Hydraulics
Prof. B.S. Thandaveswara
22.4 Most efficient Hydraulic section
For a given cross section determine the hydraulic section.
Hydraulically Best section (Hydraulically Efficient Section)
1. Rectangular Channel:
P = b + 2y
A
b=
y
A = by
P=
A
+ 2y
y
dp
= − Ay −2 + 2 = 0 , A = 2 y 2
dy
∴ by = 2 y 2
or b = 2 y
y
b = 2y
Hydraulically efficient rectangular
channel is half of a square.
2. Trapezoidal Section:
1
m
P = b + 2 y 1 + m2
A = ( b + my ) y or b =
Indian Institute of Technology Madras
A
= − my
y
Hydraulics
Prof. B.S. Thandaveswara
P=
A
− my + 2 y 1 + m 2 (For a given area of flow)
y
Differentiate with reference to y assuming A and m to be constant.
dp
= − Ay −2 − m + 2 1 + m 2 = 0
dy
Substituting for area, the above equation can be rewritten as
− y −2 ( b + my )y − m + 2 1 + m 2 = 0
+
b + my + my
= +2 1 + m 2 = 0
y
b + 2my
= y 1 + m2
2
Half the top width = side slope distance (for given side slope)
b = 2 y 1 + m − 2my = 2 y ⎡ 1 + m 2 − m ⎤
⎢⎣
⎥⎦
Substitute this value of b into the equation A and P and simplifying
P = 2 y ⎡2 1 + m2 − m⎤
⎢⎣
⎥⎦
A = y 2 ⎡ 2 1 + m2 − m ⎤
⎢⎣
⎥⎦
⎡
⎤
A
y=⎢
⎥
2
⎣⎢ 2 1 + m − m ⎦⎥
0.5
Substitute the value of y into P
P=2
A1 / 2
(2 1+ m − m)
2
(2 1+ m − m)
2
0.5
(
)
P = 2 A ⎛⎜ 2 1 + m 2 − m ⎞⎟
⎝
⎠
which is the m value that makes P least?
D.w.r to m and equate it to zero
dP
=0
dm
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
m=
tan θ =
3
= 3
3
y
1
3
= =
my m
3
∴θ = 60
∴b =
2
y
3
This means section is a half hexagon. If a semi circle is drawn with radius equal to
depth y then sides of this section are tangent to the circle.
1
m= 3
60
60
2
3
b = __ y
Half Hexagon - inscribed circle
of radius equal to depth is
tangential as shown in figure
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
Triangular Section: If θ is half angle
⎡ 2a sin θ
⎤
Area = ay sin θ ⎢∵
y = ay sin θ ⎥
2
⎣
⎦
ay sin θ a cos θ sin θ a
=
= sin 2θ
R=
2a
2
4
⎡ y
⎤
⎢⎣∵ a = cos θ ⎥⎦
a sinθ
y
a
θθ
a
Hydraulically efficient channel
Half a square on vertex
R should be max.
dR
2a
= 0,
cos 2θ = 0
dθ
4
∴θ = 45
o
o
45
45
Free surface width is equal to the diagonal
Half Square on its apex
It is a half square resting on its apex and maximum width is equal to diagonal.
Alternative derivation for Triangular Section:
A = y 2 tan θ
y = A / tan θ
P = 2 y 1 + m2
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
l
m
y
θ
= 2 y sec θ
=2
A
sec θ
tan θ
⎫⎪
dp
d ⎧⎪
A
=
sec θ ⎬
⎨2
dθ dθ ⎪⎩ tan θ
⎪⎭
⎡ sec θ tan θ
dp
sec3 θ ⎤
⎥=0
=2 A⎢
−
3/ 2
dθ
⎢⎣ tan θ
2 ( tan θ ) ⎥⎦
sec θ tan θ −
sec3 θ
2 ( tan θ )
3
=0
2 sec θ tan θ − sec3 θ = 0
∴ Solve for θ
∴ θ = 45
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
Hydrostatic Caternary (Linteria)
Equation for hydrostatic catenary is given by
y ⎡⎛ 3 2 15 4 ⎞
5
5 4
⎤
⎛3
⎞
k ⎟ φ + ⎜ k 2 + k 4 ⎟ sin 2φ −
k sin 4φ ⎥
⎜1 − k −
⎢
2k ⎣⎝ 4
864 ⎠
32 ⎠
256
⎝8
⎦
y1 = y cos φ
x1 , y1 are measured from mid point of the surface
x1 =
θ
; θο = slope angle at the point x 1 y 1 . θο varies from 0 at the
2
bottom of the curve to θο at the ends.
k= sin
φ⎤
⎡
sin ⎥
⎢
2⎦
φ = sin −1 ⎣
, θ is slope at any point ( x, y)
k
For the hydraulically efficient channel
θο = 35° 37' 7" , y= 3.5 m. Find A, R, D, Z at full depth.
Also plot the cross section of the channel
θο
solution:
2
=
35° 37 ' 7 "
= 17°48 ' 33.5 "
2
k= sin
sinφ =
k=
θο
2
sin
= 0.30585
φ
2 or k =
k
sin
φ
y1 = y cos φ
φ = 90
2
φ
cos
φ
2
A = 17.0992 m 2
P= 10.443 m
T = 6.7114 m
D = 2.5478 m
y1 (m)
φ ( in
deg )
k
φ
2
sinφ
φ
2 sin cos
2
2
φ 1
cos =
2 2k
or
sin
=
1
2 cos
φ
2
= cos 45
∴ k = 0.707
A
= 1.6374 m
P
Z = A D = 22.293 m5 / 2
R=
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
90
81.78
73.39
64.62
55.15
44.4
31.0
0
0.7071
0.6614
0.6235
0.5916
0.564
Indian Institute of Technology Madras
0.5400 0.51887
0.5
Hydraulics
Prof. B.S. Thandaveswara
Exercise: Plot the graph using the above data
x
x1
y
y1
A = 17.0992 m
T = 6.7114 m
2
P= 10.443 m
R=
A
P
= 1.6374 m
Z = A D = 22.293 m
D = 2.5478 m
5/2
Flexible Sheet: Filled with water upto rim, and held firmly at the top ends without any
effect of fixation on shape. Shape assumed under self weight of water is called
Hydrostatic Catenary.
Rounded bottom triangular section
(
)
P = 2 y ( m + cot m )
A = y 2 m 2 cot −1 m
T
−1
1
r
m
r
y
1
R=
θ0
__
cot θ0 = m
1
θ0 = cot-1m
θ0
m
A
P
T = 2r
=
r
2
=
y
2
(1 + m )
2
Hydraulically efficient sections could be derived using Lagrange Multiplier approach.
Indian Institute of Technology Madras