10 varied flow_with figures

7/29/2019 10 Varied Flow_with Figures

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10FlowinOpenChannels

10.1 Varied Flow (Non uniform flow)

bottom slope: S0 Slope of EL = Sf

Bernouilli equation:

S0 dx + y +

v2

2 g= y + dy +

v2

2 g+ d

v2

2 g

+ S dx

divided by dx :dy

dx+

d

dx

v2

2 g

= S0 Sf,

d

dxy +

v2

2 g

= S0 Sf

dE

dx= S0 Sf E specific energy( ),

dE

dx=1

Q2 b

g A3 =1 Fr2

dE

dx=

dE

dy

dy

dx

dy

dx=

S0 Sf

1 Fr2

Fr =Q

2 b

g A3, non rectangular channel

Basis equation of gradually varied flow (water surface profiles can be deduced from this

equation):

dy

dx

=

S0 Sf

1

Fr2, QChannel Shape in Fr Sf( )

This equation shows how the depth (y) changes with distance (x) in terms of bed slope

(S0) and friction (Sf).

Classification of surface profiles:

For a given discharge Q:

Sf =Q n

A R2 3

2

=

n2 Q2 p4 3

A

10 3, Fr

2=

Q2 b

g A3

Sf Fr as A y

When theflowisuniformSf= S

0and y = y

0

S0

oncey over y0

Sf

below S0

Sf< S

0when y > y

0

Sf> S

0when y < y

0

and F r2>1 when y < yc

Fr2< 1 when y > yc


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From these inequalities, we can know the sign ofdy

dx. For example the surface slope

changes for differnt slopes & Froude numbers.

The classification proceeds as follows:

S0< SC y0 > yC MildSlope M profile

S0> SC y0 > yC SteepSlope S profile

S0= SC y0 > yC CriticalSlope C profile

S0= 0 y

0> yC HorizontalSlope H profile

S0< 0 y

0> yC AdverseSlope A profile

Takins a mild slope channel as an example:

yc: critical depth y0: normal depth

Identifying 3 zones in the flow:

Zone 1: above the normal detph

Zone 2: between normal & critical depth

Zone 3: below critical depth

Zone1: y > y0> y

cS

f< S

0F

r

2 0 surface rising

Zone 2 : y0> y > yc Sf > S0 Fr

2 y

c> y S

f> S

0F

r

2>1

dy

dx> 0 surface rising

At the boundary:

Zone 1: As y then Sf Fr 0,

dy

dx S0

Hence the water surface is asymtotic to a horizontal line for it maximum.

As yy0 then Sf S0 anddy

dx 0

Hence the water surface is asymtotic to the line y = y0 . i.e. uniform flow.

Zone 2: As y y0 then Sf S0 anddy

dx 0

Hence the water surface is asymtotic to the line y = y0.


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As yyc then Fr 1 anddy

dx

This is physically impossible. In reality a very steep surface will occur.

Zone 3: As in zone 2, As yyc then Fr 1 and dydx

As y0 thendy

dx S0 the slope of channel bed.

The gradually varied flow equation is not available here.


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Example:


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Methods of solution of varied flow equation:

There are 3 forms of varied flow equation:

dH

dx

= Sf 1( )

dE

dx= S

0 Sf 2( )

dy

dx=

S0 Sf

1 Fr2

3( )

There are 2 basic methods (For prismatic channels):

1 - Direct step: distance from detph

2 - Standard Method: depth from distance

1Thedirectstepmethod:distancefromdepth

2( ) dE

dx= S

0 Sf dE= S0 Sf( ) dx x =

Ei+1 Ei

S0

S3( )

wherei & i +1representlocationalongthechannelwiththedistancex :

S: theaveragevalueofSintheinterval.Variesverygraduallyalongthechannel,ifxissmall

Thismethodistodetermineapointinthechannelwherethedepthisknownand

use that point as the starting point for computation of a distance upstream /

downstream:

S=Si+ S

i+1

2= S=

Q n

A R2 3

2

=

1

2

Q n

Ai R

i

2 3

2

+

1

2

Q n

Ai+1

Ri+1

2 3

2

Thismethodwillcalculateadistanceforagivenchangeinsurfaceheight.

3( ) finite difference( ) x =1 Fr

2

S0 Sf

y

Steps in solution are:

1. Determine the control depth as a starting point.2. Decide on expected curve and depth change if possible.3. Choose a suitable depth step by y4. Calculation term in brackets at "mean" depth (yinitial + y/2).5. Calculate x6. Repeat 4 & 5 until the appropiate distance / depth change is reached.


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Example:

A flow rate of 25 m3/s occurs in a trapezoidal channel. Bottom width = 6m, slide slopes

= 1.5:1, z = 1.5, Manning value n=0.025, bed slope = 0.0001 m/m. If the depth at a

point in the channel is 1.3m, how far (upstream / downstream) from this point will thedepth be 1.6 m?

Calculate critical depth :Q

2 b

gn A3=1 =

252 6 +2 z yc( )

9.81 6 yc +z yc2( )

3=

252 6 + 3 yc( )

9.81 6 yc +1.5 yc2( )

3 yc =1.1 m

Critical slope : A = 6 yc +1.5 yc2= 6 1.1+1.5 1.1( )

2

= 8.415 m2

Rh =A

p

=

8.415

6 +2 yc2

+z2

yc2=

8.415

6 + 2 1.1( ) 1+ 1.5( )2= 0.844

Sc = gn n2

A

b Rh4 3

= 9.81 0.025

2 8.415

6 + 3 1.1( ) 0.844 4 3= 0.007 S

0< Sc Mild Slope

Example:

A flow rate of 10 m3/s occurs in a rectangular channel 6m wide, Manning value

n=0.013, bed slope = 0.0001 m/m. If the depth at a point in the channel is 1.5m, how far

(upstream / downstream) from this point will the depth be 1.65 m?

Q =1

n A R2 3 S

0

1 2, 10 =

1

0.013 6 y

0( )6 y

0

6 +2 y0

2 3

0.001( )1 2

= 0.769 6 y0( )

6 y0

6 + 2 y0

2 3

Solving by trial-and-error, we obtain y0=1.94m

Calculate critical depth: (specific energy)

yc= q

2g3 = 10 6( )

2

9.813 = 0.66m, y0> y

c, the slope is mild. y

c< y < y

0 in zone 2

so depth 1.65 m must be upstream of depth 1.5 m. We choose y = 0.005m

x =Ei+1 EiS

0 S

, E=y +v

2

2 g, S=

Q n

AR2 3

2

2 The standard step method: depth from distance

Steps in solution are similar to the ones in the direct step. However for each x the

followings steps are performed:


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From Eq 2( ) : E = x S0 Sf( )

mean4( )

1. Assume a value of depth y (the control depth or las solution depth)2. Calculare Specific Energy EG3. Calculate Sf4. Calculate E using (4)5. Calculate Ei+1= Ei+E6. Repeat until Ei+1 = EG

Another formula of Stantard Step Method

Based on equation 3( ) : y ' = dydx

= S0Sf1 Fr

2=

S0

Q n

A R2 3

2

1 Q

2 b

g A3

5( )

UsingtrapezoidalMethodtointegratethisODequation: yi+1 = yi +dy

dx x = yi + y' x

in interval i to i +1 : y'= 0.5 yi'+ y i+1

'( ) y i+1 = y i + 0.5 y i' + yi+1'( ) x 6( )

SoifA,n, P,Q & y iareknown,y i+1 & y i+1'

canbesolvedbythefollowingsteps:

Using (5) to calculate yi

' where yi

is known form initial on previous steps.

a) Set

yi+1

'= y i

'asafirstestimate.

b) Calculate

yi+1

fromEq.(6)foraselectedx.

c) Adjust

y i+1'

= y i' from(5)using

y i+1fromstepe)

d) If

y i+1' y i

'> : tolerance repeat steps (c), (d) & (e) until

y i+1' yi

' < yi+1 y i < proceed to the next section of the channel

repeattheprocess.

Irregular channels: natural channels

+ Morecomplicatedgeometry&flowinthemainchannel&floodplain.+ Insteadof depth y, we areusing theheight h-stage (water level above a

datum).


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+ Foragivendischarge,theriverstageisknownatonesection calculatethestageattheadjacentsection.Atbothsections,aplanofthewaterwayis

available:A,P,n,etc.areknown.

+

Wetakeatrialvalueoftheunnknowstageh2andcalculateA2, v2, v22

/2g

and hence H2 (total energy) the friction slope

Sf2

=

v2

2

e2

2 R

2

can also be

calculated and from mean value

Sf1

+ Sf2

( ) 2 we obtain a total energy

differenceH2-H2.ThetestofthetrialprocessiswhetherthisvalueofH2is

equaltothepreviouscalculatedone.Ifnot,afurthertrialwillbecontinued.

H2= y

2+z

2+

v2

2

2 g= H

1+

1

2 x Sf1 + Sf2( )Eq(1)

Differencebeweentheenergyvalues:

HE =v2

2

2 g+z

2+ y

2 H

1Sf1 + Sf2

2 x Sincez

2,H

1& Sf2 = const.

dHE

dy2

=

d

dy2

y2+

v2

2

2 gSf22

x

=1 Fr2

2 1

2 x

dSf2dy

2

SinceSf2 hasapproximatedby :dSf2dy

2

3 Sf2y

2

3 Sf2R

2

we havedHE

dy2

=1 Fr22+

3 Sf2 x

2 R2

or y2=

HE

1 Fr2+

3 Sf2 x

2 R2

y2isamountbywhichthewaterlevelmustbechangedinordertomakeasmallerrorHEvanish.

Note : Inthenaturalriverthesurfacewidthbis tothewettedperimeterP

Fr2=

v2 b

g A

v2

g R


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10.2Uniformflow-Chezyequation

All forces are balanced:

dv

ds= 0

dy

ds= 0 y : water depth

Fx= F1 +w sin F2 0A0 = 0 1( )

+ Resistance force exented on the bottom & both sides:

0, A

0A

0: areaofsides & bottom( ) 0, P, l

+ No change: F1 F2 =Q V2 V1( ) = 0

+

W = A l sin =hL

12

lS

0= tan Slope of Channel bed

Inopechannelissmall tan sin = S0

HydraulicRadius: Rh =A

P P =

A

Rh

A l sin0

A

Rh l = 0

0= Rh tan = Rh S0

+

Inpipeflow,wehave0= f

v2

8

f v2

8= Rh S0 = gn Rh S0 v =

8 gn

fRh S0 = C Rh S0

C=8 gn

fChezycoefficient

Q = C A Rh S0 Chezyequation(Frechhydraulician)

Fundamentalequationforuniformflowinopenchannels


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10.3ChezyandManningCoefficient

C=R

h

1 6

nSI unit( ) n : Manning coefficient Q =

1

n A Rh

2 3 So

1 2SI unit( )

Chezy -Manningequation USunit( ) : C=R

h

1 6

n , Q=

1.49

n A Rh

2 3

So1 2

C=8gn

f, n = Rh

1 6 8gn

f

Example 1:

A rectangular channel lined with asphalt width = 6m, So = 0.0001. Calculate the depth

of uniform flow when the flowrate is 12 m3/s.

Q = 1n A Rh

2 3 So1 2; A = B y = 6 y, Rh = A

P= 6 y

6 + 2 y, n = 0.013 Smooth( )

12 =1

0.013 6 y( )

6 y

6 + 2 y

2 3

Solvingbytrialy =2.2m,

forroughasphaltn = 0.016, y = 2.55m

Example 2:

A concrete-line canal with a n-value = 0.0014, is constructed on a slope og 0.33m/kmand conveys 23 m3/s. Find uniform flow depth if the canal has a trapedoizal cross-

section with bottom width = 6m, side slope of z = 1.5

Q =1

n A Rh

2 3 So1 2

, S0= 0.33 m km = 0.00033, A = 6 y + 2

1

2 y 1.5y

= 6 y +1.5 y 2

P = 6 + 2 y2+ y z( )

2= 6 + 2y 1+ 1.5( )

2= 6 + 3.61 y, Rh =

A

P=

6 y +1.5 y 2

6 + 3.61 y

23 =1

0.014 6 y +1.5 y 2( )

6 y +1.5 y 2

6 + 3.61 y

0.00033( )

1 2

17.726 = 6 y +1.5 y 2( )6 y +1.5 y 2

6 + 3.61 y

y =1.77m


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10.4UniformLaminarFlow

Flow in a thin sheet infinite width (without side wall) considering one rectangular cross-

section with depth y0 & width b.

Rh =A

P=

b y0b

= y0

0 = Rh S0 = y0 S0

Lamin ar flow : = dv

dy

0=

dv

dy

0

v = vc = 1 r

2

R2

= vs 1

R y( )2

R2

dv

dy= vs

2 R y( ) 1( )R

2

= 2vs

R y( )R

2

dv

dy

0

= 2vs y0

v =

1

y0

vs 1 R y( )

2

R2

dy

0

y0

=vs

y0

dy0

y0

vs

y0

R2 2 R y + y 2

R2

dy

0

y0

= ...

=

vs

y0

y0

vs

y0

R2 y

0 R y

0

2+

1

3y

0

3

R2

R =y0

= vs 1

3vs =

2

3vs

0= dv

dy

0

= 2 vsy0

=

2

3

2 v

y0=

3v

y0,

0= y

0 S

0= gn y0 S0 =

3v

y0

v =

gn y02 S

0

3 =

gn y02 S

0

3 2, Re =

vy0

Re =

gn y03 S

0

3 2

Chezy form : v =gn y0

3 2 S0

3 y

0S

0= C y

0S

0 C= gn Re 3


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10.5HydraulicRadiusConsiderations

+ Hydraulic Radius Rh plays a prominent role in the equations of open-channelflows: Rh = f(y).

+ Rectangular channel:

A = By, P = B + 2y, Rh =A

P=

B y

B + 2 y

y 0 Rh 0

y Rh b

2FornarrowdeepsectionRh =

B

2

+ Even for any (non-rectangular) section, when the channel narrow & depp, thisapproximation can be applied.

+ For constant depth y = const. Rh =B y

B + 2 y, B = 0 Rh = 0

B Rh y

+ Wide shallow rectangular channel: Rh y + This approximation is also valid for non-rectangular channel if the setion is wide

& shallow. Hydraulic Radius is approximated by the mean depth.

Trapezoidal cross-setcion channel

z: Side slopes, z = tan

A = B y +z y2

, P = B + 2 y 1+z2

, Rh =A

P=

B y +z y2

B + 2 y 1+z2

dRh

dy=

1+ 2 z y B( ) 1+ y B( ) 1+z 2[ ]1+ 4 1+z

2 y B( ) 1+ y B( ) 1+z 2[ ]


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For 0 90 0 z ( ) & any y

Denominator >Numerator

dRh

dy


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Example:

Solve above problem for a trapezoidal channel with a slope z =2.

The best efficient trapezoidal channel has : Rh =

y

2

A = y 2 2 1+z2 z( ) = 2 5 2( ) y 2

Q =1

n A Rh

2 3 S0

1 2 10 =1

0.019 2 y 2 5 2( )

y

2

2 3

0.0001( )1 2

y = 2.56 m B = 2y 1+z2 z( ) = 2 2.56 5 2( ) =1.21m

10.6 Specific Energy, Critical Depth and Critical Slope- Wide

RectangularChannels:

Specific Energy : Bakhmetejf, 1912( ) E = y +v

2

2gn, E = y +

12

2gn

Q

A

2

,

Flowrateperunitofwidthq=Q

bQtotalchanneldischarge( )

v =q

y E = E = y +

12

2gn

q

y

2

q = 2 gn y2 E y 3( )


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A: q = constant, plot E versus y

Critical depth, yc, corresponding to the condition of minimum specific energy, Emin:

dE

dy=1

q2

g y 3= 0 yc =

q2

g

1

3

Critical depth, yc, is a function of only the flowrate per unit width q for a rectangular

channel:

Emin

= yc +q

2

2 g yc2=

3

2 yc yc =

2

3 E

min

E= y +q

2

2 g y 2, if y 0 E,if y E y

Emin yc as q

At y = yc

1=q

2

g yc3=

v yc( )2

g yc2=

v2

g yc= F

r

2, y > y

c

: Subcritical y < yc

: Supercritical

+ We can conclude that the critical depth condition is specified by the Frbecoming unity.

+ Ony value of E Corresponding to 2 values of y: 2 possible depths for a givenvalue of a Specific Energy. This 2 depths are called alternate depths.


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B: E = constant, plot q versus y

q = 2 g y2 E y

3( ) dq

dy=

2 g 2 y E 3 y2( )

y2 E y

3

dqdy

= 0 y 2E 3 y( ) = 0 y = 0, yc =2

3 E

Flow over a hump:

Assumption: Structure is sufifcient smotth, no additional friction loss.

B: E (1) & (2)

y1+

v1

2

2 gn

= y2+

v2

2

2 gn

+ z

q= const.


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z > hmax

dam /choke : no water overflow

flow back up gainning the elevation until obtain enough specific Energy to flow over

the dam.

Channel with restriction:

Assuming: Constriction is smooth, no addition friction loss. Qtotal = constant, q/unit as

flow passes the constriction.


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qcon =

B1

Bcon

q1, qmax =

B1

Bcon

q1, Criticalcondition:B 0

dy

dx1 Fr

2( ) < 0

Hence +IfFr1: Supercritical

dy

dx> 0 y over step

+ If there is a downward step dzdx

< 0


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dy

dx1 Fr

2( ) > 0

Hence +IfFr 0 y over step

+IfFr >1: Supercritical dy

dx< 0 y over step

Example:

Uniform Subcritical flow at a depth 1.5m in a lon rectangular channel of width 3m,

Manning value n=0.012, slope of 0.001. Calculate:

a) Minimum height of hump which can be brult on the floor of the chanel to produce

critical depth.

b) The maximum width of constriction which will produce critical depth.

(q = const)

E0= y +

1

2 gn

Q

A

2

Q =1

n A Rh

2 3 S0

1 2=

1

0.015 1.5 3( )

1.5 3

2 1.5 + 3

2 3

0.001( )1 2

= 7.83m3s

E0 =1.5 +1

2 9.817.83

1.5 3

2

=1.65m

Flowoverhump,criticaldepthoccursonceE= Emin

when yc =q

2

gn

1 3

=

7.83 3( )2

9.81

1 3

= 0.886m

Emin =3

2 yc =

3

2 0.886 = 1.33m humpheightz = E0 Emin =1.65 1.33 = 0.32m

b) E = const

yc =2

3 E

min=

2

3 E

0=

2

3 1.65 =1.1m yc =1.1 =

q2

gn

1 3

=Q /Bcon( )

2

gn

1 3

=7.83/Bcon( )

2

9.81

1 3

Bcon =7.83

1.1( )3 9.81

= 2.17m

Critical Slope:

In uniform flow, the slope of the channel bottom under critical depth condition is called

critical slope Sc. For a rectangular channel of a wide depth (Rh=yc), Sc can be obtained

as follows.


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Q =1

n A Rh

2 3 Sc

1 2=

1

n b yc yc

2 3 Sc

1 2 q =

Q

b=

1

n yc

5 3 Sc

1 2

q = gn yc3=

1

n

yc5 3 Sc

1 2 Sc =

gn n2

yc1 3

Ifslopeofachannel > Sc : steemslope

< Sc : mildslope

Example:

For a flowrate 1.5 m3/s in a rectangular channel 12m wida a water depth is 1.2m. Is the

flow a subcritical or supercritical if n = 0.017, what is Sc ot this channel? What chanel

slope would be required to produce uniform flow at a depth of 1.2m?

q =Q

B=

1.5

12= 0.125 m

2s, critical depth : yc =

q2

gn

1

3

= 0.117m y =1.2m > yc : flowissubcritic

Sc =gn n

2

yc

1 3=

9.81 0.017( )2

0.117( )1 3

= 0.0058, Channelslope : Q =1

n A R

h

2 3 S0

1 2 S0

1 2=

Q n

A Rh

2 3

S0=

Q n

A Rh

2 3

2

=

1.5 0.017

1.2 121.2 12

2 1.2+12

2 3

2

= 3.136 106 S0 < SC : mild slope

10.7SpecificEnergy,CriticalDepthandCriticalSlope-Nonrectangular

Channels:

E = y +1

2 gn

Q

A

2

= y +1

2 gn

Q

f y( )

2

,dE

dy=1+

Q2

2 gn

2

A3

dA

dy

=1+

Q

2 gn2b

A3

=1

Q2 b

gn A3

dE

dy=

0,

Q2 b

gn A3=

1 Q=

gn A3

b Fr=

Q2 b

gn A3=

v

gn A b

Critical depth: at E = Emin

Critical slope: Q =gn A

3

b=

1

n A Rh

2 3 SC1 2 SC = gn n

2A

b Rh4 3


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Example:

A flow of 28 m3/s recurs in an earth-lined canal having a base width of 3m, side slopes

z=2, Manning n-values of 0.022. Calculate the critical depth & critical slope:

Critical depth : happenswhendE

dy= 0

Q2 b

gn A3=1 =

282 3+ 2 z yc( )

9.81 3 yc +z yc2( )

3=

282 3+ 4 yc( )

9.81 3 yc + 2 yc2( )

3 yc =1.5m

Critical slope : Sc = gn n2

A

b Rh4 3

, A = 3 yc + 2 yc

2= 3 1.5 + 2 1.5( )

2= 9 m2

Rh=

A

P=

9

3+ 2 yc2+ 4 yc

2=

9

3+ 2 1.5 5=

0.93m SC=

9.81( ) 0.022( )

2

9

3 0.93( )4 3=

0.00523

10.8ControlsandtheoccurrenceofCriticaldepth:

+ In subcritical flow, a disturbance can move upstream, In practice, a controlmechanism such as a gate can make its influence felt on the flow upstream, so

that subcritical flow is subject to downstream control.

+ Conversely, supercritical flow can not be influenced by any feature downstream,and can be controlled from upstream.

+ Subcritical flow is produced by downstream control, and supercriticalflow byupstream control.

"The water doesn't know what's happening downstream".


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10.9HydraulicJump:

When a change from supercritical to subcritical flow occurs in open channel: Hydraulic

jump appears to be a complicated phenomenon: turbulence, air entrainment, ...

Assumptions: rectangular, unifor flow.

Work energy & continuity Eqs:


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Fext( )x= F

1 F

2 FX = Q V2x V1x( ) = q V2 V1( ) V = q y( )

y1

2

2

y2

2

2 FX = q

2 1

y2

1

y1

FX

=

q2

gn y1

+

y1

2

2

q2

gn y2

+

y2

2

2

= M1 M2

M=q2

gn y+

y2

2Force + MomentumEquation

InthesimplehydraulicJump: FX = 0

q2

gn y1+

y1

2

2=

q2

gn y2+

y2

2

2 V = q y( )

y2y1

= 12

1+ 1+ 8 v12

gn y1

= 12

1+ 1+ 8 Fr1

2

[ ] or y1

y2

= 12

1+ 1+ 8 v22

gn y2

= 12

1+ 1+ 8 Fr2

2

[ ]

Boundary-Equation: Work-Energy:

hL j = y1 +z1 +v1

2

2 gn

y2 +z2 +

v2

2

2 gn

Location og Hydraulic jump:

y2

y1

=

1

21+ 1+

8 v1

2

gn y

1

10.24( ) ory

1

y2

=

1

21+ 1+

8 v2

2

gn y

2

10.25( )

+ If the jump occurs on the mild slope, the flow remains at normal depth on thesteep slope and decelerates on the mild slope until the jump froms.

In this case: y < y01 < yc Table 6 Zone 3: M3


24/25

24

+ If the jump occurs on the steep slope, the slow jumps from the normal depth tothe conjugate depth y2 on the sttep slope, decelerates it until it reachs the normal

depth y02 on the mild slope.

In this case: y > y02 > yc Table 6 Zone 1: S1

In order to determine which scenarios applies, we are going to calculate the conjugate

depths across the jump corresponding to the twho normal depths.

For a given flowrate:

If y1 > y01 on steep slope, an M3 profile will take place, then the flow decelerates and

the depth will increase, so that the jump occurrs.

If y1 < y01 : impossible situation. For M3 the depth cannot decrease in dowstream and

consequently the jump must occur in the steep slope.

an incresae in the n-value in the downstream (mild slope) channel could increase the

normal depth y02 so, it would be impossible for the jump to form on the downstream:

channel roughness plays a role in positioning of hydraulic jump.

Example:

A flowrate of 15m3/s occurs in a long rectangular channel with a width of 3m, and n =

0.015. Uniform low. Verify whether a hydraulic jump must occur and its position:

y01 < yc : upstream is steep slope: S-profile.

y02 > yc : downstream is mild slope: M-profile.

Supercritical --> subcritical --> Hydraulic jump occurs

To determine whether the jump located on upstream / downstream of the channel break,we assume that it occurs on the downstream and check to see if this assumption is valid.

y1=

1

2y

21+ 1+

8 v2

2

gn y2

= ... v

2=

Q

A2

=

15

3 1.5= 3.33m /s

....=

1

2 1.5 1+ 1+

8 3.332

gn 1.5

=1.24m

As y1 > y01, conditions are right for a M3 curve to be developed.

Consequently, the jump will be located on the downstream. To find the exact position of

the jump, we are going to calculate the distance downstream from the break (control) to

the point where the depth of 1.24m occurs. Using direct step method:


25/25

x =E

i+1 E

i

S0 S

E1.24

= y +v 2

2 gn= y +

Q2

2 gn A2=1.24 +

152

2 9.81 3 1.24( )2= 2.07m

E0.75

= y +v 2

2 gn= y +

Q2

2 gn A2= 0.75 +

152

2 9.81 3 0.75( )2= 3.02m

S1.24

=

Q n

A Rh2 3

2

=

15 0.015( )

3 1.24( )3 1.24

3+ 2 1.24

2 3

2

= 0.079 , S0.75

= 0.012, sameaschannelslopeduet

S=S

1.24+ S

0.75

2= 0.01 , x =

2.07 3.02

0.0015 0.01=111.76m

The hydraulic jump is located 111.76 m downstream of the control.

10 varied flow_with figures

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