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zavuch27 [327]
3 years ago
10

A proposed embankment fill requires 7100 ft of compacted soil. The void ratio of the compacted fill is specified as 0.5. Four bo

rrow pits are available as below, which list the respective void ratios of the soil and the cost per cubic feet for moving the soil to the proposed construction site. Make the necessary calculations to determine the total cost for each pit. Assume the specific gravity of soil solids, Gs to be the same at all pits.
1. Calculate the volume of solid (Vs) = ft3 Borrow Pit A: void ratio = 1.05 and the unit cost = $ 6 /ft3
2. Calculate the total cost to pay to Pit A = Borrow Pit B: void ratio = 0.61 and the unit cost = $ 7 /ft3
3. Calculate the total cost to pay to Pit B = Borrow Pit C: void ratio = 0.81 and the unit cost = $ 4 /ft3
4. Calculate the total cost to pay to Pit C = Borrow Pit D: void ratio = 0.73 and the unit cost = $6 /ft3
5. Calculate the total cost to pay to Pit D =
Engineering
1 answer:
podryga [215]3 years ago
3 0

Answer:

1) the volume of solid (Vs) is 4733.33 ft³

2) the total cost to pay Pit A is $58220

3) the total cost to pay to Pit B is $53344.67

4) the total cost to pay to Pit C is $34269.32

5) the total cost to pay to Pit D is $49132.02

Explanation:  

given the data in the question;

Vs = ?, V = 7100 ft³, e = 0.5

1) Calculate the volume of solid (Vs)

Vs = V / ( 1 + e )

we substitute

Vs = 7100 / ( 1 + 0.5 )

Vs = 7100 / 1.5

Vs = 4733.33 ft³

Therefore, the volume of solid (Vs) is 4733.33 ft³

2) Calculate the total cost to pay to Pit A

VpitA / Vcomp = ( 1 + e_{PHA}) / ( 1 + e_{comp )

we substitute

VpitA / 7100 = ( 1 + 1.5) / ( 1 + 0.5)

VpitA / 7100 = 1.3666666

VpitA = 1.3666666 × 7100

VpitA  = 9703.33 ft³

Total cost = volume × unit cost = 9703.33 ft³ × 6 = $58220

Therefore, the total cost to pay Pit A is $58220

3) Calculate the total cost to pay to Pit B = Borrow Pit C: void ratio = 0.81 and the unit cost = $ 4 /ft3

Vpitb / Vcomp = ( 1 + e_{PHB}) / ( 1 + e_{comp )

we substitute

VpitB / 7100 = ( 1 + 0.61) / ( 1 + 0.5)

VpitB / 7100 = 1.0733333

VpitB = 1.0733333 × 7100

VpitB  = 7620.666 ft³

Total cost = volume × unit cost = 7620.666 × 7 = $53344.67

Therefore, the total cost to pay to Pit B is $53344.67

4) Calculate the total cost to pay to Pit C = Borrow Pit D: void ratio = 0.73 and the unit cost = $6 /ft3

VpitC / Vcomp = ( 1 + e_{PHC}) / ( 1 + e_{comp )

we substitute

VpitC / 7100 = ( 1 + 0.81) / ( 1 + 0.5)

VpitC / 7100 = 1.2066666

VpitC = 1.2066666× 7100

VpitC  = 8567.33 ft³

Total cost = volume × unit cost = 8567.33 × 4 = $34269.32

Therefore, the total cost to pay to Pit C is $34269.32

5) Calculate the total cost to pay to Pit D

VpitD / Vcomp = ( 1 + e_{PHD}) / ( 1 + e_{comp )

we substitute

VpitD / 7100 = ( 1 + 0.73) / ( 1 + 0.5)

VpitD / 7100 = 1.1533333

VpitD = 1.1533333 × 7100

VpitD  = 8188.67 ft³

Total cost = volume × unit cost = 8188.67 × 6 = $49132.02

Therefore, the total cost to pay to Pit D is $49132.02

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Answer:

Explanation:

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The weigth depends on the size and specific gravity.

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T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy

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T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

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T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

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1/4 * b^3 * cos(30) * L * SG  > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

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Now from mass conservation

m_3 = m_2 + m_1

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The temperature of the mixed exit stream given as

T_3m_3 = T_2m_2 +T_1m_1\\\\17 ( 3 + m_2) = 7 \times m_2 + 32 \times 3\\\\51 + 17 m_2 = 7 m_2 + 96\\\\10 m_2 = 96 - 51\\\\m_2 = 4.5 kg/min\\\\\\\\

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3 years ago
Water vapor at 100 psi, 500 F and a velocity of 100 ft./sec enters a nozzle operating at steady sate and expands adiabatically t
almond37 [142]

Answer:

a)exit velocity of the steam, V2 = 2016.8 ft/s

b) the amount of entropy produced is 0.006 Btu/Ibm.R

Explanation:

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T1 = 500f

P2 = 40 psi

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At P2 = 40psi and s1 = 1.708 Btu/Ibm.R

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Let's find the actual h2 using the formula :

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Using the first equation, exit velocity of the steam =

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Solving for V2, we have

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b) The amount of entropy produced in BTU/ lbm R will be calculated using :

Δs = s2 - s1

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At h2 = 1197.77 Btu/Ibm and P2 =40 psi,

S2 = 1.714 Btu/Ibm.R

Therefore, amount of entropy produced will be:

Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R

= 0.006 Btu/Ibm.R

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3 years ago
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