Open Channel Hydraulics Topic 3 | Specific Energy ….… · Open Channel Hydraulics Topic 3 |...

Open Channel Hydraulics

Topic 3 | Specific Energy and Control Section

Prepared by:

Tan Lai Wai et [email protected]

Learning Outcomes

At the end of this topic, 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

Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])


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

Datum

1.1 Concept of Specific Energy

g

VyE

2

2

1 2

2z

2y

fh

g

V

2

22

g

V

2

21

1y

1z


For constant Q,2

2

2gA

QyE

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,

45°0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E

y

subcritical, Fr < 1

supercritical, Fr > 1critical, Fr = 1

Emin E1= E2

y1

y2

g

V

2

21

g

V

2

22

ysuper

yc

ysub

y1

y2yc

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 energy2

2

2gA

QyE

y

A

gA

Q

y

E

d

d1

d

d3

2

c

c

TgA

Q3

2

10

y

AT

d

d

12

2

cc

c

AgA

TQdy

dA

12

c

c

gD

V

1c

c

gD

V

1Fr

Differentiating



Since

Specific Energy of Flow in Rectangular Section

qBQ 2

2

2gy

qyE

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= q20

1

2

3

4

5

6

0 10 20 30 40 50

Q (m3/s)

y (

m)

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

yEg

gAyEg

y

A

y

Q

22

d

d

d

d

12

2

cc

c

AgA

TQ

1Fr

Q

gA

A

QT c

c

c

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

gA

QyE

When y = 1.2 m, m 593.12.1381.92

102.1

2 22

2

2

2

gA

QyE


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

gA

TQ

g

Q

T

A

c

c23

Rewritten as a function of critical depth,

is usually provided


For rectangular channel, T = B and A = By

becomes

g

Q

T

A

c

c23

g

Q

B

yB c233

gB

Qyc 2

23

Since B

Qq 3

2

g

qyc (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 yEgyy

qq 322

d

d2 min

032 min cyE0d

d

y

qgives

cyE2

3min (only applies to rectangular channel)


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

Condition of So Type of slope

So = 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

2

1

3

23 1ccc

c

c SRAnT

gAQ At critical slope, So = Sc

13

2

c

c

gA

TQ

3

4

2

cc

cc

RT

gAnS


Activity 3.2

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

(a) Critical depth

(b) Critical velocity

(c) Critical slope

3

2

g

qyc

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

(a)

m 585.1481.9

253

2

2

cy


1c

c

gD

V

(for rectangular section, T = B)

(b)

1c

c

gy

V(for rectangular section, D = y)

m/s 943.3585.181.9 cc gyV

(c)

007328.0

585.124

585.14

585.181.902.0

3

4

2

cS

3

4

2

c

cc

R

gynS

3

4

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 possibledepth of flow y2 for the same specific energy and thecorresponding 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 366.181.9

53

2

3

2

g

qyc

(a) (ii)

1086.0681.9

5Fr

33

gy

q

gy

V

For rectangular section, D = y

flow lsubcritica 11086.0Fr

0001712.0

366.1

81.90044.0

3

1

2

3

1

2

3

4

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

gy

qyE

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

035.62 2

2

2

2 gy

qy

035.681.92

522

2

2

y

y

m 4789.02 y

817.44789.081.9

5Fr

33

gy

q

gy

VAt y2 = 0.4789 m,

flow calsupercriti 1817.4Fr

Activity 3.4

For a trapezoidal channel with bottom width B = 6 m and side slopez = 2, find the critical flow depth if the discharge is 17 m3/s usingtrial-and-error method.

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

z = 2y1

B = 6 m

Q = 17 m3/s

g

Q

T

A

c

c23

81.9

17

46

26 232

c

cc

y

yy

365.7

23

332

c

cc

y

yy

m 8468.0cyFrom trial-and-error,


1.3.2 Graphical Method

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

c

c DAT

A or

3

Activity 3.5

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

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


g

Q

T

A

c

c23

46.293

c

c

T

A

yc (m) Ac3/Tc

1 51.20

2 571.43

0.5 5.36

0.7 15.79

0.8 24.43

0

0.2

0.4

0.6

0.8

1

1.2

0 10 20 30 40 50 60

c

cc

c

c

y

yy

T

A

46

26323

Also,

0

0.2

0.4

0.6

0.8

1

1.2

0 10 20 30 40 50 60c

c

T

A3

yc (m)

29.46

0.84 m

m 84.0cyFrom 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

S2

yc

MildSteep

Drop

Pool

M2

yo

yo

yo

yc

controlcontrol


(c) Free overflow (sudden drop)

(d) Reservoir water flows on a steep slope

Horizontal bed

H2

yc

controlS2

yc

Reservoir

Steep 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

yc

H

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 3

H y H y H y

Subcritical yo yc

Supercritical yo yc

H Hmin

or Emin H

Eo

Submerged weir

y1 y3 yo

y2 yc

E2 Eo H

H Hmin

or Emin H

Eo

Rarelyoccur

y1 y3 yo

y2 yc

E2 Emin

H Hmin

or Emin H

Eo

Control weir

y1 y3 yo

y1 = y1y3 = 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

y2yc

E2 Emin

1 2 30

H

yo y2yc

E2 Emin

Eo

1 2 30

Hyo

y2 yc

E2 Emin

Jump

Eo

y1 yc and y3 yo

yc y3

1 2 30

H

yoy2 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 Hmin

By 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

0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E

y

Emin E2

y1,3 yo

y2

yc

Eo

HHmin

yc y2 yo

1 2 30H

y2yo yc

EoE2

Subcritical upstream yo yc

yc

0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E

y

Emin E2

y1 y3 yo

y2yc

Eo

HHmin

EGL

EGL


For Case 1,

H Hmin

H < Eo Emin

E1 E3 Eo

y2 yc

StepsUseful equations

All sections Rectangular section

1. Calculate yo and yc

Manning: Manning:

2. Calculate Hmin

3. Determine y1, y2 & y3

y1 y3 yo

E2 = Eo H

2

1

o

3

2

S

QnAR

g

Q

T

A

c

c23

3

2

g

qyc

2

1

o

3

2

o

S

qnRy

2

2

oo2gA

QyE

2o

2

oo2gy

qyE

cyE2

3min

c

cgA

QyE

2

2

min

minomin EEH

222

2

22

EgA

Qy 22

2

2

22

Egy

qy


Case 2: H Hmin

0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E

y

E2Emin

y1,3 yo

y2 yc

Eo

HHmin

0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E

y

E2Emin

y1 y3 yo

y2 yc

Eo

HHmin

1 2 30

H

yo y2yc

E2 Emin

Eo

y2 yc yo


Eo

E2 Emin

y2 yc yo


H

1 2 30

y2yc

yo

EGL

EGL


For Case 2,

H Hmin

Hmin Eo Emin

E1 E3 Eo

E2 Emin

y2 yc




Manning: Manning:

2. Calculate Hmin


y1 y3 yo

y2 = yc

2

1

o

3

2

S

QnAR

g

Q

T

A

c

c23

3

2

g

qyc

2

1

o

3

2

o

S

qnRy

2

2

oo2gA

QyE

2o

2

oo2gy

qyE

cyE2

3min

c

cgA

QyE

2

2

min

minomin EEH

0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E

y

E2Emin

y2 yc

E1E3

Hmin

1 2 30

Hyo

E2 Emin

JumpEo

y1 yc and y3 yo

yc y3

y1

y2 yc yo

y1 yo

y3 yc


y2 yc yo

y1 yo

y3 yc


EGL y1 y1 yo

y3 yo

0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E

y

E2Emin

y1 y1 yo

y2 yc

Eo

HHmin

1 2 30

H

yoy2 yc

E2 Emin

Eo

y1 yo and y3 yc

yc y3

Backwater

y1

EGL

y3 yo

Hmin

E1,3

yo

yo

Eo

HHmin

y2 yc



For Case 3,

H Hmin

H Eo Emin

E1 E3 Eo

y2 yc




Manning: Manning:

2. Calculate Hmin


y1 y3 yo

E1,3 = Emin H

3

2

g

qyc

2

1

o

3

2

o

S

qnRy

2o

2

oo2gy

qyE

cyE2

3min

minomin EEH

3,123,1

2

1,32

EAg

Qy

3,121,3

2

1,32

Eyg

qy

g

Q

T

A

c

c23

2

2

oo2gA

QyE

c

cgA

QyE

2

2

min

2

1

o

3

2

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 thechannel, 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

2

1

o

3

2

o

S

qnRy

2

1

3

2

o

oo

0075.0

015.04

10

24

4

y

yy


m 8605.081.9

4

103

2

3

2

g

qyc

4330.024

4 3

2

o

oo

y

yy

m 6804.0o y

yo yc supercritical flow

m 369.16804.081.92

4

10

6804.02 2

2

2o

2

oo

gy

qyE

m 291.18605.02

3

2

3min cyE

Step 2. Calculate Hmin

m 078.0291.1369.1minomin EEHOpen Channel Hydraulics

by Tan Lai Wai et al. ([email protected])

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 m onSo = 1/3000 and Manning n 0.01.

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

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

(c) Calculate depth of flow upstream, downstream and abovethe 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 8499.081.9

454.23

2

3

2

g

qyc

m 6.1o y

yo yc subcritical flow

/s/mm 454.23000

177.06.1

01.0

11 32

1

3

2

2

1

o3

2

o

SRy

nq

m 720.16.181.92

454.26.1

2 2

2

2o

2

oo

gy

qyE

m 275.18499.02

3

2

3min 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

22

Egy

qy

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,y1 1.778 m and y3 0.4669 m

since y1 yo and y3 yc

3,121,3

2

1,32

Eyg

qy

m 875.16.0275.1min3,1 HEE

875.181.92

454.22

1,3

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

2gy

q

22

22

2gy

q

q2q

1 2 30

EGL

y1 y3y2

yo

q

y

qmaxqo0

1

2

3

4

5

6

0 10 20 30 40 50

Q (m3/s)

y (

m)

yoy2yc

q2

min3

2Eyc

y

E

Side view

yo

y2

qo

q2

322

2

2gB

Qyc

Since B2 < Bo, q2 > qo

21

21

2

1

21

1122 ygB

Qy

g

VyE

22

22

2

2

22

2222 ygB

Qy

g

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 yc2> yc2

Therefore, y2 = yc2 or yc2 and E2 = Emin if B2 Bmax

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

Condition of upstream flow

yo

Case 1 Case 2 Case 3

B y B y B y

Subcritical yo yco

Supercritical yo yco

B2 Bmax

or Emin 2 Eo

orq qmax

y1 y3 yo

y2 yc2

E2 Eo

B2 Bmax

or Emin 2 Eo

orq qmax

y1 y3 yo

y2 yc2

E2 Emin 2 = Eo

B2 Bmax

or Emin 2 Eo

orq qmax

Control constriction

y1 y3 yo

y1 = y1y3 = y3y2 yc2

E1,3 E'min 2 Eo

E2 E'min 2

yo y2 yc2

1 2 30

yo

Eo

E2 Eo

yc2 y2 yo

1 2 30

yo

Eo

E2 Eo

Eo

1 2 30

yo

E2 Emin 2 Eo

1 2 30

yo

E2 Emin 2 Eo

Eo

yco

y2 yc2

yco y2 yc2

ycoy2yc2

y2yc2yco

1 2 30

yo

E2 Emin 2

Eo

y1 yo and y3 yc2

yco y3

Backwater

y1

y2yc2

yo

1 2 30

E2 Emin 2

Jump

Eo

y1 yc2 and y3 yo

yco y3

y1

y2yc2

Steps in Analysing Flow Through Constriction1. Calculate yo and yco

Determine 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 Bmax

When 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

0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E

y

EminoEmin 2

y1 y3 yo

y2

yco

EoE1E2E31 2 30

y2

yo

yc2

EoE2

yco

EGL

yc2

B2 or q2

B or qo

0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E

y

Emino Emin 2

y1 y3 yo

y2

yco

EoE1E2E3

yc2

B2 or q2

B or qo

1 2 30

yo

EoE2

EGL

yco

yc2

y2

Bmax or qmax

Bmax or qmax


For Case 1,

B2 Bmax

Emin 2 Eo

E2 Eo

y2 yc2

Steps Useful equations

1. Calculate yo and yco

Manning: or



y1 y3 yo

E2 = Eo

2

1

o

3

2

S

QnAR

3

2

og

qyc

2

1

o

3

2

o

S

qnRy

;2 2

o

2

oogy

qyE

2min2

3cyE

222

2

22

Egy

qy

;3

2max

2g

qyc

max

maxB

Qq


Case 2: B2 Bmax

y2 yc2 yo


y2 yc2 yo


0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E

y

Emino

y1 y3 yo

yco

Emin 2 Eo1 2 30

yo

y2yc2

Eo

yco

EGL

y2 yc2

B or qo

0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E

y

Emino

y1 y3 yo

yco

Emin 2 Eo

y2 yc2

B or qo

1 2 30

yo

EoE2

EGL

yco y2 yc2

Bmax or qmax

Bmax or qmax

E2


For Case 2,

B2 Bmax

Emin 2 Eo

E2 Emin 2 Eo

y2 yc2



Manning: or



y1 y3 yo

y2 = yc2

2

1

o

3

2

S

QnAR

3

2

og

qyc

2

1

o

3

2

o

S

qnRy

;2 2

o

2

oogy

qyE

2min2

3cyE

;3

2max

2g

qyc

max

maxB

Qq


Case 3: B2 Bmax

y2 yc2 yo


y2 yc2 yo


0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E

y

Emino

yo

yco

Emin 21 2 30

yo

Eo

yco

EGL

y2 yc2

B or qo

0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E

y

Emino

y3

yco

Emin 2

y2 yc2

B or qo

1 2 30

yo

Eo

E2Emin

EGL

yco

yc2

Bmax

Bmax or qmax

E2Emin

B2Bmax

y3

yc2 y3

Emin 2

y1

yc2y1

yo

y3

y1

Jump

Backwater

y1

B2Bmax

Emin 2


For Case 3,

B2 Bmax

Emin 2 Eo

E2 Emin 2

y2 yc2



Manning: or



y1 y3 yo ; y2 = yc2

2

1

o

3

2

S

QnAR

3

2

og

qyc

2

1

o

3

2

o

S

qnRy

;2 2

o

2

oogy

qyE

2min2

3cyE

;3

2max

2g

qyc

max

maxB

Qq

;2

maxB

Qq

21,3

2max

1,31,32 yg

qyE

3

2max

2g

qyc

;2

32min cyE


Activity 3.8

A bridge is to be built across a 50-m wide rectangular channelcarrying flow of 200 m3/s at depth 4.0 m. For reducing the span ofthe bridge, what is the minimum width of channel such that theupstream 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 177.181.9

43

2

3

2

o g

qyc

m 0.4o y

yo yco subcritical flow

/s/mm 0.450

200 3o

B

Qq

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 701.2051.43

2

3

2min2 Eyc

3

2max

2g

qyc

m 39.149.13

200

max

max

max

max q

QB

B

Qq

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 channelcarrying flow of 200 m3/s at depth 4.0 m. The construction hascaused the width of the channel to be reduced to 30-m. Determinethe 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 177.181.9

43

2

3

2

o g

qyc

m 0.4o y

yo yco subcritical flow

/s/mm 0.450

200 3o

B

Qq



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 701.2051.43

2

3

2min2 Eyc

3

2max

2g

qyc

m 39.149.13

200

max

max

max

max q

QB

B

Qq

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

22

EygB

Qy

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/sof 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 criticaldepth to occur.

(b) Calculate the depth of flow upstream, downstream and atthe 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 6121.081.9

5.13

2

3

2

o g

qyc

m 8.0o y

Since yo yco subcritical flow

/s/mm 5.12

3 3o

B

Qq


m 9792.08.081.92

5.18.0

2 2

2

2o

2o

oo

gy

qyE

m 6528.09792.03

2

3

2min2 Eyc

m 816.1652.1

3

max

max q

QB

/sm 652.16528.081.9 2332max cgyq


(b) If B2 = 1.2 m,

m 816.1max2 BB Case 3, where new qmax, i.e. qmax is required

Step 3. Calculate y1, y2 and y3

/sm 5.22.1

3 2

2

max B

Qq

m 8605.081.9

5.23

2

3

2max

22

g

qyy c

m 291.18605.02

3

2

32min

cyE


1,321,3

2o

1,32

Eyg

qy

min31 EEE

291.181.92

5.12

1,3

2

1,3

y

y

From trial-and-error, y1 1.213 m and y3 0.3489 m


Activity 3.11

Flow inside a rectangular channel of 3.0 m width has a velocity of3.0 m/s at 3.0 m depth. The channel is experiencing a step of 0.61m high at the channel bottom. What is the constriction to be madeto the channel width in order to ensure the depth of flowupstream does not change.

Given V = 3 m/s, B = 3 m, yo = 3 m, and H = 0.61 m

Thus, q = yoV = 3 3 = 9 m2/s


m 021.281.9

93

2

3

2

o g

qyc

m 0.3o y

Since yo yco subcritical flow


m 459.3381.92

93

2 2

2

2o

2

oo

gy

qyE

m 032.3021.22

3

2

3min 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 reducey1 to yo, q has to be increased, i.e. via width constriction.


0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E (m)

y (m)

E2Emin

y1 y1 yo

yc =2.021

Eo

H=0.61 m

1 2 30

0.61 m

3 myc=2.021 m

Emin3.032 m

Eo=3.459 m

y1 yo and y3 yc

yc=2.021 m y3

Backwater

y1

EGL

y3 yo

Hmin=0.427 m

E1,3

yo =3.459

0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E (m)

y (m)

E2Emin

y1 y1 yo

yc =2.021

Eo

H=0.61 m

1 2 30

0.61 m

3 myc=2.021 m

Emin3.032 m

Eo=3.459 m

y1 yo and y3 yc

yc=2.021 m y3

Backwater

y1

EGL

y3 yo

Hmin=0.427 m

E1,3

yo =3.459

E1,3

yc2


m 642.33,1min EE

m 428.2642.33

2

3

2min2 Eyc

m 278.285.11

333

max

max

q

QB

/sm 85.11428.281.9 2332max cgyq


0

2

4

6

8

10

0 2 4 6 8 10

E (m)

y (

m)

E (m)

y (m)

yc2 =2.021

Eo

H=0.61 m

1 2 30

0.61 m

3 m

Emin3.642 m

Eo=3.459 m

yc=2.021 m y3

y1=3 m

EGL

y3

Hmin=0.427 m

Emin

y1 = yo =3.459

E1,3

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, andlongitudinal slope of 0.0008 is conveying flow at 9.8 m3/s. If aconstriction is made by reducing channel width to 2.4 m, calculatedepth 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


2

1

o

3

2

S

QnAR

2

1

3

2

o

oo

0008.0

015.08.9

262

6.26.2

y.

yy


m 131.181.9

6.2

8.93

2

3

2

o

g

qyc Since yo yco subcritical flow

197.5262

6.26.2

3

2

o

oo

y.

yy

m 270.2o yThrough trial-and-error,


m 411.227.281.92

6.2

8.9

27.22 2

2

2o

2o

oo

gy

qyE

m 607.1411.23

2

3

2min2 Eyc

Bmax is when q = qmax, where Emin = Eo


m 536.1381.6

8.9

max

max q

QB

/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 < Eo

E2 = Eo

y1 = y3 = yo = 2.270 m

/sm 083.44.2

8.9 2

2

2 B

Qq

m 193.181.9

083.43

2

3

2

o g

qyc


22

22

222gy

qyE

411.281.92

083.422

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 Choking

Choking 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.


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 the unavoidable 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 -Open Channel Hydraulics

by Tan Lai Wai et al. ([email protected])

THANK YOU


Lecturers

• Dr. Tan Lai Wai ([email protected])

• Dr. Mohd Adib Mohammad Razi ([email protected])

• Dr. Hartini Kasmin ([email protected])

• Dr. Mohd. Shalahuddin Adnan ([email protected])

• Dr. Mohd Ariff Ahmad Nazri ([email protected])

• Dr. Siti Nazahiyah Rahmat ([email protected])

• Mdm. Zarina Md Ali ([email protected])

• Mdm. Noor Aliza Ahmad ([email protected])


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