Soil Question

[EASC3611]

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.

 

[fattdad]

There is no solution to your problem. It's unfortunate that the original thread it gone as I provided information on the reason that this problem can't be solved. In summary:

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!

 

[PEinc]

Is this a real job issue or is it a homework problem?

 

[EASC3611]

Real job issue...However, I don't have the background in soils coming from a hard rock Mining background.

 

[msucog]

sorry for my ramblings this morning (wifey was scheduled to go under the knife today so i was sort of out in left field). i'm guessing that even if the situation could happen, my calcs would have been wrong since i've never been proficient at doing these problems off the top of my head (even though i should be able to).

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.

 

[jdonville]

EASC,

How do you know my solution is incorrect?

Jeff

 

[EASC3611]

What was your solution? Original post is gone. I think it didn't seem right, too high or too low for some reason.

 

[jdonville]

EASC3611,

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

 

[EASC3611]

Area = 3.36ha = 336 000 000cm2
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.

 

[fattdad]

Your calculation doesn't make sense. It seems like you are looking at a volume of water that is passing through the soil mass, rather than residing in the soil mass. Liquid and Plastic limits are moisture contents that reside in the soil. Therefore, you have to have sufficient void space in the soil mass for the given volume of water. You don't.

This is a frustrating exercize and you are overlooking the obvious in your calculation.

f-d

¡papá gordo ain’t no madre flaca!

 

[dgillette]

Is 2.25 g/cc the specific gravity, i.e., the weight and volume of the solids only, or is it the dry density, equal to the weight of the solids divided by the volume of solids and voids?

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?

 

[EASC3611]

The 2.25 g/cubic centimetre applies to the dry soil and is not related to the PL.

 

[dgillette]

The parts just aren't fitting together. Where did that number come from? That is, how was it measured, and what is the definition of "bulk density" in your mind? As jdonville and fattdad are interpreting it, they are absolutely correct that your numbers are incompatible with each other. They are considering "bulk density" to be the weight of solids divided by the total volume of solids and voids, better known to geotechs as the 'dry density' or 'dry unit weight.' (I would probably have made the same interpretation if I had come into the discussion earlier.) It is possible that you actually mean the 'specific gravity,' i.e., the weight of solids divided by the volume of solids only.

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.