4.12.6 Problem 5 Calculation of Total Bed-Material Discharge Using the
Basic Power Function Relationship
Determine the bed-material discharge for the 100-year discharge for a stream with the
following data. Note that these data are not in the Fr range of Table 4.2. Compare the result
with the expanded power function application in Problem 6.
Width W = 60.96 m, Depth y = 1.829 m, Velocity V = 2.438 m/s, Q = 271.83 m3/s. Sediment
properties of D50 = 0.31 mm and size distribution factor G =1.32.
Simons et al. (1981) Equation 4.48 is qs = c s1 y Cs2 V Cs3
Use Table 4.1 and cross-interpolate to obtain the values of Cs2 and Cs3. For Cs1 in SI units
use the following Equation
Cs1 (SI) = Cs1 (English units) x 0.3048 (2-Cs2 -Cs3 )
Cross-interpolate in Table 4.1
Cs1 = 1.63 x 10-5 (English)
Cs2 = 0.45
Cs3 = 3.64
Cs1 = 1.63 x 10-5 x 0.3048(2-0.45-3.64) = 1.63 x 10-5 x 0.3048-2.09 =0.000195 (SI)
qs = 0.000195 (1.829)0.45 (2.438)3.64 = 0.0066 m2/sec
Qs = 60.96 x 0.0066 = 0.402 m3/sec
Using a specific gravity of 2.65
Qs = 2.65 x 9810/9.81 x 3600 x 24 x 0.402 /1000 = 92,000 metric-tons/day
4.12.7 Problem 6 Calculation of Total Bed-Material Discharge Using the
Expanded Power Function Relationship
Determine the bed-material discharge for the 100-yr discharge for a stream with the data
given in Problem 5. The data are repeated below.
Width W = 60.96 m, Depth y = 1.829 m, Velocity V = 2.438 m/s, Q = 271.83 m3/s, S =
.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
qs = 2123.4 (2.438)3.30 (1.829)0.468 (0.000521)0.613 = 518 metric tons/m/day
4.40
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