4.13.7 Problem 6 Calculation of Total Bed-Material Discharge Using the
Expanded Power Function Relationship
Determine the bed-material discharge for the 100-year discharge for a stream with the data
given in Problem 5. The data are repeated below.
Width W = 200 ft, Depth y = 6 ft, Velocity V = 8 ft/s, Q = 9,600 cfs, S = 0.000521. Sediment
properties of D50 = 0.31 mm and size distribution factor G =1.32.
Kodoatie et al. (1999) Equation 4.50 is qs = a Vb yc Sd
The sand size places this in a medium sand bed stream in Table 4.3.
From Table 4.3
a = 2123.4, b = 3.30, c = 0.468, d = 0.613
a (English units) = 1.1 x 0.3048(1+b+c) a = 1.1 x 0.3048(1+3.30+0.468) (2123.4) = 8.095
qs = 8.095 (8)3.30 (6)0.468 (0.000521)0.613 = 173.78 tons/ft/day
Qs = 173.78 x 200 = 34,800 tons/day
4.13.8 Problem 7 Calculate Total Bed-Material Discharge Using Yang's Sand Equation
Yang's sand equation (Equation 4.51) is as follows:
V*
VS
Vcr S
ωD50
ωD50
V*
log
+ 1.799 - 0.409 log
log C t = 5.435 - 0.286 log
- 0.457 log
- 0.314 log
-
ω
ν
ω
ν
ω
ω
Vcr
2.5
VD
=
+ 0.66 for 1.2 < * 50 < 70
VD
ω
ν
log * 50 - 0.06
ν
and
Vcr
VD
= 2.05 for * 50 ≥ 70
ω
ν
Compute the bed-material transport in tons per day.
Given:
Q = 7,820 ft3/sec
ν = 1.21 x 10-5 ft2/s
W = 341 ft
ω = 0.141 ft/sec
y = 7.58 ft
D50 = 0.00105 ft
V = 3.02 ft/sec
S = 0.00022
4.53
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