What is the amount of water required per hour for compacting loose soil being spread by a shovel and dozer at the rate of 230 m³/h for a soil having density of 1·5 gm/m³ and 8% moisture content needing 12% optimum moisture content for compaction ?
- A6900 litres/hour
- B13800 litres/hourCorrect
- C20700 litres/hour
- D27600 litres/hour
Solution
Concept
Water to be added to reach OMC
Formula
Water = (OMC − w) × dry mass of soil
Step-by-step solution
1. Mass of soil per hour = 230 m³ × 1.5 t/m³ = 345 t = 3,45,000 kg. 2. Water to be added = (12% − 8%) = 4% of that mass = 13,800 kg. 3. 13,800 kg of water = 13,800 litres per hour.
Explanation
The soil already contains 8 percent moisture. To reach the optimum moisture content of 12 percent, another 4 percent of the soil mass must be added as water. The printed unit "gm/m³" should read t/m³ (g/cm³).
Why the other options are wrong
6,900, 20,700 and 27,600 litres/hour correspond to 2%, 6% and 8% of the soil mass.
Shortcut
345 t × 4% = 13.8 t = 13,800 litres.
Common mistake
Using 12% (the full OMC) instead of the deficit of 4%.
Correct answer: B) 13800 litres/hour