Question

The sediment yield at the outlet of a river having a catchment area of 8 km\(^2\) is 6000 tons/year. If the sediment density is 1.5 g/cm\(^3\), the average erosion rate of the river basin is _________mm/yr.

Hint:

To calculate the erosion rate, first convert sediment yield into volume using sediment density, then divide by the catchment area to get the rate in units of depth per year.

Answer 1

Step 1: Understanding the given data.
Sediment yield = 6000 tons/year
Catchment area = 8 km\(^2\)
Sediment density = 1.5 g/cm\(^3\)
We need to calculate the erosion rate in mm/year. First, we need to convert the sediment yield into volume, then divide it by the catchment area to get the erosion rate.

Step 2: Converting tons to grams.
Since 1 ton = \(10^6\) grams, the sediment yield in grams per year is:

6000 tons/year = 6000 × 10^6 grams/year

Step 3: Converting sediment density to compatible units.
The density is given as 1.5 g/cm\(^3\), and we need to convert it to kg/m\(^3\):

1.5 g/cm³ = 1500 kg/m³

Step 4: Calculating the sediment volume.
The volume of sediment can be found using the formula:

Volume = Mass / Density

Volume = 6000 × 10^6 grams/year ÷ 1.5 g/cm³ = 4000 × 10^6 cm³/year

Now, convert the volume to cubic meters:

4000 × 10^6 cm³/year = 4000 m³/year

Step 5: Calculating the erosion rate.
The catchment area is 8 km² = \(8 × 10^6\) m². The erosion rate is:

Erosion rate = Volume / Area = 4000 m³/year ÷ 8 × 10^6 m² = 0.0005 m/year = 0.5 mm/year

Thus, the average erosion rate is 0.50 mm/year.