m6b water surface profiles.pdf

8/12/2019 M6b Water Surface Profiles.pdf

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M6b: Water Surface Profilesand Hydraulic Jumps

Robert PittUniversity of Alabama

andShirley Clark

Penn State - Harrisburg

Example 3.10 (Chin 2006): Constr ict ion i n Channel

A rectangular channel 1.30 m widecarries 1.10 m 3/sec of water at adepth of 0.85 m.

a) If a 30 cm wide pier is placed inthe middle of the channel, find theelevation of the water surface atthe constriction.

b) What is the minimum width of theconstriction that will not cause arise in the upstream watersurface?

a) Neglecting the energy losses between the constriction andthe upstream section, the energy equation requires that thespecific energy at the constriction be equal to the specificenergy at the upstream section:

gV

yg

V y

22

22

2

21

1 +=+

( )( ) sm

mmm

AQ

V /00.130.185.0

sec/10.1 3

11

===

( )

( ) m

m

mm

g

V y E 901.0

sec/81.92

sec/00.185.0

2 2

221

1 =+=+=

Where Section 1 is the upstream section and Section 2 isthe constricted section. Therefore y 1 = 0.85 m and:

The specific energy at Section 1 is:

Equating the specific energies at Sections 1 and 2 yields:

( )( ) ( )[ ]22

2

3

22

2

230.030.1sec/81.92

sec/10.1

2901.0

ymmm

m y

gAQ

ym−

+=+=

Which simplifies to:

901.00617.0

22

2 =+ y y

There are three solutions to this equation:

mmm y 23.0;80.0;33.02 −=



2

35.01

11

==gy

V Fr

m y 80.02 = 49.0

2

22

==gy

V Fr

Of the two positive depths, the correct one mustcorrespond to the same flow condition as upstream. At theupstream section:

The upstream flow is thereforesubcritical.

9.1;33.0 22 == Fr m yif

Subcritical

Supercritical (not possible)

b) The minimum width of constriction that does not causethe upstream depth to change is associated with the critical

flow conditions at the constriction. Under these conditions(and for a rectangular channel):

3/12

21 2

3

2

3⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ====

gq

y E E E cc

( ) 3/12

1

/

2

3

⎥⎦

⎤

⎢⎣

⎡=

g

bQ

E ( )

( )

3/1

22

23

sec/81.9

sec/10.1

2

3

901.0 ⎥⎥⎦

⎤

⎢⎢⎣

⎡=

mbm

m

and b =0.75m. If the constricted channel is less than 0.75 m,then the flow will be choked and the upstream depth willincrease.

If b is the width of the constriction that causes critical flow,then:

Prasuhn 1987

Bridge HydraulicsYarnell tested different types of model piers to obtain thefollowing equation that predicts the increase in watersurface elevation (y 3) due to the bridge:

( )( )423

23

3

156.05 α α +−+=∆Fr K KFr

y y

bridgethetodueelevationsurfacewater inincreasetheis yand

pierstheof spanthetowidth pier theof ratiothebb

where

∆

=1

α

Since the flow is subcritical, the downstream depth andFroude number would be known (from computation of thebackwater profile from some known location downstream).K is selected based on the pier geometry, as shown on the

following table.



3

Prasuhn 1987

Water Surface Slope Classifications(Figure 3.11, Chin 2006)

Hydraulic Classification of Slopes (Table 3.5, Chin 2006)

Prasuhn 1987

Mild slope because y n>y c

M – 1: above both y n and y c (above both )M – 2: between y n and y c (between th e upper two)

M – 3: below both y n and y c (below both)



Prasuhn 1987

S – 1: above both y n and y c (above both )S – 2: between y n and y c (between the upp er two)S – 3: below both y n and y c (below both)

Steep slope because y c>y n

Prasuhn 1987

Critical slope because y c=y n

C – 1: above both y n and y c (above both )There is no C – 2: not possible to be between y n and y c(they are the same)C – 3: below both y n and y c (below both)

Prasuhn 1987

Horizontal slope because y n is infinite

There is no H – 1: not possible to be above both (y n atinfinity)H – 2: between y n and y c (between both)H – 3: below both y

nand y

c(below both )

Prasuhn 1987

Adverse slope because y n is inf inite andchannel s lopes upwards

There is no A – 1: not possibl e to be above both (y n atinfinity)

A – 2: between y n and y c (between bo th) A – 3: below b oth y n and y c (below both)



Prasuhn 1987

Computation of Water Surface Profiles

x

gV

y

S S f o ∆

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ +∆

=− 2

2

α

o f S S

gV

y L −

⎥⎦

⎤

⎢⎣

⎡

+=∆

1

2

2

2α

The equation describing the shape of the water surfaceprofile in an open channel is derived from the energyequation and written in the form:

Rearranging leads to:

which describes the distance between upstream anddownstream sections.

Example 7-11 (Prasuhn 1987) Direct Step Method toCalculate Water Surface Profiles

A discharge of 800 cfs occurs in a long rectangular channelthat is 20 ft wide and has a slope of 0.0005 (= 5x10 -4). Thechannel ends in an abrupt drop-off. The Manning’s n is 0.018.Calculate the water surface profile from the drop-off to adistance at which the depth has reached 99 percent of the

normal depth.

The normal and critical depths are needed to determine thetype of curve. From the Manning equation:

( )[ ] ( )( )( ) 3/2

2

2/13/52/13/23

220018.0

0005.02049.149.1sec/800

y ft

y ft n

S AR ft Q no

+===

and the normal depth at y 2 is determined to be y n = 8.01 ft

( ) ft ft

ft cfs yc 68.3sec/2.32

/403

2

2

==

ft cfs ft

ft q /40

20

sec/800 3

==

The unit width discharge and the critical depth values are:

The water surface profile will therefore be a M-2 curve.The depth at the brink (the drop off location) will be:

ft y y cbrink 58.27.0 =≈

( ) ft ft y x c 1568.3441 ==≈∆

and the distance fromthe brink to the criticaldepth will be:



Example 7-10 (Prasuhn 1987)

o f S S

gV

yg

V y

x −

⎟⎟

⎠ ⎞

⎜⎜

⎝ ⎛ +−

⎟⎟

⎠ ⎞

⎜⎜

⎝ ⎛ +

=∆ 22

22

2

21

1

2

3/249.1 ⎟⎟

⎠ ⎞

⎜⎜

⎝ ⎛ =

av

avav f R

V nS

The distances between the downstream and upstreamsections are calculated to be:

And the friction slopes for each segment can be calculatedusing the Manning’s equation, using average n, V, and Rvalues for each reach:

The computations are for a distance upstream of the drop offuntil the depth is 0.99 (8.01 ft) = 7.93 ft.

The computations are carried out in the following table:

13,05262125.36x10 -45.134.380.2338.3245.044.42158.67.93

684047506.82x10 -45.624.170.8648.1015.214.34153.67.68

209014491.04x10 -36.523.810.7877.2375.994.00133.66.68

6415181.72x10 -37.793.410.6356.4507.043.62113.65.68

1231083.27x10 -39.712.940.3005.8158.553.1993.64.68

15----------5.51510.872.6873.63.68

X=Σ∆x(ft)

∆x(ft)

S f Vav(ft/s)

R av(ft)

∆(y+V2/2g)y+V2/2gV (ft/s)R (ft) A (ft2)y (ft)

Hydraulic Jumps A hydraulic jump is a steady, nonuniform phenomenonthat occurs in open channels when a supercritical flowencounters a deeper subcritical flow. As the supercriticalflow encounters the slower moving water, it tends to flowunder it and then spread upward, creating a large eddy orroller. This can dissipate a significant amount of energyand is desirable below a spillway or sluice gate before theflow enters a natural channel, reducing erosion potential.

Figure 3.12, Chin 2006



Characteristics of Hydraulic Jumps (Table 4.5, Chin 2000)

Prasuhn 1987

( )12

22

21

22V V Q

yb y −=− ρ γ γ

( )1812

1 21

1

2 −+= Fr y y

The forces in the flow direction consist of the hydrostaticforces:

This can be used to derive the following that can be used

to predict the conjugate (required) downstream depth ofa hydraulic jump:

The following graph can also be used to predict thedepths, the length of the jump, and the head loss in the

jump. Prasuhn 1987



Example 7-7 (Prasuhn 1987) HydraulicJump Characteristics

What is the downstream depth required for a jump to occurin a 20 ft wide rectangular channel, if the upstream depth is2 ft and the discharge is (a) 200 cfs, or (b) 640 cfs. What isthe head loss in each case?

( )( ) sec/5

202

sec/200 3

1 ft ft ft

ft AQ

V ===

( )( )62.0

2sec/2.32

sec/52

1

11

=== ft ft

ft

gy

V Fr

For 200 cfs:

The upstream flow is subcritical and a hydraulic jumpwill not occur.

For 640 cfs:

( )( ) sec/16

202

sec/640 3

1 ft ft ft

ft

A

QV ===

( )( )99.1

2sec/2.32

sec/162

1

11

=== ft ft

ft

gy

V Fr

( ) ( ) ⎟ ⎠ ⎞

⎜⎝ ⎛

−+==−+= 199.1812

1

21812

1 2221

1

2

ft y

Fr y y

ft y 72.42 =

The upstream flow is therefore supercritical, and a jump will occur. The resulting downstream depth canbe calculated:

The resulting energy loss is therefore:

( ) ( )( )( )

ft ft ft

ft ft y y y y H L 53.0

72.424272.4

4

3

21

312 =−=−=

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ +−

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ +=−=

gV

yg

V y H H H L 22

22

2

21

121

Prasuhn 1987

Combined Water Surface ProfilesWater surface profiles are the result of an obstruction to theflow or some change or changes to the channel itself, such asits slope. The transitions that may result involve distinctprofiles as illustrated in the following profiles:

Alw ays s ubcri ti cal ; the requi red dow nst ream con tr ol is p rovided b y thedownstr eam normal water depth, a friction contr ol because it is related to

the channel friction thr ough the normal depth and the Manning equation.



Prasuhn 1987

The upstream subcrit ical flow must have a downstream control, while thedownstr eam supercriti cal flow must h ave an upstream control . Therefore, thetransition curve must pass through t he critical depth at the grade break (theonly possible control for both directions).

Prasuhn 1987

This transition is s imilar to th e above example: the control w ill be the gradechange following the horizontal section (the critical depth). Depending onthe length of the horizontal s ection, the H-2 curve can either be above orbelow the normal depth of the mild slop e.

Prasuhn 1987

The transition fro m a steep to a mild slope cannot be accomplish ed by thewater profiles alone and requires a hydraulic jump . The equation describingthe ratio of y 2/y1 can be applied at the break in the grade wit h y 1 taken as theupstream normal d epth. The y 2 conjugate depth can be compared to thedownstr eam normal depth. If y 2 is greater than the downstream normaldepth, the jump must be downstr eam of the break, otherwise it is upstreamof the br eak. This is because y 2 decreases as y 1 increases for constantchannel and discharge conditions.

Prasuhn 1987

This is a more complic ated example. A jump may occur i n either section, butsince the adverse section ends with a drop off, there is the possibil ity thatno jump will occur.



Prasuhn 1987

On a mild slope, the flow wi ll be at the normal depth unti l the gate is loweredinto th e flow. Immediately thereafter, it wil l act as a control on t he upstreamsection. As the gate is lowered below the criti cal depth, it will act as a controlon the downstream section also, and a hydraulic jum p will occur. If the gate

remains above the critical depth, there will be lit tle energy loss and thedownstream depth will remain at nearly normal depth.

On a steep slope, the instant the gate enters the flow, a surge will f orm thatmoves upstr eam some distance and become station ary. The gate acts asboth a control for the upstream subcritical flow and the downstreamsupercritical flow .

m6b water surface profiles.pdf

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