Chapter 1 Additional

8/6/2019 Chapter 1 Additional

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CHANNEL GEOMETRY

Three types of channel :-

A NATURAL CHANNEL

Figure 9.2 : Natural channel


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CHANNEL GEOMETRY

B ARTIFICIAL CHANNEL

Figure 9.3 : Artificial channel


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CHANNEL GEOMETRY

C PRISMATIC CHANNEL

Uniform cross section & slope at whole channellength.

Usually artificial channel.

CHANNEL GEOMETRY ELEMENTy = Depth of water (m)

T = Top width water surface (m)B = Base width water surface (m)P = Wetted perimeter (m)A = Wetted area (m 2)R = Hydraulic radius p R = A/P (m)D = Hydraulic depth p D = A/T (m)


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CHANNEL GEOMETRIC ELEMENT

T

B

Py

Figure 9.4 : Channel geometric element


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Figure 9.5 : Channels sides slope

I SIDES SLOPE, Z

1

z

Note : If slope, U= 45 r p z = 1( z at left & right side is same )


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Figure 9.5 : Channels sides slope

II CHANNEL SLOPE, S o ( unit less )

U


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III RELATIONSHIP BETWEEN v & Q

AVQ ! . 2.5where ;

Q = Discharge or flow rate (m 3/s)v = Velocity (m/s)


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DERIVATION OF CHANNEL FORMULA

B

y1

z

1

z

31

2

L

Figure 2.6 : Derivation of channel formula


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From Figure 2.6 :

T = Top width water surface

zy2BT !

A = Wetted area

A = Area 1 + Area 2 + Area 3

? A

2zyByA

yzy21

yByzy21

A

!

!


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P = Wetted perimeter

2

22

222

22

z1yL

z1yL

yzyL

zyyL


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CHANNEL GEOMETRIC ELEMENTP = Wetted perimeter

2z1y2P

L2P

!

!

therefore ;

Note :Use this trapezoidal formula (A, T & P) to find formulae forrectangular & triangular shape.

p For Rectangular z = 0

pfor Triangular = 0


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SHAPE A T P

By B B + 2y

zy 2 2zy

By + zy 2 B + 2zy

B

y

T

zz11 y

T

y1

z

T

B

1 z

d

T

y U

2z1y2

2z1y2

)si(8

2 U U

U

2si

2 U

( Ui ra ia ) ( Ui ra ia )( Ui a gle )

To sum up .. Table 2.1 : Cha els geometric eleme ts


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Lets take abreak!!!


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Based on the figure given find :-

i) Top width water surface (T), wettedarea (A), wetted perimeter (P) &hydraulic radius (R).

ii) If Q = 2.4 m 3/s, determine the flowstate.

iii) If inclined length (L) = 50 m, find thecost to construct this channel (Givenexcavation cost = RM 3/m 3 and liningcost = RM 5/m 2)

EXAMPLE 2.1


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3 m

2 m

1 m

60 r

.... Cont

SOLUTION: Given:-

B = 3 my = 2 mt = 3 m ( channel height) )


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SOLUTION

Find z value first :

60 rz

15774.0

60tan1z

z1

60tan

!r

!@

!r

Therefore ;

(i) Top width water surface, T

m3096.5T

)2)(5774.0)(2(3T

zy2BT

!

!!


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(i) Wetted area, A

SOLUTION .... Cont

2

2

2

m3096.8A

)2)(5774.0()2)(3(A

zyByA

!

!

!

(i) Wetted perimeter, P

m6189.7P

)5774.0(1)2)(2(3P

z1y2BP2

2

!

!!


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(i) Hydraulic radius, R

SOLUTION .... Cont

(ii) State of flow

m091.1R

6159.73096.8

R

PA

R

!

!

!

gDv

Fr !


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SOLUTION .... Cont

(ii) Cont . State of flow

Find v & D first :

m/s2888.03096.8

4.2v

v

!!

!

m5650.1D

3096.53096.8

D

T

AD

!

!

!


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SOLUTION .... Cont

(ii) Cont . State of flow

Thus ;

flowcriticalSu 1 074.0

)565.1)(81.9(2888.0

gv

r

r

r

p!

!

!

(iii) Construction cost

Construction cost includes :-

(a) Excavation cost

(b) Lining cost


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SOLUTION .... Cont

(iii) Cont . Construction cost

(a) Excavation cost

3

2

2

m709.83volumeExcavation

(50) )3()5774.0()3()3(volumeExcavation

(L) ztBtvolumeExcavation

ALvolumeExcavation

!

!

!

!

Therefore ;

2129.49costExcavation

m83.709m

3costExcavation 3

3

!

v!


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SOLUTION .... Cont

(iii) Cont . Construction cost

(b) Lining cost

Therefore ;

2

2

2

m496.42areaLining

)50( (0.5774)1(2)(3)3areaLining

)L( z1t2BareaLining

PLareaLining

!

!

!

!

2482.09RMcostLining

m42.496m

5RMcostLining 2

2

!

!


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SOLUTION .... Cont

(iii) Cont . Construction cost or overall cost

Hence ;

4611.58RMcostonConstructi

2482.09RM2129.49RMcostonConstructi

costLiningcostExcavationcostonConstructi

!

!!


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W ater flows 0.8 m depth inside a 1.2 mdiameter culvert. Calculate top widthwater surface (T), wetted area (A),

wetted perimeter (P) and hydraulic radius(R).

EXAMPLE 2.2

SOLUTION:

Given:-y = 0.8 md = 1.2 mT ?? A?? P?? R??


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SOLUTION

Find Uvalue first :

1.2 m

0.8 m

EE

E

0.6 m0.2 m

E

r! U 2180

U

Where ;

r!

!E 47.19

6.02.0

sin 1


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SOLUTION .... Cont

Cont . Find Uvalue first :

Thus ;

3.8213ra iai to co vert94.218

)47.19)(2(180

! Upr! U

rr! U

Therefore ;

(i) Top wi th water surface, T

m131.1T

294.218

si)2.1(T

2 siT

!

!

U!


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SOLUTION .... Cont

(ii) Wetted area, A

2

2

2

m801.0A

94.218sin8213.382.1

A

sin8

dA

!

r!

U U!

(iii) Wetted perimeter, P

m293.2P

2)2.1)(8213.3(

P

2d

P

!

!

U!


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(iv) Hydraulic radius, R

SOLUTION .... Cont

m349.0R

293.2801.0

R

R

!

!

!

Chapter 1 Additional

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