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lilavasa [31]
2 years ago
11

The in situ moist unit weight of a soil is 17.3 kN/m^3 and the moisture content is 16%. The specific gravity of soil solids is 2

.72. This soil is to be excavated and transported to a construction site for use in a compacted fill. If the specification calls for the soil to be compacted to a minimum dry unit weight of 18.1 kN/m^3 at the same moisture content of 16%. 1) How many cubic meters of soil from the excavation site are needed to produce 2000 m^3 of compacted fill? 2) How many 20 ton truckloads are needed to transport the excavated soil?

Engineering
1 answer:
hammer [34]2 years ago
7 0

Answer:

Explanation:

The solution to given problem is attached below

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A shopaholic has to buy a pair of jeans , a pair of shoes l,a skirt and a top with budgeted dollar.Given the quantity of each pr
fredd [130]

Answer:

you might be facing some difficulty in observing this point of division between your question and me

0 0
2 years ago
Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carb
Nezavi [6.7K]

Answer:

y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02

Explanation:

Hello,

a. On the attached document, you can see a brief scheme of the process. Thus, to know the degrees of freedom, we state the following unknowns:

- \xi_1 and \xi_2: extent of the reactions (2).

- F_{O_2}^2, F_{CH_4}^2, F_{H_2O}^2, F_{HCHO}^2 and F_{CO_2}^2: Molar flows at the second stream (5).

On the other hand, we've got the following equations:

- F_{O_2}^2=50mol/s-\xi_1-2\xi_2: oxygen mole balance.

- F_{CH_4}^2=50mol/s-\xi_1-\xi_2: methane mole balance.

- F_{H_2O}^2=\xi_1+2\xi_2: water mole balance.

- F_{HCHO}^2=\xi_1: formaldehyde mole balance.

- F_{CO_2}^2=\xi_2: carbon dioxide mole balance.

Thus, the degrees of freedom are:

DF=7unknowns-5equations=2

It means that we need two additional equations or data to solve the problem.

b. Here, the two missing data are given. For the fractional conversion of methane, we define:

0.900=\frac{\xi_1+\xi_2}{50mol/s}

And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:

0.860=\frac{F_{HCHO}^2}{50mol/s}

In such a way, one realizes that the output formaldehyde's molar flow is:

F_{HCHO}^2=0.860*50mol/s=43mol/s

Which is equal to the first reaction extent \xi_1, therefore, one computes the second one from the fractional conversion of methane as:

\xi_2=0.900*50mol/s-\xi_1\\\xi_2=0.900*50mol/s-43mol/s\\\xi_2=2mol/s

Now, one computes the rest of the output flows via:

- F_{O_2}^2=50mol/s-43mol/s-2*2mol/s=3mol/s

- F_{CH_4}^2=50mol/s-43mol/s-2mol/s=5mol/s

- F_{H_2O}^2=43mol/s+2*2mol/s=47mol/s

- F_{HCHO}^2=43mol/s

- F_{CO_2}^2=2mol/s

The total output molar flow is:

F_{O_2}+F_{CH_4}+F_{H_2O}+F_{HCHO}+F_{CO_2}=100mol/s

Therefore the output stream composition turns out into:

y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02

Best regards.

7 0
2 years ago
Liquid oxygen is stored in a thin-walled, spherical container 0.75 m in diameter, which is enclosed within a second thin-walled,
nadezda [96]

Answer:

Explanation:

the solution is well stated

3 0
2 years ago
Fill in the Blank: The _____ has a very sturdy handle with good leverage for tightening nuts and bolts. It can be used to apply
guapka [62]

Answer:

D Rachet

Explanation:

I tightens it

8 0
2 years ago
A 75,000 ft3 clarifier is to be used to treat wastewater. The recycle ratio is 50%, the sludge volume index (SVI) is 125, and th
Inessa05 [86]

Answer:

11 hours approximately

Explanation:

We are to calculate mean cell residence time mcrt

= Mass of solid in reactor/mass of solid wasted in a day

Q = Qe + We

Q = 2.5

Qw = 0.5

Qe = 2.5 - 0.5

= 2 MGD

10⁶/svi

= 10⁶/125

= 8000

X = 3500

Xe = 20mg/

1MGD = 0.1337million

Mcrt = 75000x3500/[0.5*8000*10⁶+2*20*10⁶] x 0.1337

= 262500000/[4000000000+40000000} x 0.1337

= 262500000/574800000

= 0.45668 days

= 0.45668 x 24 hours

= 10.9603 hours

Approximately 11 hours

3 0
2 years ago
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