Chapter 5 drop sturcutures

CHAPTER 5: DROP STRUCTURES

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0401544 - HYDRAULIC STRUCTURES

University of Sharjah

Dept. of Civil and Env. Engg.

DR. MOHSIN SIDDIQUE

ASSISTANT PROFESSOR

LEARNING OUTCOME

After taking this lecture, students should be able to:

(1). Obtain in-depth knowledge on various types of drop

structures used in open channels and their design guide

lines

(2). Identify the suitable drop structure for various flow

conditions

(3). Apply the design guide lines for the design of selected

drop structure

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References: Novak, A.I.B. Moffat, C. Nalluri, R. Narayanan, Hydraulic Structures, 4the Ed.

CRC Press

Santosh K. G., Irrigation Engineering and Hydraulic Structures, Khanna

Pubilshers

DROP STRUCTURES (CANAL DROPS)

A drop (or fall) structure is a regulating structure which lowers the water level along its course.

The slope of a canal is usually milder than the terrain slope as a result of which the canal in a cutting at its headworks will soon outstrip the ground surface. In order to avoid excessive infilling the bed level of the downstream canal is lowered, the two reaches being connected by a suitable drop structure

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A typical canal fall 4



TYPES OF DROP STRUCTURES

Drops are usually provided with a low crest wall and are subdivided into the following types:

(i) the vertical drop

(ii) the inclined drop

(iii) the piped drop and

(iv) Farm drop structures

Note: The above classification covers only a part of the broad spectrumof drops, particularly if structures used in sewer design are included; acomprehensive survey of various types of drops has been provided,e.g. by Merlein, Kleinschroth and Valentin (2002); Hager (1999)includes the treatment of drop structures in his comprehensivecoverage of wastewater structures and hydraulics.

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(i) the vertical drop,

(a). Common (straight) drop

(b). Sarda-type fall (India)

(c). YMGT-type drop (Japan)

(d) Rectangular weir drop with raised crest (France)

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(i.a) Common (straight) drop: The common drop structure, in which the aerated free-falling nappe (modular flow) hits the downstream basin floor, and with turbulent circulation in the pool beneath the nappe contributing to energy dissipation, is shown in Fig

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The following equations fix the geometry of the structure in a suitable form for steep slopes:

where d is the height of the drop crest above the basin floor and Lj the length of

the jump. Lj=6.9(y2-y1)

A small upward step, h (around 0.5<h/y1<14), at the end of the basin floor is desirable in order to localize the hydraulic jump formation. Forster and Skrinde(1950) developed design charts for the provision of such an abrupt rise.

32/ gdqDr =

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The USBR (Kraatz and Mahajan, 1975) impact block type basin also provides good energy dissipation under low heads, and is suitable if the tailwater level (TWL) is greater than the sequent depth, y2.

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h

d/s Floor

level

Crest level

yo

Tailwaterlevel


The following are the suggested dimensions of such a structure with impact block type basin

where yc is the critical depth.

The values of Ld can be

obtained from the following

Fig

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2

456.0185.0

228.0691.0

368.4195.3406.0

368.4195.3406.0

2

2

sfd

c

t

ccc

t

cc

t

cc

f

LLL

y

L

yy

d

y

L

Ls

yy

hL

yy

dL

+=

+

−

+

=

−+−=

−+−=

Use –ve sign with d and h2

Note: sometime d is replaced with ho.

However, both are the same.

h2=RL of crest-D/S FSL

d=RL of crest-stilling basin flood level

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0.85yc

0.8yc

0.4yc

h2

h=drop height=RL of crest-D/S Floor level



Side wall height above tailwater

h

d/s Floor

level

Crest level

yo


12

Tailwaterlevel

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(i.b) Sarda-type fall (India): This is a raised-crest fall with a vertical impact, consisting of a crest wall, upstream and downstream wing walls, an impervious floor and a cistern (basin), and downstream bank and bed protection works

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Two types of crest are used; the rectangular one for discharges up to 10m3/s** and the trapezoidal one for larger discharges (see Punmia and Lal, 1977).

**(14m3/s) according to Santosh K. G.

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Longitudinal Section of rectangular crest Sarda fall

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Longitudinal Section of trapezoidal crest Sarda fall

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The following are the design criteria established by extensive model

studies at the Irrigation Research Institute in India.

1. Length of crest: Since fluming is not permissible in this type of fall, the crest length is normally kept equal to the bed width of the canal

2. Shape of crest:

Rectangular

Top width, Bt=0.55d1/2(m)

Base width, B1=(H+d)/G

where G is the relative density of the crest material (for masonry, G=2).

Trapezoidal

top width, Bt=0.55(H+d)1/2 (m).

For the base width, B1, upstream and downstream slopes of around 1 in 3 and 1 in 8 are usually recommended

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3. Design Discharge:

Rectangular : Q=1.835LH3/2(H/Bt)1/6

Trapezoidal: Q=1.99LH3/2(H/Bt)1/6

Where L is length of crest, Bt is top width, H is head of water over crest

4. Length and depth of cistern

length, Lc=5(H.HL)1/2

depth, dc=1/4(H.HL)2/3

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Where, HL= drop


5. Upstream wing walls

For trapezoidal crest, the upstream wing walls are kept segmental with radius equal to 5 to 6 times H and subtending an angle of 60o at center and then carried tangential into the berm as shown in Figure.

The foundations are laid on the impervious concrete floor itself

For rectangular crest, the approach wings may be splayed straight at an angle of 45o

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6. Upstream Protection

Brick pitching in a length equal to upstream water depth may be laid on upstream bed, sloping towards the crest at a slope of 1:10.

Drain pipes should be provided at the u/s bed level in the crest so as to drain out the u/s bed during the closure of channel.

Upstream curtain walls: 11/2” brick (~35cm) thick upstream curtain wall is provided, having a depth equal to 1/3rd of water depth

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Upstream curtain wall


7. Impervious floor downstream of the crest

Length: Total length of impervious floor can be determined by Bligh’s theory for small works and by Khosla’ theory for large works. The minimum length of flood d/s of toe of crest wall should be

Lbd=2(D1+1.2)+HL

D1=U/S FSL-BL and HL= drop

Thickness: the floor thickness can be worked out for uplift pressure (using minimum thickness of 0.4m to 0.6m) and only a nominal thickness of 0.3m is provided on the upstream side.

Note: seepage theories play key role in calculation of length and thickness of floor.

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Solution:

7. Impervious flood

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8. Downstream protection:

The d/s bed may be protected with dry brick pitching, about 20cm thick resting on 10cm thick ballast. The length of d/s pitching is given by the values of table below or 3 times the depth of downstream water, whichever is more. The pitching may be provided between two or three curtain walls. The curtain walls may be 1 ½ brick (~35cm) thick and of depth equal to ½ the downstream depth; or as given in table (minimum=0.5m)

24HL= drop height


9. Slope pitching:

After the return wing, the side of the channel are pitched with one brick of edge. The pitching should rest on a toe wall 1 ½ brick thick and of depth equal to half the downstream water depth.

The side pitching may be curtailed at angle of 45o from the end of pitching or extended straight from the end of bed pitching

10. Downstream wings. Downstream wings are kept straight for a length of 5 to 8 times (H.HL)

1/2 and may then be gradually wrapped. They should be taken up to the end of impervious floor.

All the wing walls must be designed as retaining walls, subjected to full pressure of submerged soil at their back when channel is closed. Such as wall generally has a base width equal to 1/3rd its height

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(i.c) YMGT-type drop (Japan): This type of drop is generally used in flumed sections suitable for small canals, field channels, etc., with discharges up to 1m3/s. The following are the recommended design criteria:

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The following are the recommended design criteria:

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(i.d) Rectangular weir drop with raised crest (France): SOGREAH (Kraatz and Mahajan, 1975) have developed a simple structure suitable for vertical drops of up to 7m (for channel bed widths of 0.2–1 m with flow depths (at FSL) of 0.1–0.7 m):

Hdr

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For the design of crest,

discharge, Q=CL(2g)1/2H3/2,

where C=0.36 for the vertical upstream face of the crest wall and 0.40 for the rounded upstream face (5–10 cm radius).

The crest length, L=LB-0.10 m for a trapezoidal channel and is B1 (the bed width) for rectangular channels.

For the design of cistern,

Volume of basin, V=QHdr/150 (m3),

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(ii) the inclined drop and

(a) Common chute

(b) Rapid fall type inclined drop (India)

(c) Stepped or cascade-type fall

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(ii.a) Common chute: This type of drop has a sloping downstream face (between 1/4 and 1/6,called a glacis) followed by any conventional type of low-head stilling basin; e.g. SAF or USBR type III. The schematic description of a glacis-type fall with a USBR type III stilling basin, recommended for a wide range of discharges and drop heights, is shown in Fig.

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(ii.b) Rapid fall type inclined drop (India): This type of fall is cheap in areas where stone is easily available, and is used for small discharges of up to 0.75m3/s with falls of up to 1.5 m. It consists of a glacis sloping between 1 in 10 and 1 in 20. Such a long glacis assists in the formation of the hydrualic jump, and the gentle slope makes the uninterrupted navigation of small vessels (timber traffic, for example) possible.

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(ii.c) Stepped or cascade-type fall: This consists of stone-pitched floors between a series of weir blocks which act as check dams and are used in canals of small discharges; e.g. the tail of a main canal escape. A schematic diagram of this type of fall is shown in Fig.

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TYPES OF CANAL FALLS

(iii) the piped drop.

A piped drop is the most economical structure compared with an inclined drop for small discharges of up to 50 liter/s. It is usually equipped with a check gate at its upstream end, and a screen (debris barrier) is installed to prevent the fouling of the entrance.

Types:

(a) Well drop structure

(b) Pipe fall

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(iii.a) Well drop structure: The well drop (Fig) consists of a rectangular well and a pipeline followed by a downstream apron. Most of the energy is dissipated in the well, and this type of drop is suitable for low discharges (up to 50 L/s) and high drops (2–3 m), and is used in tail escapes of small channels.

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(iii.b) Pipe fall; This is an economical structure generally used in small channels. It consists of a pipeline (precast concrete) which may sometimes be inclined sharply downwards (USBR and USSR practice) to cope with large drops. However, an appropriate energy dissipator(e.g. a stilling basin with an end sill) must be provided at the downstream end of the pipeline.

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(iv) Farm drop structures:

Farm channel drops are basically of the same type and function as those in distribution canals, the only differences being that they are smaller and their construction is simpler.

The notch fall type of farm drop structure (precast concrete or timber) consists of a (most commonly) trapezoidal notch in a crested wall across the canal, with the provision of appropriate energy-dissipation devices downstream of the fall. It can also be used as a discharge measuring structure.

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The details of a concrete check drop with a rectangular opening,

widely used in the USA, are shown in Fig. Up to discharges of about 0.5m3/s, the drop in the downstream floor level (C) is recommended to be around 0.2 m and the length of the apron (L) between 0.75 m and 1.8m over a range of drop (D) values of 0.3–0.9m.

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Solved Examples

1. Common drop structure with impact block type basin

2. Sarda type drop structure

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Example 1: Find the dimensions for a straight drop structure with a impact block type basin.

Given:

• Q = 250 ft3/S (7.08m3/s)

• Drop: h = 6.0 ft. (1.93m)

(Upstream and Downstream Channel -Trapezoidal)

• B = 10.0 ft. (3.2048)

• Z = 1V:3H

• So = 0.002 (after providing for drop)

• n = 0.030

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0.85yc

0.8yc

0.4yc

h2

h=drop height=RL of crest-D/S Floor bed level



Side wall height above tailwater

h

d/s Floor

bed level

Crest level

yo


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Solution:

Step 1. Estimate the required approach and tailwater channel elevation difference, h. This is estimated and given above as 6.0 ft.

Step 2. Calculate normal flow conditions approaching the drop to verify subcritical conditions. By trial and error using Manning’s equation

Q=1.49/n(AR2/3So1/2)

yo = 3.36 ft.,

Velocity of approach, Vo=Q/A:

Vo = 3.71 ft/s,

Froude no. =Vo/(gyo)1/2= 0.36;

Therefore, flow is subcritical.

yo=?

B=10ft

1V:ZH

i.e. (Z=3)

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Solution:

Step 3. Calculate the critical depth over the weir into the drop structure. Calculate the vertical dimensions of the stilling basin.

Start by finding the critical depth over the weir based on the unit discharge, q = Q/B = 250/10 = 25ft.2/s

ftg

qyc 69.2

2.32

253/1

23/1

2

=

=

=

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Solution:

Next calculate the required tailwater depth above the floor of the stilling basin:

y2 = 2.15yc = 2.15 (2.69) = 5.77ft.

Now the distance from the crest down to the tailwater needs to be calculated:

h2 = (h-yo) = -(6.0-3.36) = -2.64 ft. (-ve indicate vertically downward distance)

Finally, calculate the total drop from the crest to the stilling basin floor:

d = -(h2 + y2) = -(2.64 + 5.77) = -8.41 ft. (round to 8.4 ft. )

Since the nominal drop, h, is 6.0 ft., the floor must be depressed by 2.4 ft.

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Solution:

Step 4. Estimate the basin length.

( )( )

( )( )

ftLL

L

ft

y

L

yy

d

y

L

Ls

ftyyhL

ftyydL

sfd

c

t

ccc

t

cct

ccf

4.102

89.1094.9

2

89.10

69.2

26.6456.0185.0

69.269.2

41.8

69.2

26.6228.0691.0

456.0185.0

228.0691.0

26.669.269.2

64.2368.4195.3406.0/368.4195.3406.0

94.969.269.2

41.8368.4195.3406.0/368.4195.3406.0

22

2

=+

=+

=

=

+

−−

+

=

+

−

+

=

=

−−+−=−+−=

=

−−+−=−+−=

ftyyLL ccdB 3.1769.2*75.169.2*8.04.1075.18.0 =++=++=

Therefore the total basin length = 17.3ft

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Solution:

Step 5. Design the basin floor blocks and end sill.

Block height = 0.8yc = 0.8(2.69) = 2.1ft.

Block width = Block spacing = 0.4yc = 0.4(2.69) = 1.1ft.

End sill height = 0.4yc = 0.4(2.69) = 1.1ft.

Step 6. Design the basin exit and entrance transitions.

Sidewall height above tailwater elevation = 0.85yc = 0.85(2.69) =2.3 ft.

Armour approach channel above headwall length = 3yc = 3(2.69) = 8.1ft.

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Example 2

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10m


Solution:

1. Length of crest:

same as d/s bed width

2. Shape of crest: Rectangular

because Q<14m3/s

Top width, Bt=0.55d1/2(m)

Base width, B1=(H+d)/G

3. Crest Level: Applying Q formula:

Assume Bt=0.8m (It will be later recomputed with above formula)

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Solution:

2. Shape of crest:

Velocity of approach

Velocity head

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Solution:

2. Shape of crest:

Therefore

Top width, Bt=0.55d1/2=0.55(2.27) 1/2=0.825m

Base width, B1=(H+d)/G=(0.76+2.27)/2=1.5m

Keep 1.5m width of base

P

d

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Solution:

4. Length and depth of cistern

length, Lc=5(H.HL)1/2

depth, dc=1/4(H.HL)2/3

Depth of cistern, dc

HL is drop (fall)= 1.5m (given)

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Solution:

5. Upstream wing walls

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Solution:

6. Upstream Protection

Yu= water depth @u/s

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Solution:

7. Impervious flood

Lbd=2(D1+1.2)+HL

D1=U/S FSL-BL and

HL= drop

Remember !!

H/Lc ≤1/C

Lc~C.H

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Solution:

7. Impervious flood

5

Remember !!

t=h/(G-1)

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Solution:

7. Impervious flood

Use nominal thickness of 0.3m at u/s 56


Solution:

8. Downstream protection

Total depth of curtain wall below d/s bed level =Floor thickness + depth of curtain wall + 0.3m thick PCC=(0.35+0.2)+0.75+0.3≅≅≅≅1.6m

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Solution:

9. Slope pitching

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Solution:

10. Downstream wings

Wing walls are kept straight and parallel up to the end of the floor and joined to return walls

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Wing walls

Plan view


Solution:

7. Impervious flood

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Wing walls


Problem 1: Vertical drop without blocks

The volume flow rate in a rectangular channel is 5ft3/s/ft. A vertical drop is used to lower the channel 6.0ft. The flow is subcritical above and below the drop structure. Determine the dimension of the drop structure for a tailwater (d/s water) depth of 1.67ft.

Solution: q=5ft3/s/ft, d=6ft, ytw=1.67ft

Calculate Basin dimensions:

Calculate sequent depths,

Length of hydraulic jump,

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( ) ( ) ftyyL j 0.133.018.29.69.6 12 =−=−=

ftyDd

y

ftyDd

y

r

r

18.2)6()106.3(66.166.1

3.0)6()106.3(54.054.0

27.032

27.02

425.031

425.01

=×=⇒=

=×=⇒=

−

−

3106.3

−= xDr


Problem 1: Vertical drop without blocks

Basin length,

Pool depth,

End sill height,

62

( ) ftLdLDd

LBjr

B 6.1866/13106.33.1/3.427.0327.0

=

+×=⇒+=

−

ftDY rp 74.1)106.3(22.0322.0

=×==−

( ) ftyyh tw 51.067.118.22 =−=−=


Problem 3:

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LB=18.6ft

d

THANK YOU

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