Soil Question
Soil Question
(OP)
My thread was deleted for some reason. Here goes:
Find the volume of water that infiltrates a 15.0 cm thick layer of soil of surface area 3.36 ha (336 000 000cm2)as the soil is taken from its plastic limit of 23% to its liquid limit of 49%. The bulk density of the dry soil is 2.25 g/cubic centimeter.
The 2.25 g/cubic centimeter applies to the dry soil and is not related to the plastic limit.
From another user (not sure of his or her name now):
Knowing the dry density of the soil (mass of solids per unit volume), I can determine how much moisture it would take to increase the moisture content from 23% to 49%
- a difference of 26%.
26/100 * 2.25g = 0.585g/cm3
Given, I have a depth of 15 cm, I would need 0.585g/cm3 * 15cm = 8.775 g/cm2.
Now, I determine how many cm2 in 3.36 ha and multiply accordingly.
3.36ha = 336 000 000cm2
So, 336 000 000cm2 * 8.775 g/cm2 = 2 948 400 000g =
2 948 400 000cm3 because 1g = 1cm3.
Not correct solution.
Any ideas welcomed.
Find the volume of water that infiltrates a 15.0 cm thick layer of soil of surface area 3.36 ha (336 000 000cm2)as the soil is taken from its plastic limit of 23% to its liquid limit of 49%. The bulk density of the dry soil is 2.25 g/cubic centimeter.
The 2.25 g/cubic centimeter applies to the dry soil and is not related to the plastic limit.
From another user (not sure of his or her name now):
Knowing the dry density of the soil (mass of solids per unit volume), I can determine how much moisture it would take to increase the moisture content from 23% to 49%
- a difference of 26%.
26/100 * 2.25g = 0.585g/cm3
Given, I have a depth of 15 cm, I would need 0.585g/cm3 * 15cm = 8.775 g/cm2.
Now, I determine how many cm2 in 3.36 ha and multiply accordingly.
3.36ha = 336 000 000cm2
So, 336 000 000cm2 * 8.775 g/cm2 = 2 948 400 000g =
2 948 400 000cm3 because 1g = 1cm3.
Not correct solution.
Any ideas welcomed.





RE: Soil Question
For a dry density of 2.25 g/cc the amount of moisture for the plastic limit would have to be 0.52 g/cc, which means there is only 0.48 cc for the 2.25 g of soil solids to occupy. This equates to a specific gravity of soil solids equal to 4.68, which is outside any normal range for soil.
For a dry density of 2.25 g/cc the amount of moisture for the liquid limit would have to be 1.1 g/cc, which means there is insufficient volume for the soil grains - an impossible condition.
There is no answer to your problem and any attempts to solve it will lead to the wrong answer. Maybe there is something amiss with your dry density. . . . ?
f-d
¡papá gordo ain’t no madre flaca!
RE: Soil Question
RE: Soil Question
RE: Soil Question
EASC3611...you dealing with mine spoils by any chance or is this residual "everyday" surface type soils? i'm somewhat curious about this real world problem...even if it can't really happen.
RE: Soil Question
How do you know my solution is incorrect?
Jeff
RE: Soil Question
RE: Soil Question
Solution I gave was as outlined in your first post in this thread.
These are well-understood basic relations. I suggest you look at the result again and bear in mind fatdad's comments.
Are you dealing with metal mining tailings? I once dealt with tailings with a SG of solids of about 3.5 (a lot of iron), but this is a little extreme.
Suggest that you double check the Atterberg limits data.
Jeff
RE: Soil Question
Depth = 15cm
Volume of soil = 336 000 000cm2 * 15cm = 5 040 000 000cm3
PL = 23%
LL = 49%
Bulk Density = 2.25g/cm3
Volume = 336 000 000cm2 * 15cm = 5 040 000 000cm3
Mdry = 5 040 000 000cm3 * 2.25 g/cm3 = 11 340 000 000g
M23% = 11 340 000 000 * 1.29 = 13 948 200 000g
M49% = 11 340 000 000 * 1.49 = 16 896 600 000g
Mwater = M49% - M23%
Mwater = 16 896 600 000g - 13 948 200 000g
Mwater = 2 948 400 000g
Mwater = 2 948 400kg
Density of water = 1000kg/m3
Volume of Water = (2 948 400kg) / (1000kg/m3) = 2 948.4 m3
Therefore, the volume of water that infiltrates the soil is 2 948.4 m3.
Probably thinking now I should use .23 and .49, and not 1.23 and 1.49.
RE: Soil Question
This is a frustrating exercize and you are overlooking the obvious in your calculation.
f-d
¡papá gordo ain’t no madre flaca!
RE: Soil Question
For those of us calibrated in lb/cubic foot, a dry density of 2.25 g/cc is 140.4 lb/c.f., pretty close to concrete. With typical specific gravities, up to 2.7, the water content could not possibly be higher than 8 percent. (This is another way to reach fattdad's conclusion that something is amiss with the input numbers.)
If, instead, that's a specific gravity, it is unusually low. Or is that number the bulk density of porous particles?
EASC3611 - What is this material, and what is the definition of 'bulk density' that you are using?
RE: Soil Question
RE: Soil Question
If you want any help here, you will have to back up and explain what you mean by terms, rather than just restating what you said before. Otherwise, you can go back to the basic weight and volume relations and derive the solution yourself, which shouldn't be too difficult.