# Sol4 - Problem#1 Refer to the Rhine River and determine the...

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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 =p-u.2 u. =./thSf A Rh = F assume rectangular channel geometry w-d 260m-9.9m 2 Rh -— u. =«/9.81m/s -9.l99m-0.0001312 =2d+w_2-9.9m+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—m-Tnismm 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). EFL-6 =1.07 r. 0.563 ds=1 .3mm=0. 13cm According to Fig. 4. the bedforms should be antidunes / / (’EKCM) F?- 9' 37mm: I’v' {1 ,r kg': 0‘70 KB' 1- 3-0 O! K§= 0.03 CVt : S 7’ K {0 ﬂ if? “"3 ELI—Z] " “gym? I) / \$4.73 I ﬁ__ {: f; : Bme 1mm PM, 3-] C WJC’I } l__ _ {-r O 01%? gag ’I“‘L'f\lZ :mé _ x ﬁr f,» 1? T} f /[\_r TT‘vaPl-X Tu“ “~70 ~— 7 13 7 @ 8&fo. ht “#3: 1‘ : fZEB 1’9.“ W5 U}. 008\$ ‘ i 4; TQL (6459015 “1' = Ami 0, ~56 ﬂﬁcw 9m 3.1 " A t q; 1 19* ’ Tﬂ ’1‘; : OLSEO -— oi 3‘63 “1" t. 0.119? f / @ ESTWM‘E 7F } —" ﬂ__d_____h____ Isa Fouub w @F= 0.0m? [ wua— ._._ _....__..______J I T 0.03042: f‘l’ﬁcw W075 3- ‘ In: —F fry-0.05.09;— 0W 7 («room L?— —— n A I / ® "(q/~16) S t S f = S 3 3' : t I 7. 23 t m r a :7933i2e9xq-2‘) ;:;, ____ _:L Sm. A; “ th‘ﬁ :1 OIOOOOS'I‘SN \ 1,. . \. __ '- a ._ ‘2 I pitfall _.': If}. I: If ﬂw—wxxL ® F‘NS) T I (:05ng 4‘70 .- E :11. 7 " Il-qqg-(I'VQX T E 'M 1 1:, _ 11(97-5 L93 ( r ml 1 L IIT / @ mam-am Wow um as“ ANTIUUr‘Eﬁ 7,5 — — 1 'T E-‘A-Myt'w '- _ u n ___ If -— ‘ _ —-- -——._______ _ _ > “I )0 rd}. 5 0 VA?“ ww‘cﬁ DIME? 3 1| T L _ __ _ fl 3 . 1 . DVMES / [ZLWLEb 'I { r ch ® Efrem Hagar {A} Am; (JEN-qu DA A= an (ﬂo‘soﬁm (2;-..7344 ® ESWATE [15 WW «(QLTPoﬁF' SAL-f ‘5? K3: BJ70 + /.IA([,€ 5 THIS Woibb'm BHcmt) THE, {Maw N SUEDE-1M mammal; I ml) 134M“ ’EMH ﬂammm W15 DQCELQWED oval mm: mm COMBINE/u: 47w FUMOMEMT-kt Maw» 921cm. Mg YﬂEO‘CIZ/ﬁ‘b W643i If—L THE 3 AM E- {LELﬁm “(E Mﬁgﬁ ( AM“ 903163) EXLEDT \IAH Ll 3M wH In“: PM Gm (.750 D UM ﬁg ‘ Problem #2 Calculate the bedload transport by the size fraction in lb/ft-day. 1) Using DuBoys, MPM and Einstein-Brown. _ 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) Einstein-Brown: 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 = 0-3qbw10) + 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/ft-day) 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 width-depth 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 / as lb/ft-day you get a higher number even though there may not be as many individual particles in motion. / mdmmd C v.mvw.V®F néNQNNF m, Smémw mm Naman NSF F.FNF Ndnﬁhm mfhmdm 53:2 90 dev Wmmwﬁor N._.NN.mm mermdmr wxolnzo maﬁa unoimm amuég \$9? .505 £50 umoﬁum .52. “.0522 :ozoﬂu ouw EEG \$.59 \$068 \$33? amp—«w hmwcw raw—0 3N2me Eyck emu chat—.3 98%: @886 \$83 \$92 353. \$5383 a: Eggnog m: 32?. .85 5.50 emu >382 88.0 388° wESd NEE“ Nemwmm mmomﬁvood 8358.0. \$5383 38885 39.255 520 v.93? Band 853 NC F 5 0832 «58285 8 38883 as Bu ESE: 38va 553 8886 8890 mﬁmmm miwwﬁ Eggnog 2.5883 10898.0 838:3 \$qu gag—E dmuénlo Eu “2 Ed smaﬁm ammvnew Slum >mm :30..m.:_2wc_m_ 83 in 20.0 1 393 human Ema \$3 2 8.: 0. mvod A839. vnod A830.“ Rod 8 E 0.. an. my sage an. Ed Swan: an. «No 663. Ed 8mg; a 8.0 an“? 8; 8:9. 55 to En EE 3 9% EE 9 0% m3 o \$52 98 >. NE? 89 a /.}.v Rmod * m\uE moooood 5.» mwmrmmowd .3 \aﬂm «558.0 m \ I E «5.5685 cm .3 WM I S E 33 n. NE Eww < ...
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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.

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Sol4 - Problem#1 Refer to the Rhine River and determine the...

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