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HE 614 – River Engineering and Sediment Transport ANSWER 1. A wide open channel has a water depth of h = 3 m, mean current velocity u´ = 2.0 m/s, and mean bed slope = 4 x 10-4. The bed material characteristics are d50 = 350 µm, d90 = 1000 µm, sediment density ρs = 2650 kg/m3, fluid density ρ = 1000 kg/m3, fluid temperature Te = 20oC (ν = 1 x 10-6 m2/s). What are the type of bed forms and the bed form dimensions according to the method of Van Rijn? Given

h=3m

mean bed slope = 4 x 10-4

d50 = 350 µm

u´ = 2.0 m/s Te = 20oC (ν = 1 x 10-6 m2/s) Wanted

ρs = 2650 kg/m3

d90 = 1000 µm ρ = 1000 kg/m3

A. What are the type of bed forms and B. The bed form dimensions according to the method of Van Rijn? Solution

According to Van Rijin method

A. Bed form type

τ-b,c = ρg(ū/ Ċ)2

Ċ = 18log (12h/3d90) =18*log (12*3/(3*1000*10-6)) = 73.43m1/2/s

= 1000*9.81*(2/73.43)2 = 7.28N/m 2

τb,cr = 0.21N/m2 (from shields curve) T = (τ-b,c - τb,cr)/τb,cr = (7.28-0.21)/ 0.21 = 33.67 D* = d50 ((s-1)g/v2)1/3 = 350*10-6 ((2.65-1)9.81/10-12)1/3 =8.85

Fr= ū/ (gh)0.5 =2/(9.81*3)0.5 =0.37

According to Van Rijin method bed form type is function of (T, D*) or BF type = F (T, D*) = F (33.67, 8.85) this belongs between T≥25, Fr