sab2513 hydraulic chapter 5

7/31/2019 SAB2513 Hydraulic Chapter 5

1/32

1


2/32

Characteristics of Gradually Varying Flow (GVF) Flow depths vary gradually along the channel.

Changes in flow depth occurs over a long distance.

Water surface nearly horizontal

2 conditions:

(a) Backwater (b) Drawdown


3/32

Analysis and derivation of GVF Equation

Head loss is similar to uniform flow

and

Bed slope is small

Channel is prismatic

Constant roughness along the channel reach

oRSACQ oSARnQ3

21


4/32

Differential Equation of GVF


5/32

(5.1)

Differentiate with respect to x:

(5.2)

Where;

and

g

v

dx

d

dx

dy

dx

dz

dx

dE

dx

dz

dx

dH

2

2

2

2vH z E z y

g

;fS

dx

dH oS

dx

dz

dx

dy

dy

dA

gA

Q

dx

dy

gA

Q

dx

d

g

v

dx

d3

2

2

22

22


6/32

Since dA/dy = T

Rewrite Eqn. (5.2),

Rearrange the above Eqn.,

(5.3)

3

2

1gA

TQ

SS

dx

dy fo

dx

dy

gA

TQ

dx

dy

g

Q

dx

d3

22

2

dx

dy

gA

TQ

dx

dSS of

3

2


7/32

If K = conveyance at any depth y and Ko = conveyancecorresponding to normal depth yo, then:

So,

;

f

QK

S

2

2

K

K

S

So

o

f

o

o

QK

S


8/32

Rearrange the above Eqn.,

(5.4)

Eqn. (5.4) is useful in developing direct integrationtechniques and it is applicable for channel with any

geometrical shape

2

0

2

3

1

1

o

KS

Kdy

Q TdxgA


9/32

Differential Equation for Rectangular Channel

Substitute relevant parameters and simplifying Eqn. (5.4)resulted;

(5.5)

Where;

and

3

2

1

1

y

y

K

K

Sdx

dy

c

o

o

n

ARK

32

oo

QK

S


10/32

Differential Equation for Very Wide RectangularChannel (R = y)

Substitute relevant parameters and simplifying Eqn. (5.5)

resulted;

;for Mannings n (5.6)

3

310

1

1

y

y

y

y

Sdx

dy

c

o

o


11/32


;for Chezy (5.7)

3

3

1

1

o

o

c

y

ydyS

dx y

y


12/32

Computations of GVF

Computation of GVF involves basically solution of dynamiceqn. of GVF

Several methods of computation:

(a) Direct Integration method

(b) Numerical Integration method

(c) Graphical integration method

(d) Standard step method Numerical Integration method will be discussed

The GVF differential equation is written in finite differenceform


13/32

Numerical Integration Method

The unknown length of channel to be determined is dividedinto several portions called segment


14/32

Referring to the segment above:

= length of the segment

= changes of flow depth between segments, whichfixed in the calculation =

= average flow depth in each segment =y

x

y

1

2

1

ii yy

ab yy 2

1


15/32

Rectangular Channel (Eqn. 5.5)


(5.8)

Where;

and

3

2

1

1

y

y

K

K

Sx

y

c

o

o

;32

g

qyc

o

ooo

S

Q

n

RAK

32 3

23

2

2

yB

yB

n

yB

n

RAK


16/32

Very Wide Rectangular Channel (Eqn. 5.6)


For Manning; (5.9a)

Where;

3

310

1

1

y

y

y

y

Sx

y

c

o

o

3

2

gqyc


17/32

Very Wide Rectangular Channel (Eqn. 5.6)


For Chezy; (5.9b)

Where;

3

3

1

1

o

o

c

y

yy Sx y

y

32

gqyc


18/32

Differential Equation of GVF

Y


19/32

19

A very wide rectangular channel with bed slope

0.001, Mannings n 0.025 and discharge 2.5

m3/s.m. Determine the backwater curve created by

a low dam that has water depth of 2.0 m

immediately upstream of the dam. The upstream

computations may be carried out to a depth 1%

greater than the normal depth. (Use numericalintegration method in your computations and divide

the range of depth into 4 equal parts).

Example 1


20/32

20

3

10

3

1

1

yy

yy

Syx

o

c

o


21/32

21

Solution

Find yo, so from Manning Eqn.

R = y

n

SAR

Qo

32

n

SyBy

Qooo

32

n

Syq

oo3

5

53

1000

025.05.2

oy

q=Q/B

myo 5.1


22/32

22

Solution

Find yc, for numerical integration method

32

g

qy

c

3

2

81.9

)5.2(

myc 86.0


23/32

23

Solution

1% above yo, 1% of 5.101.0 oy

my 52.1

m02.0015.0 02.05.1 y

?L

oy my 52.1 my 0.2


24/32

24

m52.1 m0.2m88.1m76.1m64.1

4x 3x 2x 1x

4 3 2 1y y y y

divide the range of depth into 4 equal parts


25/32

25

Solution

From Eqn (5.9a)

where

310

3

1

1

yy

yy

S

y

xo

c

o

my 12.04

52.12


26/32

26

Solution

Or

Where and

3

10

3

1

1

001.0

12.0

y

y

y

y

x

o

c

BAx 120

3

1

yy

A c3

10

1

yy

B o


27/32

27

Solution

Or

Where and

3

10

3

1

1

001.0

12.0

y

y

y

y

x

o

c

BAx 120

3

1

yy

A c3

10

1

yy

B o


28/32

28

xy A B

2.00-1.88

1.88-1.76

1.76-1.64

1.64-1.52

L=

Create Table!!!

The length of backwater, L = m

y

ix


29/32

29

xy A B

2.00-1.88 1.94 0.912 0.576 190.1

1.88-1.76 1.82 0.895 0.475 226.1

1.76-1.64 1.70 0.871 0.341 306.9

1.64-1.52 1.58 0.839 0.159 632.9

So, L= 1355.3m

Create Table!!!

The length of backwater, L = 1355m

y

ix


30/32

30

xy A B

2.00-1.88

1.88-1.76

1.76-1.64

1.64-1.52

So, L=

Create Table!!!

The length of backwater, L = 1718m

y

ix


31/32

31


32/32

32

sab2513 hydraulic chapter 5

Documents