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If we want to build any hydraulic structure, we should consider the sedimentation (scour and deposition). Riprap Filter stone Stream channel Grond bank Toe stortenet ath laga va to reduce the risk ofuscemning of the structure Bank protection by using armor stone on bank and bottom of stream (What is the minimum size of the stone? Classification of sediment transport: 1.Bedd load, 2. Suspended load, (1+2) as Total load. Dissolved load (invisible) Suspended load (silt, clay) Saltation (bouncing) Bedload (sand, gravel, boulders) Sliding Rolling
Sediment Transport Formula: water surface zeh velocity concentration transport Uz CE UCT suspended lood Co ET VASCA bed Suspended load: current velocity sediment suspended concentration load h uc (qu = ſe*uedh = {c+u) In practice, the average suspended load qua =C* *h where qu is the suspended load per unit width ( msm) C is the average sediment concentration ppm (parts per million), volume per volume: m/million m (dimensionless), sometimes: mass per volume as mg/L. U is the average velocity (m/s) his water depth (m)
Solution of suspended load formula using Rouse parameter, C/C = [(h-z)/z)][(a/Ch-a)]* (kur) Where w/(ku) is the Rouse number Suspended load transport Equation 4= "ucdz = Uch SOLVED PROBLEMS 1. What is the sediment size corresponding to beginning of motion when the steam slope So = 0.001? Flow water depth is 1.25 m. Solution: We consider the Shields diagram with constant 0.047 or 0.056 Now, td[(y-7)d.] – yhS/[(-y)d.) = 0.047 then de = yhSo[(y) - Y)0.047] = 9.81x1.25x0.001/[[26.5-9.81)(0.047)] = 0.0156 m = 15.6 mm. Using safe factor as 2.0, we can find the minimums size of armor stone as D=2*15.6 =31.2 mm 2. Use Meyer-Peter and Muller's method to calculate qb (bed material load, in mass) in kg/m-s and quin més for a channel given the slope So = 0.01, the flow depth h=20 cm, and the grain size dso = 15 mm Solution:
Meyer Peter and Muller Method qbv/[ KG - 1) gd ] 5 = 801 - ) in which G-1=[7--7)=(26,5 - 9.81) 9.81 =1.70 and [(G-1) gd-3] =[(1.709.81 0.0153 195= 0.0075 m/s t. = yhS (-y)d.] =9.81x0 2x0.01/[(26.5 - 9.8170.015)] = 0.078 Therefore, qbv = 8[KG - 1) gd, 10* (-) = 8(0.0075) (0.078 – 0.047) 3 = 3.27x10ms 3. Calculate the sediment concentration at level of 3.00 m for the following data: Sediment diameter: 0.07. Water depth: 5.0 m Settling velocity: 0.35 cm/s Reference concentration (ppm): 270 m (million m) at reference level: a a=0.05h Bottom slope of channel: 0.0006. Karman coefficient: 0.3 Sediment specific weight: 27 kN/m2 Water specific weight: 9.81 kN/m Solution: Consider the bed load formula (in Lecture Notes, Equation 31) as C.C.={[(h - 2)/z][(a/(h - a)]} ww(kur) Where w/(kus) is the Rouse number For water depth: h=5.0 m, bottom slope. S = 0.0006 shear velocity
u= (ghs)0.5 = (9.81x5x0.0006)0.5 = 0.17 m/s. For settling velocity: Ws=0.0035 m/s, Karman coefficient: k=0.3. Rouse number = 0.0035/(0.3x0.17)=0.069 For relative concentration: C/C = {[(5 - 3/3)](0.25/(5 -0.25)]}0.069= 0.7936 C = 270x0.7936 = 214.28 m®/(million m). (parts per million). 4. Calculate the bed load material transport in a river during one year. Given data: Average velocity in the river: 0.9 m/s Average concentration in the river: 250 ppm (m”million mº). Average depth of water: 1.8 m Width of river: 22 m. Solution: Consider bed load formula as suspended load per unit width a q= I ucd: = U( C) h For average velocity: U=0.9 m/s, and average concentration: Cay – 250m/109 m) 9=0.9(m/s)x250(m/108 m)xh(m) = 4.05x 10+ m?s'm. For river width: B=22 m Q: = Bxq;= 8.91x10-2 m/s Total suspended material per year ZQ.=Q.XT = 8.91x10->x(60x60x24x365) = 280986 m 5. Calculate the daily sediment load in a nearly rectangular 50-m-wide stream with an average flow depth h = 2 m and slope So = 0.0002 when 25% of the sediment load is fine silt, 25% is very fine silt, 25% is clay, and the middepth concentration is C = 50,000 mg/1.
Solution: The shear velocity is u, = (ghSo) 5 = [(9.81) (2) (0.0002)15 = 0.0626 m/s Assume T = 10-C, Bs = 1, and <=0.4. then we have Sediment ds (mm) (m/s) Ro=/(Bsku.) a (mm) Clay 0.002 2.6x10-6 1.04x104- 4x10-6 Very fine silt 0.004 1x10-5 3.99x10-4 8x10-6 Fine silt 0.008 4.2x10- 1.68x10- 16x10-6 From the above table, we see that the Rouse numbers are very small. So, the concentration profiles are almost uniform, i.e., C = 50.000 mg/1= 50 kg/m3 anywhere. Assume the channel bed is smooth. The average velocity V= (u./s) In{u.hv)+3.25u. (0,06266x0.4)xln (0.0626) (2) (1.31 x 10^)+ (3.25) (0.0626) = 1.99 m/s Then the sediment transport rate is (as a continuity equation) Q: =VA = V W HC = (1.99) (50) (2) (50) = 9950 kg/s or = (9950 kg/s)(1 ton/1000 kg)(3600 24 s) day = 859,680 tons/day