hydraulic Chapter3

BFC21103 Hydraulics

Chapter 3. Specific Energy and Control Section

Tan Lai Wai, Wan Afnizan & Zarina Md [email protected]

Updated: September 2014

Learning Outcomes

At the end of this chapter, students should be able to:

i. Apply specific energy concept in determining critical flow conditions

ii. Analyse flow over broad‐crested weir

iii. Analyse flow through width constriction

BFC21103 Hydraulics Tan et al. ([email protected])

Specific energy (introduced by Bakhmeteff) is the energy of flow measured with respect to the channel bottom.

Bottom slope So

Datum

θ


1.1 Concept of Specific Energy

gV

yE2

2

+=

Energy line slope Sf

1 2

Water surface or hydraulic grade line slope Sw

2z

2y

fh

gV2

22

gV2

21

1y

1z


For constant Q, 2

2

2gAQ

yE +=

The concept of specific energy is useful in defining critical depth and in the analysis of flow problems.

Variation of E with y is represented by a cubic parabola,

E = y

45°E

y

subcritical, Fr < 1

supercritical, Fr > 1critical, Fr = 1

Emin E1= E2

y1

y2

gV2

21

gV2

22

ysuper

yc

ysub

y1y2

yc


For a specific E (except Emin), there are two flow depths y1 and y2, i.e.

Subcritical y

Supercritical yAlternate depths

If there is energy loss, e.g. during hydraulic jump, y1 and y2 are known as conjugate (or sequent) depths

Critical flow occurs when specific energy is minimum, Emin with yc = critical depth

Note: Negative flow depth is not possible.

1.2 Alternate Depths and Critical Depth


At minimum specific energy Emin, y = yc and A = Ac

Specific energy 2

2

2gAQ

yE +=

yA

gAQ

yE

dd

1dd

3

2

−=

cc

TgAQ

3

2

10 −=

yA

Tdd

=

12

2

=cc

c

AgATQ

dydA

12

=c

c

gDV

1=c

c

gDV

1Fr =

Differentiating


Since

Specific Energy of Flow in Rectangular Section

qBQ = 2

2

2gyq

yE +=

Rearranging ( )yEgyq −= 22

Variation of q with y is represented by the following curve,

yc

ysub

y1

y2

yc

subcritical, Fr < 1

supercritical, Fr > 1

critical, Fr = 1

ysuper

q

y

qmaxq1= q2


For a specific q (except qmax), there are two flow depths y1 and y2, i.e.

Subcritical y

Supercritical yConjugate depths

Critical flow occurs when discharge per unit width is maximum, i.e. qmax

Keeping E constant,

( )yEgAQ −= 2

( )( )yEggA

yEgyA

yQ

−−−=

22

dd

dd

12

2

=cc

c

AgATQ

1Fr =

QgA

AQ

T c

cc

2

0 −=


State of flow can be established by comparing yo with yc.

Characteristics Flow condition

Fr = 1yo = yc

Critical flow

Fr < 1yo > yc

Subcritical flow

Fr > 1yo < yc

Supercritical flow


Activity 3.1

The rate of flow in a 3‐m wide rectangular channel is 10 m3/s. Calculate the specific energy if the depth of flow is

(a) 3 m; and

(b) 1.2 m.

Given Q = 10 m3/s and B = 3 m.

When y = 3 m, m 063.33381.92

103

2 22

2

2

2

=×××

+=+=gAQ

yE

When y = 1.2 m, m 593.12.1381.92

102.1

2 22

2

2

2

=×××

+=+=gAQ

yE


1.3 Calculation of Critical Depth

Critical depth can be determined by:

i. Trial and error; or

ii. Graphically

1.3.1 Critical Depth from Trial‐and‐Error

For all channel sections, during critical flow (Emin)

13

2

=c

c

gATQ

gQ

TA

c

c23

=Rewritten as a function of critical depth,

is usually provided


For rectangular channel, T = B and A = By

becomes

gQ

TA

c

c23

=

gQ

ByB c

233

=

gBQ

yc 2

23 =

Since BQ

q = 3

2

gq

yc = (only applies to rectangular channel)


Critical depth also occurs when q is maximum

( )yEgyq −= 22

( )yEgyq −= 22 2

Differentiating q with respect to y

( )cc yEgyyq

q 322dd

2 min −=

032 min =− cyE0dd

=yq

gives

cyE23

min = (only applies to rectangular channel)


Critical slope Sc is used to categorize the type of channel slope

Condition of So Type of slopeSo = Sc Critical slope

So < Sc Mild slope

So > Sc Steep slope

Critical slope Sc can be calculated by equating Manning resistance flow equation to critical flow condition

21

323 1

cccc

c SRAnT

gAQ ==At critical slope, So = Sc

13

2

=c

c

gATQ

34

2

cc

cc

RT

gAnS =


Activity 3.2

A 4.0 m wide rectangular channel conveys water to a reservoir. If the discharge in the channel Q = 25 m3/s and Manning coefficient n = 0.02, find

(a) Critical depth

(b) Critical velocity

(c) Critical slope

3

2

gq

yc =

Given Q = 25 m3/s, B = 4.0 m, n = 0.02

(a)

m 585.1481.9

253

2

2

=×

=cy


1=c

c

gDV

(for rectangular section, T = B)

(b)

1=c

c

gyV

(for rectangular section, D = y)

m/s 943.3585.181.9 =×== cc gyV

(c)

007328.0

585.124585.14

585.181.902.0

34

2

=

⎟⎠⎞

⎜⎝⎛

×+×

××=cS

34

2

c

cc

R

gynS =

34

2

cc

cc

RT

gAnS =


Activity 3.3

(a) An infinitely wide and straight river has a discharge of 5.0 m3/s/m.Calculate:(i) Critical depth(ii) Froude number of the flow when the flow depth is 6.0 m and

determine the type of flow(iii) Critical slope of the channel if Manning coefficient n = 0.0044.

(b) Based on the river characteristics given in (a), find the possible depth of flow y2 for the same specific energy and the corresponding Froude number.

Given q = 5.0 m3/s/m, y = 6.0 m, n = 0.0044, For infinitely wide channel R ≈ y


(a) (i) m 366.181.95

3

2

3

2

===gq

yc

(a) (ii)

1086.0681.9

5Fr

33=

×===

gy

qgyV

For rectangular section, D = y

flow lsubcritica 11086.0Fr →<=

0001712.0366.1

81.90044.0

31

2

31

2

34

2

=×

===

cc

cc

y

gn

R

gynS

(a) (iii) For rectangular section, T = B


(b) Specific energy at y1 = 6 m

m 035.6681.92

56

2 2

2

21

2

11 =××

+=+=gyq

yE

The alternate depth of y1 = 6 m with E2 = E1 = 6.035 m is

035.62 2

2

2

2 =+gyq

y

035.681.925

22

2

2 =××

+y

y

m 4789.02 =y

817.44789.081.9

5Fr

33=

×===

gy

qgyV

At y2 = 0.4789 m,

flow calsupercriti 1817.4Fr →>=


Activity 3.4

For a trapezoidal channel with bottom width B = 6 m and side slope z = 2, find the critical flow depth if the discharge is 17 m3/s using trial‐and‐error method.

Given Q = 17 m3/s, B = 6 m, z = 2

z = 2y1

B = 6 m

Q = 17 m3/s

gQ

TA

c

c23

=

( )81.9

174626 232

=++

c

cc

yyy

( )365.7

233

32

=++

c

cc

yyy

m 8468.0=cyFrom trial‐and‐error,


1.3.2 Graphical Method

Critical depth of flow yc can be solved by plotting y against ccc

c DATA

or 3

Activity 3.5

For a trapezoidal channel with bottom width B = 6 m and side slope z = 2, find the critical flow depth if the discharge is 17 m3/s graphically.

Given Q = 17 m3/s, B = 6 m, z = 2


gQ

TA

c

c23

=

46.293

=c

c

TA

yc (m) Ac3/Tc

1 51.20

2 571.43

0.5 5.36

0.7 15.79

0.8 24.43

( )c

cc

c

c

yyy

TA

4626

323

++

=Also,

0

0.2

0.4

0.6

0.8

1

1.2

0 10 20 30 40 50 60c

c

TA3

yc (m)

29.46

0.84 m

m 84.0=cyFrom the graph,


1.4 Control Sections

A control section is where for a given discharge Q, the flow depth yand velocity V are fixed.

The critical depth yc is also a control point since at this section Fr = 1, effective when subcritical flow changes to supercritical flow. When supercritical flow changes to subcritical flow, a hydraulic jumps usually bypass the critical depth as control point.

A control section 'controls' the upstream or downstream flow.


Examples of control sections:

(a) Flow from a mild channel to steep channel

(b) A mild‐slope channel discharging into a pool

M2

S2yc

MildSteep

Drop

Pool

M2

yo

yo

yoyc

control control


(c) Free overflow (sudden drop)

(d) Reservoir water flows on a steep slope

Horizontal bed

H2

yc

controlS2

yc

ReservoirSteep yo

control


(e) Flow through sluice gate (f) Flow over spillway

M1

M3

ycMild

yo

control

M1

yc

control

control

Jump

Mild

yo


(g) Flow over broad‐crested weir

(h) Flow through constricted channel width

ycH

Hump

control

yc

B

control

Plan view

Constriction


1.5 Flow Over Broad‐Crested Weir

Flow in a prismatic open channel is uniform if there is no obstruction e.g. of a hydraulic structure.

If broad‐crested weir is installed, uniform flow changes to non‐uniform flow. Changes to the water surface profile is influenced by the weir height H and the flow condition before the weir (upstream flow), i.e. either supercritical or subcritical.

H

Weir

0 1 2 3

yo

yo = normal depth of flow

y1 = depth of flow just before weir

y2 = depth of flow on the weir

y3 = depth of flow just after weir


Minimum Height of Weir Hmin

Height of weir H determines the depth of flow above the weir y2, i.e. whether y2 = yc or not.

Hmin = minimum height of weir which will start to produce critical flow depth above the weir (y2 starts to change to yc)

Generally, depth of flow above the weir y2 is

If H < Hmin → y2 ≠ yc

If H = Hmin → y2 = yc

If H > Hmin → y2 = yc

Therefore, y2 = yc and E2 = Emin if H ≥ Hmin



Condition of upstream flow

yo

Case 1 Case 2 Case 3H y H y H y

Subcritical yo > yc

Supercritical yo < yc

H < Hminor

Emin + H < EoSubmerged

weir

y1 = y3 = yoy2 ≠ yc

E2 = Eo − H

H = Hminor

Emin + H = EoRarelyoccur

y1 = y3 = yoy2 = ycE2 = Emin

H > Hminor

Emin + H > EoControl weir

y1 ≠ y3 ≠ yoy1 = y1′y3 = y3′y2 = yc

E1,3 = Emin + HE2 = Emin

yo < y2 < yc

1 2 30H

y2

yo

yc

EoE2

yc < y2 < yo

1 2 30H

y2yo yc

EoE2

Eo

1 2 30

Hyo

y2=yc

E2 = Emin

1 2 30

H

yo y2=yc

E2 = Emin

Eo

1 2 30

Hyo

y2 = yc

E2 = Emin

Jump

Eo

y1′ > yc and y3′ < yo

yc y3

1 2 30

H

yo y2 = yc

E2 = Emin

Eo

y1′ > yo and y3′ < yc

yc y3

Backwater

y1

y1


Steps in Analysing Flow Over Broad‐Crested Weir

1. Calculate yo and ycDetermine state of upstream flow yo, i.e. either subcritical or supercritical by comparing with yc.If yo > yc → subcritical upstreamIf yo < yc → supercritical upstream

2. Calculate HminBy comparing height of weir H with Hmin, the condition of flow over weir can be established, i.e.If H < Hmin → Case 1If H = Hmin → Case 2If H > Hmin → Case 3

3. Determine y1, y2 and y3.


Case 1: H < Hmin

yo < y2 < yc

1 2 30H

y2

yo

yc

EoE2

Supercritical upstream yo < yc

E = y

E

y

Emin E2

y1,3 = yoy2yc

Eo

HHmin

yc < y2 < yo

1 2 30H

y2yo yc

EoE2

Subcritical upstream yo > yc

yc

E = y

E

y

Emin E2

y1 = y3 = yo

y2yc

EoH

Hmin

EGL

EGL


For Case 1,

H < HminH < Eo − EminE1 = E3 = Eoy2 ≠ yc

StepsUseful equations

All sections Rectangular section

1. Calculate yo and yc

Manning: Manning:

2. Calculate Hmin

3. Determine y1, y2 & y3

y1 = y3 = yoE2 = Eo − H

21

o

32

S

QnAR =

gQ

TA

c

c23

= 3

2

gq

yc =

21

o

32

o

S

qnRy =

2

2

oo 2gAQ

yE += 2o

2

oo 2gyq

yE +=

cyE23

min =c

c gAQ

yE2

2

min +=

minomin EEH −=

222

2

2 2E

gAQ

y =+ 222

2

2 2E

gyq

y =+


Case 2: H = Hmin

E = y

E

y

E2=Emin

y1,3 = yo

y2 = yc

Eo

H=Hmin

E = y

E

y

E2=Emin

y1 = y3 = yo

y2 = yc

Eo

H=Hmin

1 2 30

H

yo y2=yc

E2 = Emin

Eo

y2 = yc < yo


Eo

E2 = Emin

y2 = yc > yo


H

1 2 30

y2=yc

yo

EGL

EGL


For Case 2,

H = HminHmin = Eo − EminE1 = E3 = EoE2 = Eminy2 = yc




Manning: Manning:

2. Calculate Hmin


y1 = y3 = yoy2 = yc

21

o

32

S

QnAR =

gQ

TA

c

c23

= 3

2

gq

yc =

21

o

32

o

S

qnRy =

2

2

oo 2gAQ

yE += 2o

2

oo 2gyq

yE +=

cyE23

min =c

c gAQ

yE2

2

min +=

minomin EEH −=


E = y

E

y

E2=Emin

y2 = yc

E′1=E3Hmin

1 2 30

Hyo

E2 = Emin

JumpEo

y1′ > yc and y3′ < yo

yc y3

y1

y2 = yc < yoy′1 > yoy′3 < yc


y2 = yc > yoy′1 > yoy′3 > yc


EGL y1 = y1′ ≠ yo

y3 ≠ yo

E = y

E

y

E2=Emin

y1 = y1′ ≠ yo

y2 = yc

Eo

H>Hmin

1 2 30

H

yo y2 = yc

E2 = Emin

Eo


yc y3

Backwater

y1

EGL

y3 ≠ yo

Hmin

E′1,3

yo

yo

Eo

H>Hmin

y2 = yc


For Case 3,

H > HminH > Eo − EminE′1 = E′3 ≠ Eoy2 = yc




Manning: Manning:

2. Calculate Hmin


y′1 ≠ y′3 ≠ yoE′1,3 = Emin + H

3

2

gq

yc =

21

o

32

o

S

qnRy =

2o

2

oo 2gyq

yE +=

cyE23

min =

minomin EEH −=

3,123,1

2

1,3 2E

AgQ

y ′=′

+′3,12

1,3

2

1,3 2E

ygq

y ′=′

+′

gQ

TA

c

c23

=

2

2

oo 2gAQ

yE +=

cc gA

QyE

2

2

min +=

21

o

32

S

QnAR =


Activity 3.6

10 m3/s of flow is conveyed in a rectangular channel of 4 m width, n= 0.015 and So = 0.0075. If a weir with height 0.92 m is built in the channel, determine the depth of flow on the weir.

Given Q = 10 m3/s, B = 4 m, n = 0.015, So = 0.0075, and H = 0.92 m

B

y

Step 1. Determine yo and yc

21

o

32

o

S

qnRy =

21

32

o

oo

0075.0

015.0410

244

×⎟⎠⎞

⎜⎝⎛

=⎟⎟⎠

⎞⎜⎜⎝

⎛+ yy

y


m 8605.081.9410

3

2

3

2

=⎟⎠⎞

⎜⎝⎛

==gq

yc

4330.024

4 32

o

oo =⎟⎟

⎠

⎞⎜⎜⎝

⎛+ yy

y

m 6804.0o =y

yo < yc → supercritical flow

m 369.16804.081.92

410

6804.02 2

2

2o

2

oo =××

⎟⎠⎞

⎜⎝⎛

+=+=gyq

yE

m 291.18605.023

23

min =×== cyE

Step 2. Calculate Hmin

m 078.0291.1369.1minomin =−=−= EEH


Step 3. Determine y2

Since H = 0.92 m > Hmin = 0.078 m → Case 3 Hydraulic jump &y2 = yc = 0.8605 m


Activity 3.7

A rectangular channel conveys flow at yo = 1.6 m and R = 0.77 on So = 1/3000 and Manning n = 0.01.

(a) What is the minimum height of weir to control the flow in the channel?

(b) Calculate depth of flow upstream, downstream and above the weir in (a).

(c) Calculate depth of flow upstream, downstream and above the weir if the height of weir is(i) 0.4 m, and(ii) 0.6 m.


Given yo = 1.6 m, R = 0.77, So = 1/3000, n = 0.01

(a) Step 1. Determine yo and yc

m 8499.081.9454.2

3

2

3

2

===gq

yc

m 6.1o =y

yo > yc → subcritical flow

/s/mm 454.230001

77.06.101.011 32

1

32

21

o32

o =⎟⎠⎞

⎜⎝⎛×××== SRy

nq

m 720.16.181.92

454.26.1

2 2

2

2o

2

oo =××

+=+=gyq

yE

m 275.18499.023

23

min =×== cyE


m 445.0275.1720.1minomin =−=−= EEH


(b)

In (a), H = Hmin → Case 2.

For Case 2, y1 = y3 = yo = 1.6 my2 = yc = 0.8499 m

(c) (i) If H = 0.4 m < Hmin = 0.445 m → Case 1

For Case 1, y1 = y3 = yo = 1.6 m

E2 = Eo − H = 1.72 − 0.4 = 1.32 m

Step 3. Determine y1, y2 & y3.

222

2

2 2E

gyq

y =+

32.181.92

454.222

2

2 =×

+y

y

Since yo is subcritical, yo > y2 > yc, y2 = 1.032 m

Through trial‐and‐error, y2 = 1.032 m or y2 = 0.7085 m


(c) (ii) If H = 0.6 m > Hmin = 0.445 m → Case 3: Backwater

For Case 3, y2 = yc = 0.8499 m

Through trial‐and‐error, y′1 = 1.778 m and y′3 = 0.4669 m

since y′1 > yo and y′3 < yc

3,121,3

2

1,3 2E

ygq

y ′=′

+′

m 875.16.0275.1min3,1 =+=+=′ HEE

875.181.92454.2

21,3

2

1,3 =′×

+′y

y


1.6 Flow Through Constricted ChannelIf width of a prismatic channel is reduced/enlarged at a section, uniform flow changes to non‐uniform flow. Changes to the water surface profile is influenced by the width of constriction B2 and the flow condition before the constriction, i.e. either supercritical or subcritical.

B2q

B

1 2 30

q2

Plan view

2o

2o

2gyq

22

22

2gyq

q2q

1 2 30

EGL

y1 y3y2yo

q

y

qmaxqo

yoy2yc

q2

min32Eyc =

y

E

Side viewyoy2

qoq2

322

2

2 gBQ

yc =

Since B2 < Bo, q2 > qo

21

21

2

1

21

11 22 ygBQ

yg

VyE +=+=

22

22

2

2

22

22 22 ygBQ

yg

VyE +=+=

Bed elevations at 1 and 2 are the same, E1 = E2


Maximum Width of Constriction Bmax

Width of constriction B2 determines the depth of flow at the constricted section y2, i.e. whether y2 = yc2 or not.

Bmax = maximum width of constriction which will start to produce critical flow depth at the constriction (y2 starts to change to yc2)

yco or yc1 = critical depth of flow along the unconstricted section

yc2 = critical depth of flow at the constricted section.

Generally, depth of flow at constriction y2 is

If B2 > Bmax → y2 ≠ yc2

If B2 = Bmax → y2 = yc2

If B2 < Bmax → y2 = y′c2> yc2Therefore, y2 = yc2 or y′c2 and E2 = Emin if B2 ≤ Bmax



Condition of upstream flow

yo

Case 1 Case 2 Case 3B y B y B y

Subcritical yo > yco

Supercritical yo < yco

B2 > Bmaxor

Emin 2 < Eoor

q < qmax

y1 = y3 = yoy2 ≠ yc2E2 = Eo

B2 = Bmaxor

Emin 2 = Eoor

q = qmax

y1 = y3 = yoy2 = yc2

E2 = Emin 2 = Eo

B2 < Bmaxor

Emin 2 > Eoor

q > qmaxControl constriction

y1 ≠ y3 ≠ yoy1 = y1′y3 = y3′y2 = yc2

E1,3 = E'min 2 ≠ EoE2 = E'min 2

yo < y2 < yc2

1 2 30yo

EoE2 = Eo

yc2 < y2 < yo

1 2 30

yoEo

E2 = Eo

Eo

1 2 30yo

E2 = Emin 2 = Eo

1 2 30

yo

E2 = Emin 2 = Eo

Eo

ycoy2 yc2

yco y2 yc2

ycoy2=yc2

y2=yc2yco

1 2 30

yo

E2 = E′min 2

Eo

y1′ > yo and y3′ < yc2

yco y3

Backwater

y1y2=yc2

yo1 2 30

E2 = E′min 2

Jump

Eo

y1′ > yc2 and y3′ < yo

yco y3

y1y2=yc2


Steps in Analysing Flow Through Constriction

1. Calculate yo and ycoDetermine state of upstream flow yo, i.e. either subcritical or supercritical by comparing with yco.If yo > yco → subcritical upstreamIf yo < yco → supercritical upstream

2. Calculate yc2, qmax and BmaxWhen width of a channel is being constricted, yc2 can be obtained since Emin = Eo. Once Bmax is calculated, the condition of flow through the constriction can be established, i.e.If B2 > Bmax → Case 1If B2 = Bmax → Case 2If B2 < Bmax → Case 3

3. Determine y1, y2 and y3.


Case 1: B2 > Bmax

yo < y2 < yc2

Supercritical upstream yo < yco

yc2 < y2 < yo

Subcritical upstream yo > yco

E = y

E

y

EminoEmin 2

y1 = y3 = yoy2

yco

Eo=E1=E2=E31 2 30

y2yo

yc2

EoE2

yco

EGL

yc2B2 or q2B or qo

E = y

E

y

Emino Emin 2

y1 = y3 = yo

y2yco

Eo=E1=E2=E3

yc2B2 or q2B or qo

1 2 30yo

EoE2

EGL

ycoyc2

y2

Bmax or qmax

Bmax or qmax


For Case 1,

B2 > BmaxEmin 2 < EoE2 = Eoy2 ≠ yc2

Steps Useful equations

1. Calculate yo and yco

Manning: or

2. Calculate yc2, qmax and Bmax


y1 = y3 = yoE2 = Eo

21

o

32

S

QnAR =

3

2

o gq

yc =

21

o

32

o

S

qnRy =

;2 2

o

2

oo gyq

yE += 2min 23

cyE =

222

2

2 2E

gyq

y =+

;3

2max

2 gq

yc =max

max BQ

q =


Case 2: B2 = Bmax

y2 = yc2 > yo


y2 = yc2 < yo


E = y

E

y

Emino

y1 = y3 = yo

yco

Emin 2 = Eo1 2 30

yoy2=yc2

Eo

yco

EGL

y2 = yc2

B or qo

E = y

E

y

Emino

y1 = y3 = yo

yco

Emin 2 = Eo

y2 = yc2

B or qo1 2 30

yo

EoE2

EGL

yco y2 =yc2

Bmax or qmax

Bmax or qmax

E2


For Case 2,

B2 = BmaxEmin 2 = EoE2 = Emin 2 = Eoy2 = yc2



Manning: or



y1 = y3 = yoy2 = yc2

21

o

32

S

QnAR =

3

2

o gq

yc =

21

o

32

o

S

qnRy =

;2 2

o

2

oo gyq

yE += 2min 23

cyE =

;3

2max

2 gq

yc =max

max BQ

q =


Case 3: B2 < Bmax

y2 = yc2 > yo


y2 = yc2 < yo


E = y

E

y

Emino

yo

yco

Emin 21 2 30

yoEo

yco

EGL

y2 = y′c2

B or qo

E = y

E

y

Emino

y′3

yco

Emin 2

y2 = y′c2

B or qo1 2 30

yo

Eo

E2=E′min

EGL

ycoy′c2

Bmax

Bmax or qmax

E2=E′min

B2<Bmax

y′3y′c2 y′3

E′min 2

y′1

yc2y′1

yoy′3

y′1

Jump

Backwater

y′1

B2<Bmax

E′min 2


For Case 3,

B2 < BmaxE′min 2 ≠ EoE2 = E′min 2y2 = y′c2



Manning: or



y1 = y3 ≠ yo ; y2 = y′c2

21

o

32

S

QnAR =

3

2

o gq

yc =

21

o

32

o

S

qnRy =

;2 2

o

2

oo gyq

yE += 2min 23

cyE =

;3

2max

2 gq

yc =max

max BQ

q =

;2

max BQ

q =′

21,3

2max

1,31,3 2 ygq

yE′

′+′=′

3

2max

2 gq

yc′

=′

;23

2min cyE ′=′


Activity 3.8

A bridge is to be built across a 50‐m wide rectangular channel carrying flow of 200 m3/s at depth 4.0 m. For reducing the span of the bridge, what is the minimum width of channel such that the upstream water level will not be influenced by the constriction?

Given Q = 200 m3/s, yo = 4 m, B = 50 m

Step 1. Determine yo and yco

m 177.181.94

3

2

3

2

o ===gq

yc

m 0.4o =y

yo > yco → subcritical flow

/s/mm 0.450200 3

o ===BQ

q

yc is influenced by q. When q changes, yc varies as well. Therefore, at constriction where q ≠ qo, yc2 exists (calculated in Step 2).


Step 2. Calculate yc2 and Bmax

m 051.4481.92

44

2 2

2

2o

2o

2

oo =××

+=+=ygB

QyE

m 701.2051.432

32

min2 =×== Eyc

3

2max

2 gq

yc =

m 39.149.13

200

maxmax

maxmax ===→=

qQ

BBQ

q

With no energly loss, Emin = Eo, therefore

At width Bmax, E2 = Emin and q2 = qmax

Also,

/sm 90.13701.281.9 2332max =×== cgyq

rearranging gives

Since


Activity 3.9

A bridge is to be built across a 50‐m wide rectangular channel carrying flow of 200 m3/s at depth 4.0 m. The construction has caused the width of the channel to be reduced to 30‐m. Determine the depth of flow upstream, downstream and under the bridge.

Given Q = 200 m3/s, yo = 4 m, B = 50 m

Step 1. Determine yo and yco (similar to the solution in Activity 3.8)

m 177.181.94

3

2

3

2

o ===gq

yc

m 0.4o =y

yo > yco → subcritical flow

/s/mm 0.450200 3

o ===BQ

q


Step 2. Calculate yc2 and Bmax (similar to solution in Activity 3.8)

m 051.4481.92

44

2 2

2

2o

2o

2

oo =××

+=+=ygB

QyE

m 701.2051.432

32

min2 =×== Eyc

3

2max

2 gq

yc =

m 39.149.13

200

maxmax

maxmax ===→=

qQ

BBQ

q

With no energly loss, Emin = Eo, therefore

At width Bmax, E2 = Emin and q2 = qmax

Also,

/sm 90.13701.281.9 2332max =×== cgyq

rearranging gives

Since


Step 3. Determine y1, y2 and y3

m 051.4o2 == EE

Since B2 = 30 m > Bmax = 14.39 m → Case 1

At B2 = 30 m,

222

22

2

2 2E

ygBQ

y =+

051.43081.92

20022

2

2

2 =×××

+y

y

From trial‐and‐error, y2 = 0.8399 m or y2 = 3.902 m

Since yo > yco, thus yc2 < y2 < yo. Therefore, y2 = 3.902 m


Activity 3.10

A rectangular channel of 2.0 m width is required to convey 3 m3/s of flow. The normal depth is 0.8 m. At downstream of the channel, the width of the channel is to be reduced.

(a) Determine the width of the maximum constriction for critical depth to occur.

(b) Calculate the depth of flow upstream, downstream and at the constriction if the constricted width is 1.2 m.

Given Q = 3 m3/s, B = 2.0 m, yo = 0.8 m


(a) Step 1. Determine yo and yco

m 6121.081.95.1

3

2

3

2

o ===gq

yc

m 8.0o =y

Since yo > yco → subcritical flow

/s/mm 5.123 3

o ===BQ

q


m 9792.08.081.92

5.18.0

2 2

2

2o

2o

oo =××

+=+=gyq

yE

m 6528.09792.032

32

min2 =×== Eyc

m 816.1652.13

maxmax ===

qQ

B

/sm 652.16528.081.9 2332max =×== cgyq


(b) If B2 = 1.2 m,

m 816.1max2 =<BB → Case 3, where new qmax, i.e. q′max is required

Step 3. Calculate y1, y2 and y3

/sm 5.22.13 2

2max ===′

BQ

q

m 8605.081.95.2

3

2

3

2max

22 ==′

=′=g

qyy c

m 291.18605.023

23

2min =×=′=′ cyE


1,321,3

2max

1,3 2E

ygq

y ′=′

′+′

min31 EEE ′=′=′

291.181.925.2

21,3

2

1,3 =′××

+′y

y

From trial‐and‐error, y′1 = 0.8735 m or y′3 = 0.8476 m


Activity 3.11

Flow inside a rectangular channel of 3.0 m width has a velocity of 3.0 m/s at 3.0 m depth. The channel is experiencing a step of 0.61 m high at the channel bottom. What is the constriction to be made to the channel width in order to ensure the depth of flow upstream does not change.

Given V = 3 m/s, B = 3 m, yo = 3 m, and H = 0.61 mThus, q = yoV = 3 × 3 = 9 m2/s


m 021.281.99

3

2

3

2

o ===gq

yc

m 0.3o =y

Since yo > yco → subcritical flow


m 459.3381.92

93

2 2

2

2o

2

oo =××

+=+=gyq

yE

m 032.3021.223

23

min =×== cyE


m 427.0032.3459.3minomin =−=−= EEH

Step 3. Determine y1, y2 and y3

Since H = 0.61 m > Hmin = 0.427 m

→ Case 3: Backwater upstream of weir

m 642.361.0032.3min3,1 =+=+=′ HEE

In order to maintain the same specific energy and reduce y′1 to yo, q has to be increased, i.e. via width constriction.


E = y

E (m)

y (m)

E2=Emin

y1 = y1′ ≠ yo

yc =2.021

Eo

H=0.61 m

1 2 30

0.61 m

3 myc=2.021 m

Emin=3.032 m

Eo=3.459 m


yc=2.021 m y3

Backwater

y1

EGL

y3 ≠ yo

Hmin=0.427 m

E′1,3

yo =3.459

E = y

E (m)

y (m)

E2=Emin

y1 = y1′ ≠ yo

yc =2.021

Eo

H=0.61 m

1 2 30

0.61 m

3 myc=2.021 m

Emin=3.032 m

Eo=3.459 m


yc=2.021 m y3

Backwater

y1

EGL

y3 ≠ yo

Hmin=0.427 m

E′1,3

yo =3.459

E′′1,3

yc2


m 642.33,1min =′=′ EE

m 428.2642.332

32

min2 =×=′= Eyc

m 7595.085.119

maxmax ===

qQ

B

/sm 85.11428.281.9 2332max =×== cgyq


E = y

E (m)

y (m)

yc2 =2.021

Eo

H=0.61 m

1 2 30

0.61 m

3 m

Emin=3.642 m

Eo=3.459 m

yc=2.021 m y3

y1=3 m

EGL

y′3

Hmin=0.427 m

E′min

y′1 = yo =3.459

E′′1,3

B2=0.7595 mqoB=3 m

1 2 30

q2

Plan view

yc2=2.428 m


Activity 3.12

A rectangular channel of 2.6 m width with Manning n = 0.015, and longitudinal slope of 0.0008 is conveying flow at 9.8 m3/s. If a constriction is made by reducing channel width to 2.4 m, calculate depth of flow upstream and downstream of the constriction. Sketch the flow surface profile.

Given Q = 9.8 m3/s, B = 2.6 m, B2 = 2.4 m, n = 0.015, So = 0.0008


21

o

32

S

QnAR =

21

32

o

oo

0008.0

015.08.9262

6.26.2

×=⎟⎟

⎠

⎞⎜⎜⎝

⎛+ y.y

y


m 131.181.96.28.9

3

2

3

2

o =⎟⎠⎞

⎜⎝⎛

==gq

yc Since yo > yco → subcritical flow

197.5262

6.26.2

32

o

oo =⎟⎟

⎠

⎞⎜⎜⎝

⎛+ y.y

y

m 270.2o =yThrough trial‐and‐error,


m 411.227.281.92

6.28.9

27.22 2

2

2o

2o

oo =××

⎟⎠⎞

⎜⎝⎛

+=+=gyq

yE

m 607.1411.232

32

min2 =×== Eyc

Bmax is when q = qmax, where Emin = Eo


m 536.1381.68.9

maxmax ===

qQ

B

/sm 381.6607.181.9 2332max =×== cgyq

When B2 = 2.4 m, m 536.1max2 =>BB

Step 3. Calculate y1, y2 and y3

→ Case 1, where Emin2 < EoE2 = Eo

y1 = y3 = yo = 2.270 m

/sm 083.44.28.9 2

22 ===

BQ

q

m 193.181.9083.4

3

2

3

2

o ===gq

yc


22

22

22 2gyq

yE +=

411.281.92083.4

22

2

2 =××

+y

y

411.28497.0

22

2 =+y

y

Through trial‐and‐error,

m 7059.0 orm 242.2 22 == yy

Since it is subcritical upstream, m 242.22 =y

1 2 30

y2=2.242myo=2.270m

yc2=1.607m

Eo = 2.411 m

E2

yco=1.193m

EGL


1.7 ChokingChoking of flow occurs when

of a broad‐crested weir in an open channelminHH >

max2 BB < at the constricted width in an open channel

Choked conditions are undesirable in the design of culverts and other surface drainage features involving channel transitions.

i.e. when the specific energy or depth of flow immediately upstream of the weir or constriction increases or is being controlled.

BFC21103 Hydraulics Tan Lai Wai ([email protected])

Assignment #3

Q1. (a) Critical depth occurs in an open channel when the specific energy is minimum. Sketch the corresponding flow depth versus specific energy graph. From this concept, derive the general equation used to determine critical flow depth in an open channel.

(b) A rectangular channel 3.05 m wide carries 3.4 m3/s uniform flow at a depth of 0.6 m. A 0.2 m‐high weir is placed across the channel.(i) Does the weir cause hydraulic jump upstream of the weir?

Provide reason why.

(ii) Calculate the flow depth above the weir, and just upstream of the weir. Classify the surface profile of flowupstream of the weir. Sketch the resulting flow‐surface profile and energy line, showing the critical depth yc and normal depth yo.


Q2. (a) An engineer is to analyze flow in an open channel in which the channel is designed to be constricted by placing bridge embankment at both sides of the channel. Explain the consequences due to the constriction.

(b) An 8‐m wide rectangular channel is conveying flow uniformly at a rate of 18.6 m3/s and depth of 1.2 m. A temporary short span bridge is to be built across the channel in which bridge embankment is needed at both sides of the channel causing the channel to be constricted under the proposed bridge.(i) Calculate the maximum channel width under the proposed bridge

which will not cause backwater upstream.

(ii) If the channel width under the proposed bridge is 4 m due to theunavoidable condition, calculate the expected flow depth under the bridge, at just upstream and just downstream of the bridge.

(iii) If the flow depth just upstream of the proposed bridge is to be limited to 0.2 m higher than the normal depth, calculate the channel width under the bridge.

‐ End of Question ‐

THANK YOU


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