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Unformatted text preview: Problem #1:
Refer to the Rhine River and determine the following in SI units:
1. Bedform type from Liu and Albertson 2'0 =thSf =pu.2 u. =./thSf
A Rh = F assume rectangular channel geometry wd 260m9.9m 2
Rh — u. =«/9.81m/s 9.l99m0.0001312 =2d+w_29.9m+260m u.=0.109m/s
Rh = 9.199172
80 d3 . . . . 24 . .
a) = —— 1— —l Thls equation was derived assuming C D = — +1.5 , Wthh according to
d“, Rep
ﬁg. 5.2 in the text is best suited for the sand sized sediment class.
a) =130.75mm/s
a: = 0.109m/s : 0.832
a) 0.131m/s
u. ds
Re. = where, ds=d50=1.3mm Re. = 141.45
v
1 1.6V _4 6 . .
6 = 6 = 1.07x10 m d— = 12.195 >1, therefore the ﬂow is hydraulically rough.
71‘ 50
'Hygg’xﬂ'y—mTnismm Hydraﬁllcany rough
.2 ‘ I U 2 3
310 m 1 u 10 1O 10 1o"
1
10 Fig. 1. Bedform classiﬁcation according to Liu (1957) (Adopted from Julien, 1998) According to Fig. 1. the bedforms are antidunes. 2. Bedforrn type from Chabert and Chauvin To
2', = ———————
(Vs  n M
r. = 0.563
Re. = 141.45 ‘D Fig 2. Bedform classiﬁcation according to Chabert and Chauvin, 1963 (From Julien, 1998) According to Fig.2 the bedfoms are Dunes / 3. Bedform type from Simons and Richardson
Iov =stream power (lb ft/s)
10:11.84 Pa= 0.247 lb/ft2
v=1.76m/s = 5.774 ft/s
stream power=1.428 lb ft/s
d5=l .3mm According to Fig.3 the data plot off the chart, but assuming that the trends continue the bedforms
should be in transition between dunes and antidunes. / ave!
No motion 0 Incipient without motion ‘I  O hcipiom motion
I ' 0 Initial ripples
'' O lnitiat dunes
0 Initial transition
A Initial antidunes
10‘“ 10" 1o" 1 1:) Particle diameier d. tom) _!
Fig. 4. Bedform classiﬁcation from Bogardi, 1974 (From Julien, 1998). EFL6 =1.07
r. 0.563 ds=1 .3mm=0. 13cm According to Fig. 4. the bedforms should be antidunes
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EXLEDT \IAH Ll 3M wH In“: PM Gm (.750 D UM ﬁg ‘ Problem #2
Calculate the bedload transport by the size fraction in lb/ftday.
1) Using DuBoys, MPM and EinsteinBrown. _ 0.173
d% Where dS is in mm 1:0 is in lb/m and qbv is in cfs/ft width.
/
MPM: qbv = 8(7. — 1.6% 1/(G —1)gdf M DuBoys: qb‘, 2'0 2'0 — 0.0125 — 0.0196115) EinsteinBrown: qb‘, =1571'5wodx when 1:00.52 To calculate the bedload transport by size fraction the following equation must be used for
each of the above methods: N
q!) = EAR(1m = 03qbw10) + 0'4qb<45°> + 0'3q”("9°)
i=1 Ap was found from the grain size distribution of the reach.
Please see attached Excel sheet for calculations
2) Repeat calculations after using only the grain shear stress for the calculations. The grain shear stress accounted for 71% of the total, thus the shear stresses used in the
calculations will be multiplied by 0.71. Please see attached Excel sheet for calculations
3) Repeat calculations assuming a uniform dso
Please see attached Excel sheet for calculations / 4) Prepare a summary table with one equation per line and the above calculations in separate
columns. Compare and discuss the results. Below is the summary table for each of the 4 questions above. Please see the attached Excel
spreadsheet for full calculations Summary Table Grain Size Fraction Uniform d50 l Grain shear stress
edload (lb/ftday) 59,221.2 105,255. 49,2165 57,7762 .2 72,4592 122,624.7 164,643.4 116,896.8 The Duboys method is based Mcept that sediment moves in thin layers along the bed.
The applied bed shear stress must exceed the critical shear stress to initiate motion. Meyer—Peter and Muller developed a complex bedload formula based on the median sediment
size d50. This formula is most appropriate for channels with large widthdepth ratios. Einstein’s equation introduced the idea that grains move in steps proportional to their size. He
defined the bed layer thickness as twice the particle diameter. He used extensive probability
concepts to formulate a relationship for contact sediment discharge. There is no critical shields
number in this equation, which might be a reason why it has a higher estimate for transport. All three of the equations aSsume a plane bed situation, which is inaccurate because the shear
stresses that were calculated include form drag. The grain shear stress bedload yielded a smaller
transport number. The transport was only slightly smaller because the grain shear stress was the
dominant contribution at 71% of the total shear stress. The uniform d50 transport rates were all calculated to be greater than the mixed dm&d50 transport
rate. This is because the d50 sediment is much heavier and larger than the dlo, so when expressed /
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This note was uploaded on 09/16/2009 for the course CIVE 716 taught by Professor Julien during the Fall '06 term at Colorado State.
 Fall '06
 Julien

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