Question

i have the answer for the question but i dont know how to draw the phase diagram if anyone can help plzz

The moist density of a soil is 1750 kg/m3. Given w= 23% and

Gs= 2.73, determine:

a. Dry density

b. Porosity

c. Degree of saturation

d. Mass of water, in kg/m^3, to be added to reach full saturation.

and this is my answer

Given moist density =1750kg/m³

w=23%

Gs=2.73

Step1:

a. Dry desity formula is

dry density $\rho$ _d = ( moist density) / (1+water content of the soil mass()

$\rho$ _d = ( $\rho$ _moist) / (1+w)

$\rho$ _d = 1750 /(1+0.23)

$\rho$ _d = 1422.76

Dry density of the soil is 1422.76 kg/m³ .

Step 2:

b. porosity formula is n = e/(1+e)

e is void ratio

$\rho$ _d =(Gs/(1+e)) $\rho$ _w

$\rho$ _d / $\rho$ _w =Gs/(1+e)

1422.76/1000 =2.73/(1+e)

1.42276=2.73/(1+e)

1+e=2.73/1.42276

1+e=1.9188057

e=0.9188

n=e/(1+e)

n=0.9188(1+0.9188)

n=0.4788

porosity is 0.4788

Step 3:

c.Degree of Saturation is

Se=wGs

S* 0.9188= 0.23*2.73

S=0.6279/0.9188

S=0.6833

Degree of saturation is 0.6833.

Step 4:

d. Mass of water, in kg/m^3, to be added to reach full saturation is

Mass of water = S*e * $\rho$ _w

Mass of water = 0.6833*0.9188*1000

Mass of water = 627.9 kg/m³ to be added to reach full saturation.

Answer 1

Your solution for d option is not correct. It is asked to find extra water for making saturation 100%. Please find the correct solution and phase diagram in attached image.