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

Correct Answer: 0.45

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: \[ 6000 \, {tons/year} = 6000 \times 10^6 \, {grams/year} \] 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\(^3\): \[ 1.5 \, {g/cm}^3 = 1500 \, {kg/m}^3 \] Step 4: Calculating the sediment volume.
The volume of sediment can be found using the formula: \[ {Volume} = \frac{{Mass}}{{Density}} = \frac{6000 \times 10^6 \, {grams/year}}{1.5 \, {g/cm}^3} = 4000 \times 10^6 \, {cm}^3/{year} \] Now, convert the volume to cubic meters: \[ 4000 \times 10^6 \, {cm}^3/{year} = 4000 \, {m}^3/{year} \] Step 5: Calculating the erosion rate.
The catchment area is 8 km\(^2\) = \(8 \times 10^6\) m\(^2\). The erosion rate is: \[ {Erosion rate} = \frac{{Volume}}{{Area}} = \frac{4000 \, {m}^3/{year}}{8 \times 10^6 \, {m}^2} = 0.0005 \, {m/year} = 0.5 \, {mm/year} \] Thus, the average erosion rate is \(0.50\) mm/year.