QUESTIONS

  1. Give two alternate definitions of particle sphericity.

    Particle sphericity (true sphericity) is the ratio of the surface area of a sphere having the same volume as the particle to the surface area of the particle. Alternatively, particle sphericity is defined as the ratio of the diameter of a sphere having the same volume as the particle (i.e., the nominal diameter) to the diameter of a sphere circumscribing the particle.

  2. What is the difference between specific weight and specific gravity?

    The specific weight of a sediment particle is its weight per unit volume. The specific gravity of a sediment particle is the ratio of its weight to the weight of an equal volume of water. Specific weight has units of weight per unit volume. Specific gravity is dimensionless.

  3. What is standard fall velocity? What is standard fall diameter?

    The standard fall velocity of a particle is the average rate of fall that it would attain if falling alone in quiescent water of infinite extent at the temperature of 24°C.

    The standard fall diameter of a particle is the diameter of an equivalent sphere having the same standard fall velocity and specific gravity.

  4. What is the difference between sediment production and sediment yield?

    Sediment production (gross sediment production) refers to the amount of sediment eroded and removed from the source(s), whereas sediment yield refers to the actual delivery of eroded soil particles to a given downstream point. Sediment yield amounts are generally less than those produced at the upland source(s).

  5. Describe the differences between normal and accelerated erosion.

    Normal erosion is that which has been occurring at variable rates since the first solid materials formed on the surface of the earth. This type of erosion is extremely slow in most places, and is largely a function of climate, parent rocks, precipitation, topography, and vegetative cover.

    In contrast with normal erosion, accelerated erosion occurs at a much faster rate, usually through reduction of vegetative cover. Deforestation, cultivation, forest fires, and systematic destruction of natural vegetation by human activities (e.g., for surface mining and construction) result in accelerated erosion.

  6. Name four sources of sediment.

    The following are four sources of sediment: (1) sheet erosion, (2) rill erosion, (3) gully erosion, and (4) channel erosion.

  7. What is the rainfall factor R in the Universal Soil Loss Equation?

    In the Universal Soil Loss Equation, the rainfall factor R is the average value of the series of annual sums of EI products, in which E is the storm's total kinetic energy and I is its maximum 30-minute intensity. The product EI reflects the combined potential of raindrop inpact and runoff turbulence to transport dislodged soil particles.

  8. What is sediment delivery ratio?

    The sediment-delivery ratio (SDR) is the ratio of sediment yield to gross sediment production.

  9. Why is sediment delivery ratio inversely related to drainage-basin area?

    The sediment-delivery ratio is inversely related to drainage-basin area because the efficiency in transporting sediment decreases with an increase in drainage area. Therefore, the greater the drainage-basin area, the greater the chances for the downstream deposition of sediment eroded at the upland sources, and the lower the sediment-delivery ratio.

  10. Why are two formulas needed in the Dendy and Bolton approach to the computation of sediment yield?

    Reservoir sedimentation data studied by Dendy and Bolton have shown that sediment yield per unit area increases sharply in the range of mean annual runoff from 0 to 2 inches. Thereafter, sediment yield per unit area decreases exponentially. This difference in behavior is due to the different climatic conditions: in the first case, an arid to semiarid climate, with sparse vegetation, fast erosion rates, and increasing sediment yield; in the second case, a subhumid to humid climate, with established vegetation, slow erosion rates, and decreasing sediment yield per unit area. The pair of Dendy and Bolton formulas reflect this natural trend.

  11. Describe the classifications of sediment load based on: (1) predominant mode of transport, and (2) whether the particle sizes are represented on the channel bed.

    Based on the predominant mode of transport, sediment load can be classified into: (a) bed load, and (b) suspended load. Bed load is the fraction of sediment load that moves by saltation and rolling along the channel bed, primarily by action of bottom shear stresses caused by vertical velocity gradients. Suspended load is the fraction of sediment load that moves in suspension by the action of turbulence.

    Based on whether the particle sizes are represented in the channel bed, sediment load can be classified into: (a) bed-material load, and (b) fine-material load. Bed-material load is the fraction of sediment load whose particle sizes are significantly represented in the channel bed. Fine-material load, commonly referred to as wash load, is the fraction of sediment load whose particle sizes are not significantly represented in the channel bed.

  12. What are possible forms of bed roughness in alluvial channels?

    The possible forms of bed roughness in an alluvial channel are: (1) ripples, (2) dunes and superposed ripples, (3) dunes, (4) washed-out dunes or transition, (5) plane bed with sediment motion, (6) antidunes, and (7) chute and pool.

  13. What is range of applicability of the Meyer-Peter formula for bedload transport?

    The development of the Meyer-Peter formula was based on flume data with uniform bed material size in the range 3 to 28 mm. Such coarse sediments do not produce appreciable bed forms; therefore, the formula is applicable for coarse sediment transport (i.e., largely gravel) with negligible form roughness.

  14. What is the basic difference between the Colby 1957 and Colby 1964 procedures for the computation of discharge of sands?

    The Colby 1957 method requires information on the measured concentration of suspended bed material. It estimates the unmeasured bed-material discharge and adds this to the measured bed-material discharge to provide an estimate of the total bed-material discharge.

    The Colby 1964 method, on the other hand, does not require a measurement of suspended bed-material discharge, basing its prediction of total discharge of sands (i.e., total bed-material discharge) entirely on flow and sediment characteristics.

  15. What is a sediment rating curve?

    At a given stream gaging site, a sediment rating curve is an x-y plot showing water discharge in the abscissa and sediment discharge in the ordinates. Such a curve is obtained either by the simultaneous measurement of water and sediment discharge or, alternatively, by the use of sediment transport formulas.

  16. What is sediment routing?

    Sediment routing refers to the distributed and unsteady calculation of sediment production, transport and deposition in catchments, streams, rivers, canals, reservoirs, and estuaries.

  17. What is the trap efficiency of a reservoir?

    Trap efficiency refers to the ability of a reservoir to entrap the sediment that is transported by the flowing water entering the reservoir. Trap efficiency is defined as the ratio of trapped sediment to incoming sediment, in percentage.

    Trap efficiency is a function of: (1) the ratio of reservoir volume to mean annual runoff volume, and (2) the characteristics of the sediment particles.

  18. What is a debris basin?

    A debris or sediment-retention basin is a small reservoir located in an upland area with the specific purpose of trapping sediment and debris before they are able to reach the main fluvial network system.

  19. Describe two techniques to measure suspended-sediment discharge. How do they differ in the evaluation of suspended sediment concentration?

    The techniques to measure suspended sediment discharge are: (1) EDI, or equal-distance-increments, and (2) ETR, or equal-transit-rate. In the EDI method, sampling is performed at the centroids of equal-discharge increments. In the ETR method, sampling is performed at the centroids of equal-length increments.

    In the EDI method, the suspended-sediment concentration is the average obtained from several depth-integrating samples. In the ETR method, the suspended-sediment concentration is that of a composite sample encompassing several depth-integrating samples.

PROBLEMS

  1. Calculate the fall velocity of a sediment particle using Stokes' law. Assume a diameter 0.1 mm, kinematic viscosity 1 centistoke, and specific gravity 2.65.

    Using Stokes' law (Eq. 15-4), the fall velocity of the particle is:

    w = [981 cm/s2 × (0.1 mm × 0.1 cm/mm)2 × (2.65 - 1.00) / (1.00)] / [18 × 1 cs × 0.01 (cm2/s)/cs] = 0.9 cm/s.  ANSWER.

  2. Calculate the specific weight of a sediment deposit in a reservoir, after an elapsed time of 100 y, under moderate drawdown conditions. Assume the following mix of particle sizes: sand 55%, silt 30%, clay 15%.

    From Table 15-2, the specific weight of reservoir sediment deposits, after 100 y, under moderate drawdown conditions is:

    sand size:  93 + (0.0 × log 100) = 93.0 lb/ft3

    silt size:  74 + (2.7 × log 100) = 79.4 lb/ft3

    clay size:  46 + (10.7 × log 100) = 67.4 lb/ft3

    The weighted value of specific weight is:

    (93.0 × 0.55) + (79.4 × 0.30) + (67.4 × 0.15) = 85.1 lb/ft3.  ANSWER.

  3. Compute the average annual soil loss by the universal soil loss equation for a 300-ac watershed near Lexington, Kentucky, with the following conditions: (1) cropland, 250 ac, contoured, soil is Keen silt loam, slopes are 7% and 150 ft long, C = 0.15; (2) pasture, 50 acres, 75% canopy cover, 60% ground cover with grass, soil is Ida silt loam, slopes are 10% and 200 ft long.

    (1) From Fig. 15-2: R = 200; from Table 15-4: K = 0.48; from Fig. 15-3: LS = 1.0; C = 0.15; from Table 15-7: P = 0.50.

    Therefore: A1 = 200 × 0.48 × 1.0 × 0.15 × 0.50 = 7.2 tons/ac/y.

    (2) From Fig. 15-2: R = 200; from Table 15-4: K = 0.33; from Fig. 15-3: LS = 1.9; from Table 15-5: C = 0.032. No value of erosion-control-practice factor (P = 1.0) has been established for pasture.

    Therefore: A2 = 200 × 0.33 × 1.9 × 0.032 × 1.0 = 4.01 tons/ac/y.

    The average annual soil loss is:

    A = (250 ac × 7.2 tons/ac/y) + (50 ac × 4.01 tons/ac/y) = 2000 tons/y.  ANSWER.

  4. Compute the average annual soil loss by the universal soil loss equation for a 1-mi2 forested watershed near Bangor, Maine. The soil is Fayette silt loam, the slopes are 3% and 300 ft long, and the site is 80% covered by forest litter.

    From Fig. 15-2: R = 100; from Table 15-4: K = 0.38; from Fig. 15-3: LS = 0.4; from Table 15-6, assume a C value in midrange: C = 0.003. No value of P has been established for forest (P = 1).

    Therefore:

    A =  100 × 0.38 × 0.4 × 0.003 × 1.0 = 0.0456 tons/ac/y.

    The average annual soil loss is:

    A =  1 mi2 × 640 ac/mi2 × 0.0456 tons/ac/y = 29.2 tons/y.  ANSWER.

  5. Using the Dendy and Bolton formula, calculate the sediment yield for a 25.9-km2 watershed with 5 cm of mean annual runoff.

    The watershed area is:  A = 25.9 km2 × 0.3863 mi2/km2 = 10 mi2

    The mean annual runoff is:  Q = 5 cm × 0.3937 in/cm = 1.97 in

    Using Eq. 15-13a:

    S = 1280 × 1.970.46 [1.43 - (0.26 × log 10)] = 2045.7 tons/mi2/y

    The sediment yield is:

    S = 2045.7 tons/mi2/y × 10 mi2 = 20,457 tons/y.

    Converting to SI units:

    S = 20,457 tons/y × 2000 lb/ton × 0.4536 kg/lb × 0.00981 kN/kg

    S = 182,060 kN/y.  ANSWER.

  6. Determine whether a particle of 2-mm diameter is at rest under a 3-m flow depth and 0.0002 channel slope. Assume a specific gravity 2.65 and kinematic viscosity 1 centistoke.

    The dimensionless shear stress (Eq. 15-7) is:

    τ* = (γ d So) / [(γs - γ) ds] = (d So) / [((γs/γ) - 1 ) ds]

    τo = (3 m × 0.0002) / [((2.65/1.00) - 1) × 2 mm × 0.001 m/mm] = 0.182

    The shear velocity (Eq. 15-16) is:

    U* = (τo / ρ)1/2 = (g d So)1/2 = [9.81 m/s2 × 3 m × 0.0002]1/2

    U* = 0.07672 m/s

    The boundary Reynolds number (Eq. 15-15) is:

    R* = U* ds / v = (0.07672 m/s × 2 mm × 0.001 m/mm) / [1 cs × 10-6 (m2/s)/cs] = 153.4

    Using Fig. 15-6, the point for T*= 0.18 and R* = 153.4 plots above the Shields curve. Therefore, the 2-mm diameter particle is at motion under a flow depth of 3 m and channel slope 0.0002.  ANSWER.

  7. Determine the form of bed roughness that is likely to prevail under the following flow conditions: mean velocity 3 ft/s, flow depth 8 ft, channel slope 0.0002, and mean particle diameter 0.3 mm.

    Using Eq. 15-14, the bottom shear stress is:

    τo = γ d So = 62.4 lb/ft3 × 8 ft × 0.0002 = 0.1 lb/ft2

    The stream power is:  τoV = 0.1 lb/ft2 × 3 ft/s = 0.3 ft-lb/s/ft2

    Using Fig. 15-8, with median particle size d50 = 0.3 mm, and stream power τoV = 0.3 ft-lb/s/ft2, the form of bed roughness is dunes.  ANSWER.

  8. Given the following flow characteristics: flow depth 9 ft, mean velocity 3 ft/s, channel slope 0.00015, mean particle diameter 0.4 mm, mean channel width 250 ft. Calculate the bed material transport rate by the Duboys formula.

    Using Eq. 15-14, the bottom shear stress is:

    τo = γ d So = 62.4 lb/ft3 × 9 ft × 0.00015 = 0.08424 lb/ft2

    From Fig. 15-10, with d50 = 0.4 mm: τc = 0.02 lb/ft2

    ψD = 57 ft3/lb/s.

    Using Eq. 15-21:

    qs = 57 ft3/lb/s × 0.08424 lb/ft2 × (0.08424 - 0.02) lb/ft2

    qs = 0.30846 lb/s/ft

    Therefore: Qs = 0.30846 lb/s/ft × 250 ft = 77.115 lb/s.  ANSWER.

    Qs = (77.115 lb/s × 86,400 s/d) / (2000 lb/ton) = 3331 tons/d.  ANSWER.

  9. Given the following flow characteristics: flow depth 3 ft, mean velocity 5 ft/s, energy slope 0.009, mean particle diameter 1.0 in., mean channel width 30 ft. Calculate the bedámaterial transport rate (in tons per day) by the Meyer-Peter formula.

    The discharge per unit width is: q = 5 ft/s × 3 ft = 15 ft2/s.

    Using Eq. 15-22:

    qs = [(39.25 × 152/3 × 0.009) - (9.95 × 1 in / (12 in/ft))]3/2

    qs = 1.515 lb/s/ft

    Therefore: Qs = 1.515 lb/s/ft × 30 ft = 45.45 lb/s.  ANSWER.

    Qs = (45.45 lb/s × 86,400 s/d) / (2000 lb/ton) = 1963 tons/d.  ANSWER.

  10. Given the following flow characteristics: flow depth 5 ft, mean velocity 4 ft/s, mean channel width 180 ft, measured concentration of suspended bed material discharge 200 ppm. Calculate the total bed-material discharge (in tons per day) by the Colby 1957 method.

    Using Fig. 15-11, the uncorrected unmeasured sediment discharge is: qu' = 26 tons/d/ft.

    Using Fig. 15-12, the relative concentration of suspended sands is:

    Cr = 1050 ppm. The availability ratio is: Cm /Cr = 200 / 1050 = 0.19.

    Using the mean line of Fig. 15-13, the correction factor is 0.52.

    Therefore, the corrected unmeasured sediment discharge is:

    qu = 0.52 × 26 tons/d/ft = 13.52 tons/d/ft.

    The discharge per unit width is: q = 4 ft/s × 5 ft = 20 ft3/s/ft.

    Using Eq. 15-23:

    qs = [0.0027 × 200 (mg/L or ppm) × 20 (ft3/s/ft)] + 13.52 tons/d/ft

    qs = 24.32 tons/d/ft.

    Qs = 24.32 tons/d/ft × 180 ft = 4378 tons/d.  ANSWER.

  11. Given the following flow characteristics: flow depth 5 ft, mean velocity 3 ft/s, median bed material size 0.3 mm, mean channel width 225 ft, water temperature 70°F, wash load concentration 300 ppm. Calculate the discharge of sands by the Colby 1964 method.

    Using Fig. 15-14, the uncorrected discharge of sands for 1 ft depth, mean velocity 3 ft/s and particle size 0.3 mm is: 14.5 tons/d/ft. Likewise, the uncorrected discharge of sands for 10 ft depth is 16.2 tons/d/ft. By logarithmic interpolation, for 5 ft depth, the uncorrected discharge of sands is 15.7 tons/d/ft.

    From Fig. 15-15: k1 = 0.9; k2 = 1.0; and k3 = 1.0.

    Using Eq. 15-24:

    qs = [1 + (0.9 × 1.0 - 1) × 1.0) × 15.7 = 14.13 tons/d/ft

    Qs = 14.13 tons/d/ft × 225 ft = 3179 tons/d.  ANSWER.

  12. A reservoir is to be built with a total storage capacity of 50 hm3. The contributing drainage basin is 800 km2, and the mean annual runoff at the site is 200 mm. Assume well-graded sediment deposits with average specific weight 1400 kg/m3. (a) How long will it take for the reservoir to lose 20% of its storage volume? (b) How long will it take for the reservoir to fill up with sediment? Estimate sediment yield by the Dendy and Bolton formula.

    The mean annual runoff is 200 mm. The mean annual volume inflow to the reservoir is:

    I = 200 mm × 800 km2 × 0.001 m/mm × 106 m2/km2 × 10-6 hm3/m3 = 160 hm3

    The contributing drainage basin in square miles is:

    A = 800 km2 × 0.3863 mi2/km2 = 309 mi2.

    The mean annual runoff in inches is:

    Q = 200 mm × 0.03937 in/mm = 7.874 in.

    Using the appropriate Dendy and Bolton formula (Eq. 15 -13b), the annual sediment yield is:

    S = 1965 × e-0.055 × 7.874 (1.43 - 0.26 log 309) = 997.3 tons/mi2/y

    S = 997.3 tons/mi2/y × 2000 lb/ton × 0.4536 kg/lb × 0.3863 mi2/km2

    S = 349,505 kg/km2/y

    The annual sediment inflow expressed in reservoir deposit volume is:

    IS = (349,505 kg/km2/y × 800 km2) / (1400 kg/m3) = 199,717 m/y

    IS = 199,717 m3/y × 10-6 hm3/m3 = 0.2 hm3/y

    (1)(2)(3)(4)(5)(6)(7)
    Interval

    i    		 Reservoir
    capacity
    (hm3)    		 Accumulated
    volume
    (hm3)    		 C /I
    ratio    		 Average C /I
    in interval    		 Trap
    efficiencyTi
    (%)    		 Number
    of years
    to fill
    Ni
    0 50 0 0.132 - - -
    1 40 10 0.250 0.281 95 53
    2 30 20 0.188 0.219 93 54
    3 20 30 0.125 0.157 90 56
    4 10 40 0.063 0.094 87 57
    5 0 50 0.000 0.032 69 72
    Sum 292

    The number of years to fill each increment of 10 hm3 of reservoir volume is:

    Ni = 10 hm3 / [(0.2 hm3/y) × (Ti /100)] = (5000/Ti ) y.

    In 53 y the reservoir will lose 20% of its storage volume.  ANSWER.

    In 292 y the reservoir will completely fill with sediment.  ANSWER.

  13. A reservoir is to be built with a total storage capacity of 120 hm3. The contributing drainage basin is 425 km2, and the mean annual runoff at the site is 45 mm. Assume coarse sediment deposits with average specific weight 13 kN/m3. (a) How long will it take for the reservoir to lose 80% of its storage volume? (b) How long will it take for the reservoir to fill up with sediment? Estimate sediment yield by the Dendy and Bolton formula.

    The mean annual runoff is 45 mm. The mean annual volume inflow to the reservoir is:

    I = 45 mm × 425 km2 × 0.001 m/mm × 106 m2/km2 × 10-6 hm/m3

    I = 19.125 hm3

    The contributing drainage basin in square miles is:

    A = 425 km2 × 0.3863 mi2/km2 = 164.2 mi2

    The mean annual runoff in inches is: Q = 45 mm × 0.03937 in/mm = 1.77 in

    The sediment yield by the Dendy and Bolton formula (Eq. 15-13a) is:

    S = 1280 × (1.77)0.46 (1.43 - 0.26 log 164.2) = 1421 tons/mi2/y

    S = 1421 tons/mi2/y × 2000 lb/ton × 4.45 N/lb × 0.3863 mi2/km2

    S = 4,885,497 N/km2/y = 4885.5 kN/km2/y.

    The annual sediment inflow expressed in reservoir deposit volume is:

    IS = (4885.5 kN/km2/y × 425 km2) / (13 kN/m3) = 159,718 m3/y

    IS = 159,718 m3/y × 10-6 hm3/m3 = 0.16 hm3/y

    (1)(2)(3)(4)(5)(6)(7)
    Interval

    i    		 Reservoir
    capacity
    (hm3)    		 Accumulated
    volume
    (hm3)    		 C /I
    ratio    		 Average C /I
    in interval    		 Trap
    efficiencyTi
    (%)    		 Number
    of years
    to fill
    Ni
    0 120 0 6.275 - - -
    1 96 24 5.020 5.648 100 150
    2 72 48 3.765 4.393 100 150
    3 48 72 2.510 3.138 100 150
    4 24 96 1.255 1.883 100 150
    5 0 120 0.000 0.613 100 150
    Sum 750

    The number of years to fill each increment of 24 hm3 of reservoir volume is:

    Ni = 24 hm3 / [(0.16 hm3/y) × (Ti /100)] = (15000/Ti ) y

    In 600 y the reservoir will lose 80% of its storage volume.  ANSWER.

    In 750 y the reservoir will completely fill with sediment.  ANSWER.

  14. Derive the conversion factor 0.0027 in Eq. 15-33.

    Assume: Qs (tons/d) = C Cs (mg/L) × Q (ft3/s)

    Qs = Cs Q [(mg/L)-(ft3/s) × 103 L/m3 × 10-6 kg/mg]

    Qs = Cs Q [10-3 (kg/m3)-(ft3/s) × 0.028317 m3/ft3 × 86,400 s/d] / [0.4536 kg/lb × 2000 lb/ton]

    Qs = 0.0026968 Cs Q [tons/d]

    Therefore: C = 0.0026968; or C = 0.0027.  ANSWER.

  15. Calculate the suspended sediment discharge (in tons per day) for the following cases: (1) suspended sediment concentration 100 ppm, water discharge 1200 ft3/s, and (2) suspended sediment concentration 80,000 ppm, and water discharge 5000 ft3/s.

    (1) Qs = 0.0027 × 100 × 1200 ft3/s

    Qs = 324 tons/d.  ANSWER.

    (2) From Table 15-8, for Cs = 80,000 ppm: Cs (mg/L) = 1.06 Cs (ppm)

    Therefore: Qs = 0.0027 × 1.06 × 80,000 × 5000 ft3/s

    Qs = 1,144,800 tons/d.  ANSWER.

  16. Derive the unit conversion factor C in the following formula: Qs = C Cs Q, in which Qs is given in kilonewtons per day, Cs in milligrams per liter, and Q in cubic meters per second.

    Qs (kN/d) = C Cs (ppm) Q (m3/s)

    Qs = Cs Q [ppm-(m3/s) × 1 (mg/L)/ppm] = Cs Q [(mg/L)-m3/s]

    Qs = Cs Q [(mg/L) -m3/s × 103 L/m3 × 10-6 kg/mg] = Cs Q [10-3 kg/s]

    Qs = Cs Q [10-3 kg/s × 0.00981 kN/kg × 86,400 s/d]

    Qs = 0.8476 Cs Q [kN/d]

    Therefore: C = 0.8476.  ANSWER.

  17. Calculate the suspended sediment discharge (in kilonewtons per day) for a suspended sediment concentration of 150 ppm and a flow of 68 m3/s.

    Using the formula derived in Problem 15-16:

    Qs = 0.8476 Cs (ppm) Q (m3/s) = 8645 kN/d.  ANSWER.

    Alternatively: Qs = 150 ppm × 68 m3/s = 10,200 pm-(m3/s)

    Qs = 10,200 [ppm-(m/s) × 1 (mg/L)/ppm] = 10,200 [(mg/L)-m3/s]

    Qs = 10,200 [(mg/L)-m/s] × 103 L/m3 × 10-6 kg/mg = 10.2 kg/s

    Qs = 10.2 kg/s × 0.00981 kN/kg × 86,400 s/d = 8645 kN/d.  ANSWER.

  18. Calculate the suspended sediment discharge (in kilonewtons per day) for a suspended sediment concentration of 22,000 ppm and a flow of 155 m3/s.

    From Table 15-8, for Cs = 22,000 ppm: Cs (mg/L) = 1.02 Cs (ppm)

    Cs = 22,000 × 1.02 = 22,440 mg/L.

    Using the formula derived in Problem 15-16:

    Qs = 0.8476 Cs (ppm) Q (m3/s) = 2,948,122 kN/d.  ANSWER.

    Alternatively: Qs = 22,000 ppm × 155 m3/s = 3,410,000 ppm-(m3/s)

    Qs = 3,410,000 [ppm-(m3/s) × 1.02 (mg/L)/ppm] = 3,478,200 [(mg/L)-m3/s]

    Qs = 3,478,200 [(mg/L)-m3/s] × 103 L/m3 × 10-6 kg/mg = 3478.2 kg/s

    Qs = 3478.2 kg/s × 0.00981 kN/kg × 86,400 s/d

    Qs = 2,948,067 kN/d.  ANSWER.

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