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Hidraulika Saluran Terbuka
Energi spesifik
Rumus Bernoulli
Pressure in an OCF
Curved channel
Specific energy
• Specific energy E is defined as:
Average SE:
Rectangular Channel:
Dimensionless specific energy
Change in flow depth
Change in flow depth
Change in width
Specific Energy
In a channel with constant discharge, Q
2211 VAVAQ ==
2
2Q
yE +=V
yE
2
+= where A=f(y)22gA
yE +=g
yE
2+= where A=f(y)
Consider rectangular channel (A=By) and Q=qB
2
2
2gy
qyE +=
A
B
y
3 roots (one is negative)
q is the discharge per unit width of channel
4
5
6
7
8
9
10
y
Specific Energy: Sluice Gate
2
2
2gy
qyE +=
1
sluice gate
y1 EGLq = 5.5 m2/sy2 = 0.45 mV2 = 12.2 m/s
E2 = 8 m
0
1
2
3
4
0 1 2 3 4 5 6 7 8 9 10
E
2
21 EE =
y2
y1 and y2 are ___________ depths (same specific energy)
Why not use momentum conservation to find y1?
E2 = 8 m
alternatealternate
Given downstream depth and discharge, find upstream depth.
2
3
4
y
Specific Energy: Raise the Sluice Gate
2
sluice gate
y1EGL
0
1
2
0 1 2 3 4
E
y
2
2
2gy
qyE +=
1 2
E1 = E2
y2
as sluice gate is raised y1 approaches y2 and E is minimized: Maximum discharge for given energy.
4
Specific Energy: Step Up
Short, smooth step with rise ∆y in channel
Given upstream depth and discharge find y2
3
4
0
1
2
3
0 1 2 3 4
E
y
∆y
1 2E E y= + D
0
1
2
0 1 2 3 4
E
y
Increase step height?
Non-uniform flows
definition
Gradually Varied Flow
2 2
1 21 2
2 2o f
V Vy S x y S x
g g+ + D = + + D
Energy equation for non-uniform, steady flow
12 yydy −=( )2 2
2 12 1
2 2o f
V VS dx y y S dx
g g
æ ö= - + - +ç ÷è ø
2
2f o
Vdy d S dx S dx
g
æ ö+ + =ç ÷è ø
PP
AA
T
dy
y
dy
dxS
dy
dxS
g
V
dy
d
dy
dyof =+
+
2
2
( )2 12 2
o fS dx y y S dxg g
= - + - +ç ÷è ø
Gradually Varied Flow
2
3
2
3
2
2
22
2
2
22Fr
gA
TQ
dy
dA
gA
Q
gA
Q
dy
d
g
V
dy
d−=
−=⋅
−=
=
dxS
dxS
Vddy=+
+2 Change in KE
Change in PEChange in KEChange in PE
[ ] fo SSFrdx
dy−=− 21
dy
dxS
dy
dxS
g
V
dy
d
dy
dyof =+
+
2
dy
dxS
dy
dxSFr of =+− 21
Change in PEChange in PE
We are holding Q constant!
21 Fr
SS
dx
dy fo
−
−=
Gradually Varied Flow
21 Fr
SS
dx
dy fo
−
−= Governing equation for
gradually varied flow
• Gives change of water depth with distance along channel
• Note• Note
– So and Sf are positive when sloping down in direction of flow
– y is measured from channel bottom
– dy/dx =0 means water depth is constant
yn is when o fS S=
Surface Profiles
• Mild slope (yn>yc)
– in a long channel subcritical flow will occur
• Steep slope (yn<yc)
– in a long channel supercritical flow will occur– in a long channel supercritical flow will occur
• Critical slope (yn=yc)
– in a long channel unstable flow will occur
• Horizontal slope (So=0)
– yn undefined
• Adverse slope (So<0)
– yn undefinedNote: These slopes are f(Q)!
Normal depth
Steep slope (S2)
Hydraulic Jump
Sluice gate
Steep slope
Obstruction
Surface Profiles
21 Fr
SS
dx
dy fo
−
−= S0 - Sf 1 - Fr2 dy/dx
+ + +
- + -
- - + 0
1
2
3
4
0 1 2 3 4
E
y
yn
yc
More Surface Profiles
S0 - Sf 1 - Fr2 dy/dx
1 + + +
2 + - -2
3
4
yy
21 Fr
SS
dx
dy fo
−
−=
2 + - -
3 - - + 0
1
2
0 1 2 3 4
E
y
yn
yc
Direct Step Method
xSg
VyxS
g
Vy fo ∆++=∆++
22
2
2
2
2
1
1
VV−+−
2
2
2
1
energy equation
of SS
g
V
g
Vyy
x−
−+−
=∆22
21
21
solve for ∆x
1
1
y
qV =
2
2
y
qV =
2
2
A
QV =
1
1
A
QV =
rectangular channel prismatic channel
Direct Step Method
Friction Slope
2 2
4/3f
h
n VS
R=
2
8f
h
fVS
gR=
Manning Darcy-Weisbach
SI unitsh
2 2
4 / 32.22f
h
n VS
R=
h
English units
Direct Step
• Limitation: channel must be prismatic (so that velocity is a function of depth only and not a function of x)
• Method• Method– identify type of profile (determines whether ∆y is + or -)
– choose ∆y and thus yn+1
– calculate hydraulic radius and velocity at yn and yn+1
– calculate friction slope yn and yn+1
– calculate average friction slope
– calculate ∆x
Direct Step Method
=y*b+y^2*z
=2*y*(1+z^2)^0.5 +b
=A/P
of SS
g
V
g
Vyy
x−
−+−
=∆22
2
2
2
1
21
=(G16-G15)/((F15+F16)/2-So)
A B C D E F G H I J K L M
y A P Rh V Sf E Dx x T Fr bottom surface
0.900 1.799 4.223 0.426 0.139 0.00004 0.901 0 3.799 0.065 0.000 0.900
0.870 1.687 4.089 0.412 0.148 0.00005 0.871 0.498 0.5 3.679 0.070 0.030 0.900
=Q/A
=(n*V)^2/Rh^(4/3)
=y+(V^2)/(2*g)
of SS −
Standard Step
• Given a depth at one location, determine the depth at a second location
• Step size (∆x) must be small enough so that changes in water depth aren’t very large. Otherwise estimates of water depth aren’t very large. Otherwise estimates of the friction slope and the velocity head are inaccurate
• Can solve in upstream or downstream direction
– upstream for subcritical
– downstream for supercritical
• Find a depth that satisfies the energy equation
xS
g
VyxS
g
Vy fo ∆++=∆++
22
2
2
2
2
1
1
control
control
H = E + z0
Hitungan profil muka air
metode integrasi numerik
• Kecepatan rerata :
Debit
2/13/21IR
nV =
2/13/21IAR
nQ =
Landai energi:
n
3/42
22
RA
QnI f =
312
3/42220
1 −−
−−
−
−=
ATgQ
RAQnI
dx
dy
3
2
0
1gA
TQ
II
dx
dy f
−
−=
Integrasi numeris
• Deret Taylor: yn+ = yi + (dY/dx) dx
xdx
dy
dx
dyyy
ii
ii
+
+=
++
121
1
xffyy iii )( 121
11 ++ ++=
32
3/42220
/1 gATQ
RAQnIf
−
−=
−
Direct step method
• zl + yl + v12/2g = z2 + y2 + v2
2/2g + hf
• zl – z2 = I0 ∆x
• hf = If ∆x
xIgVygVyxI f ∆++=++∆ // 2
221
2
2
121
10
fII
gVygVyx
−
+−+=∆
0
2
121
1
2
221
2 )/()/(
fII
EEx
−
−=∆
0
12 )(
Direct step method
VQ 2
2
2 2g
VQ 1
2
1 2g
1 2
h a
z1
z1
z2
z2
y2
y1
latihan
• Suatu aliran segi empat dengan lebar B = 2 m
mengalirkan air dengan debit Q = 1 m3/s.
Kedalaman air pada dua titik yang berdekatan
adalah 1,0 dan 0,95 m. Apabila koefisien adalah 1,0 dan 0,95 m. Apabila koefisien
Manning n = 0,02 dan kemiringan dasar
saluran I0 = 1 : 2500, hitung jarak antara kedua
tampang tersebut.
Standard step method
• Berdasarkan nilai yj awal yang diketahui, dihitung nilai fi dari pers. (12.43c).
• Pertama kali dianggap fi+1 = fi.
• Hitung nilai yi+l dari pers. (12.42) dengan menggunakan nilai fi+2 yang diperoleh dalam langkah 2 atau nilai fi+1 yang diperoleh dalam langkah di atas.
• Hitung nilai baru y dengan menggunakan nilai f yang dihitung • Hitung nilai baru yi+l dengan menggunakan nilai fi+l yang dihitung dari nilai yi+1 dari langkah 3.
• Apabila nilai yi+1 yang diperoleh dalam langkah 3 dan 4 masih berbeda jauh, maka langkah 3 dan 4 diulangi lagi.
• Sesudah nilai yi+1 yang benar diperoleh, dihitung nilai yi+2 yang berjarak x dari yi+1.
• Prosedur di atas diulangi lagi sampai diperoleh nilai y di sepanjang saluran.
Non-Uniform Open Channel Flow
Let’s evaluate H, total energy, as a function of x.
H = z+ y + α v2 / 2g( )
dH dz dy α dv2 dH
dx=
dz
dx+
dy
dx+
α
2g
dv2
dx
Where H = total energy headz = elevation head,αv2/2g = velocity head
Take derivative,
Replace terms for various values of S and
So. Let v = q/y = flow/unit width - solve for
dy/dx, the slope of the water surface
–S =−So +dy
1−q
2
3
since v = q / y–S =−So +
dx1−
gy3
since v = q / y
1
2g
d
dxv2[ ]=
1
2g
d
dx
q2
y2
= −
q2
g
1
y3
dy
dx
Given the Froude number, we can simplify and
solve for dy/dx as a fcn of measurable
parameters
Fr2 = v2 / gy( )dy
dx=
So − S
1− v2 / gy=
So − S
1 − Fr2dx=
1− v2 / gy=
1 − Fr2
where S = total energy slope
So = bed slope,
dy/dx = water surface slope
*Note that the eqn blows up when Fr = 1 and goes to
zero if So = S, the case of uniform OCF.
Uniform Depth
Yn > Yc
Mild Slopes where - Yn > Yc
Now apply Energy Eqn. for a reach of length L
y1 +v1
2
2g
= y2 +
v22
2g
+ S −So( )L
y1 +v1
2
− y2 +
v22
L =
y1 +2g
− y2 +
2g
S − S0
This Eqn is the basis for the Standard Step Method
Solve for L = ∆x to compute water surface profiles
as function of y1 and y2, v1 and v2, and S and S0
Backwater Profiles - Mild Slope Cases
∆x
Backwater Profiles - Compute Numerically
Compute
y3 y2 y1
Routine Backwater Calculations
1. Select Y1 (starting depth)
2. Calculate A1 (cross sectional area)
3. Calculate P1 (wetted perimeter)
4. Calculate R1 = A1/P1
5. Calculate V1 = Q1/A1
6. Select Y2 (ending depth)6. Select Y2 (ending depth)
7. Calculate A2
8. Calculate P2
9. Calculate R2 = A2/P2
10. Calculate V2 = Q2/A2
Backwater Calculations (cont’d)
1. Prepare a table of values
2. Calculate Vm = (V1 + V2) / 2
3. Calculate Rm = (R1 + R2) / 2
4. Calculate Manning’sS =nVm
2
2
Energy Slope Approx.
5. Calculate L = ∆X from first equation
6. X = ∑∆Xi for each stream reach
S =1.49Rm
23
L =
y1 + v1
2
2g
−
y2 + v2
2
2g
S − S0
Application - 100 Year Floodplain
Main Stream
Tributary
C
DQD
QC
Bridge
Floodplain
Main Stream
Cross Sections
Cross Sections
A
B
QA
QB
Bridge Section
The Floodplain
Top Width
Floodplain Determination
The Woodlands
� The Woodlands planners wanted to design the community to withstand a 100-year storm.
� In doing this, they would attempt to minimize any changes to the existing, undeveloped floodplain as development proceeded through time.
HEC RAS (River Analysis System, 1995)
HEC RAS or (HEC-2)is a computer model designed for
natural cross sections in natural rivers. It solves the
governing equations for the standard step method,
generally in a downstream to upstream direction. It can
Also handle the presence of bridges, culverts, and
variable roughness, flow rate, depth, and velocity.
HEC - 2
Orientation - looking downstream
Multiple Cross Sections
River
HEC RAS (River Analysis
System, 1995)
HEC RAS Bridge CS
HEC RAS Input Window
HEC RAS Profile Plots
3-D Floodplain
HEC RAS Cross Section
Output Table
AA
Critical Flow
T
dy
y
Find critical depth, yc
2Q
yE +=
0=dy
dE
Arbitrary cross-section
A=f(y) dA
0
1
2
3
4
0 1 2 3 4
E
y
yc
PP
AAy
T=surface width
22gA
QyE +=
TdydA =
3
2
1
c
c
gA
TQ=
A=f(y)
2
3
2
FrgA
TQ=
22
FrgA
TV=
dA
AD
T= Hydraulic Depth
013
2
=−=dy
dA
gA
Q
dy
dE
Critical Flow:
Rectangular channel
yc
T
Ac
3
2
1
c
c
gA
TQ=
qTQ = TyA cc =
cTT =
3
2
33
32
1
cc gy
q
Tgy
Tq==
3/12
=
g
qyc
3
cgyq =
Only for rectangular channels!
Given the depth we can find the flow!
Critical Flow Relationships:
Rectangular Channels
3/12
=
g
qyc cc yVq =
=
g
yVy
cc
c
22
3
V
because
forceinertial Kinetic energy
g
Vy
c
c
2
=
1=gy
V
c
cFroude numberFroude number
velocity head =g
Vy cc
22
2
=
2
c
c
yyE += Eyc
3
2=
forcegravity
forceinertial
0.5 (depth)0.5 (depth)
g
VyE
2
2
+=
Kinetic energy
Potential energy
Critical Flow
• Characteristics
– Unstable surface
– Series of standing waves
Occurrence
Difficult to measure depth
0
1
2
3
4
0 1 2 3 4
E
y
• Occurrence
– Broad crested weir (and other weirs)
– Channel Controls (rapid changes in cross-section)
– Over falls
– Changes in channel slope from mild to steep
• Used for flow measurements
– ___________________________________________Unique relationship between depth and discharge
Broad-crested Weir
H
P
yc
E
3
cgyq = 3
cQ b gy=
3/12
=
g
qyc
Broad-crestedweirc cQ b gy=
Eyc
3
2=
3/ 2
3/ 22
3Q b g E
æ ö=
è ø
Cd corrects for using H rather than E.
weir
E measured from top of weirE measured from top of weir
3/ 22
3dQ C b g H
æ ö=
è ø
Hard to measure yc
Broad-crested Weir: Example
• Calculate the flow and the depth upstream.
The channel is 3 m wide. Is H approximately
equal to E?
y
E
0.5
yc
Broad-crestedweir
yc=0.3 m
Solution
How do you find flow?____________________
How do you find H?______________________
Critical flow relation
Energy equation
Hydraulic Jump
• Used for energy dissipation
• Occurs when flow transitions from
supercritical to subcriticalsupercritical to subcritical
– base of spillway
• We would like to know depth of water
downstream from jump as well as the
location of the jump
• Which equation, Energy or Momentum?
Hydraulic Jump!
Basic equation
• Continuity:
• Momentum:
• Energy:
Conjugate depth
• Conjugate depth equation
• Dimensionless:• Dimensionless:
Energy loss
• Energy loss:
• Dimensionless:• Dimensionless:
Exercise
• A hydraulic jump takes place in a 0.4 m wide
laboratory channel. The upstream flow depth
is 20mm and the total flow rate is 31 l/s. The
channel is horizontal, rectangular and smooth. channel is horizontal, rectangular and smooth.
Calculate the downstream flow properties and
the energy dissipated in the jump. If the
dissipated power could be transformed into
electricity, how many 100 W bulbs could be
lighted with the jump? [Woro, 35345]
answer
ρgQ∆H = 1000 x 9.8 x 31 x 10-3 x 0.544 = 167 W
Types of hydraulic jump
Length of hydraulic jump
• Rule of thumb: L = 6 d2 4 < Fr < 20
• Harger et al. (1990)
Application – 1
• Considering a dissipation basin at the
downstream end of a spillway, the total
discharge is Q 2000m3/s. The energy
dissipation structure is located in a horizontal dissipation structure is located in a horizontal
rectangular channel (25m wide). The flow
depth at the downstream end of the spillway
is 2.3 m. Compute the energy dissipation in
the basin.
solution
• Upstream flow condition:
– Flow depth d1 = 2,3 m
– Flow velocity V1 = 34,8 m/s
– Upstream Froude number = V1/(g d1)1/2 = 7,3– Upstream Froude number = V1/(g d1) = 7,3
– Conjugate depth d2 = 22,7 m
– Flow velocity V2 = 3,52 m/s
– Fr2 = 0,24
– Energy dissipated= 40,7 m
– Power generated = ρgQH = 796 x 106 W
Application – 2 • Considering a hydraulic jump in a horizontal
rectangular channel located immediately upstream
of an abrupt rise, estimate the downstream flow
depth d3 for the design flow conditions d1 0.45 m,
V1 10.1 m/s. The step height equals zo 0.5 m.
solution
WEIRS AND GATES
Weir (bendung)
• A weir is an obstruction on a channel bottom
over which the fluid must flow.
• Weir provides a convenient method of
determining the flowrate in an open channel determining the flowrate in an open channel
in terms of a single depth measurement.
Sharp crested weir
(bendung ambal tajam)
Geometry of a sharp crested weir
rectangular triangular trapezoidal
Flow rate of a sharp crest weir
• Assume: velocity profile
upstream is uniform,
pressure within the
nappe is atmospheric
• Fluid flow pass the
sharp crest flows
horizontal over the
plate with non uniform
velocity profile
Sharp crested weir – flow rate
Rectangular sharp crested weir – flow rate
Rectangular weir – flow rate
Triangular weir
Triangular weir – flow rate
Nappe position
Broad crested weir
Broad crested weir
Broad crested weir
Application
• Water flows in a rectangular channel of width b = 2 m with flowrate between Qmin = 0.02 m3/s and Qmax = 0.60 m3/s. This flowrate is to be measured by using (a) a rectangular sharp-crested weir, (b) a triangular sharp-crested weir with θ=90º, or (c) a broad-crested weir. In all cases the bottom of a triangular sharp-crested weir with θ=90º, or (c) a broad-crested weir. In all cases the bottom of the flow area over the weir is at a distance Pw = 1 m above the channel bottom. Plot a graph of
Q= Q(H) for each weir and comment on which weir would be best for this application.
Solution
Underflow Gates
Underflow Gates
Underflow Gates
Underflow Gates
Contoh
• Water flows under the sluice gate shown in
Fig. The channel width is b = 20 ft, the
upstream depth is y1= 6 ft, and the gate is a =
1.0 ft off the channel bottom. Plot a graph of 1.0 ft off the channel bottom. Plot a graph of
flowrate, Q, as a function of y3.
Solution
Solution
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