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Elodia
21 day ago
5

A 0.91 m diameter corrugated metal pipe culvert (n = 0.024) has a length of 90 m and a slope of 0.0067. The entrance has a squar

e edge in a headwall. At the design discharge of 1.2 m3 /s, the tailwater is 0.45 m above the outlet invert. Determine the head on the culvert at the design discharge. Repeat the calculation for head if the culvert is concrete.

Engineering
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The in situ moist unit weight of a soil is 17.3 kN/m3 and the moisture content is 16%. The specific gravity of soil solids is 2.
Mrrafil [318]

Answer:

Explanation:

Given that,

Moisture content w = 16%

The in situ moist unit weight of soil: γ(in situ) = 17.3 kN/m³

Specific gravity of the soil

G(s) = 2.72

Minimum dry unit weight of the soil

γd(compacted) = 18.1 kN/m³

Moisture content remains the same

w = 16%

Question:

How much cubic meters of soil are necessary from the excavation site to create 2000 m³ of compacted fill?

Now, let's work out the in situ dry unit weight γd(in-situ) with the formula:

γd(in-situ) = γ(in-situ) / [1 + (w/100)]

γd(in-situ) = 17.3 / [1 + (16/100)]

γd(in-situ) = 17.3 / (1 + 0.16)

γd(in-situ) = 17.3 / 1.16

γd(in-situ) = 14.89 kN/m³

To find out the Volume of soil to be excavated (Vex):

Let the Volume to be excavated = V

We will use the relationship:

V = V(fill) × γd(compacted) / γd(in situ)

Given that, V(fill) = 2000 m³

V(fill) is the compacted fill volume.

Therefore, we have:

V = V(fill) × γd(compacted) / γd(in situ)

Vex = 2000 × 18.1 / 14.89

Vex = 2423.34 m³

Thus, the soil volume to be excavated equals 2423.34 m³.

3 0
3 months ago
The water level in a large tank is maintained at height H above the surrounding level terrain. A rounded nozzle placed in the si
pantera1 [306]

Answer:

Explanation: Kindly refer to the attached files for a detailed solution process.

Take note that the diagram file is the first one and is positioned accordingly.

8 0
2 months ago
Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727C (1341F). (a) What is the proeutectoid phase? (b) How
pantera1 [306]

Answer:

a) The phase before eutectoid is commonly referred to as cementite, with the chemical formula Fe₃C.

b) The total mass of ferrite obtained is 0.8311 kg.

The total cementite mass equals 0.1689 kg.

c) The total cementite mass accounts for 0.9343 kg.

Explanation:

Provided:

1 kg of austenite

a carbon content of 1.15 wt%

Cooled below 727°C

Questions:

a) Identify the proeutectoid phase.

b) Calculate the mass of total ferrite and cementite, Wf =?, Wc =?

c) Determine the mass of both pearlite and the proeutectoid phase, Wp =?

d) Create a schematic to illustrate the resulting microstructure.

a) The proeutectoid phase is referred to as cementite with the formula Fe₃C.

b) To find the total mass of formed ferrite:

W_{f} =\frac{C_{cementite}-C_{2} }{C_{cementite}-C_{1} }

With:

Ccementite = composition of cementite = 6.7 wt%

C₁ = composition of phase 1 = 0.022 wt%

C₂ = overall composition = 1.15 wt%

Inserting the values yields:

W_{f} =\frac{6.7-1.15}{6.7-0.022} =0.8311kg

For the total mass of cementite:

W_{c} =\frac{C_{2}-C_{1}}{C_{cementite}-C_{1} } =\frac{1.15-0.022}{6.7-0.022} =0.1689kg

c) The mass of pearlite:

W_{p} =\frac{6.7-1.15}{5.94} =0.9343kg

d) The diagram illustrates the different compositions: (pearlite, proeutectoid cementite, ferrite, eutectoid cementite)

6 0
4 months ago
A distância entre duas retas reversas é a medida de um segmento orientado que *?
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Fatec – SP) Let A be a point on line r, which is contained within plane α. It holds true that: a) there is exactly one line that is perpendicular to line r at point A. b) there exists one unique line, not lying in plane α, that is parallel to line r. c) there are infinitely many distinct planes that are parallel to plane α and contain line r. d) there are infinitely many distinct planes that are perpendicular to plane α and contain line r. e) there are countless distinct lines contained within plane α that are parallel to line r.
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