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If we want to build any hydraulic structure, we should consider the sedimentation (scour and deposition). Riprap Fiter stone Stream channel Graded bank 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) Saltation (bouncing) Sliding Toe is strengthened with large rock to reduce the risk of undermining of the structure Suspended load (silt, clay) Bedload (sand, gravel, boulders) Rolling
Sediment Transport Formula: zuh water surface velocity concentration transport JKE- U₂ C₂ UCT suspended lood Z=0 EXE VAVA bed current sediment suspended concentration load. velocity 9ss = Suspended load: h 11c (q=fe*u, dh = [c*u.) In practice, the average suspended load qu=C*U*h where que is the suspended load per unit width (m³/s/m) C is the average sediment concentration ppm (parts per million), volume per volume: m³/million m³ (dimensionless); sometimes: mass per volume as mg/L. U is the average velocity (m/s) h is water depth (m).
Solution of suspended load formula using Rouse parameter, C/C₂= [(h-z)/z)][(a/(h-a)]ww (kut) Where w/(ku) is the Rouse number. Suspended load transport Equation q=fuCdz = UCzth 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, To/[(y₁ - y)ds] = yhSo/[(x - y)d] =0.047 then ds=yhSo/[(-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 qem (bed material load, in mass) in kg/m-s and quin m²/s 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/[ (G − 1) gd,j] 05 = 8(t. - t)LS G-1 (1-1)/y=(26,5-9.81)/9.81 =1.70 [(G – 1) gd*] 5 =[(1.70)(9.81)(0.0153)]5=0.0075 m3s t. = yhSo/[(-y)d.] =9.81x0.2x0.01/[(26.5-9.81)(0.015)] = 0.078 T₂ Therefore, qb = 8[(G – 1) gd3]03 (t. – T«e)15 = 8(0.0075) (0.078 – 0.047)15 = 3.27×10- mis 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 m³/(million m³) at reference level: a = 0.05h Bottom slope of channel: 0.0006, Karman coefficient: 0.3 Sediment specific weight: 27 kN/m³ Water specific weight: 9.81 kN/m³. Solution: Consider the bed load formula (in Lecture Notes, Equation 31) as C/C₁ {[(h-z)/z][(a/(h-a)]}(ku) Where w/(ku:) is the Rouse number For water depth: h= 5.0 m, bottom slope. S = 0.0006 shear velocity in which, and
u₁= (ghS)0.5 = (9.81x5x0.0006) 0.5 = 0.17 m/s. For settling velocity: w₁=0.0035 m/s, Karman coefficient: k = 0.3. Rouse number = 0.0035/(0.3x0.17)= 0.069 For relative concentration: C/C₂= {[(5-3)/3)][(0.25/(5 -0.25)]30.069-0.7936 C= = 270x0.7936=214.28 m³/(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 (m³/million m³), Average depth of water: 1.8 m Width of river: 22 m. Solution: Consider bed load formula as suspended load per unit width q=uCdz = U(Ca) h For average velocity: U=0.9 m/s, and average concentration: Cav = 250 m³/106 m³) q=0.9(m/s)x250(m³/106 m³)xh(m) = 4.05x 104 m³/s/m. Q=Bxq, = 8.91x10-³ m³/s Total suspended material per year ZQ₁ =QxT = 8.91x10-³x(60x60x24x365) = 280986 m³ 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. For river width: B= 22 m,
Solution: The shear velocity is u.=(ghSo)5 = [(9.81) (2) (0.0002)]05 = 0.0626 m/'s Assume T = 10-C, ß. = 1, and x = 0.4, then we have Sediment ds (mm) (m/s) Ro=co/(B,xu.) a (mm) Clay 0.002 2.6x10-6 1.04×10 4x10-6 Very fine silt 0.004 1×10-5 3.99x10 8x10-6 Fine silt 0.008 4.2x10-5 1.68x10-3 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./k) In(u.h/v)+3.25u, = (0.0626/x0.4)xln (0.0626) (2)/(1.31×106)+(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 x 24 s)/day = 859,680 tons/day