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andreev551 [17]
3 years ago
13

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
Engineering
2 answers:
belka [17]3 years ago
8 0

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%

amm18123 years ago
8 0

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

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1.Mohamed Siddiqui appeals his convictions for fraud and false statements to a federal agency, and obstruction in connection with a federal investigation.   Siddiqui challenges the district court's admission into evidence of e-mail and foreign depositions.

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Alexeev081 [22]

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Explanation:

Hello!

To solve this exercise follow the steps below

1. we will call 1 the initial state, 2 the steam that enters and 3 the final state

2. We find the quality of the initial state, dividing the mass of steam by the total mass.

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3 Find the internal energy in the three states using thermodynamic tables

note:Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

u1=IntEnergy(Water;x=0,6(quality);P=200kPa) =1719KJ/kg

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m2=the mass of the steam that has entered.

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(m1)(u1-u3)=(m2)(u3)-(m2)(u2)

m2=m1\frac{u1-u3}{u3-u2} =10\frac{1719-2529}{2529-2808} =29kg

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