River conservation and rehabilitation eah 416 - USMredac.eng.usm.my/EAH/document/River conservation and rehabilitation...SS dx 22 Specific ... 30 10 50.419 15.833 1.200 50.492 0.030

RIVER CONSERVATION AND REHABILITATION

EAH 416PUAY HOW TION

PC LAB AEROSPACE

AFTER COMPLETING THIS COURSE,YOU SHOULD BE ABLE TO

• Write and understand the governing equations for steady flow

• Predict water surface profile (wsp)

• Understand the required boundary conditions to solve for wsp

• Solve numerically for water surface profile by using :

• Spreadsheets

• Programming language : fortran/matlab (or other languages)

• Commercial software : HEC-RAS (or others)

Governing Laws

Conservation of mass

Continuity equation

Conservation of energy

Bernoulli equation

Conservation of momentum

Momentum equation

State of material Equation of state

Flow classification

Based on space

Uniform flow

Non uniform flow

Based on time

Steady flow

Unsteady flow

0t

0t

0x

0x

flow variables, e.g. velocity, depth, cross section areas..

THE MANY FORMS OF GOVERNING EQS..

0u v w

x y z

0

A Q

t x

22 2

2 2 2

1 1 xyxx xzx

u uu uv uw Pg

t x y z x x y z

Continuity equation (per unit volume) 1-D depth averaged continuity equation

Momentum equation (per unit volume),

, hydraulic radiusbxQ uQ hgA gA R

t x x gR

1-D depth averaged momentum equation

Some examples…………

In x-direction,

HOW DO WE CHOOSE THE SUITABLE FORM OF GOV. EQ ?

• The flow direction :

• Mainly in one direction – use 1D equation

• Mainly in two-directions – use 2D equation

• Flow is complex, all directions – use 3D equation

• What phenomena are we interested in?

• Hydraulic jump- at least 2D equation

• Turbulent flow structure – 3D equation

• Wave breaking on surface – 3D equation

• Nature of the flow

• Steady flow : use steady equation

• Unsteady flow : use unsteady equation

• Study purpose and accuracy needed

• To reduce complexity of the solution, assumptions

are made and governing equations are

simplified. Assumptions made must not sacrifice

accuracy and cause lost of important flow

characteristics.

HOW DO WE CHOOSE THE SUITABLE FORM OF GOV. EQ ?

OPEN- CHANNEL STEADY FLOW

• Characteristics of open channel flow

• Existence of surface

• If the flow changes gradually,

hydrostatic pressure

is adequate (see figure)

• Basic equation

• Continuity equation, energy

equation, momentum equation

Flow classification

Based on space

Uniform flow

(a)

Non uniform flow

(b)

Based on time

Steady flow

(c)

Unsteady flow

(d)

Steady non-uniform flow

(b)+(c)

Gradually varied flow

Rapidly varied flow

Unsteady non uniform flow

(b)+(d)

Gradually varied

unsteady flow

Rapidly varied unsteady flow

GRADUALLY VARIED FLOW

• DERIVATION OF CONTINUITY AND ENERGY EQUATION

• FLOW SURFACE PROFILE CLASSIFICATION

• UNIFORM FLOW RESISTANCE LAWS

DERIVATION OF CONTINUITY EQUATION FOR OPEN CHANNEL FLOW

…… Eq (1)

…… Eq (2)

Based on the law of mass

conservation, Eq.(1)=Eq.(2)

From derivation (see handout), we have the continuity

equation for unsteady flow,

0A Q

t x

For steady flow, the first term is zero

0

constant

Q

x

Q

MOMENTUM EQUATION

Momentum equation (unsteady flow)

Neglecting turbulence stresses,

sin Reynold stressess byQ uQgA gA

t x x gR

s

o f

yQ uQgA gA S S

t x x

Derivation

skipped

ENERGY EQUATION

ENERGY AT TWO SECTIONS (EXPRESSED IN HEAD) FOR UNSTEADY FLOW

ENERGY AT TWO SECTIONS (EXPRESSED IN HEAD) FOR STEADY FLOW

By principle of energy conservation,

Rearranging,

Change in total head Friction loss

divide by Dx,

2 2 2

2 2 2f

V V Vz d z dz d d d d h

g g g

2

2f f

Vd z d h S x

g

D

2 2

Specific energy ,

cos2 2

o

o

H

V VH d h

g g

Channel slope o

o

S

dzS

dx

Friction slope f

f

S

SgR

o

o f

dHS S

dx

2 2

Specific energy ,

cos2 2

o

o

H

V VH d h

g g

22

2 22 2

2 2 3

2

3

2

3

2 2

3 3

cos cos2 2

1 2

2

2cos

2

2cos

2

o

o

dH d V dh dh V

dx dx g dx g dx

d d Q d Q dAV Q

dx dx A dx A A dx

Q A A dh

A x h dx

dH dh Q A A dh

dx dx g A x h dx

dh Q A Q A

dx g A h g A x

constant constant

is function of and , therefore

h x

A x h

dA A dx A

dx x dx h

dh A A dh

dx x h dx

rearrange

2 2

3 3

2

3

2

3

cos

cos

o

o f

o f

o f

dHS S

dx

dh Q A Q AS S

dx h xgA gA

Q AS S

xdh gA

dx Q A

hgA

from o

o f

dHS S

dx

Determination of flow profile of steady flow

Case 1 : uniform channel

…. Eq. (3)

2

3

2

3cos

o f

Q AS S

xdh gA

dx Q A

hgA

Where definition is used.o

dzS

dx

If we assume 0 , 1,

Eq.(3) can be transformed into the following form,

fS

For uniform channel . Therefore, 0 and A A

A Bh Bx h

Definition of normal depth, oh

0dh

dx

2

3

Therefore, numerator =0

0o f

Q AS S

xgA

2

3

Definition of critical depth,

Therefore, denominator=0

cos 0

ch

dh

dx

Q A

hgA

Other definition of critical depth,

1) 0 (minimum specific energy, )

2) Froude number=1

c

o

o

h

dHH

dx

Definition of critical slope ic

for a rectangular channel where physical shape and roughness n are

known, for an arbitrary discharge Q, there exists a slope where the

flow is normal flow (dh/dx=0) and the depth is equivalent to critical

depth, this slope is called the “critical slope”

in the case wide rectangular channel , R=h

Classification of slope

wide rectangular channel, using Manning’s n

10/3

3

1 /

1 /

o

o

c

h hdhS

dx h h

Steep slope

Mild slope

Critical slope

steep slope mild slope

METHOD OF CALCULATION

1) Standard Step Method - all kinds of channel : varying width

(river), fixed width (prismatic

channel)

For river : - the bottom bed of a river is

often uneven, so it is not practical to

measure water depth, h.

So, we measure the total value of z+d

from the datum that we set (here d=h cos

theta). Theta is the angle between the

bed and the datum line

By energy principle

2 2 2

2

2

2 2 2

2

where cos2

f

f f

f

V V Vz d z dz d d d d h

g g g

Vd z d h S x

g

Vd Z S x Z z d z h

g

D

D

2 2

1 2

2 1

1 where

2 2 2f f f f

V VZ Z S x S S S

g g

D

2 2

2 12 2

f

V VZ Z S x

g g

D

Known

(obtained from

control section)

i) Make a guess for Z2

ii) Calculate V2 using continuity equation

iii) Calculate from Manning’s eq.

2 2 2 2

1 1 2 21 2 4/3 4/3

1 2

1 1

2 2f f f

V n V nS S S

R R

1 2

1 1 2 2

Q Q

V A V A

Cross section data A2 is needed !

1) Standard Step Method – a) for non prismatic channel

When standard step method is applied on river,

a) With non-prismatic channels, the value of n, A

and R vary and have to be found from cross-

sectional data obtained from survey

b) The solution doesn’t give information on flow

depth ! Only the elevation of water surface

above a datum , which is the value of Z=z+d

c) Mostly used for river channel to determine the

water surface profile. For a river, the channel

bed is uneven, therefore the depth of water is

different across a section. Therefore it is

meaningless to determine water depth and the

elevation of water surface Z is more useful.

Therefore, R and A data for all sections

have to be provided from survey data

unknown

If guess is

incorrect,

re-guess

If guess is correct,

LEFT=RIGHT

LEFT RIGHT

EXAMPLE 1

A broad crested rectangular weir forms a control point in a river channel. At a distance of 10m upstream the weir, the

rectangular channel has an average bed elevation of 48.895m above ordinance datum (OD), a width of 10 m and the depth

of flow is 1.49m when the discharge is 19m3/s.

Upstream of the control point the width, area and roughness of the channel vary along its length. Cross-sections of the channel

have been surveyed and plotted at 10m intervals upstream the weir. For the water levels shown in column 3 of Table 1, the

appropriate values A, n and R are shown in columns 4,7,8.

Assume energy coefficient = 1.0, and the bed slope is 1 in 400. It is thought that the normal depth in the channel is about

1.3m.

There is some concern that the weir may cause the river to overtop its bank in the 100m length of the channel nearest to the

weir, particularly at chainage 80m where the bank dips to 50.7m OD.

Calculate the elevation of the water surface and determine if this is a problem.

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11)

Chainage

(m)

DL

(m)

Z

(m) OD

A

(m2)

V

(m/s)

H=Z+V2

/2g

n

(s/m1/3)

R

(m)

Sf Sfave H2=H1+

Sfave

*DL

0 (weir)

10 (CP) 10 50.385 15.111 1.257 50.466 0.035 1.144 0.00162 - -

20 10 50.411 16.350 1.162 50.480 0.035 1.158 0.00136 0.00149 50.480

30 10 50.419 15.833 1.200 50.492 0.030 1.149 0.00108 0.00122 50.492

40 10

50 10

60 10

70 10

80 10

90 10

100 10 50.499

Table 1

1) Check discharge

2) Check critical depth

Since the actual depth upstream of the weir is 1.49m, h=1.49m>ho=1.3m >hc=0.72m,

The slope is a mild slope and the water surface profile is of type M1 (backwater curve)

The calculation to obtain the elevation is shown in Table 1.

Calculation for row 2 (chainage 10m)

1) V1=Q/A1=19.00/15.111=1.257 m/s (column 5)

2) H1=Z1+V1^2/(2g)=50.385+1.257^2/(2g) = 50.466 (column 6)

3) Sf = (V1*n1)2/R14/3=(1.257x0.035)2/(1.144)4/3=0.00162 (column 9)

Since chainage 10m is a control point, calculation stop there.

Calculation for row 3 (chainage 20m)

3) Guess the value of Z2, (Z2=50.411)

4) Calculate H (column 6)=50.480, calculate Sf (column 9)

5) Calculate Sfave =1/2(Sf1+Sf2)=1/2(0.00162+0.00136)=0.00149

6) Calculate H2 (column 11) = H1+Sfave*DL=50.466+0.00149*(10)=50.480

7) Compare H2 from Step 6 with H2 from Step 4. If same, then repeat for next chainage.

If not same, then guess Z2 in step 3 and repeat calculation until H2 in Step 6 same with H2 in Step 4.

1) Standard Step Method – b) for prismatic channel

2 2 2

2

2

2 2 2

2

where is called the specific energy2

where is the slope of the bed

Therefo

f

ff

f

f fo o

V V Vz d z dz d d d d h

g g g

Vd z d h S x

g

Vd z H S x H d

g

dH dz dzS S S S

dx dx dx

D

D

2 1

2 1

re,

fo

fo

H H S S x

H H S S x

D

D

2 1 foH H S S x D

Given data Calculate from Manning’s eq.

2 2

2 22 2 2 cos

2 2

V VH d h

g g

1) Guess h2

2) We can calculate A2

3) Calculate V2 since V2=Q/A2

4) So, H2 can be estimated

Known from

control section

1) Standard Step Method – b) for prismatic channel

1 2

2 2 2 2

1 1 2 2

4/3 4/3

1 2

1

2

1

2

f f fS S S

V n V n

R R

V2 from (3) is used

R2 can be calculated

from (1)

RIGHTLEFT

If guess is

in correct

re- guess h2

If guess is correct

LEFT=RIGHT

PRACTICE 1

• PREDICT THE FLOW PROFILE FOR

STEADY FLOW USING STANDARD

STEP METHOD

hcx

y

h(x)

Control point

A prismatic, rectangular channel is 6m

wide.

The channel has bed slope of 1 in 800

and Manning’s n is 0.017 s m-1/3.

The channel terminates in a vertical drop

so that the flow falls freely into a lower

reservoir. The channel is designed to

carry 35m3/s at normal depth of 2.34m

Assume the flow passes the critical depth

near the drop. Calculate the elevation of

the water surface in the channel using

specific energy head (use standard step

method).

And determine:

a) The distance at which the water

surface is within 10mm of the normal

depth.

Hint : use dx=100m, since theoretically

the water surface approaches the

normal depth at infinity.

PRACTICE 2 The jet emerging from an underflow

vertical sluice gate contract to a vena

contracta just downstream of the gate

where the depth is D=0.85m.

Downstream of the sluice gate where the

flow is unaffected the corresponding

normal depth is 2.60m. The channel has

a uniform rectangular cross-section of

7.00m wide, Manning’s roughness n of

0.030s/m1/3 and bed slope of 1 in

200m. Assuming energy correction

coefficient is 1.00, use the standard step

method to determine the surface water

profile and the location of the hydraulic

jump.

Fr <1.7 1.7 2.0 2.5 3.0 4.0 5.0 7.0 14.0 20.0

Lj Undular

jump

4.0h2 4.4h2 4.8h2 5.3h2 5.8h2 6.0h2 6.2h2 6.0h2 5.5h2

Lj=length of jump

h2=subsequent depth after jump

2) Direct Step Method – for prismatic channel only

21

o fS Sdh

dx Fr

Prismatic

channel

2) Direct Step Method – for prismatic channel only

2

2

1

1

o f

ave

o f ave

S Sdh

dx Fr

Frx h

S S

D D

1) Predetermine delta h

A Bh hh

n

D

n= number of

calculation point

between h1*

and h2*

2) Terms here

calculated using h(i)

and h(i+1)( 1) (i)h i h h D

3) Location of h(i+1)

can be determined

x

y

hA

Control point

hB

Known depth (e.g. normal depth)

hi

In direct step method, the distance DL required for the water depth to change by a

fixed amount of Dh are calculated.

This method is preferred by many for prismatic channel since it afford direct solution

without the need to guess values and undertake several iteration

PRACTICE 3

In Practice 1 the 6m wide rectangular channel ended in a vertical drop into a reservoir. An

alternative arrangement is to be investigated in which the channel would end in a ramp or

spillway with a slope of 1:10. Assume that this part of channel is long enough for the flow to

become established at the normal depth before it encounters any backwater from the

reservoir. At the end of the channel the depth will be constant at 4.755m as dictated by the

water surface level in the reservoir. Take Q=35m3/2, energy coefficient as 1 and n=0.017.

Determine the elevation of the water surface profile in the channel.