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scoundrel [369]
3 years ago
12

A site is underlain by two layers of normally consolidated clayey sand. The unit weights of the top layer are 19 and 21 kN/m2 an

d the layer is 6 meters thick. The unit weights of the bottom layer are 20 and 22 kN/m, and the layer is 8 meters thick. Below the bottom layer lies bedrock. The water table is located 2 meters below the ground surface. The top layer soil has a friction angle of 38 degrees, a cohesion intercept of 20 kPa, and an undrained strength of 160 kPa. The bottom layer soil has a friction angle of 33 degrees, a cohesion intercept of 10 kPa, and an undrained strength of 120 kPa. Point A is located 9 meters underground. All layers have a lateral stress ratio of 0.5.
Required:
a. Draw the profile neatly, add dimensions, and draw a tree on the ground surface.
b. Determine the geostatic vertical effective stress and the geostatic horizontal effective stress at point A.
c. Draw the Mohr Circle to determine the effective stress that acts at point A on a plane inclined 10 degrees counter-clockwise from the horizontal plane. Make sure the Mohr Circle is a well-drawn circle
Engineering
1 answer:
Ivan3 years ago
5 0

Answer:

A) attached below

B) Geostatic vertical effective stress ( бv )

= 119.33 KN/m^2

Geostatic horizontal effective stress ( бn )

= 59.66 KN/m^2

C) attached below

Explanation:

attached below is a detailed solution

A) attached below

B) Determine the geostatic vertical effective stress and the geostatic horizontal effective stress at point A

Geostatic vertical effective stress ( бv )

= 119.33 KN/m^2

Geostatic horizontal effective stress ( бn )

= 59.66 KN/m^2

C) attached below

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A closed system consisting of 4 lb of a gas undergoes a process during which the relation between pressure and volume is pVn 5 c
gayaneshka [121]

Answer:

V1=5<u>ft3</u>

<u>V2=2ft3</u>

n=1.377

Explanation:

PART A:

the volume of each state is obtained by multiplying the mass by the specific volume in each state

V=volume

v=especific volume

m=mass

V=mv

state 1

V1=m.v1

V1=4lb*1.25ft3/lb=5<u>ft3</u>

state 2

V2=m.v2

V2=4lb*0.5ft3/lb=  <u> 2ft3</u>

PART B:

since the PV ^ n is constant we can equal the equations of state 1 and state 2

P1V1^n=P2V2^n

P1/P2=(V2/V1)^n

ln(P1/P2)=n . ln (V2/V1)

n=ln(P1/P2)/ ln (V2/V1)

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3 0
3 years ago
4.2 A vapor compression refrigeration machine uses 30kW of electric power to produce 50 tons of cooling. What is
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Answer:

5.833

Explanation:

Coefficient of Perfomance (COP) is the ratio of refrigeration effect to power input.

COP=\frac {RE}{P} where RE is refrigeration effect and P is power input

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6 0
3 years ago
A cylindrical insulation for a steam pipe has an inside radius rt = 6 cm, outside radius r0 = 8 cm, and a thermal conductivity k
goldfiish [28.3K]

Answer:

heat loss per 1-m length of this insulation is 4368.145 W

Explanation:

given data

inside radius r1 = 6 cm

outside radius r2 = 8 cm

thermal conductivity k = 0.5 W/m°C

inside temperature t1 = 430°C

outside temperature t2 = 30°C

to find out

Determine the heat loss per 1-m length of this insulation

solution

we know thermal resistance formula for cylinder that is express as

Rth = \frac{ln\frac{r2}{r1}}{2 \pi *k * L}   .................1

here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity

so

heat loss is change in temperature divide thermal resistance

Q = \frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}

Q = \frac{(430-30)*(2 \pi * 0.5 * 1}{ln\frac{8}{6} }

Q = 4368.145 W

so heat loss per 1-m length of this insulation is 4368.145 W

4 0
3 years ago
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