Question

1. A soil core sampling tube of 4 cm diameter, 12 cm length and initial mass of 0.525 kg (sample only), was dried at 105o C and had final mass of 0.490 kg. Determine dry bulk density, porosity and water content. 10 pts

Answer 1

Answer:

porosity = 0.07 or 7%

dry bulk density = 3.25g/cm3]

water content =

Explanation:

bulk density = dry Mass / volume of  sample

dry mass = 0.490kg = 490g

volume = πr2h = 3.142 * 2 *2 *12 = 150.8cm3

density = 490/150.8 = 3.25g/cm3

porosity = $\frac{wet mass - dry mass }{wet mass}$ = $\frac{0.525 - 0.49}{0.525}$ = 0.07 or 7%

water content =  $\frac{wet mass - dry mass}{wet mass}$ = 7%

Answer 2

Answer:

Given diameter 4cm=0.04m,lenght=0.12m

Iniatial mass m1=0.525kg final mass m2=0.490kg

dry bulk density =?

Porosity=?

Water content=?

Water content

Mass of water=0.525-0.490=0.035kg

m=Mw/Ms=0.935)0.525=0.0667 Pr 6.67%

Dry bulk density

Recall Density= mass/volume

Volume =πr^2h=π*(0.02)^2*0.12=7.54*10^-3m^3

Dry density=0.490/(7.54*10^-3)=64.97kg

Porosity

n=e/1+e

Were void ratio e=0.0667*2.6=0.1734

Porosity=0.1734/1+0.1734=0.15