite large and a very long and deep stilling basin is required. They can be

adopted for all tail water conditions. However, they can be used only when the river bed

consists of a strong and stiff rock.

The bucket type energy dissipators are basically of the following three types.

1. Solid roller bucket 2. Slotted roller bucket

3. Ski-jump (or flip or trajectory ) bucket.The solid and slotted roller buckets are used where the tail water rating curve

(TWRC) is above the jump height curve (JHC). Both these types of buckets remain

submerged in the tail water, and hence these are also called submerged buckets.

A ski-jump bucket is used where the tail water rating curve (LWRC) is below the

jump height curve (JHC). The river bed in this case should consist of very strong rock.

(i) Solid roller bucket: This type of bucket shown in fig. 2.20 performs well when deeply

submerged.

Fig. 2.20

It consists of a circular bucket type apron with a concave profile of considerable radius and a

lip which deflects the high velocity flow away from the stream bed. The dissipation of

surplus energy in the water is accomplished by the diffusion of the water jet in a large mass

of water and by overcoming the boundary resistance of the scooped bed and material. The

sheet of water is deflected upward by the bucket lip forming two elliptical rollers, an

anticlockwise roller or high boil in the water surface in the bucket and a clockwise ground

roller below the bucket. As a result of this action considerable energy dissipation occurs

before the jet reaches the channel bottom.The different parameters of the bucket can be decided as below.


24/41

57

Bucket Radius

The bucket radius may be taken as the lowest of the following:

(a) R.S. Varshneys curve.

The radius of the bucket can be read from the curve exhibited in figure 2.21.

(b) The following relation has been found to give a satisfactory value for the radius of the

bucket.

R = 0.6 ( ' )H Hd (2.14)

where

H = Fall from the crest of the spillway to bucket invert in m.

Hd = Head over the crest in m.

Fig. 2.21 Relationship of Bucket Radius & Head

(c) Ven Te Chows Formula:

R = 0.305 x 10p (2.15)

wherep =V H

H

d

d

+ +

+

6 4 4 88

3 6 19 5

. .

. .

V= Velocity at the toe of spillway in m/sec.

(d) R.S. Varshneys dimensionsless formula:

FR

D1 10 09 198= +. . (2.16)

(e) R.S. Varshney & M.L. Bajajs Formula:F1 = 13.0 R

1/419.50 (2.17)


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58

Bucket Lip

In case of roller buckets, a 45 bucket exit angle is most effective.

Invert

Positioning of the bucket invert needs a major decision and is mainly governed by the

tail water levels, rock strata and cost of excavation and concreting. The desirable tail water

depths range from 1.3 to 1.4 times the conjugate depth y2

The Pressures on the Solid Roller Bucket: are in the neighbourhood of hydrostatic head

measured as the difference in the tail water elevation and the elevation of the point in bucket

under reference. These can be determined with the help of design chart (Fig. 2.22) given by

R.S. Varshney.

(ii) Slotted or dentated bucket: - High position of the river bed downstream of roller

bucket and unsymmetrical gate operation reduces to a great extent the usefulness of the solid

Fig. 2.22 Pressure on solid roller bucket-after Varshney

roller bucket. The dentated roller bucket substantially reduces the high boil and violent

ground roller. This type of bucket can also be tried for flows where tail water depth is less

thany2 curve.

Various designs are available for slotted roller bucket. The U.S.B.R. design

(Fig. 2.23) may be adopted for design, where necessary.

Fig. 2.23 Slotted Bucket U.S.B.R Design


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59

(iii) Ski Jump Bucket: This type of energy dissipator is suitable where the stream bed is

composed of firm rock and the tail water depth is less than required for the formation of

hydraulic Jump. Fig. 2.24 shown the definition sketch of skijump bucket.

Fig. 2.24 ski-jump Definition sketch

Bucket Invert

The invert level is decided mainly from structural design point of view. If power

houses are to be sited below the skijump bucket, then the invert should be fixed higher than

power house top. In certain cases it happens that the tail water depth is lower than y2, but the

bucket invert is lower than tail water levels. In such cases the bucket invert is decided so as to

provide a concrete cover of 1.5 to 3m over the bed rocks. The pure shijump or flip. as it is

quite often called, is provided such that bucket lip is always higher than maximum tail waterlevel.

Bucket Radius

All the relations given for the solid roller bucket are applicable in this case also.

Entrance and Exit Slopes

For entrance slope the steepest spillway slope that should be used is 4 vertical to 1horizontal. The exit angle is an important factor in determining the length of the trajectory.Theoretically, if friction, air retardation etc. are negelected, the following formula can be

used to evaluate the horizontal component of the jet trajectory

xV

g= 0

22sin

(2.18)

(For notations refer figure. 2.24) The jet trajectory height is given by the relation,

y = h sin2 (2.19)


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60

A judicious selection of exit angle is necessary. An exit angle of 30 to 35 will make a goodchoice.

The following formula by R.S. Varshney may be used to evaluate bucket exit angle

in degrees.

FR

H130 000091 8= +. (2.20)

(This relation was developed for cases F1< 8. However it can be used for F1 > 8.0 also after

neglecting minus sign).

In cases when maximum tail water elevation is below the bucket lip, the shape of the

lip may be flat for ease in construction. The shape of the lip in case of ski jump buckets,

where the tail water is slightly above the bucket lip needs careful attention in design. In suchcases high sub-atmospheric pressures occur at the downstream of lip. These subatmospheric

pressures can be controlled by making aeration arrangements below the lip. Another simpler

and effective way of sub luing these negative pressures is to provide a curved or a sloping lip

as shown in figure 2.25 (after S.N. Gupta and R.S. Varshney).

Fig. 2.25

Dynamic Pressures on Spillway and Bucket

Dynamic force on the spillway and bucket is exerted by the force of falling water.

There is a continuous change of velocity from section to section, which results, according to

Newton's second law of motion, in dynamic force on the structure. This force can be evaluted

after neglecting friction on spillway surface and approach velocity. The resulting force is

given by the relation

F = Q ( V2 - V1 ) (2.21)


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61

where is the mass density of water.Q is the discharge and

(V2 - V1) is the change in velocity.

Forces in x and y direction can be written as

Fx = Q (V2x - V1x) (2.22)Fy = Q (V2y - V1y) (2.23)

where suffixesx ory, denote the corresponding components in x ory direction. The resultant

ofFx and Fy forces is the net dynamic force on the spillway face. Pressures on slip bucket can

be found by D.B. Gummensky's formula

hV

gRh

d= +

2

1

where hd= design head at any point on the bucket in m of waterV= velocity of flow in m/sec.

R = radius of curvature in m.

h = hydrostatic head over the point under reference.

2.7 Spillway Crest Gates

For a gated-spillway crest gates are provided on the crest of the spillway. By

installing the crest gates, additional storage equal to the volume of water stored between thecrest level and the top of gates is made available. On the other hand, for an ungated spillway,

the useful storage in the reservoir is only upto the crest level of the spillway.

When the flood occurs, the gates are opened (or lifted up) so that the full spillway

capacity is available for discharging the flood. However, during the periods of the low flows,

the gates can be kept closed and an increase in the reservoir level is permitted. These low

flows, if required to be discharged downstream, are passed through the dam outlets or

sluiceways.

Crest gates can be provided on all types of spillways, except the siphon spillway for

which the gates are not required. In siphon spillways, the rise of water level above the fullreservoir is relatively small as compared to that in other types of spillways, and, therefore,

there is no need of gates. Gates installed on the spillways of the earth and rockfill dams

require extra precaution, because any operational failure of the gates may lead to overtopping

of the dam, resulting in its failure and catastrophe.

Various types of gates commonly used in preactice are described below:

1. Flash boards 2. Stoplogs and Needles

3. Vertical lift gates 4. Tainter (or radial) gates

5. Roller gates 6. Drum gates7. Bear-trap gates.


29/41

62

Hydraulic Design of Ogee Spillway & Bucket Type Energy Dissipator

Design Example 2.2

Design ogee spillway with the following data:

1. Height of spillway crest from the river bed 100 m

2. No. of spans 5

3. Length of each span (clear) 12.5 m

4. Thickness of each pier 3 m

5. Downstream slope of spillway glacis 0.8 horiz to 1 vert.

6. Tail water curve is below y2 curve

7. Rock conditions good

8. Discharge (design) 8500 cumec

Design

A. Head over crest and coefficient of discharge

Clear waterway = 5 x 12.5 = 62.50 m

Intensity of discharge =8500

62 5.

= 136 m3/sec/m

Let us assume a coefficient of discharge 2.1, hence

Q = CL H3/2

= 2.1 x 62.5 x H3/2

or 8500 = 131 H3/2

or H3/2

= 64.9

H = 16.2 m.

The maximum coefficient of discharge is 2.21, if not affected by other hydraulic

parameters like submergence, velocity of approach etc. To get correct value of Cd, the effects

of different parameters have to be worked out.

(i) Effect of approach depth

P = height of spillway from river bed = 100 m

P

H= =

100

16 26 2

..

As this is more than 4.0, there is no effect of approach depth and coefficient of

discharge may be taken as 2.21.

(ii) Effect of head due to velocity of approach

Velocity of approach =)2.16100)(345.62(

8500

++


30/41

63

=8500

74 5 116 2. .

= 0.98 metre/sec

Head due to velocity of approach

= ( . ).

0 982 9 8

2

=0 98

20

.=0.049 m

This is very small and is neglected.

(iii) Effect of tail water conditions

In this case d+ hd = 100 + 16.2 = 116.2 m

and Hd = 16.2 m

25.72.16

2.116==

+

d

d

H

dh

This is more than 1.7; the discharge coefficient is not affected by tail water

conditions.

(iv) Effect of heads other than design head

The crest is designed for full design discharge to avoid sub-pressures along spillway

surface. At discharge higher than design head, some negative pressures may be expected, but

they will be within allowable limits. At discharges less than the design discharge, the

coefficient of discharge will be reduced proportionately according to (H/Hd) ratio.

(v) Effect of upstream face slope

The upstream face of the dam is proposed to be kept vertical. A batter of 1 in 10 will

be provided from stability considerations in lower part. This batter being small will have no

effect on coeffecient of discharge.

Effective Length of Spillway

Due to piers and abutments, there will be reduction in the discharging length or

effective length will be less than the actual length due to end contractions.

Cut water (90/ nosed piers are proposed to be used and also rounded abutments. The

values ofKp = 0.01 and Ka = 0.1. Hence effective length of spillway will be


31/41

64

Le = L - 2 (N.Kp + Ka) Hd

= 62.5 - 2 (4 x 0.01 + 0.1) Hd

= 62.5 - 0.28 Hd

Since the effective length is less than net clear span of the spillway, a design headequal to 16.5 m is assumed. Hence

Le = 62.5 - 0.28 x 16.5 = 62.5 - 4.62 = 57.88 m

Hence Q = 2.21 x 57.88 x 16.53/2

= 128.0 x 67 = 8570 cumec > 8500 cumec

Hence alright

The crest profile will be designed forHd= 16.5 m

Downstream Profile

The profile recommended by Waterways Experiment Station Vicksburg Mississippi

U.S.A of U.S. Army is

x

H

y

Hd d

=

1 85

2

.

or x1.85 = 2 Hd0.85y

or y =x

H

x x

d

1 85

0 85

1 85 1 85

2 2 108 216

.

.

. .

. .=

=

The calculated coordinates for the downstream profile are

x (m) y (m) x (m) y (m)

0

1

2

3

4

5

7

9

10

0

0.046

0.167

0.354

0.603

0.905

1.710

2.684

3.240

12

14

16

18

20

22

23.5

4.575

6.02

7.88

9.74

11.85

14.35

15.73


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65

Tangent Point Coordinates

The slope of the downstream glacis = 0.8 horizontal to 1 vertical. Hence dy/dx =

(1.0/0.8) = 1.25.

Differentiating equation of the downstream profile.

25.16.21

85.1 85.0==

x

dx

dy

or x0.85 =125 216

18514 6

. .

..

=

or x = 23.5 m

y = 15.73 m

Upstream Profile

The upstream profile is given by the equation

y =0 724 0 270

0126 0 4315

1 85

0 85

. ( . ). .

.

.

x H

HHd

d

d

++

Hd0.85

x (x + 0.270 Hd)0.625

Since Hd = 16.5 m, hence

y =0 724 4 45

10 22 08 1233 4 45

1 850 625. ( . )

.. . ( . )

..x x

++ +

The coordinates of the profile are found out as under

x x + 4.45 (x+4.55)1.85

0724

108

.

.

(x+ 4.45)1.85

(x + 4.45)0.625

1.233(x+

4.45)0.625

y

-0.5-1.0

-2.0

-3.0

-4.0

-4.45

3.953.45

2.45

1.45

0.45

0

12.739.90

5.29

1.99

0.23

0

0.8500.664

0.354

0.133

0.015

0

2.3602.160

1.750

1.262

0.607

0

9.912.66

2.16

1.56

0.75

0

0.0200.064

0.274

0.653

1.345

2.080

The upstream prefile extends upto

x = -0.270 Hd= -4.45 metrey = 0.126 Hd= 2.08 metre


33/41

66

The nappe coordinates in the design, forH/Hd= 10, are as below:

x entre of span Along piersx/Hd

m y/Hd y (m) y/Hd y (m)

-1.0-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.81.0

1.2

1.4

1.6

1.8

-16.8-13.2

-9.9

-6.6

-3.3

0.0

3.3

6.6

9.9

13.216.5

19.8

23.1

26.4

29.7

-0.941-0.932

-0.913

-0.890

-0.855

-0.805

-0.735

-0.647

-0.539

-0.389-0.202

-0.015

0.266

0.521

0.860

-15.5-15.4

-15.1

-14.7

-14.1

-13.3

-12.1

-10.7

-8.9

-6.4-3.34

0.25

4.4

8.6

14.2

-0.950-0.940

-0.929

-0.930

-0.925

-0.779

-0.651

-0.545

-0.425

-0.285-1.121

0.067

0.286

0.521

0.729

-15.7-15.5

-15.3

-15.3

-15.2

-12.9

-10.7

-9.0

-7.0

-4.7-2.0

1.1

4.7

8.6

12.0

The velocities and depths of water (with and without air) on the spillway face can also

be evaluated from the figures 1.16 and 1.17.

( q =8500

74 5.= 114 say 115 cumec/m)

h (in m) head below

crestvelocity in m/sec.

ddepth of solid water

(m)

d'= depth of water

and air mixture (m)

10

20

30

40

50

60

70

80

90

20.9

24.0

27.4

28.7

32.8

36.0

38.4

41.0

44.2

5.5

5.0

4.2

4.0

3.5

3.2

3.0

2.8

2.6

5.90

5.50

4.75

4.60

4.10

3.70

3.65

3.60

3.55

Provide a 1.5 m free-board over the water depth with air entrained.


34/41

67

Shape of Pier

90 cut water nosed pier is hydraulically efficient and gives minimum contractioncoefficient. This shape of pier is adopted in design. The design of cut water is shown in figure

2.26.Thickness of pier = t = 3 m

Radius of cut water = 1.6t = 4.8 m

Spillway discharge rating curve

With the help of figure 1.11 the values of coefficients of discharge for heads other

than design head can be evaluted

Fig. 2.26

and discharge calculated from the equation

Q = Cd(62.5 - 0.20 H ) H3/2

Energy dissipation-Ski jump bucket

Since foundation rock is good and the tail water level is lower than y curve (i.e. curve

giving values of depths for formation of jump at different discharges), ski-jump bucket is an

ideal and economical solution.

Bucket invert

The invert for ski -jump can be at any elevation, with the proviso that a minimumcover of 1.5 m of concrete is required over good rock. The bucket can be provided in the

body of the dam at a high elevation. This will increase impact on the downstream. The bucket

invert in this case is provided at the river bed level assuming sound rock to be 20 m below the

river bed.

Bucket radius

(a) From Varshney's curve: The vertical distance between bucket and upstream

pool elevation is 114.5 m. The bucket radius from Varshney's curve in figure 2.21 words outto be 23 m


35/41

68

(b) From the relation given in equation 2.14 we get

R = 0.6 100 165 . = 24.4 m(c) To evaluate radius of bucket by Ven Te Chow formula, the velocity at the toe

of the spillway has to be determined.

VT = 2 05g H Hd( . ) = 2 9 81 1165 8 25. ( . . ) = 19 62 108 25. . = 46 m/sec.

From figure 2.18 we get by extrapolation VA/TT = 0.98.

Hence VA = 46 x 0.98 = 45.1 m/sec.

Using Ven Te Chow's relation, we have

p =451 6 4 165 4 88

36 165 19 5

. . . .

. . .

+ + +

=1.975

R = 0.305 x 10p

= 0.305 x 101.975

=29 m.

(d) R.S. Varshney's formula we have,

FR

D11

0 09 198=

+. .

q =8500

74 5.= 114 m3/sec/m

D1 =114

451.

= 2.53 m

F1 =451

9 81 2 539 05

.

. ..

=

9 05. =0.09R

2 53196

..+

or R = 29.5 m.

(e) From R. S. Varshney and M. L. Bajaj's formula we get

9.05 = 13.0 R1/4

-19.5 R = 23.3 m

The least of the above values viz 23.0 m is adopted. The bucket will be circular.

Exit angle

The usual exit angles are 30 to 35. An angle of 30 is selected as bucket lip angle.Checking from R. S. Varshney's formula (equation 2.20), we have

9.05 = -0.000091.23

1165.3 + 8

= 41


36/41

69

Lip shape

The lip will be a sloping one with 2 horizontal and 1 vertical slope in a vertical

distance of 1.5 metre, and thereafter it would meet the rock surface at slope of 45 .

Fig. 2.27 Ogee spillway- Hydraulic profiles

Fig. 2.28 Ogee spillway- Hydraulic profiles


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70

Length and height of trajectory

Assuming a depth of 2.5 m in the bucket invert

Velocity = 45.1 m/sec

Depth = 850074 5 451. .

=2.52 m hence okay

x = kV

g

2 2sin where k is constant which accounts

for air resistance = 0.95 approximately.

x =0 95 45 1 45 1. . .x x s i n 60

9.81= 170 metre

The skijump will rise to height

y = kVo

g

2

2sin

2 = 0.95

451 451

2 9 81

. .

.

x

xx (1/4)

=0 95 45 1 45 1

8 9 81

. . .

.

x x

x

= 24.8 say 25 metre

Pressure on bucket

The maximum pressure on the bucket (at theinvert) is given by Gumeneskys formula(equ. 2.29).

Design head hd = hgR

V

+1

2

= 5.212381.9

1.451.45

+

= 10.0 x 2.5

= 25.0 m of water i.e 25.0 t/sq m

By Balloffets formula

hd = h +

22

12 R

hR

g

V

= 2.5 +

2

23

5.2231

81.92

1.451.45

= 2.5 + 103 (1-0.797) = 2.5 x 0.203

= 2.5 + 21 = 23.5 m of water

The bucket shall be designed for a pressure of 25m head of water.

The hydraulic design of spillway is shown in figure 2.28.


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Evaluation of Dynamic Force on the Spillway

Discharge intensity = 144 cumec/m

Applying Bernoullis theorem on u.s. and the bucket entrance (Section 1) and bucket

invert (Section 2), we get,

116.5 = 8.65 + d1 cos 51.1 +g

V

2

21 = d2 +g

V

2

22 (I)

But V1d1 = 114 = V2d2

Substituting for V1 =1

114

dand V2 =

2

114

din equation (I)

we have

116.5 = 8.65 + d1 0.6225 +1

1.62.19

114114

d

and

116.5 = d2 +2

262.19

114114

d

Solving by trial, we get

d1 = 2.5 m and d2 = 2.4

V1 = 45.7 m / sec V2 = 47.5 m/sec

Let Fx and Fy be the components of forces of water on curved section 1-2 and F1 and

F2 be the hydrostatic force at the respective sections (Fig 2.29)

Fig 2.29 Dynamic force on spillway --- definition sketch


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P1 = x 1 x 2.5 2 x cos 51.5 = 6.25 x 0.31125 = 1.94 tP2 = x 1 x 2.4

2= 2.88 t

Weight W of the water flowing in this section

W = 1 x 515360

. x 2 x 23 x 2.45 = 50.7 t

Applying equations 2.20 and 2.21 we have

P1 0.6225 - P2 +Fx =114

9 81.( 47.5 - 45.7 x 0.6225 )

or Fx = 223.67 t

SimilarlyFx = P1 sin 60 - W = 11.63 [ - (-47.5 sin 60) ]

= 529.4 t

Resultant force = F x 2 + Fy2 = 57.5 t

and acts at an argle = tan -1529 4

223 67

.

.= 62.1


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PROBLEMS

* What are the essential requirements of a spillway ? How would you select a suitable

site for the spillway ?* What are the factors that affect the spillway capacity ? How would you fix the

spillway capacity ?

* Describe the various component part of the spillway Explain their functions.

* How would you classify the spillway based on the purpose and based on the control ?

* Discuss the working and uses of a free overfall spillway. why this type of spillwaysare not common ?

* Explain the design of an ogee-shaped spillway . How would you fix the d /s and u / s

profiles?

* How would you compute the discharge over an ogee- shaped spillway? Discuss thevarious factors which affect the coefficient of discharge.

* Where would you prefer a chute spillway over an ogee- shaped spillway ?

* Where do you provide a side-channel spillway ?

* What is a spillway ? What are its functions? Describe in Brief the various types of

spillways.

* Write short notes on .

* Discuss the characteristics of a hydraulic jump. Differentiate between the tail water

rating curve ( TWRC ) and the jump height curve ( JHC ). How would you select the

most suitable type of energy dissipating device for different relative positions of thetwo curves ?

* What are the different types of energy dissipating methods used below the spillways?

* What are different types of 1.8 stilling basins for the horizontal aprons? Compare

these with U.S.B.R basins.

* Discuss various bucket-type energy dissipators. Where would you provide each type?

* What are different types of spillway gates?

Numerical

** An overflow spillway with u / s face vertical is to be designed for a flood of 2100

cumecs . The level of the spillway crest is at R.L of 130.00 and the river bed is at R.L of

100.00. The end walls of the spillway are 79 m apart, and there are 4 round- nosed piers, each

I m wide . Determine the total head over the crest. Take C = 2.20 and Ka = 0.10. Also draw

the spillway profile. [ Ans: 5.53 m ]


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** Design a suitable stilling basin at the toe of a spillway from the following data.

Discharge intensity (q) = 24.6 cumecs / m

H. F. L in the reservoir = 100.00

Spillway crest level = 95.00

River bed level = 59.00Tail water level = 69.00

Coefficient of velocity = 0.90

Adopt the tail water depth 5% greater than the sequent depth. Assume Lb / y2 = 4.30

[ Ans.V1 = 25.5 m / s, y1 = 0.97 m, F1 = 8.23 , y 2 = 10.82 m, Lb = 48.85 m, Floor depression

= 1.36 m ]

** The crest level of an ogee- shaped spillway is at R.L of 323.00 and the maximum

reservior level is 334.00 Calculate the maximum discharge when the flow takes place through5 vents of the effective width of 12 m each . Take C = 2.20.

*** In order to dissipate the energy below a spillway, it is proposed to form a

hydraulic jump If the depth of flow changes from 1.0 m to 4.0 m in the hydraulic jump,

determine the initial Froude number and the discharge intensity. [ Ans: 3.16 , 9.90 cumecs /

m]

*** The crest of an ogee- shaped spillway is at a height of 10m above the river bed. If

the head over the crest is 3m, find the loss of energy when a hydraulic jumps forms at the toe. Take C = 2.20. [Ans: 6.12 k Nm / kN ]

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energy dissipator