Skip to main content

Moist soil 20 kg numerical

Short answer

Water content 21.2%, dry density 1500 kg/m³, degree of saturation 71.6%, porosity 44.4%.

A mass of moist soil is 20 kg and its volume is 0.011 m³. After oven drying, the mass reduces to 16.5 kg. Assume G = 2.70. Determine water content, dry density, degree of saturation and porosity.

Given

M=20M = 20 kg, V=0.011V = 0.011 m³, dry mass Ms=16.5M_s = 16.5 kg, G=2.70G = 2.70, ρw=1000\rho_w = 1000 kg/m³.

Solution

Mass of water: Mw=20−16.5=3.5M_w = 20 - 16.5 = 3.5 kg.

Water content

w=MwMs=3.516.5=0.212=21.2%w = \frac{M_w}{M_s} = \frac{3.5}{16.5} = 0.212 = 21.2\%

Dry density

ρd=MsV=16.50.011=1500 kg/m3\rho_d = \frac{M_s}{V} = \frac{16.5}{0.011} = 1500\ \text{kg/m}^3

Void ratio (from ρd=Gρw1+e\rho_d = \dfrac{G\rho_w}{1+e}):

e=Gρwρd−1=27001500−1=0.80e = \frac{G\rho_w}{\rho_d} - 1 = \frac{2700}{1500} - 1 = 0.80

Degree of saturation

S=wGe=0.2121×2.700.80=0.716=71.6%S = \frac{wG}{e} = \frac{0.2121 \times 2.70}{0.80} = 0.716 = 71.6\%

Porosity

n=e1+e=0.801.80=0.444=44.4%n = \frac{e}{1+e} = \frac{0.80}{1.80} = 0.444 = 44.4\%

Answer: w=21.2%w = 21.2\%, ρd=1500\rho_d = 1500 kg/m³, S=71.6%S = 71.6\%, n=44.4%n = 44.4\%.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Soil Mechanics chapter 2 questions →