PE_2009,
The 5 kg is the total mass of the soil plus initial water.
Dry mass of soil = 5000g / 1.03 = 4854g
Initial mass of water = .03 * 4854 = 146g (Or 5000 - 4854)
Total water req'd = .12 * 4854 = 582g
Additional water = 582 - 146 = 436g
Please let me know, what's wrong in following solution. Some authors including Chelapati calculate amount of water to be added with following method
Total mass of soil is 5,000g which contains 3% water. To bring this soil mass to 12% water content, amount of water to be added = 5000 x (12%-3%) = 450g
The water content is a ratio of the mass of water (mw) to the mass of soil (ms), not the total mass (mt) of 5000g. i.e. w = mw / ms. => initial mw = .03ms
5000g = mw + ms. You must isolate the ms before proceeding. => 5000g = .03ms + ms => 5000g = 1.03ms => ms = 4854g
For the initial water content: mw = 5000 - 4854 = 146g For the final water content: mw = .12*4854 = 582g
Additional water required = 582 - 146 = 436g
ms should be used in the short cut instead of mt: 5000/1.03*(.12-.03) = 437g Close enough to 436g. See above for derivation of 1.03.
Bottom line: general short cut = mt/(1+w)*(Δw)
