Calculate suspended-sediment transport through time

Name the quantity crossing the section

Water carrying suspended particles passes a stream section. To calculate the suspended sediment transported, we need both water discharge and sediment concentration. A high concentration alone does not establish a large mass rate if very little water passes. Conversely, a modest concentration carried by a large water volume can transport considerable mass.

The investigation supplies six fictional one-hour intervals. At flow multiplier 1, the discharges are 10, 20, 40, 40, 20 and 10 m³/s. In the aligned case, concentrations are 100, 200, 400, 400, 200 and 100 mg/L. Each value is held constant within its interval. The step plots encode that assumption; they are not continuous sensor observations from a named storm.

Each supplied concentration is assumed to represent the discharge-weighted cross-section concentration for its bin. It is not a single unverified point sample, nor a turbidity reading. The model assumes no measurement error. Actual suspended-sediment studies need flow measurements and representative sediment observations as conditions change. USGS introduction to suspended-sediment measurement.

Convert the units before multiplying

One cubic meter contains 1,000 liters, and one kilogram contains 1,000,000 milligrams. Therefore 1 mg/L equals 0.001 kg/m³. Multiply discharge Q in m³/s by concentration C in mg/L and by 0.001 to obtain the mass rate in kg/s.

For the third aligned interval, Q = 40 m³/s and C = 400 mg/L. The rate is 40 × 400 × 0.001 = 16 kg/s. The cubic meters cancel after converting concentration, leaving kilograms per second. A one-hour bin lasts 3,600 seconds, so the transported mass is 16 × 3,600 = 57,600 kg.

This is a mass passing the section during the interval. It is not a concentration and does not tell you how much sediment is stored in the river. Keep the time interval attached to the result whenever reporting it.

Build the event ledger

Calculate all six aligned mass rates: 1, 4, 16, 16, 4 and 1 kg/s. Multiplying each by 3,600 seconds gives 3,600, 14,400, 57,600, 57,600, 14,400 and 3,600 kg. Their sum is 151,200 kg, or 151.2 metric tonnes. One metric tonne is 1,000 kg.

On paper, draw six equal-width boxes and write Q, C, rate and mass inside each. Add the masses cumulatively below the boxes. After three bins the accumulated load is 75,600 kg; after six it is 151,200 kg. The steepest cumulative increases correspond to the largest mass rates, not merely to the highest plotted concentration in isolation.

The six-hour water volume is the sum of Q × 3,600 seconds, giving 504,000 m³ at multiplier 1. Dividing total mass by total water gives 0.3 kg/m³, or 300 mg/L. This is the event’s flow-weighted concentration for the aligned case, distinct from the arithmetic average of the six concentration values.

Isolate a controlled scaling comparison

Keep the aligned timing and double the flow multiplier. Every Q doubles while C stays unchanged. Each mass rate, each bin mass, total water and total load therefore doubles. Total load becomes 302,400 kg, but the flow-weighted concentration remains 300 mg/L because both mass and water volume doubled.

This is a controlled calculation, not evidence that a real river’s concentration would remain unchanged as discharge doubles. The browser does not model particle entrainment or sediment supply. Its value is that the paired arithmetic and assumptions remain visible.

Transfer to another interval

A paper-only interval lasts 30 minutes, with discharge 15 m³/s and representative concentration 80 mg/L. Find the mass rate, water volume and suspended mass before checking.

Check the sediment-load transfer

The rate is 15 × 80 × 0.001 = 1.2 kg/s. Thirty minutes is 1,800 seconds, so water volume is 27,000 m³ and suspended mass is 2,160 kg, or 2.16 metric tonnes. The shorter interval changes accumulated mass, not the supplied instantaneous rate.