BFC21103 Hydraulics Chapter 5. Hydraulic Structures

BFC21103 Hydraulics

Chapter 5. Hydraulic Structures

Tan Lai Wai, Wan Afnizan & Zarina Md [email protected]

Updated: September 2014

Learning Outcomes

At the end of this chapter, students should be able to:

i. Recognise the use of various types of hydraulic structures;

ii. Estimate the rate of flow through/over various hydraulic structures; and

iii. Design USBR stilling basin type III

BFC21103 Hydraulics Tan et al. ([email protected])


Hydraulic structures are artificial or natural structures that can be usedto divert, restrict, stop, or manage the flow of fluid. The structures canbe made from a range of materials such as large rocks, concrete, woodentimbers or even tree trunks.

One of hydraulic engineering works is to design and analyze hydraulicstructures. Hydraulic structure is constructed to:

a. Control flow,

b. Maintain water level,

c. Control erosion, or

d. others


Examples of hydraulic structures are:

a. Sluice gate,

b. Spillway,

c. Weir,

d. Energy dissipator structures, and

e. Constrictor


Sluice gates

Spillways


Weirs

Energy dissipators

5.1 Sluice Gate


Sluice gates are used to control and measure rate of flow in open channels (in hydropower plants, and irrigation channel).

Eo = initial specific energy a = height of gate openingyo = depth of flow upstream y1 = depth of flow downstreamy2 = depth of flow further downstream

Hydraulic jump

a y1

Q y2

yoEo

Sluice gate


The flow under the vertical sluice gate is considered as rectangular orifice as long as the height of the gate opening a is smaller relative to specific energy Eo and the flow further downstream y2 does not affect the flow under the sluice gate y1, i.e. (yo y1) > y2

The minimum depth area (maximum flow velocity) immediate downstream of the sluice gate is known as vena contracta.

The depth of flow at vena contracta y1 = a, where = depth correction factor (0.624 to 0.648).


Flow under Sluice Gate

Rate of flow under the sluice gate can be derived from the Bernoulli's theorem, i.e.

g

Vy

g

p

g

Vy

g

po

22

21

11

2oo

Since the free surface is under atmospheric pressure, po = p1 = 0 N/m2

Upstream velocity Vo is very small compared to downstream velocity V1, thus Vo 0. Therefore,

g

Vyy

2

21

1o

Velocity of flow under sluice gate 1o1 2 yygV

Rate of flow under sluice gate 1o2 yygabCQ d

where, b = width of gate, and Cd = coefficient of discharge (0.596 - 0.607)


If downstream control occurs, y1 submerges completely, and the rate of flow is based on (yo y2).

Rate of flow under sluice gate is

1o2 yygabCQ d

2o2 yygabCQ d

if (yo y1) > y2

if (yo y1) y2

Checking for CASE 1

Checking for CASE 2

(memorize)!!


A sluice gate is built in a rectangular channel with bottom width bof 2.5 m. The flow and gate characteristics are yo = 2 m, = 0.625, Cd = 0.610, and a = 0.5 m. Calculate the discharge under the sluice gate if y2 = 1.8 m.

Activity 5.1



5.2 Spillway

Spillway is built in a reservoir to allow the flow of water to safely move downstream when the reservoir is full.

A spillway is shaped as a rectangular concrete channel that connects the upstream and downstream regions of a weir.

The best design of a spillway is following the lower nappe of free fall of flow (e.g. ogee spillway) and the best shape of the spillway surface is parabolic with inverted curve at downstream.

Q

Inverted curve

Falling water nappe

Spillway

CrestH


Sembrong dam spillway


Ogee spillway is also known as the overflow spillway and is widely used on gravity, arch and buttress dams. It is the most extensively used spillway to safely pass the flood flow out of reservoir.

Ogee is shaped based on the underside curve of the nappe over a sharp-crested weir

Discharge coefficient Cd for ogee spillway is about 20% higher than Cd for weir.

In high overflow spillway (H1 > Hd), the discharge corresponds to the maximum designed capacity of the spillway, and the crest of ogee spillway rises up from point A (sharp-crested crest) to maximum rise of 0.115H1 then falls in parabolic form.

A

H1Hd 0.115H1


5.3 Weir

A weir is an overflow structure built across an open channel to measure the rate of flow.

The weir offers a simple and reliable method of measuring the stream flow.

where, H1 = height of flow surface to the crest of weir, P = height of weir, and Q = rate of flow

Q

H1

P

Sharp-crested weir

Nappe


Weirs can be classified into various types, based on the width of their crests and the openings of the weirs.

Based on Width of Crest Based on Opening of Weir

Broad-crested Rectangular notch

Sharp-crested Trapezoidal notch

Triangular notch

Weir has broad crest when the thickness of the crest is more than 60% of nappe's thickness.

Weirs can also be categorised into submerged and non-submerged weirs.



5.3.1 Rectangular Weir

Rectangular weirs are divided into suppressed weirs and contracted weirs.

A weir is suppressed when the weir opening spans the full width of a channel.

Discharge Q over a sharp-crested, fully ventilated, and free-flowing suppressed weir is given as

where, Cd = coefficient of discharge (Cd = Cv Cc)

L = span or width of weirH1 = head of flow over the weir

2

3

123

2LHgCQ d

Note :-

Suppressed = uncontracted

Unsupressed = contracted

Rectangular – uncontracted weir


Checking (memorize)!!

51 P

H

The discharge coefficient, Cd is not directly given Use Rehbock formula

P

HCd

1075.0611.0

(Rehbock formula)

201 P

H2

3

1

106.1

H

PCd

The discharge coefficient has smooth transition between the two equations for 20 > H1/P > 5.

where, H1 = head of flow over weirP = height of weir

205 1 P

H15.110.1 dC

CASE 1

CASE 2

CASE 3


A suppressed rectangular weir 0.75 m high and 1.5 m long is used to discharge water from a tank under a head of 0.5 m. Estimate the discharge of the weir.

Activity 5.2

0.5 m

0.75 m

1.5 m



Contracted weir has weir length that is smaller than the width of the channel. Due to the presence of the end contractions, the effective length of weir Le is smaller than the actual length of weir L.

H1

P

B

LLe


24

Contracted weir can be contracted at both ends (number of contraction n = 2) or only one end (n = 1). For suppressed weir, n = 0.

Two end contraction

One end contraction

Zero end contraction

n = 2 n = 1 n = 0


The discharge of a free-flowing rectangular contracted weir is given as

2

3

123

2HLgCQ ed

where, L = length of weirLe = L 0.1nH1

n = number of end contractionsH1 = head of flow over the weirCd= coefficient of discharge

2

1

1

1

55.0100451.0

607.0PH

H

HCd

Bazin formula

Rectangular – contracted weir

IF the discharge coefficient, Cd is not directly given Use BAZIN formula


A rectangular weir 0.75 m high and 1.5 m long is contracted at both ends (n = 2). Estimate the discharge if the weir is discharging water from a tank under a head of 0.5 m.

Activity 5.3



5.3.2 Triangular Weir

Discharge over a triangular weir with vertex angle , under a head H1 is given as

2tan2

15

82

5

1

HgCQ d

where, Cd = coefficient of discharge (= 0.58)

H1

B

Triangular weir


Calculate the discharge of flow through a triangular notch with vertex angle of 60 and a head H1 of 0.5 m. Assume coefficient of discharge to be 0.58.

Activity 5.4


Given = 60, H1 = 0.5 m, and Cd = 0.58

2tan2

15

8 251

HgCQ d


5.3.3 Trapezoidal Weir

Discharge through a trapezoidal weir with side slope z : 1, is a combination of flow through a suppressed rectangular weir with length L, and a triangular weir with vertex angle of 2, where z = tan , i.e.

)tan5

4(2

3

21

2

3

1 HLHgCQ d

H1

L

Normal trapezoidal weir


2

3

123

2LHgCQ d

The trapezoidal weir with side slope 1(H) : 4(V) is known as Cippolettiweir. The discharge through a Cippoletti weir is given as

where, Cd = 0.63

trapezoidal weir - Cippoletti

H1

L

1

4


Activity 5.5

A Cippoletti weir with width 0.5 cm is installed at a section of a channel. Calculate the discharge when the head over the crest is 0.25 cm. Use Cd = 0.63.


Cippoletti weir with sides 1(H) : 4(V), L = 0.5 cm, H1 = 0.25 cm, Cd = 0.63

H1

L

04.14

4

1tan 1

2

3

123

2LHgCQ d


5.4 Energy Dissipator Structure

Energy dissipator structures are commonly built at the lower end of a spillway to reduce the kinetic energy of flow. These structures are built to reduce damages caused by high kinetic energy flows.

When designing an energy dissipator for a specific location, one must considers factors such as the site location, the dissipator structures, and the discharge through the dissipator. An example of energy dissipator isUSBR stilling basin type III.

Stilling basin type III is very stable with a steep jump front and less wave action downstream than stilling basin type II or free hydraulic jump. It is recommended for discharges up to 18.58 m2/s per basin width with Fr = 4.5 to 17, and velocity of flow V1 = 15.2 m/s to 18.3 m/s.


Type of stilling basin (US Dept of the Interior, Bureau of Reclamation - USBR, 1984)

Stilling basin type I

Stilling basin type II

Flow

Flow


Stilling basin type III

Stilling basin type IV

Flow

Flow


Stilling basin type V

Stilling basin type VI

Flow

Flow


Stilling basin type VII

Stilling basin type VIII

Flow


Stilling basin type IX

Stilling basin type X

Flow

Flow


USBR stilling basin type III


USBR Stilling basin type III

There are 3 sets of blocks, i.e. A. chute blocks, B. baffle piers, and C. end sill.

Block C: End sill

Block B: Baffle piers

Block A: Chute blocks

h1

w1

s1

L2

L

w3

s3

h3

t3

1z3 z4

1 h4


Block A Block B Block C

Sizing for USBR stilling basin type III

h1 = y1

s1 = y1

w1 = y1

L2 = 0.8y2

L = refer to the chart of length of jump on horizontal floor

where,

h3 = (0.168Fr1 + 0.63)y1

s3 = 0.75h3

w3 = 0.75h3

t3 = 0.2h3

z3 = 1.0

z4 = 2.0

11

4 118

Fryh

1

1BV

Qy


y1

V1

y2

Ho

H1

22 o

11

HHgV 2

1

1

2 Fr8112

1

y

y


0.0

1.0

2.0

3.0

4.0

0 2 4 6 8 10 12 14 16 18 20

1

11

gy

VFr

2y

L

Length of jump on horizontal floor


Steps in sizing for USBR stilling basin type III

Step 1. Find V1, y1, Fr1, y2

Step 2. Determine length of basin L

Step 3. Determine h1, s1, w1 and quantity of chute blocks (block A)

11

A blockof No.ws

B

Step 4. Determine L2, h3, s3, w3, t3 and quantity of baffle piers (block B)

33

B blockof No.ws

B

Step 5. Determine h4 for end sill (block C)


Activity 5.6

An 8 m-wide weir of a dam is discharging 80 m3/s of flow with H1 = 30 m, and Ho = 5m. Design an USBR stilling basin type III downstream of the spillway.





Assignment #5

Q1. Explain the functions of the following hydraulic structures:(a) sluice gate(b) weir(c) spillway(d) energy dissipator structures

Q2. A 1.5 m wide sluice gate with an opening of 10 cm is discharging water at a rate of 3 m/s. Calculate the discharge if the depth of flow behind the sluice gate is 3 m and the downstream depth is 2 m. Use coefficient of discharge Cd = 0.585.

Q3. A trapezoidal weir with side slope 1:4 and width of opening 0.5 m is used to measure flow rate of a river. Calculate the discharge if the head above the weir is 25 cm and Cd = 0.675.

- End of Question -

THANK YOU


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BFC21103 Hydraulics Chapter 5. Hydraulic Structures

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