18.1 The Hydraulic Exponent for Uniform – Flow Computation.

Hydraulics
Prof. B.S. Thandaveswara
18.1 The Hydraulic Exponent for Uniform – Flow
Computation.*
Assuming the conveyance K as a function of the depth of flow y*, it may be expressed
as
K 2 = C0 y N
(1)
in Which C0 is a coefficient and N is known as the “hydraulic exponent for uniform –
flow".
*This is strictly applicable to sections which are wide and are described by the
exponential equation
Taking logarithms on both sides of above equation and then differentiating with respect
to y, it may be written as
d
N
( ln K ) =
dy
2y
(2)
Now, taking logarithms on both sides of Eq. (2) and then differentiating this equation
with respect to y under the assumption that Resistance factor is independent of y, the
expression for N may be obtained. [See Box]
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
K 2 =C0 y N
K 2 =C0 y N
Taking logarithm on both sides
2 ln K = ln Cο + N ln y
2 ln K = ln C0 + N ln y
Differentiating with respect to y
d
N
(ln K) =
dy
2y
Consider Manning formula
2 ⎞2
⎛
1
K 2 = ⎜ AR 3 ⎟
⎜⎜ n
⎟⎟
Differentiating with respect to y
d
N
(ln K) =
dy
2y
Consider Chezy formula
1 ⎞2
⎛
K 2 = ⎜ C AR 2 ⎟
2
⎡ ⎡1⎤
⎤
2 ln K = 2 ⎢ln ⎢ ⎥ + ln A + ln R ⎥
3
⎣ ⎣n⎦
⎦
Differentiating with respect to y
1
⎡
⎤
2 ln K = 2 ⎢ln C + ln A + ln R ⎥
2
⎣
⎦
Differentiating with respect to y
2 dA 1 dR
d
2
+
( ln K ) =
dy
A dy R dy
⎝
2
d
d
1
4
( ln K ) = ⎡⎢ 2 ln + 2 ln A + ln R ⎤⎥
dy
dy ⎣
n
3
⎦
⎡ 1 dA 2 1 dR ⎤
d
+
( ln K ) = ⎢
⎥
dy
⎣ A dy 3 R dy ⎦
equating the right hand side
N T 2 T 2 A dP
= +
2y A 3 A 3 p dy
T 2 T 2 dP
= +
- R
A 3 A 3 dy
N=
⎜⎜
⎝
⎠
2y ⎡
dP ⎤
⎢ 5T - 2R
⎥
3A ⎣
dy ⎦
⎟⎟
⎠
⎡ 1 dA 1 1 dR ⎤
d
+
( ln K ) = ⎢
⎥
dy
⎣ A dy 2 R dy ⎦
=
T 1 p dA
A dP
+
A 2 R dy 2RP 2 dy
T T 1 dP
= +
A 2A 2P dy
equating the R.H.S
N
2y
=
3 T 1 dP
2 A 2P dy
N=
y⎡
A dP ⎤
⎢3T ⎥
A⎣
P dy ⎦
N=
y⎡
dP ⎤
⎢3T - R
⎥
A⎣
dy ⎦
* dA
≈T
dy
* These are the general equation for the hydraulic exponent N.
* This is strictly applicable only to section which are wide and are described by
the exponential equation
For a trapezoidal channel section having a bottom width b and side slopes 1 on m, the
expression for A, T, P and R may be obtained from Table. Substituting them in equation
in the Box and simplifying, it results in
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
⎡ ⎛y ⎞
1+ m2
⎜
⎟
⎢
b
⎡
⎤
′
1+
2y
10
8
⎠
N= ⎢
- ⎢ ⎝
⎥
3 ⎣ 1+ y′ ⎦ 3 ⎢ 1+ 2 ⎛ y ⎞ 1+ m2
⎜ b⎟
⎢⎣
⎝
⎠
my n
in which y′ =
b
⎤
⎥
⎥
⎥
⎥⎦
This equation indicates that the value of N for the trapezoidal section is a function of m
and y / b. For values of m = 0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, and 4.0, a family of curves for
N versus y /b may be constructed (Fig). These curves indicate that the value of N varies
within a range of 2.0 to 5.0.
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
Logarithmic plot of "K" as ordinate against the depth as abscissa will appear as straightline then
log ( K1/K 2 )
log ( y1/y 2 )
The hydraulic exponent is equal to twice the slope of the tangent to the curve at the
N=2
given depth. When the cross section of a channel changes abruptly with respect to
depth, the hydraulic exponent will change accordingly.
Indian Institute of Technology Madras