Open Channel Hydraulics

1

Hydraulics

Dr. Mohsin Siddique

Assistant Professor

Steady Flow in Open Channels

� Specific Energy and Critical Depth

� Surface Profiles and Backwater Curves in Channels of Uniform sections

� Flow over Humps and through Constrictions

� Hydraulics jump and its practical applications.

� Broad Crested Weirs and Venturi Flumes

Flow over Humps and through

Constrictions

Flow Over Hump

� Hump:

is a streamline construction provided at the bed of the channel.

It is locally raised bed.

Let’s examine the case of hump in a rectangular channel. We will neglect the head loss.

Flow Over Hump

� For frictionless two-dimensional flow, sections 1 and 2 in Fig are related by continuity and energy:

� Eliminating V2 between these two gives a cubic polynomial equation for the water depth y2 over the hump.

1 1 2 2

2 2

1 21 2

2 2

v y v y

v vy y Z

g g

=

+ = + +

2 23 2 1 12 2 2

2

12 1

02

2

v yy E y

g

vwhere E y Z

g

− + =

= + −

y2y1

y3Z

V1V2

1 2 3

This equation has one negative andtwo positive solutions if Z is nottoo large.It’s behavior is illustrated by E~yDiagram and depends uponwhether condition 1 is Subcritical(on the upper) or Supercritical(lower leg) of the energy curve.

B1=B2

Flow Over Hump

� The specific energy E2 is exactly Zless than the approach energy E1, andpoint 2 will lie on the same leg of thecurve as E1.

� A sub-critical approach, Fr1 <1, willcause the water level to decrease atthe bump. Supercritical approachflow, Fr1>1, causes a water-levelincrease over the bump.

� If the hump height reachesZmax=Zc=E1-Ec, as illustrated in fig, theflow at the crest will be exactlycritical (Fr =1).

� If Z = Zmax, there are no physicallycorrect solutions to Eqn. i.e., a humptoo large will “choke” the channeland cause frictional effects, typically ahydraulic jump.

1 2Super-Critical Approach

These hump arguments are reversed if the channel has a depression (Z<0): Subcriticalapproach flow will cause a water-level rise and supercritical flow a fall in depth. Point 2 willbe |Z| to the right of point 1, and critical flow cannot occur.

Z

Zmax

Flow Over Hump

Damming Action

y1=yo, y2>yc, y3=yoy1=yo, y2>yc, y3=yo

y1=yo, y2=yc, y3=yo y1>yo, y2=yc, y3<yo

ycy1

y3

Z

Z=Zc

y1

Z<<Zc

y2 y3

Z

Z<Zc

y2y1

y3

Z

Z>Zc

Afflux=y1-yo

y3

ycyo

Z

y1

Flow Over Hump

� As it is explained with the help of E~y Diagram, a hump of any height “Z”would cause the lowering of the water surface over the hump in case ofsubcritical flow in channel. It is also clear that a gradual increase in theheight of hump “Z” would cause a gradual reduction in y2 value. That heightof hump which is just causing the flow depth over hump equal to yc is knowas critical height of hump Zc.

� Further increase in Z (>Zc) would cause the flow depth y2 remaining equalyc thus causing the water surface over the hump to rise. This would furthercause an increase in the depth of water upstream of the hump which meanthat water surface upstream of the hump would rise beyond the previousvalue i.e y1>yo. This phenomenon of rise in water surface upstream withZ>Zc is called damming action and the resulting increase in depth upstreamof the hump i.e y1-yo is known as Afflux.

Flow Through Contraction

� When the width of the channel is reduced while the bed remains flat, the dischargeper unit width increases. If losses are negligible, the specific energy remains constantand so for subcritical flow depth will decrease while for supercritical flow depth willincrease in as the channel narrows.

B1 B2

y2yc

y1( )

1 1 1 2 2 2

2 2

1 21 2

1 2

2 2 2 2 2 2

2 2

1 1

'

2 2

Using both equations, we get

2Q=B y v =B y

1

Continuity Equation

B y v B y v

Bernoulli s Equation

v vy y

g g

g y y

B y

B y

=

+ = +

−

−

Flow Through Contraction

� If the degree of contraction and the flow conditions are such that upstream flow is subcritical and free surface passes through the critical depth yc in the throat.

ycyc

y1

( )

3/ 2

2

2sin

3

2 12

3 3

1.705

c c c c c c

c

c

Q B y v B y g E y

ce y E

Therefore Q B E g E

Q BE in SI Units

= = −

=

=

=

B1 Bc

y2yc

y1

Example # 11.3

� In the accompanying figure, uniform flowof water occurs at 0.75 m3/s in a 1.2mwide rectangular flume at a depth of0.6m.

� (a). Is the flow sub-critical or super-critical.

� (b). If a hump height of Z=0.1 m is placedin the bottom of flume, calculate thewater depth over the hump. Neglect thehead loss in flow over the hump.

� (c). If the hump height is raised toZ=0.2m, what then are the water depthsupstream and downstream of hump.Neglect head loss over hump.

y1

y2 y3

Z

E.L

H.G.LV12/2g V2

2/2g

Q = 0.75 m3/secB = 1.2 my1 = 0.6 mq = Q/B = 0.625 m3/sec/m

Example # 11.3Solution

� (a)

� (b) First calculate Zc

� Z<Zc therefore y2> yc

� (C) Z>Zc therefore y2= yc

2 2

330.625

9.81

0.341

c

qy

g

m y

Flow is subcritical

= =

= <

∴

2 2

1 22 2

1 2

2 2

2 12 2

2 1

2

Applying Bernoulli's Equation

2 2

2 2

0.46

q qy Z y

gy gy

q qy y Z

gy gy

y m

+ = + +

+ = + −

=

2 2

1 2 2

1

1

2 2

3 2 2

3

3

Applying Bernoulli's Equation

2 2

0.665

2 2

0.2

c

c

c

c

q qy Z y

gy gy

y m

q qy Z y

gy gy

y m

+ = + +

=

+ = + +

=

( ) ( )

cmZ

Z

gZ

g

gy

qyZ

gy

qy

c

c

c

ccc

2.14

512.0655.0

341.02

625.0341.0

6.02

625.06.0

22

2

2

2

2

2

2

2

2

=

+=

++=+

++=+

(c).

Problem 11.54

� A rectangular channel 1.2 m widecarries 1.1 m3/sec of water inuniform flow at a depth of 0.85m.If a bridge pier 0.3m wide is placedin the middle of this channel, findthe local change in water surfaceelevation. What is the minimumwidth of the constricted channelwhich will not cause a rise inwater surface upstream.

� Given that

Q = 1.1 m3/sec

B1 = 1.2 m

q1 = 0.92 m3/sec/m

yo = 0.85 m

B2 = B1-0.3 = 0.9 m

B10.3m

2 2

1 22

2

'

2 2o

o

c c c

Bernouli s Equation

q qy y

gy gy

Energy Equation

Q B y v Byv

+ = +

= =

B2=B1-0.3

Problem 11.54

2

1sin 0.912

20.606

3

2.473 / sec

1.1

0.744

o

o

c

c c

c c c

c

qce E y m

gy

y E m

V gy m

Therefore

Q B y v

B m

= + =

= =

= =

= =

=

2 2

1 22

2

2

22

2

2

2 2

0.912

.38 & 0.785

o

o

q qy y

gy gy

qy

gy

y o m m

+ = +

+ =

=

Broad Crested Weirs and

Venturi Flumes

Broad Crested Weirs and Venturi Flumes

� Flow Measurement in Open Channels

� Temporary Devices� Floats

� PitotTube

� Current meter

� Salt Velocity Method

� Radio Active Tracers

� Permanent Devices� Sharp Crested Weir/Notch

� Broad Crested Weir

� Venture Flume

� Ordinary Flume

� Critical Depth Flume

Broad Crested Weirs and Venturi

Flumes are extensively used for

discharge measurement in open

channel.

Broad Crested Weirs and Critical

flumes are based and worked on

the principle of occurrence of

critical depth.

Broad Crested Weir

� A weir, of which the ordinary dam isan example, is a channel obstructionover which the flow must deflect.

� For simple geometries the channeldischarge Q correlates with gravityand with the blockage height H towhich the upstream flow is backed upabove the weir elevation.

� Thus a weir is a simple but effectiveopen-channel flow-meter.

� Figure shows two common weirs,sharp-crested and broad-crested,assumed. In both cases the flowupstream is subcritical, accelerates tocritical near the top of the weir, andspills over into a supercritical nappe.For both weirs the discharge q perunit width is proportional to g1/2H3/2but with somewhat differentcoefficients Cd.

Broad Crested Weir

1

act d

3/ 22

d

3/ 22

d

Velocity of approach =Q/By

H= Head over the crest

B= Width of Channel

Since Q =C

V1.7C

2g

V3.09C

2g

act

act

V

Q

Q B H in SI

Q B H in FPS

=

∴ = +

= +

L

22

2

2 22

2

1

2 3

32

3/ 22

Applying Energy Equation ignoring h

VH+Z+

2g 2

For Critical flow 2 2

2VH+

2g 2 2

V2

3 2g

:

V2

3 2g

V1.7

2g

cc

c c

c c

c

c cc c c

VZ y

g

V y

g

V V

g g

V g H

V BVSince Q By V B V

g g

BQ g H

g

Q B H i

= + +

=

∴ = +

= +

= = =

∴ = +

= +

3/ 2

2V3.09

2g

n SI

Q B H in FPS

= +

Z>Zc

Vcy1

Broad Crested Weir

Coefficient of Discharge, Cd also called Weir Discharge Coefficient Cw

� Cw depends upon Weber numberW, Reynolds number R and weirgeometry (Z/H, L, surface roughness,sharpness of edges etc). It has beenfound that Z/H is the mostimportant.

� The Weber number W, whichaccounts for surface tension, isimportant only at low heads.

� In the flow of water over weirsthe Reynolds number, R isgenerally high, so viscous effectsare generally insignificant. ForBroad crested weirs Cw dependson length for. Further, it isconsiderably sensitive to surfaceroughness of the crest.

Z>Zc

Vc

Venturi Flume

Ordinary Flume

� An ordinary flume is the one in which a stream line contraction of width is providedso that the water level at the throat is drawn down but the critical depth doesn’toccur.

B1 B2

y2yc

y1

1 1 1 2 2 2

2 2

1 21 2

2 2 2 2 2 2

2 2

1 1

'

2 2

Using both equations, we get

2Q=B y v =B y

1

Continuity Equation

B y v B y v

Bernoulli s Equation

v vy y

g g

gH

B y

B y

=

+ = +

−

H = y2-y1

Venturi Flume

Critical Depth Flume (Standing Wave Flume)

� A critical depth flume is the one in which either the width is contracted to suchan extent that critical depth occurs at the throat or more common both ahump/weir in bed & side contractions are provided to attain critical depth withhydraulic jump occurrence at d/s of throat.

B1 B2

ycy1

Z

Hvc

1 1 1 2 2 2

22

1

2 c c

'

2 2

Using both equations, we get

Q=B y v

cc

Continuity Equation

Q B y v B y v

Bernoulli s Equation

vvZ H Z y

g g

= =

+ + = + +

V1

Problem: 12.66

� A broad crested weir rises 0.3m above the bottom of channel. With ameasured head of 0.6m above the crest, what is rate of discharge per unitwidth? Allow for velocity of approach.

1

3/ 22

3/ 22

3

0.3

0.6

???

As we know that;

1.72

1.72

1; using Trial and Error

= q =0.505 / sec/

act d

act d

act

Z m

H m

y Z H

q

VQ C B H

g

QQ C B H

By g

Since B

Q m m

=

=

= +

=

= +

= +

=

Take Cd=0.62

Problem: 12.67

� A broad crested weir of height 0.6m in a channel 1.5m wide has a flow over itof 0.27m3/sec.What is water depth just upstream of weir?

1

3

3/ 22

1

3/ 22

1

1

1

0.6

0.6

1.5

0.27 / sec

0.62

As we know that;

1.72

0.270.27 1.7 0.62 1.5 0.62

1.5 2

Solving above equations reults

y 0.905

act d

Z m

H y

B m

Q m

Cd

QQ C B H

By g

x x yy g

m

=

= −

=

=

=

= +

= − +

=

Hydraulic jump and its practical

applications.

Hydraulic jump

Hydraulic jump formed on a spillway modelfor the Karna-fuli Dam in Bangladesh.

Rapid flow and hydraulic jump on a dam

Hydraulics Jump or Standing Wave

� Hydraulics jump is local non-uniform flow phenomenon resulting from thechange in flow from super critical to sub critical. In such as case, the waterlevel passes through the critical depth and according to the theorydy/dx=infinity or water surface profile should be vertical. This off coursephysically cannot happen and the result is discontinuity in the surfacecharacterized by a steep upward slope of the profile accompanied by lot ofturbulence and eddies. The eddies cause energy loss and depth after thejump is slightly less than the corresponding alternate depth. The depthbefore and after the hydraulic jump are known as conjugate depths orsequent depths.

y

y1

y2

y1

y2

y1 & y2 are called conjugate depths

Classification of Hydraulic jump

Classification of hydraulic jumps: (a) Fr =1.0 to 1.7: undular jumps; (b) Fr =1.7 to 2.5: weak jump; (c) Fr =2.5 to 4.5: oscillating jump; (d) Fr =4.5 to 9.0: steady jump; (e) Fr =9.0: strong jump.

Classification of Hydraulic jump

� Fr1 <1.0: Jump impossible, violates second law of thermodynamics.

� Fr1=1.0 to 1.7: Standing-wave, or undular, jump about 4y2 long; lowdissipation, less than 5 percent.

� Fr1=1.7 to 2.5: Smooth surface rise with small rollers, known as a weakjump; dissipation 5 to 15 percent.

� Fr1=2.5 to 4.5: Unstable, oscillating jump; each irregular pulsation creates alarge wave which can travel downstream for miles, damaging earth banksand other structures. Not recommended for design conditions. Dissipation15 to 45 percent.

� Fr1=4.5 to 9.0: Stable, well-balanced, steady jump; best performance andaction, insensitive to downstream conditions. Best design range. Dissipation45 to 70 percent.

� Fr1>9.0: Rough, somewhat intermittent strong jump, but good performance.Dissipation 70 to 85 percent.

Uses of Hydraulic Jump

� Hydraulic jump is used to dissipate or destroy the energyof water where it is not needed otherwise it may causedamage to hydraulic structures.

� It may be used for mixing of certain chemicals like in caseof water treatment plants.

� It may also be used as a discharge measuring device.

Equation for Conjugate Depths

1

2

F1 F2y2y1

So~0

1 2 2 1

1

2

( )

Resistance

g f

f

Momentum Equation

F F F F Q V V

Where

F Force helping flow

F Force resisting flow

F Frictional

Fg Gravitational component of flow

ρ− + − = −

=

=

=

=

Assumptions:

1. If length is very small frictional resistance may be neglected. i.e (Ff=0)

2. Assume So=0; Fg=0

Note: Momentum equation may be stated as sum of all external forces is equal to rate of change of momentum.

L

Equation for Conjugate Depths

Let the height of jump = y2-y1

Length of hydraulic jump = Lj

2 1

1 1 2 2 2 1

1 1 1 2 2 2

1 2 ( )

( )

Depth to centriod as measured

from upper WS

.1

Eq. 1 stated that the momentum flow rate

plus hydrostatic force is the same at both

c c

c

c c

F F Q V Vg

h A h A Q V Vg

h

QV h A QV h A eqg g

γ

γγ γ

γ γγ γ

− = −

− = −

=

+ = + ⇒

2 2

1 1 2 2

1 2

sections 1 and 2.

Dividing Equation 1 by and

changing V to Q/A

.2c c m

Q QA h A h F eq

A g A g

γ

+ = + = ⇒

2

;

Specific Force=

: Specific force remains same at section

at start of hydraulic jump and at end of hydraulic

jump which means at two conjugate depths the

specific force is constant.

m

Where

QF Ahc

Ag

Note

= +

( )

( )( )

2 2 2 2

1 21 2

1 2

2 22 2

1 2

1 2

22 2

2 1

1 2

2

2 12 1 2 1

1 2

Now lets consider a rectangular channel

2 2

.32 2

1 1 1

2

1

2

y yq B q BBy By

By g By g

y yq qeq

y g y g

qy y

g y y

or

y yqy y y y

g y y

∴ + = +

+ = + ⇒

− = −

−= − +

Equation for Conjugate Depths

2

2 11 2

1 1 2 2

2 2

1 1 2 11 2

3

1

22

1 2 2

1 1 1

2

22 21

1 1

2

1

.42

Eq. 4 shows that hydraulic jumps can

be used as discharge measuring device.

Since

2

2

0 2 N

y yqy y eq

g

q V y V y

V y y yy y

g

by y

V y y

gy y y

y yF

y y

y

y

+ = ⇒

= =

+ ∴ =

÷

= +

= + −

( )

2

1

212 1

1 1 4(1)(2)

2(1)

1 1 82

N

N

F

yy F

− ± +=

= − ± +

( )

( )

212 1

221 2

Practically -Ve depth is not possible

1 1 8 .52

1 1 8 .52

N

N

yy F eq

Similarly

yy F eq a

∴ = − + + ⇒

= − + + ⇒

Location of Hydraulic Jumps

� Change of Slope from Steep to Mild

Hydraulic Jump may take place

1. D/S of the Break point in slope y1>yo1

2. The Break in point y1=yo1

3. The U/S of the break in slope y1<yo1

So1>ScSo2<Sc

yo1y2

yc

Hydraulic Jump

M3y1

Location of Hydraulic Jumps

� Flow Under a Sluice Gate

So<Sc

yo yc ysy1 y2=yo

L Lj

Location of hydraulic jump where it starts isL=(Es-E1)/(S-So)

Condition for Hydraulic Jump to occurys<y1<yc<y2

Flow becomes uniform at a distance L+Lj from sluice gate whereLength of Hydraulic jump = Lj = 5y2 or 7(y2-y1)

Problem

35

Problem 11. 87

� A hydraulic Jump occurs in a triangular flume having side slopes 1:1. The flow rate is 0.45 m3/sec and depth before jump is 0.3m. Find the depth after the jump and power loss in jump?

� Solution

Q= 0.45 m3/sec

y1=0.3m

y2=?

2 2

1 1 2 2

1 2

2

1 2

/ 3

0.858

0.679

2.997

Q QA hc A hc

A g A g

hc y

y m

E E E

E

Power Loss Q E

Power Loss Kwatt

γ

+ = +

=

=

∆ = −

∆ =

= ∆

=

1:1

T=2y

hc=y/3

Problem 11. 89

� A very wide rectangular channel with bed slope = 0.0003 and roughness n =0.020 carries a steady flow of 5 m3/s/m. If a sluice gates is so adjusted as toproduce a minimum depth of 0.45m in the channel, determine whether a hydraulicjump will form downstream, and if so, find (using one reach) the distance from thegate to the jump.

� Solution

So<Sc

yo ycys=0.45m y1 y2=yo

L Lj

Problem 11. 89

1/32

2/ 3 1/ 2

2

1 2

2 2

4/ 3

1

1.366 0.45

( )

c

o

o

m

m

s

o

qy Super Critical Flow

g

AQ R S

n

y y

y f y

V nS

R

E EL

S S

= = > ⇒

=

≈

=

=

−=

−

Thank you

� Questions….

� Feel free to contact:

39