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KengaRu [80]
3 years ago
13

A site is underlain by a soil that has a unit weight of 118 lb/ft3. From laboratory shear strength tests that closely simulated

the field conditions, the total stress parameters were measured to be C total = 250 lb/ft2 and φ total = 29°. Estimate the shear strength on a horizontal plane at a depth of 12 ft below the ground surface at this site in lbs/ sq ft
Engineering
1 answer:
Rzqust [24]3 years ago
7 0

Answer: the shear strength at a depth of 12 ft is 1034.9015 lb/ft²

Explanation:

Given that;

Weight of soil r = 118 lb/ft³

stress parameter C = 250 lb/ft²

φ total = 29°

depth Z = 12 ft

The shear strength on a horizontal plane at a depth of 12ft

ζ = C + δtanφ

where δ = normal stress

normal stress δ = r × z = 118 × 12 = 1416

so

ζ = C + δtanφ

ζ = 250 + 1416(tan29°)

ζ = 250 + 1416(tan29°)

ζ = 250 + 784.9016

ζ = 1034.9015 lb/ft²

Therefore the shear strength at a depth of 12 ft is 1034.9015 lb/ft²

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A steam power plant with a power output of 230 MW consumes coal at a rate of 60 tons/h. If the heating value of the coal is 30,0
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A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3
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Answer:

Q = -68.859 kJ

Explanation:

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mass co_2 = 1 kg

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Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

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R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

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V_1 = \frac{m RT_1}{P_1}

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V_1 = 0.5415 m3

V_2 = \frac{m RT_2}{P_2}

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V_1 = 0.1033 m3

WORK DONE

W =P_{avg}*{V_2-V_1}

w = 586*(0.1033 -0.514)

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\Delta U  = m *c_v*{V_2-V_1}

\Delta U  = 1*0.657*(584-298)

\Delta U  =187.902 kJ

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Q = \Delta U  +W

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Q = -68.859 kJ

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