V*
= 0.0706 / 0.043 = 1.64
ω
ωD50
= 0.043 x 0.00032 / 1.16 x 10 -6 = 11.86
ν
log C t = 5.435 - 0.286 log 11.86 - 0.457 log 1.64 +
(1.799 - 0.409 log 11.86 - 0.314 log 1.64) log ( 0.00472 - 0.00059)
= 1.95
C t = 101.95 = 89 ppm by weight
Qs = 221.3 x 89 x 3600 x 24/106 = 1700 metric-tons/day
4.13 SOLVED PROBLEMS FOR SEDIMENT TRANSPORT (ENGLISH)
4.13.1 Introduction
The following example problems illustrate the application of concepts and equations for
sediment transport.
4.13.2 Problem 1 Suspended Sediment Concentration Profile
From data observed on the Missouri River, calculate the vertical suspended sediment
concentration profile for the D50 sediment size in mg/l.
Given:
y o = 14.10 ft
V = 4.164 ft / s
S = 0.000210
Temp = 72.5F (22.5C)
D50 = 0.255 mm (0.000838 ft)
ω (D 50 ) = 0.1109 ft / s
υ = 1.017 x 10 -5 ft 2 / s
c a = 775,000 mg / l (48.3 lb / ft 3 )
From Equation 4.12:
Z
y - y
a
ω
c
= o
and z =
yo - a
βκV*
ca y
4.42
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