Sediment Transport Functions – Sample Calculations

The following sample calculations were the basis for the algorithms used in the HEC-RAS sediment transport functions. They were computed for a single grain size, however they were adapted in the code to account for multiple grain sizes.

Ackers-White Sediment Transport Function
by Ackers-White (ASCE Jour. Of Hyd, Nov 1973)

Input Parameters

Temperature, F	T = 55	Average Velocity, ft/s	V = 2
Kinetic viscosity, ft²/s	$\begin{array}{l}\nu\end{array}$ = 00001315	Discharge, ft³/s	Q = 5000
Depth, ft	D = 10	Unit Weight water, lb/ft³	$\begin{array}{l}\gamma _w\end{array}$ =62.385
Slope	S = 0.001	Overall d₅₀, ft	d₅₀ = 0.00232
Median Particle Diamter, ft	d_si = 0.00232
Specific Gravity of Sediment,	s = 2.65

Constants

Acceleration of gravity, ft/s²

g = 32.2

Solution

Note

Ackers-White required the use of d₃₅ as the representative grain size for computations in their original paper. In the HEC-RAS approach, the median grain size will be used as per the 1993 update. The overall d₅₀ is used for the hiding factor computations.

Hiding Factor from Profitt and Sutherland has been added for this procedure, but will be included as an option in HEC-RAS.

Computations are updated as per Acker's correction in Institution of Civil Engineers Water Maritime and Energy, Dec 1993.

Dimensionless grain diameter,

$\begin{array}{l}\displaystyle \displaystyle d_{gr} = d_{si} \left[ \frac{g \cdot (s-1)}{\nu ^2} \right] ^{\frac{1}{3}}\end{array}$

$\begin{array}{l}\displaystyle d_{gr} =15.655\end{array}$

Shear velocity u

$\begin{array}{l}\displaystyle \displaystyle u_{star} = \sqrt{g \cdot D \cdot S}\end{array}$

$\begin{array}{l}\displaystyle u_{star} =0.567\end{array}$

Sediment size-related transition exponent n,

$\begin{array}{l}\displaystyle n= \left| \begin{array} 1 1 & if \; d_{gr} \leq 1 \\ (1-0.056 \cdot log(d_{gr})) & if \; 1< d_{gr} \leq 60 \\ 0 & if \; d_{gr} > 60 \end{array}\end{array}$

$\begin{array}{l}\displaystyle n=0.331\end{array}$

Initial motion parameter A,

$\begin{array}{l}\displaystyle A= \left| \begin{array} 1 ( \frac{0.23}{\sqrt{d_{gr}}} +0.14) & d_{gr} \leq 60 \\ 0.17 & otherwise \end{array}\end{array}$

$\begin{array}{l}\displaystyle A = 0.198\end{array}$

Sediment mobility number F_gr

$\begin{array}{l}\alpha =10\end{array}$ (assumed value used in HEC6 and SAM) $\begin{array}{l}\alpha =10\end{array}$

$\begin{array}{l}\displaystyle \displaystyle F_{gr} = \frac{u_{star}^n}{\sqrt{g \cdot d_{si} \cdot (s-1)}} \cdot \left( \frac{V}{\sqrt{32} \cdot log \left( \alpha \cdot \frac{D}{d_{si}} \right)} \right) ^{1-n}\end{array}$

$\begin{array}{l}\displaystyle F_{gr} = 0.422\end{array}$

Hiding Factor HF,

Shield's Mobility Parameter $\begin{array}{l}\theta\end{array}$ ,

$\begin{array}{l}\displaystyle \displaystyle \theta = \frac{u_{star}^2}{g \cdot (s-1) d_{50}}\end{array}$

$\begin{array}{l}\displaystyle \theta =2.612\end{array}$

$\begin{array}{l}\displaystyle dRatio = \left| \begin{array} 1 1.1 & if \; \theta \leq 0.04 \\ (2.3-30 \cdot \theta ) & if \; 0.04 < \theta \leq 0.045 \\ (1.4-10 \cdot \theta ) & if \; 0.045 < \theta \leq 0.095 \\ 0.45 & otherwise \end{array}\end{array}$

$\begin{array}{l}\displaystyle dAdjust=d_{50} \cdot dRatio\end{array}$

$\begin{array}{l}\displaystyle dAdjust=1.044 \times 10^{-3}\end{array}$

$\begin{array}{l}\displaystyle \displaystyle HFRatio = \frac{d_{si}}{dAdjust}\end{array}$

$\begin{array}{l}\displaystyle DFRatio=2.222\end{array}$

$\begin{array}{l}\displaystyle HF = \left| \begin{array} 1 1.30 & if \; HFRatio \geq 3.7 \\ (0.53 \cdot log(HFRatio)+1 & if \; 0.075 < HFRatio \leq 3.7 \\ 0.40 & otherwise \end{array}\end{array}$

$\begin{array}{l}\displaystyle HF=1.184\end{array}$

Adjust Sediment Mobility Number for Hiding Factor

$\begin{array}{l}\displaystyle F_{gr} =HF \cdot F_{gr}\end{array}$

$\begin{array}{l}\displaystyle F_{gr} =0.5\end{array}$

Check for too fine sediment based on F_gr and A,

$\begin{array}{l}\displaystyle \displaystyle Check= \frac{F_{gr}}{A}\end{array}$

Sediment transport function exponent m,

$\begin{array}{l}\displaystyle m= \left| \begin{array} 1 ( \frac{6.83}{d_{gr}}+1.67) & if \; d_{gr} \leq 60 \\ 1.78 & otherwise \end{array}\end{array}$

$\begin{array}{l}\displaystyle m = 2.106\end{array}$

Check for too fine sediment based on m,

$\begin{array}{l}\displaystyle Check= \left| \begin{array} 1 0 & if \; m>6 \\ Check & otherwise \end{array}\end{array}$

$\begin{array}{l}\displaystyle Check \;= 2.522\end{array}$

Sediment transport function coefficient C,

$\begin{array}{l}\displaystyle C= \left| \begin{array} 1 10^{2.79 \cdot log(d_{gr})-0.98(log(d_{gr}))^2 -3.46} & if \; d_{gr} \leq 60 \\ 0.025 & otherwise \end{array}\end{array}$

$\begin{array}{l}\displaystyle C \;=0.0298\end{array}$

Transport parameter G_gr,

$\begin{array}{l}\displaystyle \displaystyle G_{gr} =C \cdot \left( \frac{F_{gr}}{A}-1 \right) ^m\end{array}$

$\begin{array}{l}\displaystyle G_{gr} =0.072\end{array}$

Sediment flux X, in parts per million by fluid weight,

$\begin{array}{l}\displaystyle \displaystyle X= \frac{G_{gr} sd_{si}}{D \left( \frac{u_{star}}{V} \right)^n}\end{array}$

$\begin{array}{l}\displaystyle X=6.741 \times 10^{-5}\end{array}$

Sediment Discharge, lb/s

$\begin{array}{l}\displaystyle \displaystyle G= \gamma _w QX\end{array}$

$\begin{array}{l}\displaystyle G \;=21.027\end{array}$

Sediment Discharge, tons/day

$\begin{array}{l}\displaystyle \displaystyle G_s = \frac{86400}{2000} \cdot G\end{array}$

Check to make sure particle diameter and mobility functions are not too low,

$\begin{array}{l}\displaystyle G_s = \left| \begin{array} 1 G_s & if \; Check >1 \\ 0 & otherwise \end{array}\end{array}$

$\begin{array}{l}\displaystyle G_s \;=908\end{array}$

Engelund Hansen Sediment Transport Function

by Vanoni (1975), and Raudkivi (1976)

Input Parameters

Temperature, F	T = 55	Average Velocity, ft/s	V = 5.46
Kinematic viscosity, ft²/s	$\begin{array}{l}\nu\end{array}$ = 0.00001315
Depth, ft	D = 22.9	Unit Weight water, lb/ft³	$\begin{array}{l}\gamma _w\end{array}$ = 62.385
Slope	S = 0.0001
Median Particle Diamter, ft	d_si = 0.00232	Channel Width, ft	B = 40
Specific Gravity of Sediment	s = 2.65

Constants

Acceleration of gravity, ft/s²

g=32.2

Solution

Bed level shear stress $\begin{array}{l}\tau _o\end{array}$ ,

$\begin{array}{l}\displaystyle \tau _o = \gamma _w \cdot D \cdot S\end{array}$

$\begin{array}{l}\displaystyle \tau _o = 0.143\end{array}$

Fall diameter d_f,

$\begin{array}{l}d_f = \left| \begin{array} 1 (-69.07 \cdot d_{si}^2 +1.0755 \cdot d_{si} +0.000007) & if \; d_{si} \leq 0.00591 \\ (0.1086 \cdot d_{si}^{06462} ) & otherwise \end{array}\end{array}$

$\begin{array}{l}\displaystyle d_f = 2.13 \times 10^{-3}\end{array}$

Sediment discharge lb/s,

$\begin{array}{l}\displaystyle \displaystyle g_s =0.05 \cdot \gamma _w \cdot s \cdot V^2 \cdot \sqrt{\frac{d_f}{g \cdot (s-1)}} \cdot \left[ \frac{\tau _o}{(\gamma _w \cdot s - \gamma _w ) \cdot d_f} \right] ^{\frac{3}{2}}\end{array}$

$\begin{array}{l}\displaystyle g_s \;=32.82\end{array}$

Sediment discharge ton/day,

$\begin{array}{l}\displaystyle \displaystyle G_s = g_s \cdot \frac{86400}{2000}\end{array}$

$\begin{array}{l}\displaystyle G_s =1418\end{array}$

Laursen-Copeland Sediment Transport Function

by Copeland (from SAM code, 1996)

Input Parameters

Temperature, F	T = 55	Average Velocity, ft/s	V = 5.46
Kinematic viscosity, ft²/s	$\begin{array}{l}\nu\end{array}$ = 0.00001315	Discharge, ft³/s	Q=5000
Depth, ft	D = 22.90	Unit Weight water, lb/ft³	$\begin{array}{l}\gamma _w\end{array}$ = 62.385
Slope	S = 0.0001	84% Particle diameter, ft	d₈₄ = 0.00294
Median Particle Diamter, ft	d_si = 0.00232
Specific Gravity of Sediment	s = 2.65

Constants

Acceleration of gravity, ft/s²

g=32.2

Solution

Note

the difference between the final result presented here and the result in SAM is due to the method for determining fall velocity. Rubey is used here, whereas SAM computes a
value based on a drag coefficient determined from Reynolds number. Calculation routine taken from SAM. Because the grain distribution is reduced to standard grade sizes representing each present grade class, the d₈₄ will equal the standard grade size, d_si, in this procedure.

$\begin{array}{l}\displaystyle d_{84} = d_{si}\end{array}$

Grain-related hydraulic radius R

$\begin{array}{l}\displaystyle \displaystyle R'= \frac{0.0472 \cdot V^{\frac{3}{2}} \cdot (3.5 \cdot d_{84} ) ^\frac{1}{4}}{(g \cdot S )^{\frac{3}{4}}}\end{array}$

$\begin{array}{l}\displaystyle R'=14.189\end{array}$

$\begin{array}{l}R' = 15.248\end{array}$

Grain-related bed shear stress $\begin{array}{l}\tau '_b\end{array}$ ,

$\begin{array}{l}\displaystyle \tau '_b = R' \cdot \gamma _w \cdot S\end{array}$

$\begin{array}{l}\displaystyle \tau '_b =0.095\end{array}$

$\begin{array}{l}\displaystyle \tau _b = D \cdot \gamma _w \cdot S\end{array}$

$\begin{array}{l}\displaystyle \tau '_b =0.143\end{array}$

$\begin{array}{l}\displaystyle \tau '_b = \left| \begin{array} 1 \tau '_b & if \; \tau '_b < \tau _b \\ \tau _b & otherwise \end{array}\end{array}$

$\begin{array}{l}\displaystyle \tau '_b =0.095\end{array}$

$\begin{array}{l}\displaystyle \displaystyle u'_* = \sqrt{\frac{\tau'_b \cdot g}{\gamma _w}}\end{array}$

$\begin{array}{l}\displaystyle u'_* =0.222\end{array}$

$\begin{array}{l}\displaystyle RRP = \left( \frac{d_{si}}{R} \right) ^{1.16667}\end{array}$

$\begin{array}{l}\displaystyle RRP=2.187 \times 10^{-5}\end{array}$

Dimensionless bed shear stress $\begin{array}{l}\tau ^*_b\end{array}$ ,

$\begin{array}{l}\displaystyle \displaystyle \tau ^*_b = \frac{\tau '_b}{\gamma _w \cdot (s-1) \cdot d_{si}}\end{array}$

$\begin{array}{l}\displaystyle \tau ^*_b = 0.398\end{array}$

Shield's parameter for course grains $\begin{array}{l}\theta ^*\end{array}$ ,

$\begin{array}{l}\displaystyle \theta ^* =0.647 \cdot \tau ^* _b + 0.0064\end{array}$

$\begin{array}{l}\displaystyle \theta ^* = \left| \begin{array} 1 0.02 & if \; \theta ^* < 0.02 \\ \theta ^* & otherwise \end{array}\end{array}$

$\begin{array}{l}\displaystyle \theta ^* =0.264\end{array}$

Critical shear stress, $\begin{array}{l}\tau _{cr}\end{array}$

$\begin{array}{l}\displaystyle \tau _{cr} = \left| \begin{array} 1 [\theta ^* \cdot \gamma _w \cdot (s-1) \cdot d_{si} ] & if \; \tau^*_b \leq 0.05 \\ [0.039 \cdot \gamma _w \cdot (s-1) \cdot d_{si} ] & otherwise \end{array}\end{array}$

$\begin{array}{l}\displaystyle \tau _{cr} = 9.315 \times 10^{-3}\end{array}$

Shear stress mobility parameter $\begin{array}{l}TFP\end{array}$ ,

$\begin{array}{l}\displaystyle \displaystyle TFP = \frac{\tau ' _b}{\tau _{cr}} -1\end{array}$

$\begin{array}{l}\displaystyle TFP=9.214\end{array}$

Fall velocity $\begin{array}{l}\omega\end{array}$ ,

Use Rubey's equation, Vanoni p. 169

$\begin{array}{l}\displaystyle \displaystyle F_1 = \sqrt{\frac{2}{3} + \frac{36 \cdot \nu ^2}{g \cdot d_{si}^3 \cdot (s-1)}} - \sqrt{\frac{36 \cdot \nu ^2}{g \cdot d_{si}^3 \cdot (s-1)}}}\end{array}$

$\begin{array}{l}\displaystyle F_1 = 0.725\end{array}$

$\begin{array}{l}\displaystyle \omega = F_1 \cdot \sqrt{(s-1) \cdot g \cdot d_{si}}\end{array}$

$\begin{array}{l}\displaystyle \omega =0.255\end{array}$

Particle velocity ratio SF,

$\begin{array}{l}\displaystyle \displaystyle SF= \frac{u_*}{\omega}\end{array}$

$\begin{array}{l}\displaystyle SF = 0.870\end{array}$

Particle velocity ratio parameter $\begin{array}{l}\Psi\end{array}$ ,

$\begin{array}{l}\displaystyle \Psi = \left| \begin{array} 1 [7.04 \cdot 10^{15} \cdot (SF)^{22.99} ] & if \; SF \leq 0.225 \\ (40.0 \cdot SF) & if \; 0.225 < SF \leq 1.0 \\ (40 \cdot SF ^{1.843} ) & if \; SF >1.0 \end{array}\end{array}$

Sediment transport G_s, tons/day

$\begin{array}{l}\displaystyle G_s =0.432 \cdot \gamma _w \cdot Q \cdot RRP \cdot TFP \cdot \Psi\end{array}$

$\begin{array}{l}\displaystyle G_s = 945\end{array}$

Meyer-Peter Muller Sediment Transport Function

by Vanoni (1975), and Schlichting's Boundary Layer Theory, 1968

Input Parameters

Temperature, F	T = 55	Average Velocity, ft/s	V = 5.46
Kinematic viscosity, ft²/s	$\begin{array}{l}\nu\end{array}$ = 0.00001315	Discharge, ft³/s	Q=5000
Depth, ft	D = 22.90	Unit Weight water, lb/ft³	$\begin{array}{l}\gamma _w\end{array}$ = 62.385
Slope	S = 0.0001	84% Particle diameter, ft	d₈₄ = 0.00294
Median Particle Diamter, ft	d_si = 0.00232	Channel Width, ft	B = 40
Specific Gravity of Sediment	s = 2.65

Constants

Acceleration of gravity, ft/s²

g=32.2

Solution

Shear velocity u,

$\begin{array}{l}\displaystyle u_* = \sqrt{g \cdot D \cdot S}\end{array}$

$\begin{array}{l}\displaystyle u_* = 0.272\end{array}$

Shear Reynold's number, $\begin{array}{l}R_s\end{array}$ ,

$\begin{array}{l}\displaystyle \displaystyle R_s = \frac{u_* \cdot d_{90}}{\nu}\end{array}$

$\begin{array}{l}\displaystyle R_s = 63.189\end{array}$

Schlichting's $\begin{array}{l}B\end{array}$ coefficient, $\begin{array}{l}BCoeff\end{array}$

$\begin{array}{l}\displaystyle BCoeff = \left| \begin{array} 1 (5.5 + 2.5 \cdot ln(R_s)) & if \; R_s \leq 5 \\ \begin{bmatrix} 0.297918+24.8666 \cdot log(R_s) -22.9885 \cdot (log(R_s))^2 ... \\ +8.5199 \cdot (log(R_s))^3 -1.10752 \cdot (log(R_s))^4 \end{bmatrix} & if \; 5< R_s \leq 70 \\ 8.5 & otherwise \end{array}\end{array}$

Friction factor due to sand grains f',

$\begin{array}{l}\displaystyle \displaystyle f' = \left( \frac{2.82843}{BCoeff -3.75+2.5 \cdot ln \left( 2 \cdot \frac{D}{d_{90}} \right)} \right)^2\end{array}$

$\begin{array}{l}\displaystyle f'=9.565 \times 10^{-3}\end{array}$

Nikaradse roughness ratio $\begin{array}{l}RKR\end{array}$ ,

$\begin{array}{l}\displaystyle \displaystyle RKR = \sqrt{\frac{f'}{8}} \cdot \frac{V}{\sqrt{g \cdot D \cdot S}}\end{array}$

$\begin{array}{l}\displaystyle RKR = 0.695\end{array}$

Sediment discharge lb/s,

$\begin{array}{l}\displaystyle \displaystyle g_s = \left( \frac{(RKR)^{frac{3}{2}} \cdot \gamma _w \cdot D \cdot S -0.047 \cdot (\gamma _w \cdot s -\gamma _w ) \cdot d_{si}}{0.25 \cdot \left( \frac{\gamma _w}{g} \right) ^{\frac{1}{3}} \left( \frac{\gamma _w \cdot s- \gamma _w}{\gamma _w \cdot s} \right) ^{\frac{2}{3}}} \right)^{frac{3}{2}} \cdot B\end{array}$

$\begin{array}{l}\displaystyle g_s =7.073\end{array}$

Sediment discharge ton/day,

$\begin{array}{l}\displaystyle \displaystyle G_s = g_s \cdot \frac{86400}{2000}\end{array}$

$\begin{array}{l}\displaystyle G_s =306\end{array}$

Toffaleti Sediment Transport Function

by Vanoni, for single grain size

Input Parameters

Slope,	S = 0.0001	Temperature, F	T = 55
Hydraulic Radius, ft	R = 10.68	viscosity, ft²/s	$\begin{array}{l}\nu\end{array}$ = 0.00001315
Width, ft	B = 40	Median Particle Size, ft	d_si = 0.00232
Velocity, ft/s	V = 5.46	65% finer Particle Size, ft	d₆₅ = 0.00257
		Fraction of Total Sediment	p_i = 1
		Unit Weight of Water, lb/ft³	$\begin{array}{l}\gamma _w\end{array}$ = 62.385

Constants

Acceleration of gravity, ft/s²

g=32.2

Solution

Nikaradse Roughness Value, using d₆₅, as per Einstein, 1950, p.

$\begin{array}{l}\displaystyle k_s = d_{65}\end{array}$

$\begin{array}{l}\displaystyle k_s = 2.57 \times 10 ^{-3}\end{array}$

Grain-related shear velocity as per Einstein, 1950, p. 10

$\begin{array}{l}\displaystyle u'_* = \frac{V}{\left( 5.75 \cdot log \left( 12.27 \cdot \frac{r'}{k_s} \right) \right)}\end{array}$

Check $\begin{array}{l}u'_* =0.169\end{array}$

Check for hydraulically rough or smooth grains…

Check $\begin{array}{l}u'_* =0.169\end{array}$

Check for Transitional regime

**Note**

Einstein's method for determining u' was compared with Toffaleti's graphical approach. Results showed that the two methods are in acceptable agreement, with differences on the order of less than 3%. Einstein's approach was selected for its established reputation and its relative simplicity.

Toffaleti coefficients, A and k₄,

$\begin{array}{l}\displaystyle \displaystyle A_{factor} = \frac{\left( 10^5 \cdot \nu \right)^{\frac{1}{3}}}{10 \cdot u'_*}\end{array}$

$\begin{array}{l}\displaystyle A_{factor} =0.54\end{array}$

$\begin{array}{l}\displaystyle A= \left| \begin{array} 1 \left( 9.5987 \cdot A_{factor}^{-1.5445} \right) & if \; A_{factor} \leq 0.5 \\ \left( 39.079 \cdot A_{factor}^{0.481} \right) & if \; 0.5 < A_{factor} \leq 0.66 \\ \left( 221.85 \cdot A_{factor}^{4.660} \right) & if \; 0.66 < A_{factor} \leq 0.72 \\ 48 & f \; 0.72 < A_{factor} \leq 1.3 \\ \left( 22.594 \cdot A_{factor}^{2.872} \right) & if \; A_{factor} > 1.3 \end{array}\end{array}$

$\begin{array}{l}\displaystyle A =29.065\end{array}$

$\begin{array}{l}\displaystyle \displaystyle k_{4Factor} = \frac{\left( 10^5 \cdot \nu \right) ^{\frac{1}{3}}}{10 \cdot u'_*} \cdot 10^5 \cdot S \cdot d_{65}\end{array}$

$\begin{array}{l}\displaystyle k_{4Factor} = 0.014\end{array}$

$\begin{array}{l}\displaystyle k_4 = \left| \begin{array} 1 \left( 1.0 \right) & if \; k_{4Factor} \leq 0.25 \\ \left( 5.315 \cdot k_{4Factor}^{0.1.205} \right) & if \; 0.25 < k_{4Factor} \leq 0.35 \\ \left( 0.510 \cdot k_{4Factor}^{-1.028} \right) & if \; k_{4Factor} > 0.35 \end{array}\end{array}$

$\begin{array}{l}\displaystyle k_4 = 1\end{array}$

$\begin{array}{l}\displaystyle Ak_4 = A \cdot k_4\end{array}$

Check for too low values for the product $\begin{array}{l}Ak_4\end{array}$ ,

$\begin{array}{l}\displaystyle Ak_4 = \left| \begin{array} 1 16 & if \; Ak_4 < 16 \\ Ak_4 & if \; Ak_4 \geq 16 \end{array}\end{array}$

$\begin{array}{l}\displaystyle Ak_4 = 29.065\end{array}$

More Coefficients,

$\begin{array}{l}\displaystyle T_T = 1.10 \cdot (0.051+0.00009 \cdot T )\end{array}$

$\begin{array}{l}\displaystyle T_T = 0.062\end{array}$

$\begin{array}{l}\displaystyle n_V =-.1198+0.00048 \cdot T\end{array}$

$\begin{array}{l}\displaystyle n_V = 0.146\end{array}$

$\begin{array}{l}\displaystyle c_z =260.67-0.667 \cdot T\end{array}$

$\begin{array}{l}\displaystyle c_z = 223.985\end{array}$

Fall Velocity for Medium Sand from Toffaleti Tables at 55 degrees F,

$\begin{array}{l}\displaystyle w_i = 0.340\end{array}$

$\begin{array}{l}\displaystyle \displaystyle z_i = \frac{w_i \cdot V}{c_z \cdot R \cdot S}\end{array}$

$\begin{array}{l}\displaystyle z_i = 7.76\end{array}$

$\begin{array}{l}\displaystyle z_i = \left| \begin{array} 1 (1.5 \cdot n_v ) & if \; z_i < n_v \\ z_i & otherwise \end{array}\end{array}$

$\begin{array}{l}\displaystyle z_i = 7.76\end{array}$

Empirical Relationship for $\begin{array}{l}g_{ssLi}\end{array}$ ,

$\begin{array}{l}\displaystyle \displaystyle g_{ssLi} = \frac{0.600 \cdot p_i}{\left( \frac{T_T \cdot Ak_4}{V^2} \right)^{\frac{5}{3}} \cdot \left( \frac{d_{si}}{0.00058} \right) ^{\frac{5}{3}}}\end{array}$

$\begin{array}{l}\displaystyle M_i = 2.948 \times 10^{-10}\end{array}$

Concentration,

$\begin{array}{l}\displaystyle \displaystyle C_{Li} = \frac{M_i}{43.2 \cdot p_i \cdot (1+n_{_V}) \cdot V \cdot R ^{0.756 \cdot z_i -n_{_V}}}\end{array}$

$\begin{array}{l}\displaystyle C_{Li} = 1.425 \times 10^{-18}\end{array}$

Check for unrealistically high concentration and adjust M_i if necessary,

$\begin{array}{l}\displaystyle \displaystyle C_{2d} = C_{Li} \cdot \left( \frac{2 \cdot d_{si}}{R} \right) ^{-0.756 \cdot z_i}\end{array}$

$\begin{array}{l}\displaystyle C_{2d} =75.536\end{array}$

$\begin{array}{l}\displaystyle C_{Li} = \left| \begin{array} 1 C_{Li} & if \; C_{2d} < 100 \\ \frac{100}{\left( \frac{2 \cdot d_{si}}{R} \right) ^{-0.756 \cdot z_i}} & if \; C_{2d} \geq 100 \end{array}\end{array}$

$\begin{array}{l}\displaystyle C_{Li} = 1.425 \times 10^{-18}\end{array}$

$\begin{array}{l}\displaystyle \displaystyle M_i = C_{Li} \cdot \left[ 43.2 \cdot p_i \cdot (1+ n_{_V} ) \cdot V \cdot R^{0.756 \cdot z_i - n_{_V}} \right]\end{array}$

$\begin{array}{l}\displaystyle M_i = 2.948 \times 10^{-10}\end{array}$

Bed Load Transport,

$\begin{array}{l}\displaystyle \displaystyle g_{sbi} = M_i \cdot (2 \cdot d_{si} ) ^{(1+ n_{_V} -0.756z_i )}\end{array}$

$\begin{array}{l}\displaystyle g_{sbi} =30.555\end{array}$

Lower Layer Transport,

$\begin{array}{l}\displaystyle \displaystyle g_{ssLi} = M_i \cdot \left[ \frac{ \left( \frac{R}{11.24} \right) ^{(1+ n_{_V} -0.756 \cdot z_i)} - (2 \cdot d_{si} ) ^{(1+ n_{_V} -0.756 \cdot z_i)}}{1+ n_{_V} -0.756 \cdot z_i} \right]\end{array}$

$\begin{array}{l}\displaystyle g_{ssLi} = 6.473\end{array}$

Middle Layer Transport,

$\begin{array}{l}\displaystyle \displaystyle g_{ssMi} = M_i \cdot \frac{\left( \frac{R}{11.24} \right)^{0.244 \cdot z_i} \cdot \left[ \left( \frac{R}{2.5} \right) ^{1+ n_{_V} -z_i} - \left( \frac{R}{11.24} \right)^{1+ n_{_V} -z_i} \right]}{1+ n_{_V} -z_i}\end{array}$

$\begin{array}{l}\displaystyle g_{ssMi} = 5.674 \times 10^{-1}\end{array}$

Upper Layer Transport,

$\begin{array}{l}\displaystyle \displaystyle g_{ssUi} = M_i \cdot \frac{\left( \frac{R}{11.24} \right)^{0.244 \cdot z_i} \cdot \left( \frac{R}{2.5} \right) ^{0.5 \cdot z_i} \cdot \left[ R^{(1+ n_{_V} -1.5z_i)} - \left( \frac{R}{2.5} \right)^{1+ n_{_V} -1.5z_i} \right]}{1+ n_{_V} -1.5z_i}\end{array}$

$\begin{array}{l}\displaystyle g_{ssUi} = 1.72 \times 10^{-15}\end{array}$

Total Transport per Unit Width,

$\begin{array}{l}\displaystyle g_{si} = g_{sbi} + g_{ssLi} + g_{ssMi} + g_{ssUi}\end{array}$

$\begin{array}{l}\displaystyle g_{si} 37.027\end{array}$

Total Transport,

$\begin{array}{l}\displaystyle G= g_{si} \cdot B\end{array}$

$\begin{array}{l}\displaystyle G = 1481 tons/day\end{array}$

Yang Sediment Transport Function

by Yang, from ASCE Journal of Hydraulics, Oct 1973, Dec 1984

Input Parameters

Temperature, F	T = 55	Average Velocity, ft/s	V = 5.46
Kinematic viscosity, ft²/s	$\begin{array}{l}\nu\end{array}$ = 0.00001315	Discharge, ft³/s	Q=5000
Hydraulic Radius, ft	R = 10.68	Unit Weight water, lb/ft³	$\begin{array}{l}\gamma _w\end{array}$ = 62.385
Slope	S = 0.0001
Median Particle Diamter, ft	d_si = 0.00232
Specific Gravity of Sediment	s = 2.65

Constants

Acceleration of gravity, ft/s²

g=32.2

Solution

Shear Velocity, ft/s,

$\begin{array}{l}\displaystyle \displaystyle u_* = \sqrt{g \cdot R \cdot S}\end{array}$

$\begin{array}{l}\displaystyle u_* = 0.185\end{array}$

Particle Fall Velocity, ft/s,

Use Rubey's equation, Vanoni p. 169

$\begin{array}{l}\displaystyle \displaystyle F_1 = \sqrt{\frac{2}{3} + \frac{36 \cdot \nu ^2}{g \cdot d_{si}^3 \cdot (s-1)}} - \sqrt{\frac{36 \cdot \nu ^2}{g \cdot d_{si}^3 \cdot (s-1)}}\end{array}$

$\begin{array}{l}\displaystyle F_1 = 0.725\end{array}$

$\begin{array}{l}\displaystyle \displaystyle \omega F_1 \cdot \sqrt{(s-1) \cdot g \cdot d_{si}}\end{array}$

$\begin{array}{l}\displaystyle \omega =0.255\end{array}$

Shear Reynold's Number,

$\begin{array}{l}\displaystyle \displaystyle R_s = \frac{u_* \cdot d_{si}}{\nu}\end{array}$

$\begin{array}{l}\displaystyle R_s = 32.717\end{array}$

Critical Velocity, ft/s,

$\begin{array}{l}\displaystyle V_{cr} = \left| \begin{array} 1 \omega \cdot \Bigg( \frac{2.5}{log(\frac{u_* \cdot d_{si}}{\nu}) -0.66} +0.66 \Big) & if \; 0< R_s < 70 \\ (\omega \cdot 2.05) & if \; R_s \geq 70 \end{array}\end{array}$

Log of Concentration,

$\begin{array}{l}\displaystyle log \cdot C_t = \left| \begin{array}1 \begin{bmatrix} 5.435-0.286 \cdot log (\frac{\omega \cdot d_{si}}{\nu}) -0.457 \cdot log (\frac{u_*}{\omega}) ... \\ + (1.799-0.409 \cdot log( \frac{\omega \cdot d_{si}}{\nu})-0.314 \cdot log(\frac{u_*}{\omega})) \cdot log (\frac{V \cdot S}{\omega} -\frac{V_{cr} \cdot S}{\omega}) \end{bmatrix} & if \; d_{si} < 0.00656 & Sand \\ \begin{bmatrix} (6.681-0.633 \cdot log (\frac{\omega \cdot d_{si}}{\nu}) -4.816 \cdot log (\frac{u_*}{\omega}) ) ... \\ + (2.784-0.305 \cdot log( \frac{\omega \cdot d_{si}}{\nu})-0.282 \cdot log(\frac{u_*}{\omega})) \cdot log (\frac{V \cdot S}{\omega} -\frac{V_{cr} \cdot S}{\omega}) \end{bmatrix} & if \; d_{si} \geq 0.00656 & Gravel \end{array}\end{array}$

$\begin{array}{l}\displaystyle logC_t =1.853\end{array}$

Concentration, ppm

$\begin{array}{l}\displaystyle C_t =10^{logC_t}\end{array}$

$\begin{array}{l}\displaystyle C_t =71.284\end{array}$

Sediment Discharge, lb/s

$\begin{array}{l}\displaystyle \displaystyle G= \frac{\gamma _q \cdot Q \cdot C_t}{1000000}\end{array}$

$\begin{array}{l}\displaystyle G = 22.235\end{array}$

Sediment Discharge, tons/day

$\begin{array}{l}\displaystyle \displaystyle G_s = \frac{86400}{2000} \cdot G\end{array}$

$\begin{array}{l}\displaystyle G_s = 961\end{array}$