Non Uniform Flow

8/6/2019 Non Uniform Flow

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Part II: Non-uniform Flow

Changing conditions takesplace over a long distance

Gradually varied flowRapidly varied flow

Changing of conditionstakes place over a short

distance

Reference books1. Hydraulics in Civil and Environmental Engineering, Andrew Chadwick and John Morfett,

2nd Ed, E & FN Spon, 1995.

2. Introduction to Fluid Mechanics, Robert W. Fox, Alan T. McDonald, P. J. Pritchard, 6th

Ed, John Willey & Sons, 2004.


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Occurs when there is a sudden change in geometry,

Flow over a sharp-crested weirs,

Flow through regions of rapidly varied cross-section,e.g. venturi flumes and broad-crested weirs.

A sudden change in flow regime

Normally associated with hydraulic jump phenomena, i.e.

flow with high speed and low depth is rapidly changed to

low speed and high depth.

1. Rapidly varied flow:

length)sticcharacteri:(/ LgLVFr= (6.1)

In open channel flow, flow regime is defined by Froude number:


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1.1 Characteristics of rapidly varied flows:

The use of energy and momentum principles

Main task: to determine of water depth

Bernoulli equation and continuity equation

Specific energy E

Specific discharge q

Alternate depths

Water surface profile changes suddenly and has curvature,

Pressure distribution departs from hydrostatic distribution,

The assumption of parallel streamline does not apply.

1.2 Methods of analysis for rapidly varied flows:


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1.3 Energy considerations

Example: In a rectangular channel, fluid flows over a section inwhich the bed gradually rises by z. The upstream depth y1 and

discharge Q are known. What is the depth of the flow at position 2?

Assuming no energy loss between (1) and (2), apply the energy Eq:

y

V

g y

V

g z11

2

2

2

2

2 2+ = + + (6.2)

datum

(1) (2)

Figure 6.1

p2/=0p1/=0


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Apply the continuity equation between points (1) and (2):

q V y V y= =1 1 2 2 (6.3)

where q (m3/s/m) is the specific discharge (discharge per unit width).

yq

gyy

q

gyz

1

2

12 2

2

222 2

+ = + +

2 2 2 023

2

2

1

2

1

2

2gy y g z gy

q

y

q+ + =( )

(6.4)

3 solutions for y2. Which one(s) is(are) correct?

Substitute (6.3) into (6.2):

Cubic equation


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To answer the above question, we use the concept of specific energy

H z yV

gz y

Q

gAz E= + + = + + = +

2 2

22 2

E yV

gy

Q

gA= + = +

2 2

22 2

(6.5)

The energy at a location is given by:

Channel bedwhere

E: specific energy=energy between energy line (EL) and channel bed.

For a rectangular channel: Q/A = bq/by = q/y

y

z

Water surface

datum

gV 2/

2

EL

E yq

gy= +

2

22

(6.6)

Eq. (6.6) gives two asymptotes:

y E 0 y E y

(Please note the definition of E: E = local water depth + velocity head)


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Using Eq (6.6), Eq. (6.4) can be written as E E z1 2= + (6.7)

Point A is defined by E1 corresponds to location 1,

Location 2 in the channel have E2 and is on the specific energy curve.

Possible B1 and B2 whose depths are known as alternate depths To arrive at B2 from A requires that E2 < E1- z at some intermediate

points. This means energy loss in a frictionless system impossible.

Therefore the flow depth at (2) must be B1

on the specific energy curve.

alternate depths

Figure 6.2Eq. (6.6)


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Solution:

Q = 8 m3/s; b = 5 m;

q = Q/b = 8/5 = 1.6 m2/s

Example 6.1:

The discharge in a rectangular channel of width 5m is 8 m3/s. The

normal depth is 1.25 m. Determine the depth of flow above a sectionin which the bed gradually rises by 0.2 m. Use graphical methods to

find the solution.

zEE += 21

222 /13.0)2/( yygyqyE +=+=Specific Energy

00.20.40.60.8

11.21.41.61.8

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

E (m)

y

(m)myyE 33.1/13.0

2

111 =+=

From energy Eq.,

mzEE 13.112 ==

E1=1.33

z

y2

E2=1.13

From the figure we get: y2=1m


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Example 6.2

Discharge in a rectangular channel of width 5m is 8 m3/s. The normal

depth is 1.25 m. Determine the flow depth where the section contractsby 1.0 m. Use graphical and numerical methods to find the solution.

smAQV /28.1)25.15/(8/11

===

mgVyE 33.12/2111 =+=

2222 /2)4/(8/ yyAQV ===

222

2222 /204.02/ yygVyE +=+=

21 EE =From energy Eq. 0

0.5

1.0

1.5

2.0

0 0.5 1.0 1.5 2.0

y2=1.18m

E1=1.33m=E

2

E (m)

y(m)

we get y2=1.18m


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Example 6.3


normal depth is 1.25 m. Determine the depth of flow where the

section contracts by 1.0 m and the bed gradually rises by 0.2 m.

Use graphical methods to find the solution.

Plan view

smQV /28.1)25.15/(8/ 11 === mgVyE 33.12/2

111 =+=

2222 /2)4/(8/ yyAQV ===222

2222 /204.02/ yygVyE +=+=

or 1221 zEEzEE =+=From energy equation, we have:

(1)

Plot the specific energy curve for E2,

the water depth corresponding

to E2 is y2.

z =

y2y1

Elevation


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0

0.5

1.0

1.5

2.0

0 0.5 1.0 1.5 2.0

Eq (1)

E2

z=0.2

y2=0.83 m

E1=1.33m

E (m)

y(m)


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1.4 Subcritical, critical and supercritical flow

E y

q

gy= +

2

22 (6.6)

Eq (6.6) is plotted in (a) and Eq(6.8) is plotted in (b).

The specific volume flow rate q can be expressed as a function of y,From Eq. (6.6),

)(2 2 yEgyq = (6.8)

(q is constant)

(E is constant)

subcritical

supercritical

(a) (b)

Figure 6.3Eq(6.6)

Eq(6.8)Critical point

12


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Derivation of the critical depth for a rectangular channel

The critical depth yc can be evaluated from Eq(6.6) by setting thederivative of E with respect to y equal to zero:

01/3

2

==gy

qdydE

2

2

2gy

qyE +=

0Fr112

2

== gy

V

From , to let E have an extremum value,

Since q=Vy

For a rectangular channel, we have 3gy

q

gy

VFr ==

1Fr =

Thus, for critical depth, Fr = 1.

(6.9)

Solving for depth in Eq (6.9) we have: ( ) cygqy ==3/12/ (6.11);

and 2/3 cc yE = (6.12)

(6.10)

c cV gy= (6.11a)


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A generalized derivation of the critical depth

For a generalized cross-section, the specific energy can be writtenin terms of the total discharge Q and the cross-sectional area A:

2

22

22 gA

Q

yg

V

yE +=+=

013

2

==dy

dA

gA

Q

dy

dE

(6.13)

The minimum-energy is obtained by differentiating (6.13) with

respect to y, i.e.

(6.14)

For incremental changes in depth. The change in area is

BdydA = (6.15)


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Substituting (6.15) to (6.14), we get:

2 3

1 / 0Q B gA = (6.16)

The second term in Eq(6.16) is the Froude number squared,

2 3 /Fr // /

Q A VQ B gA

gA B gA B= = = (6.18)

The ratio A/B is the hydraulic depth (B is the surface width).

For a rectangular channel, A/B=y, Eq (6.18) is simplified to (6.10).

Therefore, at the critical depth, we have:

2 3

max / 1c cQ B gA =

At the critical depth, Q is maximum and E is minimum.

(6.17)


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1. For a given constant discharge (Figure 6.3a)

The specific energy curve has a minimum value Emin at yc,

For any other E, there are two possible depths called

alternate depths, one is called subcritical and the other is

supercritical.

For supercritical flow, y < yc(Fr > 1)

For subcritical flow, y > yc (Fr < 1)

2. For a given constant specific energy (Figure 6.3b)

The discharge is maximum at depth yc;

For all other discharges, there are two possible depths for

any particular value of E.

1.5 Some more discussion to Figure (6.3):


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Summary of critical, subcritical and supercritical flows

Critical Flow (requires )

requires Eminimum or qmaximum;

For rectangular channels:

23 / ,c y q g= min 3 / 2,cE y=

Subcritical Flow:

Fr < 1,

Water velocity < wave velocity

Disturbances travel upstream and downstream

Upstream water levels are affected by downstream control

c cV gy=

1/ == gyVFr

V: water velocity in channel

gy : celerity of long wave when there is a disturbance in channel


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Slope classification: mild, steep and critical slopesA mild slope: on which uniform flow is subcritical:

A steep slop: on which uniform flow is supercritical:

A critical slope: on which uniform flow is critical:

where y0

is the uniform depth.

0 0 c(or Fr1 or S >S )cy y 1,Water velocity < Wave velocity

Disturbances travel downstream

Upstream water levels are unaffected by downstream control


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Example 6.4


normal depth is 1.25 m. Determine the depth of flow where the

section contracts by 1.0 m and the bed gradually rises by 0.3 m.

Use graphical methods to find the solution.

z =

y2y1

z=0.3m

..

6.15/8/ 11 === bQq

curve)(plot33.1/1305.0)2/( 2112

1

2

111 myygyqyE =+=+=

2/ 22 == bQq curve)(plot/204.0)2/(2

22

2

2

2

222 yygyqyE +=+=

1.03m0.3-1.33or 1221 ===+= zEEzEE

From energy equation, we have:(1)


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Corresponding to E2=1.03m, there are no crossing points on the

specific energy curve. Therefore, there are no solutions. This result

suggests that the depth y2 should be the critical depth.

From the specific energy curve, we get:

ycritical = 0.741m and E2=Emin = 1.112m. (refer to Figure a)

From E1=E2+z E1new=1.412m (refer to Figure b)

From the specific energy curve at location (1), we get the new water

depth at location (1):

y1new=1.34m

Therefore, at the contraction, the water depth is y2 = 0.74m

And the original water depth is increased to y1new = 1.34m.


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0

0.5

1.0

1.5

2.0

0 0.5 1.0 1.5 2.0

(a)E

2=1.03

No crossing pointswith the energy curve

No solution.Therefore, y

2should

be at critical depth

z=0.3

E1=1.33m

E (m)

y(m)

E2~y2

E1~y1


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0

0.5

1.0

1.5

2.0

0 0.5 1.0 1.5 2.0

(b)

E1=1.412m

E2=1.03m

E1=1.33m

y1=1.2m (original depth)

y1=1.34m (new depth)

q1=1.6m

2/s

at location (1)

q2=2m

2/s

at location (2)

z=0.3

Move by hMove by h

Emin

=1.112m

Emin

=1.112m

z=0.3

ycritical

=0.735 m

E (m)

y(m)


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Example 6.5

The discharge in a trapezoidal channel is 30 m3/s. The channel has a

base width of 4 m and side slopes of 1:2. Determine the critical

water depth corresponding to the flow rate in the channel, n=0.022.

1

2

B

yc

4m

Solution:

(1) At critical water depth,

13

2

== gA

BQFr

cccc yyyBysmQ )24(2/)4(A,44B,/30With3

+=+=+==

1)24(81.9

)44(3033

2

=+

+

cc

c

yy

ymyc 4.1=


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(2) Critical slope:

2/3 1/ 2h c

AQ R S

n=Using Eq. (5.6),

At critical flow, y = yc

2/13/230 chc

c SRn

A=

(1)

(2)

Using n = 0.022, ,)24( ccc yyA +=c

cchc

y

yyR

524

)24(

+

+=

Substituting into (2) with yc = 1.4m

.......=cS


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Example 6.6


normal depth is 1.25 m. Determine the depth of flow immediately

prior to, above and after a section in which the bed gradually rises

by 0.5 m as shown.

Solution:

We define: y0 = original depthy1 = immediately prior to the hump

y2 = depth after the hump

y3 = depth above the hump

y3y0=y1

y2

(3)

z=


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Write the energy Eq between (0) and (3):

zEE += 30mzmysmBQq 5.0,25.1,/6.15/8/For 0

2=====

2 2

0 0 0/(2 ) 1.33 E y q gy m= + =

(1)

(2)mE 83.05.033.13 ==

From the figure, we can see that for hump higher that 0.37m, therewould be no crossing points (no solution) between E3 and specific

energy curve, i.e. the fluid does not have enough energy to pass over

a hump higher than 0.37m.

To pass over a hump higher than 0.37m, water dams up (known as

damming action or backwater) before the hump until it acquires

sufficient energy to pass over, after which the hump acts as a control.

The minimum energy required is the Emin (or Ec) at the critical point.

(3)curve)(plot/1305.0)2/(Plot222

yygyqyE +=+=


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The depth of water at a control for a known flow rate is given by (6.11):

( )1/ 3 32 2 3/ 1.6 / 9.81 0.64c y q g m y= = = =The minimum energy for the flow to pass the hump is:

myE c 96.02/64.032/3min ===

(4)

(5)

Since , the flow prior to the hump is subcritical.cyy >0

The E1 prior to the hump is obtained by usingmzEE 46.15.096.0min1 =+=+=

From the figure, corresponding to E1=1.46m, we have y1 = 1.39m

or 0.34m. However, because the flow prior to the hump is subcritical,It can only increase its specific energy by increasing its depth since

there is no control point prior to the hump. Thus y1 = 1.39m.

(Note: y1 can also be obtained using )46.1)2/( 21211 =+= gyqyE

(6)


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Assuming no energy loss from (1) to (2), i.e. E1 = E2,

2

2 222

1.46 2

qy Egy= + =

From the figure, we get y2 = 1.39m or 0.34m.

Since a control usually causes the flow to change from subcritical to

supercritical or vice versa, thus the depth y2 = 0.34m is expected

downstream of the hump as shown in the figure.

(7)

Damming action


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0

0.5

1.0

1.5

2.0

0 0.5 1.0 1.5 2.0

0.34m

y1=1.39m

E1=1.46m = E2

z=0.5m

q=1.6m2/s

y0=1.25m

yc=0.64m

E3=0.83m(a)

Emin

=0.96m

No crossing pointsNo solution.

zmax

=0.37m

E0=1.33m

E (m)

y(m)


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1.6 Flow characteristics over humps and contractions

y

E

y1 y2

y2

y2

y2


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Example 6.7

The discharge in a 6 m wide rectangular concrete channel (n =

0.013) is 50 m3/s. Determine:

(a) The critical water depth and critical slope of the channel

(b) Types of flow if the flow at different sections of the channel has

water depths of 3.72, 1.92 and 1.55 m.(c) The Froude Number of the flow corresponding to flow depths of

3.72, 1.92 and 1.55 m

(d) The normal water depths and the type of slopes if the channelslopes are 0.001 and 0.04 respectively

Solution:

013.06,/50:Given3

=== m, nbsmQ

(a) 2 23 3/ (50 / 6) / 9.8 1.92 ,c y q g a m= = =

,

2/13/2

ch SRn

A

Q = ,byA c= )2/( cch ybbyR += 0258.0=cS

From Eq. (5.6),


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(b) At section of:

y=3.73m, since y > yc, subcritical flow

y=1.92m, since y = yc, critical flow

y=1.55m, since y < yc, supercritical flow

(c) The Froude number: , V=Q/(yb) =50/(yb)gyVFr /=At y=3.73m, V=2.24m/s, Fr = V/(gy)1/2 = 0.37

At y= 1.92m, V=4.34m/s, Fr = V/(gy)1/2 = 1

At y=1.55m, V=5.76m/s, Fr = V/(gy)1/2 = 1.38

(d) The normal water depths: Since

With A=by, , Q = 50m3/s, S0 = 0.001 and 0.004, )2/( ybbyRh +=

> Sc = 0.0258Steep slope0.76m < yc0.04

< Sc = 0.0258Mild slope2.71m > yc0.001

Slope typeNormal depth yS0

5.6)(Eq,2/1

0

3/2SR

nQ h=


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2. Hydraulic Jumps

A hydraulic jump is a steplike increase in fluid depth in an openchannel flow. Commonly seen below weirs and sluice gates.

Supercriticalflow (upstream)

Hydraulic

jump

Hydraulic jump

length

Subcriticalflow

C id i f ifi


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Consideration of specific energy:

Upstream:Supercritical flow, y1 < yc,

Specific energy: E1,

E

Es

y

y1

y2

yc

V1

V2

y1

y2

E2

Emin

E1

Downstream:Subcitical flow, y2 > ycSpecific energy: E2,

Energy difference: Es = E1 E2 > 0So after a hydraulic jump, there is energy loss


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Formed when transit from supercritical flow to subcritical flow

2.1 Hydraulic jump-formation and characteristics

Formation:

Transition from supercritical to subcritical flows is rapid,

Length of hydraulic jump: about 7 times downstream depth,

The transition involves large energy loss due to turbulence,

Energy loss can be found in terms of y1, y2 and upstream Fr,

Hydraulic jump can be used to dissipate water energy,

such as in a spillway of a dam to reduce damage to riverbed.

Characteristics:


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2.2 Downstream or sequent depth for a rectangular channel

The basic equations involved are:

Continuity equation

Momentum equation

Assumptions:

Ignore boundary friction (due to a short length of hydraulic jump),

The channel bed slope is very small,

Pressure distributions at (1) and (2): hydrostatic


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Momentum equation: F Mx =

F F M M 1 2 2 1 =

(6.19)

F M F M 1 1 2 2+ = + = (6.20)

where M, the pressure force plus momentum flux, is constant for

any given flow rate q.y

y1

y2

p2

p1 F +MIgnore friction

Initial depth

Sequent depth

(1)

(2)

A i h d t ti t (1) d (2) h


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Assuming hydrostatic pressure at (1) and (2), we have:

2

1 1 / 2,F gy= F gy22

2 2=

/

1 1 ,qV= M qV2 2=

qgy

y qgy

y

2

1

1

2 2

2

2

2

2 2+ = + (6.21)

= +q

gy

y2 2

2(for rectangular channel) (6.22)

Momentum function

( )1/ 2

22 11

11 8 1 ,2

yFry

= +

( )

[ ]

y

y

Fr1

2

2

21 21

2

1 8 1= + /

(6.23)

(6.24)


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2.3 Energy loss

h E E E yq

gyy

q

gyL= = = + +

1 2 1

2

2 2

2

22 2

1 2

( ) ( )

Since FrV

gy

q

gy

2

2 2

3= = And use Eqs. (6.23) and (6.24):

(6.25)

h Ey y

y yL= =

( )2 13

1 24

(6.26)

Energy loss increases dramatically as the relative height of the jump

increases.

Example 6 8


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Example 6.8

Determine the depth y4 immediately upstream of the hydraulic jump

that will form in the situation shown below if q = 1.6 m2/s.

From Example 6.6, we have:

y1 = 1.39m, y2 = 0.34m, yc = y3 = 0.64m

y0 = 1.25m

Damming action

y5=y0

y4

y3=yc


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37.025.181.9

25.1/6.1/

2

5

5

52 =

===

gy

yq

gy

VFr

( )1/ 2

242

5

11 8 1 ........

2

yFr

y

= + =

y4 can be calculatedAs y5 = 1.25m,

Energy loss: .......4

)(

54

345 ==

yy

yyhL

Assuming y5 = y0 = 1.25m,

2 4 Classification of hydraulic jumps:


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2.4 Classification of hydraulic jumps:

The actual structure of a hydraulic jump depends on Fr


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2.5 Some further discussion on hydraulic jumps

The relations for hydraulic jumps developed in the present

unit are based on a rectangular and horizontal channel,

Hydraulic jumps in other channel configurations are similar

to those for rectangular channels,

The expression of sequent depth and energy loss in other

channel configurations are somewhat different from jumps

in rectangular channels.

3 Analysis of Gradually Varied Flows


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3. Analysis of Gradually Varied Flows

Tasks

Deduce the trend of water surface change (classification ofsurface profiles)

Calculate water levels and velocity along the course of the

channel (quantitative evaluation) Analysis method

Energy equation

Continuity (mass conservation) equation

Characteristics of gradually varied flow Water depth and velocity change gradually,

Flow is non-uniform,

Water surface changes smoothly and continuously, Friction loss along the channel is not negligible.

3 1 The equations for gradually varied flow


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3.1 The equations for gradually varied flow

Assuming the change in the total energy is due to frictional lossesover some distance x, the energy grade line (EGL) and hydraulic

grade line (HGL) can be drawn as shown below.

datum

Friction slope Sf

Conservation of the energy gives:


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Conservation of the energy gives:

S x y

V

g S x y

V

gf0 11

2

2

2

2

2 2 + + = + +

S x S x y yV

g

V

gf0 2 12

2

1

2

2 2 = + +

(7.1)

(7.2)

The differential form by choosing x small dx:

S dx S dx dy d V

gf0

2

2= + + ( )

(7.3)

ddy

Vg

ddy

QgA

QgA

dAdy

Q BgA

( ) ( )

2 2

2

2

3

2

32 2

= = = (7.4)Since ,

A: the cross-sectional area; Q: discharge (m3/s).

(7.5)320

/1 gABQ

SS

dx

dy f

=

Substitute (7.4) into (7.3) and rearrange:

E (7 5) i lid f i h


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(7.6)

Eq. (7.5) allows the water surface profiles of gradually varied flow

to be deduced, y is measured vertically from the channel bottom,

the slope of the water surface dy/dx is relative to channel bottom.

For a rectangular channel, since Froude number is defined as

FrV

gy=

Eq. (7.5) can then be rearranged as

Eq. (7.5) is valid for any cross section shapes

2

0

2

0

1/1 r

ff

F

SS

gyV

SS

dx

dy

=

=

3 1 1 A note on the friction slope


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Strictly speaking, the slope term in Mannings Eq (5.6) is the frictionslope Sf, since for steady uniform flow, S0 = Sf.

Rearranging Mannings equation to find Sfgives:

Sn Q P

Af=

2 2 4 3

10 3

/

/(7.8)

Eq. (7.8) implies that Sfdecreases as the water depth y increases,

which in combination with (7.6) means that

y y S Sf< >0 0

y y S Sf>

3 2 Calculation of Gradually Varied Flow Profile


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Task: calculate the distance between two cross sections

Resources: channel slope S0, channel properties, water depth y1 at

section 1 and water depth y2 at section 2 are known

S x y

V

g S x y

V

gf0 11

2

2

2

2

2 2 + + = + +

S x E S x E f0 1 2 + = +

x

E E

S S

E

Sf=

=

1 2

0

(7.11)

Sf can be calculated using Mannings equation:

S

n Q P

Af=

2 2 4 3

10 3

/

/

(7.12)

From Eq. (7.1)

3.2 Calculation of Gradually Varied Flow Profile

3.2.1 Direct step method:

How?

Example 7.1


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The water depth of a gradually varying flow ( q = 1.6 m2/s) in a

wide rectangular channel (n = 0.015) is 0.34 m right behind a humpand 0.38 m at a location further downstream. What would be the

distance between the two locations?

(7.11)

Sf: Slope of water surface (Eq. (7.12)

S0: Slope of channel bed (Eq. (5.6)


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q2


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3.2.2 Numerical methods:


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Task: calculate yi (i=1 to n)

Resources: channel properties;

Method: An iterative procedure

3

2

2

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2

0

11'

gA

bQ

AR

QnS

F

SS

dx

dy

yh

r

f

=

== (7.13)

Using Taylors expansion:

xyyxdx

dyyy iii +=+=+ '1 (7.14)

From Eq.(7.5), we can get

y0

y1 y2 yi yi+1x x

. . .

where and)''(2

1' 1++= ii yyy

=

3

22

3/201/'

ihii

igA

bQ

RA

QnSy

3.2.2 Numerical methods:

The Iterative Procedure:


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1. Use Eq. (7.13) to calculate y'i where yi is known either as initial

point or from a previous cycle of this calculation

2. Set y 'i+1 = y ' i as a first estimate

3. Use current values of y ' i and y ' i+1 to calculate yi+1 from Eq.

(7.14) for a selected x

4. Find a revised estimate of y ' i+1 from Eq. (7.13) using the yi+1value from step 3

5. If the new y ' i+1 value is not close enough to the previouslycalculated value, repeat step 3, 4 and 5 using the latest estimate

of yi+1 found in step 4

6. Once the iteration procedure has yielded successive estimates ofy ' i+1 and yi+1 within acceptable limits of accuracy, proceed to

the next section of channel and repeat the process

7. Terminate the process when the desired reach has been covered

3 3 Water surface profiles and their classifications


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3.3 Water surface profiles and their classifications

Take a channel with mild slope as an example. The surface profile may

occupy the three regions shown below and the sign of dy/dx can be

found for each region:

y0 yc

(a) Region 1 for the mild slope:


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(a) Region 1 for the mild slope:

Since y>y0 > yc, we have Sf< S0 from Eq. (7.8)

Since y > yc, we have Fr2 < 1 from Eq. (7.10)

From Eq. (3.2.6) we get dy/dx is positive.

The asymptotic behaviour of the free surface M1:

00Frand,As Sdy/dxSy f

hence the water surface is asymptotic to a horizontal line.

0,As 00 dy/dxSSyy f

hence the water surface is asymptotic to the line y = yn.

(b)Region 2 on a mild slope:


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For the M2 profile, dxdyyy c /,as

This is physically impossible. This is because as the fluid entersa region of rapidly varying flow and Eq. (7.6) is no longer valid.

y y S Sf 0 0dy dx/ 012 rc Fyy

(c)Region 3 on a mild slope:

positive.is/1Frand, 200 dxdySSyyy fc >>>>

For the M3 profile, dxdyyy c /,as

This is physically impossible. In practice, an hydraulic jump willform before y = yc.

Note: above discussion is for mild slope. For steep, critical, horizontal

and adverse slopes, the surface profiles are given in next page.


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Non Uniform Flow