chapter 1 additional
TRANSCRIPT
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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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CHANNEL GEOMETRIC ELEMENT
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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CHANNEL GEOMETRIC ELEMENT
Figure 9.5 : Channels sides slope
II CHANNEL SLOPE, S o ( unit less )
U
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CHANNEL GEOMETRIC ELEMENT
III RELATIONSHIP BETWEEN v & Q
AVQ ! . 2.5where ;
Q = Discharge or flow rate (m 3/s)v = Velocity (m/s)
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CHANNEL GEOMETRIC ELEMENT
DERIVATION OF CHANNEL FORMULA
B
y1
z
1
z
31
2
L
Figure 2.6 : Derivation of channel formula
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CHANNEL GEOMETRIC ELEMENT
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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CHANNEL GEOMETRIC ELEMENT
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
!
!
!