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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])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
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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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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)
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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)
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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
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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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
(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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
(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,
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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,
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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.
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
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(c) Free overflow (sudden drop)
(d) Reservoir water flows on a steep slope
Horizontal bed
H2
yc
controlS2
yc
Reservoir
Steep yo
control
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(e) Flow through sluice gate (f) Flow over spillway
M1
M3
ycMild
yo
control
M1
yc
control
control
Jump
Mild
yo
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(g) Flow over broad-crested weir
(h) Flow through constricted channel width
yc
H
Hump
control
yc
B
control
Plan view
Constriction
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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
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Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
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Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
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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.
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Subcritical upstream yo yc
Eo
E2 Emin
y2 yc yo
Supercritical upstream yo yc
H
1 2 30
y2yc
yo
EGL
EGL
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
For Case 2,
H Hmin
Hmin Eo Emin
E1 E3 Eo
E2 Emin
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
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
Subcritical upstream yo yc
y2 yc yo
y1 yo
y3 yc
Supercritical upstream yo 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
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Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
For Case 3,
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
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
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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.
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Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Step 2. Calculate Hmin
m 445.0275.1720.1minomin EEH
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
(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
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Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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.
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
For Case 1,
B2 Bmax
Emin 2 Eo
E2 Eo
y2 yc2
Steps Useful equations
1. Calculate yo and yco
Manning: or
2. Calculate yc2, qmax and Bmax
3. Determine y1, y2 & y3
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
Case 2: B2 Bmax
y2 yc2 yo
Supercritical upstream yo yco
y2 yc2 yo
Subcritical upstream yo yco
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
For Case 2,
B2 Bmax
Emin 2 Eo
E2 Emin 2 Eo
y2 yc2
Steps Useful equations
1. Calculate yo and yco
Manning: or
2. Calculate yc2, qmax and Bmax
3. Determine y1, y2 & y3
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
Case 3: B2 Bmax
y2 yc2 yo
Supercritical upstream yo yco
y2 yc2 yo
Subcritical upstream yo yco
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
For Case 3,
B2 Bmax
Emin 2 Eo
E2 Emin 2
y2 yc2
Steps Useful equations
1. Calculate yo and yco
Manning: or
2. Calculate yc2, qmax and Bmax
3. Determine y1, y2 & y3
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
;2
maxB
21,3
2max
1,31,32 yg
qyE
3
2max
2g
qyc
;2
32min cyE
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
yc is influenced by q. When q changes, yc varies as well. Therefore, at constriction where q qo, yc2 exists (calculated in Step 2).
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
(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
Step 2. Calculate yc2 and Bmax
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
(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
m 8605.081.9
5.23
2
3
2max
22
g
qyy c
m 291.18605.02
3
2
32min
cyE
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Step 1. Determine yo and yco
m 021.281.9
93
2
3
2
o g
qyc
m 0.3o y
Since yo yco subcritical flow
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
m 459.3381.92
93
2 2
2
2o
2
oo
gy
qyE
m 032.3021.22
3
2
3min cyE
Step 2. Calculate Hmin
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.
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Step 1. Determine yo and yco
2
1
o
3
2
S
QnAR
2
1
3
2
o
oo
0008.0
015.08.9
262
6.26.2
y.
yy
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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,
Step 2. Calculate yc2 and Bmax
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
m 193.181.9
083.43
2
3
2
o g
qyc
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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.
Open Channel Hydraulics by Tan Lai Wai et al. ([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.
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])
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])
Open Channel Hydraulics by Tan Lai Wai et al. ([email protected])