Environmental Engineering Reference
In-Depth Information
Because the energy slope is considered to be the same for both parts, the energy from a unit element of
water transmits only in one direction: to the left wall, to the right wall, or to the channel bed. Thus, from
the view of energy the whole flow section can be divided into three parts, as shown in Fig. 5.20. The
energy concentrated on the left wall
ab
(for a unit thickness in the flow direction) is from the potential
energy of water body
abf
,
and the turbulent kinetic energy created by
ab
finally dissipates into thermal
energy in the region
abf.
Similarly, the flow region
bcef
and the bottom
bc
combine together as one group
in energy transmission, and the region
cde
and the right wall
cd
as another. No energy is exchanged
across the faces
bf
and
ce
.
The energy components are not limited in number. For the area
gh
on a unit boundary perimeter (Fig. 5.20),
the energy transmitted is taken from the water body with a volume
£
.
If the total wetted perimeter is
P
,
pa
must equal to the whole volume of the flow, i.e.
A = P
E,
or
A
E
R
(5.31)
P
Hence, the hydraulic radius itself does have a clear physical meaning. From the mechanism of the
energy transformation of flow, the turbulent energy created on the area of a unit boundary element comes
from the potential energy of a water volume,
R
; the turbulent kinetic energy created there finally dissipates
into thermal energy in the same volume of water. The hydraulic radius,
R,
represents the water volume
possessing this part of the energy (Einstein, 1934). Division or summation of the hydraulic radius only
implies the division or summation of the volume of the water body, an understandable process.
In common alluvial streams, the roughness coefficients of the side walls and the bed are different, and
the turbulence intensities there are different too. The energy taken from the flow by a unit area of the side
wall is not equal to the one on the bed, hence the hydraulic radius
, R
w
, corresponding to the bank resistance
is not equal to
R
b
, corresponding to the bed resistance, instead, the two are
A
W
R
(5.32)
w
2
h
b
A
R
b
(5.33)
B
Here
A
w
is the sum of areas
abf
and
cde
in Fig. 5.20, and
A
b
is the area
bcef
.
The method of calculating the bank and bed resistances of a rectangular flow section according to the
principle of division of energy is given as follows:
The following equations describe the relations between the flow energy and the partitioning of the
hydraulic radius
A
AA
w
b
AUAU AU
ww
bb
A
RAP
/
w
w
w
w
2
h
RAPAB
/
/
(5.34)
b
b
b
b
1
UR
2/3
1/2
J
b
b
n
b
1
2/3
1/2
U
R
J
w
w
n
There are 8 variables in the 6 equations:
A
w
,
A
b
,
U
w
,
U
b
,
P
w
,
P
b
,
n
w
,
and
n
b
Usually
n
w
can be estimated
from the bank composition, especially for rigid banks. We need one more equation to solve the problem.
Different methods have been proposed but only the Einstein and Chien (1958) method is introduced for
w
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