1

Flow under pressure through a fixed cross section: e.g.

orifices, nozzles, short pipes, gates

Open flow through an undetermined cross section: e.g.

weirs, spillways, drop structuresMain tasks:Find the discharges Q;Maintain water levels

Hydraulic Structures and Flow Measurements

Type 1

Type 2

Applications of the structures:•Irrigation•Water supply•Drainage•Sewage treatment•Hydropower engineering

Hydraulic Structures(gates, weirs, Spillways, Culverts)

2

Reference Books

1. Hydraulics in Civil and Environmental Engineering, 2nd Ed, Andrew

Chadwick and John Morfett, E & FN SPON, 1994

2. The Hydraulics of Open Channel Flow, Hubert Chanson,

Eliane Wigzell, 1999.

3. Civil Engineering Hydraulics, 4th Ed, C. Nalluri and R. E. Feathrstone,

Blackwell Science, 2001.

3

1. Orifices and sluiceways

2. Flow under gates

3. Flow over weirs

• Sharp crested (thin plate) weirs

• Long-based (broad-crested) weirs

4. Spillways

5. Culverts

Contents

xhAcQ ×××= Constant

General Relation for Q:

Our main task:to find the values of c for various kinds of structures

4

1. Orifices and sluiceways

Fig. 1 Notations for the orifice problem

22

22

11

21

22z

p

g

vz

p

g

v ++=++γγ

(9.1)Q

2

h

1

Area A

Apply Bernoulli Eq from (1) to (2):

Since p1 = p2 = patm, Z1 -Z2 = h,

and V1 = 0

or (9.2)gVh 2/22= ghVV jet 22 ==(9.1)

5

Q can then be obtained as:

ghAAVQ jet 2== (9.3)

where A is the area of the orifice.

There are two reasons:

1) the size of the jet is smaller than the size of opening;

2) the viscous shear effect between the edge of the orifice and the water.

Experimental results showed that actual

Q is smaller than Q from Eq. (9.3).

Vena contracta

6

Fig. 2 Development of the vena contracta.

Fig. 3 Velocity distribution due to viscous shear loss.

Assumed V distribution

Actual V distribution

Vena contracta

Diameter dis minimum

A2A1

Typical value:A1/A2=0.64

7

A discharge coefficient, c, is introduced to Eq.(9.3) due to the combined effect of the above two factors:

(9.4)

1) A is the area of the opening

2) c is determined by experiments, ranging from 0.6 to 0.68for small, round, and square orifices discharging into air.

2) c is larger with smaller diameters and large heads.

3) when the jet exits into the downstream water, h will be the difference between the two levels.

4) For actual structures, when orifice cannot be considered infinitely thin but, rather, is a short tube with length of l, the discharge coefficients for short tubes of various materials and configurations are given in Fig. 4.

ghcAQ 2=

8

Fig. 4 Discharge coefficients for short tubes.

c

l/d

c: discharge coefficientl: length of the tubed: diameter of the tube

Different types of pipes

9

A practical application of this concept is for sluiceways under dams.

Fig. 5 A sluiceway under a concrete gravity dam.

Use: removal of water from the reservoir;

Materials: concrete or steel plates coated, or

steel pipe;

Locations: reservoir control considerations.

10

Example 1. A drain pipe located at the bottom of a concrete storage tank, 5m below the water level. The discharge through the pipe is Q = 0.5 m3/s.Determine the pipe diameter D.

Solution:

D=?

5m

0.5m

concrete storage

ghcAQ 2=Using Eq. (9.4),

4/ ,5 ,/5.0 23 DAmhsmQ π===

c depends on l/D, using trial and error:

Assume pipe is smooth with sharp entrance:

Try D =0.1m, l/D=5, from Fig.4 → c =0.825 → Q =0.06m3/s

Try D =0.25m, l/D=2 → c =0.85 → Q =0.413m3/s

Try D =0.275m, l/D=1.8 → c =0.851 → Q =0.5m3/s

11

2. Flow under gates

Fig. 6 Notations for flow under gates.

)(2

2

10 HHgbac

hgcAQ

−=

∆=(9.5)

A: area of the opening

b: width of the gate,

perpendicular to the paper

c: discharge coefficient;

a: gate opening.

H2

H0

a H1

gV 2/21

By using Bernoulli Eq and

introducing discharge coefficient c:

12

1) H2 does not influence the flow;

2) H1 = ??

3) a << H0

aH

HgbacQ

•+=

ψ0

02 (9.6)

where c and ψ are determined by experiments, as shown in Fig. (7).

2.1 Comments on Eq.(9.5):

Note: Eq. (9.6) is valid only if the downstream water level is not

influencing the flow.

Eq. (9.5), Q can be rewritten as:

Can be related with gate opening 1H aψ=2 2 20 1 0H H H− ≈

13

Fig. 7

Discharge coefficient

for flow under gates.

0

2

4

6

8

10

0.595 0.600 0.605 0.610 0.615

0.61 0.62 0.63 0.64 0.65 0.66 0.67

c

ψ

c

H0/a

ψ

14

Example 2A 1.8 m wide vertical gate on top of a spillway withholds a 1.2 m deep water. Determine the discharge under the gate if it is raised by 0.3 m.

15

How to judge?

Method (1): calculate the conjugate depth of H2 using the formula

with y2 = H2; y1 is the depth immediately in front of the jump.

Method (2): using Fig. 8.

e.g. Try H0/a = 10, H2/a =3;

and H0/a = 10, H2/a =6;

2.2 Check the influence of downstream water level on Eq(9.6)

[ ]1812

22

21 −+= Fr

yy (9.7)

(1) No influence: When a hydraulic jump exists, i.e., the conjugate depth

of H2 is equal or greater than H1.

When y1 ≥ H1, not influenced.

16Fig. 8 The range of downstream influence on flow under gate

a

H2

aH /0

17

(2) Influenced outflow – Two cases

(a) When Absolute downstream control. 210 HHH ≤−

Q is based on h = H0 -H2,

2Q cA gh= (9.9)

Where A: area of the opening

c: discharge coefficient;

18

kQQretarded= (9.8)

where Q is obtained from Eq. (9.6) and k can be found in Fig. 9

Fig. 9 Partial downstream control of flow under gates

e.g.

For H2/a=9 and

H0/a=15,

k = 0.68

(b) Partial downstream control:

19

Example 3. A 3 m wide vertical gate discharges into a pool in which the water level is 1.5 m. The upstream water level is 2.4 m and the gate opening is 0.3 m. Determine the discharge through the structure.

20

Definition: All barriers on the bottom of the channel that cause the flow toaccelerate to pass through can be considered as weirs.

Function: Flow measuring and flow control

Types: Sharp crested weirs

─Materials: plastic, metal plate,

─Shapes: full width or contracted rectangular, vee shape etc.

Broad crested weirs

─Materials: concrete, wood,

─Shapes: Ogee, flat top etc.

3 Flow over weirs

21

Fig.11 (b) Contracted Rectangular weir

Fig.11 (a) Full-width weir

22Fig 11 (c) Vee weir Fig. 11 (d) Broadcrested weir

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The crest: the bottom edge of the opening over which the water flows;

The crest height: the height of the crest over the bottom of the channel;

The nappe: the overfalling stream of water;

Broadcrested weirs: crest thickness is more than 0.6 nappethickness;

Sharp crested weirs: crest thickness less than 0.6 nappethickness;

What to study: relationship between water depth above the weir and Q;

Methods: Bernoulli equation plus experiments.

Some definitions

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Materials: plastic or metal plate of a suitable gauge;

Geometry: rectangular or triangular depending on the applications;

Installation: the plate is vertical and spans the full width of the channel,and the weir is incorporated into the top of the plate.

Rectangular weirs

Two types:1. ‘Uncontracted’ or full-width weirs;2. A ‘contracted’ weir (crest width is less than channel width).

Weir operation: based on the gauged depth to estimate the discharge.

Estimation of discharge: (1) experimental methods.(2) numerical methods.

An idealized relationship Q ~ h is developed using Bernoulli’s Eq and then

be modified to take account of the differences between ideal & real flows.

3.1 Sharp crested (thin plate) weirs - uncontracted

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3.1.1 The rectangular weir equation: un-contracted

Apply Bernoulli equation along streamline A-A.

At Station 1, the total energy :

g

uz

g

pH

2

21

11

1 ++=ρ

(9.10)

At Station 2 as the underside

of the overspilling jet is

exposed to the atmosphere,

the pressure is taken as Patm:

g

uzH

2

22

22 +=

26

Assuming no losses occur between Station 1 and Station 2,

21 HH =

g

uz

g

uy

22

22

2

21

1 +=+ withg

pzy

ρ1

11 +=

Therefore 2/121

212 )]2

(2[g

uzygu +−=

u2 is a function of y1 and z2, i.e. u2 varies with depth above weir crest.

At Station 2, the discharge through an elemental strip of width b and depth δδδδz is

zbg

uzygzbuQideal δδδ 2/1

21

212 )]2

(2[ +−==

(9.11)

Or: ∫ +−= bdzg

uzygQ ideal

2/121

21 )]2

(2[

27

∫∫ +−==1

1

0)]

2(2[0

2/121

212

hh

ideal zbg

uzygzbuQ δδ

−

+=2/32

1

2/321

1 222

32

g

u

g

uhgb

if u12/2g is negligible compared with h1, Eq (9.12) can be reduced to:

(9.12)

2/312

32

hgbQ ideal = (9.13)

For integration, the limiting values of z:

Lower limit: z= 0 if the datum is raised to the same level as weir crest.

Upper limit: z2 = h1 ( rather drastic assumption! Physically impossible)

─ This is the idealised discharge equation for rectangular weirs.

28

Reasons for modification to Eq. (9.13):

(1) Vena contracta phenomenon ⇒ actual Q should be less than Qideal.

(2) Pressure distribution in the flow passing over the weir is not atmospheric;

(3) Viscous effects: two effects ⇒ non-uniform U distribution and loss of energy between (1) and (2).

3.1.2 Modifications to un-contracted rectangular weir equation

Pressure distribution

Contraction or vena contracta

29

d idea lQ C Q=

The values of Cd:

•Should be determined experimentally

•It is not strictly a constant,

•It is a function of:

─Reynolds number,

─weir types,

─Ratio between height above the weir crest and the weir height.

(9.14)

In practical applications, we introduce an experimentally

determined coefficient Cd, i.e.

30

Other empirical relations which incorporate Cd for un-contracted weirs under free discharge conditions:

Valid for: 30 mm < h1 < 750 mm, b > 300 mm, Ps > 100 mm, and h1 < Ps.

3/ 2 3/1 1

22 (0.602 0.083 / )( 0.0012)

3s

sQ b g h P h m= + + (9.15)

smhPhgbQ s /)001.0)(/153.01(562.0 32/311 ++= (9.16)

1. Rehbock formula:

2. White’s formula:

Valid for: h1 > 20 mm, Ps> 150 mm, and h1 < 2.2Ps.

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1. The accuracy of Q depends, to a discernible extent, on the sitting of the gauging station for measuring the upstream head (h1). It is recommend that this should be at least 2.67Ps upstream of the weir, to avoid undue drawdown effects.

2. The British Standard (BS 3680) recommendation is that the station should be between 4h1 and 5h1 upstream of the weir.

Requirements of the measurement location for h1:

32

Example 4A 40cm wide channel is equipped with a full-width rectangular weir whosecrest is 20cm above the channel bed. What will be the discharge according to the Rehbock formula and White’s formula, if the water depth above the weir crest are 0.1m, 0.15m and 0.2m? Compare the results.

Solution: b = 0.4m, PS = 0.2m, h1 = 0.1m, 0.15m and 0.2m.

33

34

35

(a) The Hamilton-Smith formula

)/1(616.0 1 sd PhC −=

Valid for: B > (b+4h1), h1/B < 0.5, 75 mm < h1 < 600mm, Ps > 300 mm,

and b > 300 mm (B = channel width).

(b) The Kindsvater-Carter formulaThis formula uses concept of “effective” head and width, he and be, where

and

(9.17)

(9.18)

3.1.3 Sharp crested (thin plate) weirs - contracted

• For contracted rectangular weirs, Eq. (9.13) for Qideal is applicable.

• But Q will be affected by the vertical and lateral contraction of the jet.

Two formulae for estimating Cd are given below:

he khh += 1 be kbb +=

36

The discharge formula is

Where Ce depends on b/B and h1/Ps, i.e.

(9.19)

kh and kb are experimentally determined quantities which allow for the effects of viscosity and surface tension to be considered.

It has been found kh = constant = 0.001 m.

Ce must be determined empirically. British Standard (BS 3680) gives

charts for the determination of Ce and kb.

)/( 232 32/3 smhbgCQ eee=

{ }Se PhBbfC /,/ 1=

37

•All of the above equations assumed that downstream depth is too low to impede the free discharge over the weir. This is known as ‘modular flow’.

•In engineering applications, modular flow operations may not be possible

sometimes.

(9.20)

where Q is the free discharge which corresponds to the upstream depth h1,

and y3 is the downstream water depth.

3.1.4 ‘Submergence’ effects

( )[ ] 385.0 2/313 /1/ hyQQS −=

If an estimate of the discharge Qs of a submerged weir has to be made,

Villemonte’s formula may be used:

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As vee weirs have much better sensitivity than rectangular weirs at low discharge, they are normally used to measure small discharges.

Fig. 14 Vee weir

2

21 / 21

1 2[ 2 ( ) ]2

i d e a lQ u b z

ug y z b z

g

δ δ

δ

=

= − +

Theory and assumptions are the same as for the rectangular weir.

The discharge through an elementary strip across the weir is:

3.1.5 Vee weirs

h1

Ps

z2

b δz

θ

Methods of analysis:

22 t a n ( / 2 )b z θ=As ,

∫ +−=1

0)2/tan(2)]

2(2[ 2

2/12

121

h

ideal zzg

uzhgQ δθ

39

The approach velocity, u1, is always negligible for vee weirs in view of the small discharges for which they are designed. Therefore

∫ −=1

0)2/tan(2)](2[ 2

2/121

h

ideal zzzhgQ δθ

2/51)2/tan(2

158

hg θ=

•Cd is a function of Reynolds number, h1, the weir height and the angle θ.

•Cd must be determined experimentally.

•The BS 3680 gives magnitudes of Cd for a wide range of weirs.

•The 90o weir is probably the most widely used with a Cd of 0.59 may be

used as a first approximation for this angle.

ideald QCQ =

(9.21)

(9.22)

Values for Cd:

40

Example 5A series of tests have been conducted to a vee weir of 90o. Determine the discharge coefficient based on the following measured data.

25.124.924.925.525.625.025.425.1Q(L/s)

0.20.1980.1960.2040.210.1950.2050.201h1(m)

5 / 21

82 ta n ( / 2 )

1 5dQ C g hθ=

5 / 21

5 / 21

0 .4 2 3 3 /8

21 5

dQ

C Q hg h

= =

0.64 0.536 0.574 0.62 0.604 0.5940.5650.587Cd

Averaging: Cd = 0.59

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1) Large and generally more heavily constructed (concrete). 2) Usually designed for use in the field, and to handle large discharges. 3) An ideal long-based weir has the following characteristics:

• Cheaply and easily fabricated, and easily installed;• Possesses a wide modular range;• Produces a minimum afflux (i.e. increase in upstream depth due to

the installation of the weir); 0.869• Requires a minimum of maintenance.

3.2 Long-based (broad-crested) weirs

Long-based weir hydraulicsFrom Open Channel section, we know:•a local increase in the velocity of flow,•a reduction in elevation of water surface, •normally critical flow will be produced.

Long-based weirs:

42

Assuming no loss, total available energy on the upstream side:

In most hydraulic structures, P is significant and the kinetic energy can normally be neglected. For free fall, the discharge will occur under critical flow condition. The depth yc over the weir (Refer to Open Channel Section)

(9.23)

(9.24)

3.2.1 Discharge equations for long-based weirs

gVhE 2/211 +=

3/23/2 1hEyc ≈=

h1

V12/2g

P

yc

Figure 15

datum

long-based weirs

2

1

43

21 / 2c ch y v g= +

1 / 21 12 ( ) 2 / 3c cv g h y g h= − =

2/31

2/31 2384.0

32

32

bhghg

bbyvQ ccideal ===

Introducing a discharge coefficient c:

2/312 bhgcQ = (9.25)

3 / 21Q M b h= (9.26)

(9.25a)

Eq (9.25) may be solved in metric terms by combining c with (2g)1/2 ⇒

where h1 is water depth and M is metric weir coefficient (Table 3.1).

Note that various relations for c can be obtained from the references books.

Velocity vc can be obtained using Bernoulli Eq between and :1 2

44

Table 3.1 Collection of Metric Weir Coefficients for various Long-Based Weirs determined by laboratory experiments.

45

Table 3.1 Collection of Metric Weir Coefficients for various Long-Based Weirs determined by laboratory experiments (continued)

46

Example 6.

A weir of 10m wide with a cross-section similar to the case No. 4 of Table

3.1. The upstream water head is 45 cm. Determine the discharge.

3 / 2 3 / 2 31 1.77 10 0.45 4.5 /Q M bh m s= = × × =

10m,b and ,45.0With 1 == mh

From Table 3.1, we have M = 1.77

Solution:

47

3.2.2 Influence of downstream water levelWhen downstream water level exceeds a certain value, it may influence the discharge over the weir. There are two conditions:

(1) ,2/12 hPH +< a hydraulic jump will form over a weir.

•For free hydraulic jump → no downstream influence to Q;

•For submerged hydraulic jump → Q is influenced by downstream

h2

48

(2) 2 1 / 2,H P h> + upstream water level is elevated. Q ∝ (h1−h2)

h1 in Eq. (9.25) or (9.26) should be replaced by the water level difference h1−h2 shown in the figure.

In both cases, the discharge coef will be reduced. The actual ranges of various flow patterns over the weirs depends on the shape of the weir. For one typical weir shape, the ranges of free overfall, free orsubmerged hydraulic jump, or subcritical overflow are shown next.

49Figure 16. Ranges of flow type over weirs

50

Example 7

A broadcrested weir is designed to have a crest height of 5m. The

maximum upstream head over the crest should not exceed 2.185m.

The height of the weir is to be selected such that the downstream water

level does not influence the upstream conditions and therefore does

not retard the flow during critical floods. What is the maximum allowable downstream water level under these conditions?

Solution: P = 5m, H1 = 2.185m → H1/P = 2.185/5=0.437

From Figure 16, to avoid influence from downstream, corresponding

to H1/P = 0.437, we should choose H2/P = 0.8

→ H2max= 0.8P = 5m