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02.05.2023 06:08
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A soil had a liquid limit of 44, a plastic limit of 21, and a shrinkage limit of

A soil had a liquid limit of 44, a plastic limit of 21, and a shrinkage limit of 14. In the summer, the in situ water content was 18%. (a) Compute the plasticity index of the soil. (b) Calculate the liquidity index of the in situ soil. (c) In which consistency state was the in situ soil
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english7750
english7750
4,6(52 marks)

(a) 23

(b)-0.13

(c) In which consistency state was the in situ soil? Plastic limit = 1.1

Explanation:

(a) Compute the plasticity index of the soil.

The formula for plasticity index is given as

Plasticity Index = Liquid limit - Plastic limit

Liquid limit = 44

Plastic limit = 21

Plastics Index = 44 - 21

= 23

(b) Calculate the liquidity index of the in situ soil.

Liquidity Index = Water content - Plastic limit/Liquid limit - Plastic Limit

Water content = 18%

Plastic limit = 21

Liquid limit = 44

Liquidity Index =18 -21/44 - 21

- 3/23

= -0.1304347826

≈ -0.13

(c) In which consistency state was the in situ soil

The consistency state of a soil tells us how firm the soil is.

Consistency state/index = Liquid limit - Water content/ Plastic Index

= Liquid limit - Water content /Liquid limit - Plastic Limit

Water content of the in situ soil = 18%

=44 - 18/44 - 23

= 26/23

= 1.1304347826

= 1.1

Since the consistency index of this situ soil is above 1, it means the soil is at plastic limit.

izzzzz49
izzzzz49
4,4(90 marks)

a)\Delta P= 2666.66 Pa

b)\Delta P= 3200 Pa

Explanation:

Given that

Diameter ,d= 0.15 mm

We know that pressure difference is given as

\Delta P=\dfrac{4\sigma }{d}

Now by putting the values

When surface tension 0.1 N/m  :

The surface tension ,\sigma=0.1\ N/m

\Delta P=\dfrac{4\times 0.1 }{0.15}\times 10^3\ Pa

\Delta P= 2666.66 Pa

When surface tension 0.12 N/m  :

The surface tension ,\sigma=0.12\ N/m

\Delta P=\dfrac{4\times 0.12 }{0.15}\times 10^3\ Pa

\Delta P= 3200 Pa

a)\Delta P= 2666.66 Pa

b)\Delta P= 3200 Pa

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