Hidraulika Saluran Terbuka

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