Dear respected Nadar,

Weight of Original soil sample =Wet Weight WW= 42.5 g

Dry Weight of soil sample = DW= 31.0 g

Weight moisture content = WW - DW = 42.5 - 31.0 = 11.5 g

% moisture content = (WW- DW)/ DW *100%

= (11.5)/31 * 100 %= 37 %

To recondition the dry soil sample to a soil sample having 50 % moisture content we have:

New % moisture content required = 50 %

Dry weight of soil sample = DW2= 31 g

Wet weight of the soil sample that would give 50 % moisture content = WW2

So we have

% Moisture content = ( WW2- DW2)/ DW *100 %

50 % = (WW2-31)/31 * 100 %

0.5 = (WW2-31)/31

WW2-31=0.5*31

WW2=46.5 g

So the amount of water needed to be added to the dry soil sample to recondition it in order to have 50 % moisture content we have

Amount of water needed = WW2- DW2 = 46.5 g - 31 g = 15.5 g.

Best regards.

Moisture is a key limitation on the productivity of your soil. Three main factors affect soil moisture content:

• how well your soil can absorb water;

• how well your soil can store moisture; and

• how quickly the water is lost or used.

Here is the procedure of increasing water content after compaction:

1. Find an appropriate mold (maybe plastic pipe for cylinder or wood made for cubic)

2. Start with an initial water content which matches with saturation ratio and dry density target .

3. Compact your sample till calculated height (still you are not on your targeted water content)

4. Calculate target water content on the wet side of optimum density and matches with saturation ratio-dry density and pour half of it on the top. let it be isolated for 1-3 days depending on your soil type. ours was cohesive clay and we kept it for 2.5 days.

4. Reverse the sample and pour other half of water on the bottom. Keep it isolated for the same time.

5. Now your sample is almost homogeneous in water content percentages.

6. Run a water content test from top, bottom and middle of every sample after you ran your test as a report.

you may see some error according to change in height, expansion, non homogeneous distribution and ... You have to verify them with measuring height after reaching water content as well as water content tests.

Dear Mr. Nadar Khokhar,

I don't understand very well your input data and your question but I'll give you an answer according to what I suppose you wish to know. If you had a wet sample weighing 42.5 g but after drying the sample weigh was 31.0 g, it means that evaporated water was 11.5 g, that is 27.05% of the wet sample (not 37%). If you wish to bring the wet soil to 50% humidity, the final wet sample will weigh 54.1 g (from 27.05*100/50). It means that you should add 11.6 g water to the 42.5 g initial soil sample. The final sample of 54.1 g contains 50% water but this is an increase from the initial moisture of 27.05% water. I suppose 37% is an error or I maybe didn't catch your question...

OK, one must be pretty specific, and I am more used to using the dried weight of 31 g. But some prefer expressing as wet weight. Once the soil is dried, don't expect to be able to rewet it without some issues. By dry weight, a 31 g dry soil sample would have 15.5 g water to be 50% by weight (1.5 x 31 g). This assumes the soil can absorb that much. There is also % by volume that is sometimes discussed. To save time, I found and added a link that should explain all this in more detail.

http://www.cfr.washington.edu/classes.esrm.410/moisture.htm

Dear Dr. Hansen, thank you for this link, I'll take a look. I did not mean dry soil would be re-moistened. I just indicated the amount of water to be added to the wet soil sample to reach 50% humidity. Why drying the soil in the experiment and add water to re-moist it?... I supposed Mr. Nadar wished to know how to moisten up to 50% water the already wet soil sample. Regards.

Dear respected Nadar,

What we need to do is to recondition the soil sample to achieve the % moisture content required. The water content need to be added to the dry matter of the soil sample to obtain the 50 % moisture content is 15.5 g.

Best regards.

Dear respected Nadar,

Weight of Original soil sample =Wet Weight WW= 42.5 g

Dry Weight of soil sample = DW= 31.0 g

Weight moisture content = WW - DW = 42.5 - 31.0 = 11.5 g

% moisture content = (WW- DW)/ DW *100%

= (11.5)/31 * 100 %= 37 %

To recondition the dry soil sample to a soil sample having 50 % moisture content we have:

New % moisture content required = 50 %

Dry weight of soil sample = DW2= 31 g

Wet weight of the soil sample that would give 50 % moisture content = WW2

So we have

% Moisture content = ( WW2- DW2)/ DW *100 %

50 % = (WW2-31)/31 * 100 %

0.5 = (WW2-31)/31

WW2-31=0.5*31

WW2=46.5 g

So the amount of water needed to be added to the dry soil sample to recondition it in order to have 50 % moisture content we have

Amount of water needed = WW2- DW2 = 46.5 g - 31 g = 15.5 g.

Best regards.

Interesting!

Dear respected Usman,

Thanks for the good comment about my answer.

Best regards.