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EoEN '. E €93on H ﬁwqommxx Problem #1
Consider the Rhine River data given to determine the velocity proﬁle as well as other hydraulic parameters. Given values: ﬂow depth, h = 9.9m channel width, w=260m slope of EGL, Sf=13.12cm/km = 00001312an Assumptions:
 rectangular channel geometry Cl?”  water temperature is 20°C .'.y= QSﬁN/m3, u 21.0 x 10'6 mz/s
 ys= 26000N/m3 Determine the following parameters: a) shear velocity in m/s A
/R,, = F assume rectangular channel geometry w  d 260m  9.9m ’ 2d+w _ 29.9m+260m
,, =9.199m l R
R u. =x/9.81m;’52 9_199m0.0001312 u. = 0.109m/ s b) von Karman constant
V = i ln 2 + C u, K The shear velocity was calculated for each point as well as the distance from the channel bed, 2
(m). The ratio of velocity to shear velocity was plotted as a function of lnz (Fig 1).
The slope of the linear best fit line was 3.875. Therefore, =3.87 =0.2581 / i
K
K Assignment #3 Jaime Geode [— Velocity Profile (based on log law)
25
20 7  — 1
15 ‘1 ' ' ""—
y = 3.875x + 11.361
.2 R2 = 0.9361
10 _, 7
51 a . _
0 1—— —‘r‘ _7_ '——1 —r— —r—' '1— —'
1 5 1 0 5 0 0.5 1 1 5 2 2 5
In (2)
c) shear stress in Pascals
To 2 7Rth .
10 = 9810N.9.199m1.312><10‘4
T —1184£—1184Pa
0 ' m2 ‘ /
d) mean ﬂow velocity in m/s
— vAh .
V = 7, where h=9m (see attached spreadsheet calculatlon)
I7 =1.76 m/s 01',
[email protected][email protected] ‘2‘!
ll
/—'\
[NJ )[email protected] ¢g\:’”1\( ' Assignment #3 Jaime Goode e) Froude number Fr = 0.185 thus, ﬂow is subcritical f) DarcyWeisbach friction factor /
V _ 2
ui f
u?
f = 8. $2
f = 0 03 /
g) Manning n
1 Z l
V = — R 3 S?
n J
1 3 l
n _ : R 3 S} f
C = 51.24 m°'5/s / i) Momentum correction factor __ 1 2 z 1 2
'8— AV: JV dA_ hV: Zthi W7 ,3=1.046 (see attached spreadsheet for calculations) / / / at a j) Energy correction factor 1 3 1 3
a = v dA E v, dh.
AVE Al M Z ’ a =1 . 12 (see attached spreadsheet for calculations) k) laminar sublayer thickness 6:11.6v u. /
5 =1.07x10'4m 1) grain shear Reynolds number Re, — “* 'dS where, ds=d50=l.3mm — V
Re. =141.45 m) Is the ﬂow hydraulically smooth of rough for this bed material?
The boundary is hydraulicallyrough if z>68 and turbulent forces dominate.
The boundary is hydraulically smooth if z<8/3 and Viscous forces dominate. 5=1.07x10'4m
d10=4.0x10'4 m, (150:1 .3x10‘3m, and (190:1 .0x10'2m 65=3.42x104m 
This is less than dS so the ﬂow is hydraulically rough.
Also, Re*>70. Determine the following for the median particle diameter
d50=1.3mm (very coarse sand)
11) dimensionless particle diameter dr = d,v[(G:1)g]§ V d. = 32.88 / o) Shields parameter __ To
(7‘. "7".)0'5 r. = 0.563 / Ti p) critical shear stress in Pa
According to Fig 7.7 in the text using d5=I .3mm, and using the general rule of thumb discussed in class, I". =1Pa:dI =1mm Tc =1.3Pa / or, z'c = no (y, — ym )d, , where rte is determined from the Shields diagram (Fig 7.6), using d¥
=33.88, which corresponds to a 1., 50.04 q) angle of repose for the bed material
Using ﬁg 7.2 and assuming that the particles are closer to “very rounded” than “very angular,” ¢=31° / r) critical Shields parameter
T C C _ (7.?  Md;
ac = 0.048 7* / s) critical shear velocity in m/s 24.6 = 0.032 mfs / t) settling velocity in mm/s 3 . . . . 24 . .
a) = EQHI — d J— I] Thls equatlon was derlved assumlng C D = — +1.5, Wthh accordmg to d 72 Re p ﬁg. 5.2 in the text is best suited for the sand sized sediment class.
a) =130.75mm/s /
u)Reynolds number of the particle cod,
Rep = U Rep = 169.97 / v) is the flow around the particle laminar 0r turbulent?
The ﬂow around the particle is turbulent because the Reynolds number of the particle is greater than 70. w) is the particle in motion? /
According to the Shields diagram (ﬁg 7.5 in the text) the particle is in motion. R6. = 141.45 and
r. = 0.563 , which is above the Shields curve, indicating that the particle is in motion. x) is the shear velocity greater or smaller than the settling velocity? u. =0.109m/s; a): 0.131m/s
u.<co / y) is the critical shear velocity smaller or greater than the settling velocity?
um = 0.032; a) = 0.13 m/s u.c< a) Problem #2 A sharp bend in the Fall River, Colorado was considered. The goal was to examine the stability
and incipient motion of a 4mm particle with an angle of repose of 37 degrees placed on the bed
of the 10 m wide channel, assuming a channel bed slope, 90, of 0.074 degrees. The particle
stability was calculated using 2 different methods, 1) Stevens and Simons’ method (Section 7.3),
which considers the sum of moments on the particle; and 2) the Brooks method (Section 7.4),
which is a simpliﬁed method of particle stability analysis that is based on the sum of forces. Each
method was used to determine the critical shear stress required to bring the particle to incipient
motion, as a function of side slope, 91, and streamline deviation, 7». For each method, the lift to
drag ratio was assumed to be one. cos(/?. + 19) M+N ,/1— 2
(~—)—a‘9+sin(/t+6)
Nnotan¢ ,8 = tan—1 SFO=_%M__ /
t + 1— 2
_ w mm) ‘Wsﬂ
771 770 1+(M/N) The application of this method required an initial assumed value of the bed shear stress. Based
on the general relationship between critical shear stress and particle diameter, an initial value of
4 Pa was selected. In order to determine the critical shear stress as a function of, 61, 7t, and [3 (the
angle from the weight component down the side slope that the particle will tend to move), the
goal seek function in Excel was used to iterate on the shear stress until the safety factor equaled
one. Because the safety factor is a ratio of the passive to active moments acting on the particle, a safety factor of one meant that these moments were equal and the particle was at incipient
motion. The Brooks method is a simpliﬁed form of particle stability and applies the sum of forces on the
particle as a function of bed slope, side slope and streamline deviation angle. This method
requires the calculation of the critical shear stress from the Shields method and diagram for particles on horizontal surfaces. In this problem the calculated 10:2.43 Pa. The Brooks equation
is as follows: 2
sin 91 sin/1 + sin 60 cosl +1_ sin2 60 + sin2 91
(1+H,d2)tan¢ sinz ¢ For the sake of comparison, the Lane approach to particle stability analysis was also applied.
This relationship is very simple and is best applied to straight channels because it assumes that
the streamlines do not deviate from the direction of ﬂow. Therefore, because the channel
considered in this problem was a sharp bend, this method is not appropriate for this case. E: 1_ sin261
TC sin2¢ Results and Discussion:
The results from each method were plotted as the critical shear stress as a function of both the side slope angle and the streamline deviation angle (ﬁg 1 and ﬁg.2). 16L. sin 61 sin A + sin 90 cos 1
= _————— + 1,. (1+H,d2)tan¢ Steven & Simons method of particle stability analysis 3.5 2.5 N
I. Critical shear stress 1: (Pa)
2'}. _. J 0 5 1o 15 20 25 30 35 40
Side slope 61 (degrees) Fig. 1 Brooks method for particle stability analysis 2.5 Critical shear stress (Pa)
01 _n
___l 0.5  O 5 1O 15 20 25 3O 35 40
Side slope angle (degrees) Fig. 2. 1) Are both methods in general agreement? Both methods are in general agreement as far as the trends in the side slope angle relative to the
critical shear stress. However, the simpliﬁed method of particle stability analysis provided by
Brooks, calculated lower values of the critical shear stress for different side slope angles. The
Brooks method relies on the critical shear stress determined for horizontal surfaces and so is
sensitive to the initial critical shear stress determined from the Shields diagram. On the other
hand, the Stevens and Simons’ method assumes an initial value for the critical shear stress and
requires iterations of the critical shear stress, such that the ratio of the passive to the active
moments, called the safety factor equals one (SFo =1). This means that at incipient motion, the
forces tending to move the particle from the bed are matched with the passive component of the
weight that tends to hold the particle in place. Both methods account for deviation of the streamlines. However, the Stevens and Simons’
method is more robust because it accounts for the angle at which the particle will tend to move
down the slope, [3. However, when the streamline deviation is highly negative on low angled side
slopes, the Steven’s and Simons method must change because at negative streamline deviations,
the drag force is largely a passive force. This changes the safety factor formula used in this
exercise, so the critical shear stress could not be calculated for low side slopes and largely
negative streamline deviations. A plot of the results for each method including the Lane method for 7t=0 (Fig.3) indicated that
for lower side slope angles the Stevens and Simons’ method predicted higher values of critical shear stress, but the methods showed moderate convergence as the side slope angle approached
the angle of repose. The Lane and Brooks methods were consistent because they are both based
on the Shields diagram. Therefore, the sum of forces method is somewhat conservative and
under predicts the critical shear stress for lower side slope angles. Comparison of Methods for K =0
3.5
3 ,
2.5 7 _ , ,
1?
$
3
e 2 +
‘6
36
d.)
5
E 1.5 <
.9
'E
1 + W —
+ Brooks
0 5 0 Lane
' + Steven and Simons
o 7"— é“ —_l— t —l_‘ T— T—l“
0 5 10 15 20 25 30 35 40
Side slope angle (degrees)
i... .__r
F1g 3. 2) Are particles more stable with streamlines going up or down the side slope? According to the plots of each method (Fig 2; Fig 3), the values for critical shear stress for a
given side slope angle are greatest for K =30. Because k is defined as the deviation of the
streamlines from the horizontal and positive downward, positively directed stream lines
contribute to the net active force on the particle tending to push the particle down the slope.
Therefore, the critical shear stress for a given side slope angle is smaller when the streamlines
are directed down the side slope (X = positive), than when the streamlines are directed up the side
slope (7. = negative). When X is negative the streamlines are directed up the slope, which also
means that the drag force is directed up the slope. Therefore, the drag force becomes a passive
force, which tends to stabilize the particle. When 9» is zero, the streamlines are parallel to the bed
slope so that the drag force is also parallel to the bed slope. In this case particle stability is
controlled by only the bed slope and the side slope. This mechanism describes the phenomenon
of bank erosion of the outside of meander bends and deposition on point bars. 3) Does the critical shear stress depend more on the side slope angle or the streamline
deviation? According to the plot of the Stevens and Simon method, as the side slope increased from 0 and
30 degrees, the critical shear stress decreased by approximately 2 Pa for all values of 9». As 9» increased or decreased from 0 to either 30 or 30 degrees, respectively, the largest difference in
critical shear stress was slightly less than 1 Pa. Therefore, the critical shear stress depends more
on the side slope angle than the streamline deviation. As the side slope approaches the angle of
repose, the effect of the deviation of the streamlines from downstream on the particle stability
decreases. Problem #3:
The goal of this problem was to design stable channel cross section geometry for a clay channel
and estimate the maximum permissible velocity in the channel given a discharge of 2000 cfs. b / Given that the clay had a density of 2,000 kg/m3, an initial value of 1.15 m/s from Table 7.3 was used to estimate the channel geometry. Also, a side slope angle, 01:340 was selected from Table
7.4 and a Manning’s n=0.025 was selected from Table 7.2. In order to calculate the channel dimensions to pass the maximum velocity through the channel,
the following calculations were performed in excel using Manning’s equation. Iterations were
performed on the ﬂow depth and velocity. The goal seek function would not converge until the
correct velocity was used to iterate on the ﬂow depth. Problem #3
Maximum permissible velocity
Stable Channel Design __ —
—
— 2000
Q cms m —
mm x 80
A z— 1 ~148ia
.. Table 7.4 / ' tang] ‘ '
_1.482561
0.025 , 1 a L
v = —R 3302
n A" Permissible
velocity, v
m/s 0.80962859 Hydraulic
Radius, R Bottom sectional Wetted
Flaw depth, h width, b Area, Perimeter,
m m m2 P m m
69.877 24.266 ...
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 Fall '06
 Julien

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