ch. 8 varied flow in open channels ch8-1... · 2019. 9. 3. · • rapidly varied flow ... → flow...

Seoul National University

Ch. 8 Varied Flow in Open Channels

8-1 Concept of Non-uniform Flows


Contents

8.1 Specific Energy

8.2 Critical Flow

8.3 Hydraulic Jump


Class objectives Introduce basic concept of the steady non-uniform flows Apply the work-energy equation to the non-uniform open channel flow

problems Derive equations of the hydraulic jump

3


Steady non uniform flow Steady non-uniform flow

– The velocity, and depth are different at the different points

– Important concepts are

• Specific energy 비에너지

• Specific force (or momentum) 비력

• Rapidly varied flow 급변류

• Gradually varied flow 점변류

4


- The specific energy was first introduced by Bakhmeteff in 1912.

- The specific energy concepts permit determination of how water surface

and velocity vary with change in flow cross-section.

- Specific energy (비에너지): total head above the channel bottom (not

from datum) = distance between the channel bottom and the energy line

22 1=2 2V QE y y

g g A = + +

8.1 Specific Energy

5

(10.13)


1. Specific energy

2

2 3

12

2 ( )

qE yg y

q g y E y

= +

= −

For a wide channel of rectangular cross section, specific energy is given

by the two-dimensional flow rate, q

(10.14)

6


1) Specific energy diagram: plot of E versus y with q constant

- With a given flowrate, E is increasing as depth decreases or increases

from the critical value, critical depth, yc, in which E has the minimum value

2) q-curve: plot of q versus y with E constant

Use diagram to solve Eq. (10.13)

2

2 3

12

2 ( )

qE yg y

q g y E y

= +

= −

7


2. E-diagram/q-diagram

( )

2

22

12

2

qE yg y

qE y yg

= +

− =

2 ( )q y g E y= −

8→ 3 alternate depths

2asymptotes 0; 0E y y→ − = =


Critical depth

0 0dE dqordy dy

= =

23 30 : ,c c

dE qq gy ydy g

= = =

min min320 :

3 2c

cydq y E or E

dy= = =

(10.15)

(10.16)

• Eq. (10.15) shows that critical depth is dependent on flowrate only.

• Critical flow can be used as means of flowrate measurement.

→ critical-depth meters (broad-crested weir)

9


E-diagram

minE

cy


In both specific energy diagram and q-curve, there exist two different

depths for a given specific energy or unit discharge

→ Alternate depths: subcritical flow depth

대응수심 supercritical flow depth

Alternate depths

11


Consider two types of local changes in flow cross section- Channel hump (step) or broad-crested weir- Channel constriction

3. Use of specific energy

12


Channel hump: The structure is sufficiently smooth and streamlined so that no additional friction losses are introduced.→ Energy line remains parallel to the channel bottom.

As flow moves from upstream channel over the hump, the distance from the channel bottom to the energy line, i.e. specific energy, E,decreases even though the total energy remains the same.It appears that the depth of flow will either increase or decrease as flow moves onto the hump depending on whether the upstream flow is supercritical or subcritical.

1) Channel hump (broad-crested weir)

13

2 21 1 2 2

1 22 2 Lp V p Vz z h

g gγ γ+ + = + + +


1) Case I: upstream flow is subcriticalWork-energy eq. from channel bottom is

So the new specific energy is just the initial specific energy minus the step dimensions, E2= E1-∆z

→ the specific energy decreases, even though total energy unchanged.

2 21 2

1 1 2

1 2

2 2(specific energy)

V VE y z yg g

E z E

= + = ∆ + +

= ∆ +

14


In terms of the total energy given by the Bernoulli equation, the depth at the step is given by

For a rectangular channel section,

2 22

2 1 1 222 2

V Qy E z E zg gA

= − ∆ + = − ∆ +

2 22

2 1 1 222 2

V qy E z E zg gy

= − ∆ + = − ∆ +

If the upstream flow is subcritical, then the decrease in E will cause a decrease in depth of flow.→ This result in a drop in water surface over the hump→ The flow velocity and the velocity head must increase.

15


• If the hump were made high enough (∆z = E1 – Emin), the critical depth

would occur over the hump.

→ Broad-crested weir

• Further, if the hump were made even higher, the flow would not have

enough specific energy to flow over the hump and the hump would

become a dam. The depth at the hump will still remain the critical depth.

→ Flow would back up, gaining enough elevation to flow over the hump at

minimum energy and critical depth.16

2E

cycy


Broad-crested weirs:

For real weir flow the relation between flowrate and head is

Combine (10.15) and (14.23)

17

3 3/23 3/2 3/2

min2 2 20.577 23 3 3c

q gy g E gE gE = = = = ×

3/22 23w

q C gH= (14.23)

(10.15)

3/213w

ECH

=

2

132 2

w

c

PC fH

VE E P

g

=

= − +

Re,We,

Broad-crested weirs

0.08 0.5w

HL

<


2) Case II: upstream flow is supercritical

If the upstream flow is supercritical, the decrease in E will cause a

increase in depth over the hump.

18

2 21 2

1 1 2

1 2

2 2(specific energy)

V VE y z yg g

E z E

= + = ∆ + +

= ∆ +

2 22

2 1 1 222 2

V Qy E z E zg gA

= − ∆ + = − ∆ +

1E


2) Channel constriction

When contraction occurs, specific energy, E is constant, but unit discharge (q) increases.

If the upstream flow is subcritical, a decrease in flow depth occurs in the constriction.

If the upstream flow is supercritical, a depth increase occurs in the constriction. 19

Q, q0 Q, qc

00

cc

Q Qq qB B

= > =

q∆


• If the channel constriction is made sufficiently narrow, the critical depth

would occur in the constriction.

→ Venturi flume, Parshall flume

20


• If the constriction is narrowed further, the flow will be unable to pass

through with the specific energy available and the constriction

becomes a choke. For subcritical flow upstream, flow will then back

up much as occurred with the hump, creating an unconventional dam,

until sufficient specific energy is built up to pass through the

constriction at the critical depth. If the supercritical flow exists

upstream, a hydraulic jump will form and the flow will back up, gaining

enough elevation to flow through the constriction at minimum energy

condition.

21


8.2 Critical Flow

3,c c cq V y q gy= =

c cV gy=

• Critical velocity may be obtained from combining the two equations.

• This equation is similar to another equation of open channel flow.

→ speed of propagation of a small gravity wave on the surface of a

liquid of depth y → Ch. 6 (pp. 200-203)

c gy=

(a)

(b)

22

1. Critical velocity


Supercritical flow

Subcritical flow

Critical flow

23

• In supercritical flow, V > Vc, small

gravity waves will not propagate

upstream but rather will be swept

downstream.

• In subcritical flow, V < Vc, small

gravity waves will propagate

upstream.


• In subcritical flow, Fr < 1

• In supercritical flow, Fr > 1

VFrgy

=

Supercritical flow

Subcritical flow

Critical flow

24

1VFrgy

= >

1VFrgy

=


2 233

min

101) 1.4632.2

3 2.19 .2

c

c

qy ftg

E y ft

= = =

= =

( )

( ) ( )

2

2

2

2 22 22 2

2

2) At the dpeth of 0.5ft (supercritical flow)100.5 6.71

2 0.5

10 1.556.712

6.68

E ftg

y yg y y

y

= + =

= + = +

=

2 233

153) 1.9132.2c

qy ftg

= = =

6.68

1.460.5

6.71

[Example 1]Water flows in a rectangular channel of width 5 ft. 1) When discharge is 50 ft3/s find the critical depth and minimum specific energy. 2) When depth of flow is 0.5 ft, find the specific energy and the alternate depth. 3) If discharge increases to 75 ft3/s, what is the critical depth?

25


[Example 2] A uniform flow of 100 ft3/s occurs in a rectangular channel of 5 ft width and 5 ft depth. A smooth rise of 0.5 ft is formed in the channel bed. 1) Determine the depth of flow above step. 2) What is the height of the step that will produce a critical flow?

The discharge per unit width is

The critical depth is

Initial flow depth is larger than critical depth, so the flow is subcritical.

2100 20 /5

Qq ft sb

= = =

2.32

2

3 min2.32 , 1.5(2.32) 3.48cqy ft E ftg

= = = =

3.4826


1) The initial energy available for flow,

Applying the Bernoulli equation across the step and ignoring losses,

2) Required ∆z to cause critical depth at the step: Critical depth will occur at the step when the available energy E1 is equal to Emin.

( )( )

ftggy

qyE 25.552

2052 2

2

2

2

1 =+=+=

2 2

1 22 21 2

2 2 22 22 2

2

2

2 26.21 6.210.5 4.75

1.35 4.45 choose

5

4.454.7

. 5

5

2

q qy z ygy gy

y E yy y

y or ft ftE ft

+ = ∆ + +

= + + → = + =

= →=

ftzzzEE 77.1,48.325.5,min1 =∆∆+=∆+= 27


• If uniform flow were to occur at critical depth, the slope of the channel bottom would be called the critical slope Sc.

• For a rectangular channel of great width,

• Sc is obtained by equating the flowrate of (10.5) and (10.15)

• S > Sc → steep slope → will produce supercritical uniform flow• S < Sc → mild slope → will produce subcritical uniform flow

h cR y=

2. Critical slope

3 5/3 1/2

2

2 1/3

c c c

cc

uq gy y Sn

gnSu y

= =

=

28


2 233

12.5 501) 1.69 12.532.2 4

4 subcritical flow

c

c

q Qy ft qg B

y ft y

= = = ← = = =

= > →

2 2

2 1/3 2 2

2) For wide rectangular channel, (y = 1.69ft) c occurs at the slope32.2 (0.017) 0.00351.49 1.6

ritical unifor

9

m flow

cc

gnSu y

×= = =

×

( )

2/3 1/20

2/31/2

0 3

0

2/

1 0.017 1500 0.015981.49 4 404 40

40 2 4

3)Tocompute uni

subcritcal u

form flow,use Manning Eq.

0.00025 nifrom flo5 mild l e ws op

h

c

h

u Sn

nS Q

Q AR

Au R

S S

=

= = × × + ×

= → →

=

<

IP 10.6 (p. 455)For a flowrate of 500 cfs in a rectangular channel 40 ft wide, the water depth is 4 ft. 1) Is this flow subcritical or supercritical? 2) If n = 0.017, what is the critical slope of this channel for this flowrate? 3) What channel slope would be required to produce uniform flow at a depth of 4 ft?

29

← uniform flow ?


For upstream channel, flow is supercritical uniform flow For downstream channel, flow is subcritical uniform flow

Critical slope

30

𝑦𝑦𝑐𝑐


31

EMV 유속계Head Tank

700

6560

펌프

600

For given Q,adjust thechannelslope critical utocreate t niforhe m flow


For non-rectangular channels, specific energy equation is given as

The critical depth relationship can be found by

3. Critical depth in non-rectangular channels

2

3

20 1 02

dE Q dAdy g A dy

= → + − =

(10.18)22 2

21 1 ( )2 2 ( ) 2

Q Q QE y y y A yg A g A y g

− = + = + = +

(A)

32

3

2

dA gAdy Q

→ =


For non-rectangular channels,

Substituting this into (A) results in the critical depth relationship as

The critical flow can be defined by the Froude number as

2

3 1Q bgA

=

dA bdy= (B)

(10.19)

2

3 1( / )cQ b VFrgA g A b

= = = (10.20)

Top width

33

3gAQb

→ =


The critical slope (for critical uniform flow) can be derived by following the same procedure as for the wide rectangular channel

A, b, and Rh are all functions of depth and dependent on the section shape. Critical slope is thus a function of section shape.

(10.20)

32/3 1/2

h cgA uQ AR Sb n

= =

2

2 4/3ch

gn ASu bR

=

34


IP 10.7 (pp. 457-8)

A flow of 28 m3/s occurs in an earth-lined canal having a base width of 3 m, side slopes z = 2 and Manning n-value of 0.022. Calculate the critical depth and critical slope.

Solution22

3 2 3

(28) (3 4 )19.81 (3 2 )

1.50

c

c c

c

yQ bgA y yy m

× += =

× +

=

35


Area and hydraulic radius corresponding to the critical flow conditions are

2 2 2

2

2 4/3

2

2 4/3

3 2 3 1.50 2 1.50 9.09.0 0.93

3 2 5 1.502 3 2 2 1.50 9.0

9.81 0.022 9.0 0.005231 9.0 0.93

c c

h

ch

A y y mAR mP

b B zy m

gn ASu bR

= + = × + × =

= = =+ ×

= + = + × × =

=

× = = ×

36


(1) Mild slope to steep slope channel

The critical depth can be expected in the situation where a long channel

of mild slope is connected to a long channel of steep channel.

Far up the former channel, uniform subcritical flow at normal depth,

y01, will occur, and far down the latter a uniform supercritical flow at a

smaller normal depth, y02, will occur.

In a reach of varied flow, depth must pass through the critical.

4. Occurrence of critical depth

37


(2) Steep slope to mild slope channel

When a long channel of steep slope discharges into to a long

channel of mild channel, normal depth will occur upstream and

downstream from the point of slope change.

However, critical depth will not be found near this point.

The hydraulic jump will form whose location will be dictated by the

details of slopes, roughness, channel shapes, and so forth.

The critical depth will be found within the hydraulic jump.

38


(3) Free outfall

A long channel of mild slope is concluded to the brink

The critical depth occurs a short distance (3~4 yc) upstream from the

brink

The brink depth (yb) is 71.5% of the critical depth (Rouse, 1970)

39


Hydraulic jump occurs, when the flow moves from the depth lower than critical depth (y1) to the depth which is deeper than critical depth (y2 ). That is, flow change from super-critical to sub critical

8.3 Hydraulic Jump

40

1. Hydraulic jump


When channel is wide rectangular, there are two conjugate depths (공액수심) exist, and from this conjugate set of depths we can estimate the height of the jump.Ignore the friction (short distance), then momentum equation becomes.

1 2 2 12 2

1 2

2 12 22 2

1 2

1 2

( )

( )2 2

(A)2 2

F F q V Vy y q qq

y yy yq q

gy gy

ρ

γ γ ρ

− = −

− = −

+ = +

22 1

1 1

81 1 12

y Vy gy

= − + +

41

Force per unit width

Specific force = force per unit width per specific weight


From (A), the specific force (비력) is defined as a function of depth→ M-curve

2 2

2q yMgy

= +

• A set of conjugate depths (공액수심) exists.

• Conjugate depth can be found by passing a vertical line since a vertical line is a line of constant M.

• The hydraulic jump can take place only across the critical depth.

2. Specific force

2

30 1c

dM qdy gy

= → =

q constant

42


Conjugate depth (공액수심): same pressure and momentum relation.

Alternate depth(대응수심): same specific energy Left figure is specific force, right figure is specific energy From the left hand side, find conjugate depth and then plug

into the right hand side, then we can figure out how much energy is changed

3. Conjugate and alternative depths

43

q constantq constant


Because of eddies (rollers), air entrainment, and flow decelerations in the hydraulic jump, large head losses are to be expected. In the horizontal channel, the loss of energy is given as

• When the channel is very wide or rectangular, the unit discharge expression is (q / y),

+−

+=∆

gVy

gVyE

22

22

2

21

1

( )

−+−=∆ 2

221

2

2111

2 yygq

yyE

4. Energy loss by hydraulic jump

44


( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

2

1 2 2 21 2

2 21 2 1 2 2 1

1 2 2 21 2

2 21 2 2 1

1 2 2 21 2

21 2

2 11 2

2 22 1 2 1

2 11 2

32 1

1 2

1 12

4

4

14

24

4

qE y yg y y

y y y y y yy yy y

y y y yy yy y

y yy y

y y

y y y yy yy y

y yE

y y

∆ = − + −

+ −

= − +

+ −= − +

+

= − − + − +

= −

−∆ =

45


Undular jump (1


Stable jump (4.5


Ogee-Crested Spillways (Q = 150, 200, 250 l/min)

발표자프레젠테이션 노트수정


Half-circle dam

Broad-crested weir

Trapezoidal dam

발표자프레젠테이션 노트수정

슬라이드 번호 1ContentsClass objectivesSteady non uniform flow8.1 Specific Energy1. Specific energyUse diagram to solve Eq. (10.13)2. E-diagram/q-diagramCritical depthE-diagramAlternate depths3. Use of specific energy1) Channel hump (broad-crested weir)슬라이드 번호 14슬라이드 번호 15슬라이드 번호 16Broad-crested weirs슬라이드 번호 182) Channel constriction 슬라이드 번호 20슬라이드 번호 218.2 Critical Flow슬라이드 번호 23슬라이드 번호 24슬라이드 번호 25슬라이드 번호 26슬라이드 번호 27슬라이드 번호 28슬라이드 번호 29Critical slope슬라이드 번호 313. Critical depth in non-rectangular channels슬라이드 번호 33슬라이드 번호 34슬라이드 번호 35슬라이드 번호 364. Occurrence of critical depth슬라이드 번호 38슬라이드 번호 398.3 Hydraulic Jump슬라이드 번호 412. Specific force3. Conjugate and alternative depths4. Energy loss by hydraulic jump슬라이드 번호 455. Classification of Hydraulic jump슬라이드 번호 47슬라이드 번호 48슬라이드 번호 49

ch. 8 varied flow in open channels ch8-1... · 2019. 9. 3. · • rapidly varied flow ... → flow...

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