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Naddika [18.5K]
2 years ago
15

A shrinkage limit test is performed on a soil. The initial mass and volume of the soil are: V1=20.2cm^3 , while the final mass a

nd volume are M2=24g and V2=14.3cm^3 . Note that in the initial state the soil is saturated, whereas in the final state the soil is completely dry.
Calculate:
a. the shrinkage limit SL of the soil.
b. the void ratio at the SL.
c. Gs of the soil solids.
d. the initial void ratio.
Engineering
1 answer:
love history [14]2 years ago
3 0
C. Gs of the soil solids
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Answer:

Flow velocity

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Explanation:

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The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 0.126m, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 352k and 98 kPa, respectively

solution

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2 years ago
A 150-lbm astronaut took his bathroom scale (aspring scale) and a beam scale (compares masses) to themoon where the local gravit
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b) W = 150 lbf

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7 0
3 years ago
The unit weight of a soil is 14.9kN/m3. The moisture content of the soil is17% when the degree of saturation is 60%. Determine:
Serggg [28]

Answer:

a) 2622.903 N/m^3

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c)4.878811765

Explanation:

Find the void ratio using the formula:

y = \frac{G_{s}*y_{w} + w*G_{s}*y_{w} }{1+e} ....... Eq1

Here;

G_{s} is specific gravity of soil solids

y_{w} is unit weight of water = 998 kg/m^3

w is the moisture content = 0.17

e is the void ratio

y is the unit weight of soil = 14.9KN/m^3

Saturation Ratio Formula:

w*G_{s} = S*e  ..... Eq2

S is saturation rate

Substitute Eq 2 into Eq 1

y = \frac{(\frac{S*e}{w}) * y_{w} + S*e*y_{w}  }{1+e}

14900 = \frac{3522.352941*e + 598.8*e }{1+e} = \frac{4121.152941*e}{1+e}\\\\ e= 1.38233

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y_{s} = \frac{(G_{s} + e)*y_{w}  }{1+e} \\=\frac{(4.878811765 + 1.38233)*998  }{1+1.38233}\\\\= 2622.902571 N/m^3

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The process of making a function call itself is known as recursion. With the use of this strategy, complex problems can be reduced to more manageable, simpler ones. Recursion might be a little challenging to comprehend. Experimenting with it is the most effective way to learn how it functions.

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