Environmental Engineering Reference
In-Depth Information
White and Colebrook established an accurate good interpolation by
taking
z
0
=
0
.
03
k
s
+
0
.
01
δ
. For this transition region the equation reads:
5
.
75
u
∗
log
12
y
k
s
+
v
¯
=
(2.25)
0
.
03
δ
The last equation can be transformed to the uniform flow equation of the
de Chézy (see Section 2.6).
In the horizontal direction, the velocity distribution is influenced
by the
B
/
y
ratio. For a non-wide canal (
B/y <
5), the velocity distribution
is three-dimensional. When the
B/y
ratio is larger than 5, the velocity
distribution becomes almost two-dimensional with the exception of a
small region near the vertical sidewalls.
Corrections in view of the average velocity
The velocity distribution in the vertical and width direction of a canal
incorporates the fact that the velocity is not the same in all the points of
the cross section. For that reason, the velocity in a point (
v
A
) is not equal
to the average velocity
v
¯
=
Q/A
, but it can be written as
v
A
=¯
v
±
v
.
The discharge
Q
is by definition equal to the average velocity
v
times the
total area
A
, but it is also the sum of the volumes of water passing the
small areas
A
with a velocity
v
A
.
¯
v
A
A
(
vA
Q
=
=
v
¯
±
v
)
A
=
vA
¯
±
=
(
±
(
vA
)
By definition:
Q
0
The total energy in a cross section comprises the potential energy,
the pressure energy and the kinetic energy; the latter is expressed by the
velocity head
v
2
/2
g
. The velocity head for the average velocity differs
from the summation of the velocity head for each point and, therefore,
it should be multiplied by the coefficient of Coriolis (
α
) to obtain the
total head.
The mass of water flowing with a velocity
v
a
through a small area
A
is
ρv
a
A
. The kinetic energy passing that area per unit of time is
the product of the mass and the velocity squared: ½
ρv
a
A
and the total
¯
vA
) and hence:
=
kinetic energy for the whole cross section is: ½
ρv
A
A
.
The total kinetic energy for the whole cross section can be expressed
as
αρA
(
v
3
/
2
g
).
Equating this quantity with ½
ρv
A
A
results in
¯
(
v
A
∗
(
v
)
3
A
A
)
¯
v
±
α
=
=
(
v
3
¯
∗
A
)
(
v
3
¯
∗
A
)
(
v
3
A
¯
±
¯
v
2
vA
+
¯
vv
2
A
±
v
3
∗
3
3
A
)
=
α
v
3
¯
∗
A
Search WWH ::
Custom Search