Hydraulics
Prof. B.S. Thandaveswara
12.1 Critical flow depth computations
One of the important aspects in Hydraulic Engineering is to compute the critical depth if
discharge is given.
Following methods are used for determining the critical depth.
(i) Algebraic method.
(ii) Graphical method.
(iii) Design chart.
(iv) Numerical method. Bi section method/ Newton Raphson method.
(v) Semi empirical approach - a method has introduced by Strarb.
12.1.1 Algebraic method
In this method the algebraic equation is formulated and then solved by trial and error.
The following example illustrates the method.
1. Consider a trapezoidal channel:
2.
A = ( b + myc ) yc
D=
( b + myc ) yc
( b + 2myc )
Zc =
Q
= constant = C1 = known
g
⎧⎪ ( b + myc ) yc ⎫⎪
C1 = ( b + myc ) yc ⎨
⎬
⎩⎪ ( b + 2myc ) ⎭⎪
1/ 2
(1)
C12 ( b + 2myc ) = ( b + myc ) y3c
3
leads to
y6c + py5c + qyc4 + ry3c + syc + t = 0
in which the cons tan ts p, q, r,s and t are known.
Solve this by polynomial or by trial and error method.
It would be easier to solve the equation (1) by trial and error procedure.
After obtaining the answer check for the Froude number which should be equal to 1.
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
Example:
Consider a Rectangular channel and obtain the critical depth for a given discharge.
Solution:
Area = b y
∴ Z =
2
y3/
c =
D =
A by
=
=y
T
b
Q
= b y y1/ 2
g
Q
b g
⎛ Q ⎞
yc = ⎜
⎜ b g ⎟⎟
⎝
⎠
2/3
2
⎛ q ⎞
= ⎜
=
⎜ g ⎟⎟
⎝
⎠
3
q2
g
12.1.2 Trial and error method
For a given trapezoidal channel obtain the critical depth by trial and error method.
Solution:
For trapezoidal channel
⎡( b + myc ) yc ⎤⎦
A D=⎣
( b + 2myc )1/ 2
3/ 2
⎛ ( b + my ) y ⎞ 3 Q 2
= constant
Squaring ⎜⎜
⎟⎟ y =
g
⎝ ( b + 2my ) ⎠
For a given b, m, Q, select a value of yc
3
Assume b = 6 m, m = 2m, Q = 12 m3 / s
(
)
3
6 + 2yc y3c
6 + 4yc
( 3 + yc )3 y3c =
3 + 2yc
=
Solve for yc
144
= 14.679
9.81
36
= 3.6697
9.81
Assume a value of yc and compute A D and compare with the value obtained by
yc
1.2
0.5
0.8
0.65
0.70
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A
D
A D
23.708
1.339
6.170
3.10
3.94
Remarks
too high
low
high
Q
.
g
Hydraulics
Prof. B.S. Thandaveswara
Remarks column indicate that the values are high or low when compared to the given
value. The improvement is done till it converges.
In the above table yc lies between 0.65 and 0.70.
This could be improved further by selecting the values in between these two.
12.1.3 Graphical method
For natural channels and complicated channels, the graphical method is adopted. A
curve is generated assuming different values of yc and Z. The value of
Q
is computed
g
and yc is obtained from the chart. A one meter diameter culvert carries a discharge of
0.7 m3/s. Determine the critical depth.
T
d0 y
θ
⎡
⎤
1 ⎢ θ − sin θ ⎥
D= ⎢
⎥ do
g ⎢ sin θ ⎥
2 ⎦
⎣
2 (θ − sin θ )
1.5
Z=
⎡ θ⎤
32 ⎢sin ⎥
⎣ 2⎦
0.5
d 0.5
0
Knowing the value of d0 for different values of depth A and D could be obtained from the
table.
Example:
A one meter diameter pipe carries a discharge of 0.7 m3/s. Determine the critical depth.
Zc =
Q
0.7
=
= 0.2235
g 3.132
Construct a graph of yc Vs Z and obtain the value of yc
From the graph yc = 0.4756
From the design chart determine the critical depth for a circular channel of 0.9 m
diameter. Discharge 0.71 m3/s.
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
Solution:
Z=
Z
d 02.5
0.71
= 0.22669
9.81
= 0.29499
yc
= 0.56,
d0
ψ=
α Q2
g
=
1.0 (17 )
yc = 0.49527 m
2
9.81
⎡ 29.5 ⎤
yc = 0.81 ⎢ 0.75 1.25 ⎥
⎣2 6 ⎦
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( from table )
= 29.5
0.27
−
6
= 0.86 m
30 ( 2 )
Hydraulics
Prof. B.S. Thandaveswara
12.1.4 Graphical Procedure
Straub proposed several semi empirical equations to obtain the critical depth. The
advantage of this is a quick estimation of the critical depth. However, the equations are
non homogenous.
yc
Z=A D
Graph showing variation of section factor
with critical depth for a given pipe of
diameter do
yc
y
__
or __c
b
d0
A D
A D
_____
or _____
2.5
2.5
b
d0
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
Reference:
Straub W.O, Civil Engineering, ASCE, 1978 Dec, pp 70 - 71 and Straub 1982.
Table: Semi empirical equations for the estimation of yc (Straub, 1982) MKS units
Channel type
Equation for yc in terms of
ψ = α Q2 / g
1/ 3
⎛ψ ⎞
⎜ 2⎟
⎝b ⎠
b
Rectanglar
1
Ψ
⎛
⎞
0.81⎜ 0.75 1.25 ⎟
⎝m b
⎠
m
b
Trapezoidal
l
⎛ 2Ψ ⎞
⎜ 2⎟
⎝m ⎠
m
0.27
−
b
30m
Range of applicability
Q
0.1 < 2.5 < 4.0
b
Q
For 2.5 < 0.1
b
use equation for rectangular channel
0.20
TRIANGULAR
y
( 0.84cΨ )0.25
y = cx2
Parabolic
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x
y = cx 2
Hydraulics
Prof. B.S. Thandaveswara
⎛ 1.01 ⎞ 0.25
⎜⎜ 0.26 ⎟⎟ Ψ
⎝ d0 ⎠
d0
yc = 0.053
d 0.3
0
yc = [ m ]
Circular
Range of applicability
Q0.52
0.02 ≤
yc
≤ 0.85
do
Q = m3s −1 , d 0 = [ m ]
y
⎛ψ
0.84b0.22 ⎜ 2
⎞
⎟
⎝a ⎠
x
b
0.25
Range of applicability
y
0.05 ≤ c ≤ 0.85
2b
a = major axis
b = minor axis
a
Elliptical
y
1/ ( 2m +1)
⎛ m3ψ c 2m −2 ⎞
⎜⎜
⎟⎟
4
⎝
⎠
1
____
m-1
y = cx
x
Exponential
Example:
b = 6.0 m, m = 2, Q = 17m3 / s determine yc
Solution:
From table
ψ
⎛
⎞
yc = 0.81⎜ 0.75 1.25 ⎟
⎝m b
⎠
where ψ =
0.27
−
b
30m
for 0.1 <
Q
b 2.5
< 4.0
α Q2
g
Q
=
17
⎛ 29.5 ⎞
yc = 0.81⎜ 0.75 1.25 ⎟
⎝2 6
⎠
0.27
= 0.19,
b
6 2.5
It is in the range of the equation. Substituting the appropriate values,
The value of
2.5
1(17 )
ψ=
= 29.5
9.8
2
Indian Institute of Technology Madras
−
6
= 0.86 m
30 ( 2 )
1/ m −1
y = cx ( )
Hydraulics
Prof. B.S. Thandaveswara
Problem:
Non rectangular channel involves trial and error solution.
Obtain the critical depth for the trapezoidal channel of bottom width 6 m with a side
slope of 2.5: 1, which carries a discharge of 20 m3/s.
1
m yc
1
m
6m
Solution:
Trial and error procedure
A= ( b+my ) y = ( 6 + 2.5 yc ) yc
T = b+2my=6 + 5yc
D=
A ( 6 + 2.5yc ) yc
=
T
6 + 5yc
Zc =
Q
=A D
g
V 2 Q2 / A 2
20* 20
=
=
=
2g
2g
[6 + 2.5 yc ]2 19.62
⎧ ( 6 + 2.5yc ) yc ⎫
( 6 + 2.5yc ) yc ⎨
⎬
⎩ 6 + 5yc
⎭
0.5
yc = ?
vc = gyc = ?
Solution of Algebraic or Transcendental Equations by the Bisection Method
In the algebraic expression F(x) =0, when a range of values of x is known that contains
only one root, the bisection method is a practical way to obtain it. It is best shown by an
example.
The critical depth in a trapezoidal channel is to be determined for given flow Q and
channel dimensions.
1−
Q 2T
gA3
=0
The formula must be satisfied by some positive depth yc greater than 0 (a lower bound)
and less than, an arbitrarily selected upper bound say, 10 m.
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
T is the free surface width b + 2myc . The interval is bisected and this value of yc tried. If
the value is positive, then the root is less than the midpoint and the upper limit is moved
to the midpoint and the remaining half bisected, etc.
This method gives the solution very quickly.
T
1
F(x)
m
m
1y
0
100
b
Trapezoidal
Newton Raphson Method is discussed elsewhere.
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Bisection